Section 2 - Springer978-94-009-5770-1/1.pdf · winding is jWLl + -. - and the resonant frequency f=...

Section 2

SOLUTIONS

SOLUTIONS

1. Resonant frequency fr = 1/2rrv(LC) = 1126 kHz.

Q = w,L/R = 2rrfrL/R = 177.

Impedance Z = R + j I wL - _1_) = R [1 + j (WL __ 1_)~. \' wC R wCR~

:. the angle rp and magnitude IZI of Z are given by:

tan rp = [': - w~RJ and IZI = Rv(1 + tan2 rp),

i.e. ( w Wr) tanrp=Q --- . Wr W

( W)2 _ tan rp(W )_ 1 = 0 wr Q wr

W tan rp J[ (tan rp)] and Wr = 2 Q + 1 + 2 Q ; since at resonance W = Wr and tan rp

= 0, and only the positive sign has meaning. At the upper half-power frequency W = Wl, tan rp = 1.

:= 21Q + J~ + 4~21 At the lower half-power frequency W = W2, tan rp = - 1.

:: = - 2~ + J [1 + 4~l Wl

In this case, Q = 177, sofi = - = 1129 kHz 2rr

and W2 12 = -= 1122 kHz. 2rr

2. [max = 344 rnA so [max/V(2) = 243 rnA. Bandwidth at [max/V(2) = 16 kHz.

:. Q = frI(16 X 103) = 884 X 103/(16 X 103) = SS.2S, R = S/[max = S X 103/344 = 14.S3 n, --

175

176 PROBLEMS IN ELECTRONICS WITH SOLUTIONS

L = RQ/wr = 14.53 X 55.25/(2rr X 884 X 103) H = 144.5 J..lH,

C= 1/4rr2f,. 2L = 1/ 47T2 X (884 X 103)2 X 144.5 X 10-6 F, = 224 J..lJ..lF.

3. Let the resistance, inductance and capacitance be R ohms, L henrys and C farads respectively.

The impedance of the parallel combination

z = [(R + jwL) (~)] =!:... [ 1 - jR/wL ]

R + jwL + _.1_ CR 1 + j (WL __ 1_) ]wC R wCR

At what is often taken as resonance, Z is a pure resistance and

[:~]= t~- W,~RJ. w =J(_1 - R2)= 27Tt,

r LC L2 ,..

Thus =-1 J( 1 ,f,. 27T 88 X 10-6 X 375 X 10-12

. H 4 82 )

882 X 10-12 Z

= 876.4 kHz.

4. V= I/(l/jwL + jwC+ l/R).

V has a maximum value equal toIR when wC= l/wL, i.e. when w = I/V(LC) = W, say.

V drops 3 dB from its maximum value when wC - l/wL = ± I/R and the corresponding angular frequencies are WI and W2,

where W2/W, - Wr/W2 = -1/w,CR

and wtlwr - Wr/Wl = + l/wrCR.

By comparison with the solution to Problem 1 the Q factor is seen to be w,CR = R/w,L.

5. Using the solution to Problem 1 it is seen that

Q = Wr /(Wl - W2)'

Q ~ Wr/2(Wl- wr) ~ wr/2(wr - W2)'

In this case Q ~ 27T X 106/2(2rr X 5 X 103) = 100.

SOLUTIONS 5-8 177

It is easily shown that the parallel resonant impedance Z = L/CR, i.e.

Z = Q/w"C= 100/(21T X 106 X 200 X 10-12) n = 79.6 kn.

6. The impedance-frequency curve for the circuit will have a maximum value of R at some frequency. The bandwidth is the difference in hertz between the two frequencies at which the impedance is R/V(2) and can be shown* to be 1/21TCR.

Here bandwidth = 250 X 103 Hz, C = 50 X 10-12 F

soR = 12740 n.

7. In reducing circuits such as (a) and (b) to that of (c) the rules that determine the values of Zp, Zs and Mare:

(1) Zp = impedance measured between primary terminals of actual circuit when secondary is opened.

(2) Zs = impedance measured by opening secondary of actual circuit and determining the impedance between these open points when the primary is open-circuited.

(3) M is determined by assuming a current I flows in the primary circuit. The voltage which appears across an open-circuited secondary is then ±jwML

For circuit (a) applying the above rules:

Zp = jW(LI + 4,), Zs = jW(L2 + 4,), jwL,,/ = jwMI.

coefficient of coupling = Lm/V[(L1 + Lm){L2 +Lm)] .

For circuit (b):

Zp = (C1 + c",)/wC1c"" Zs = (C2 + c",)/wC2c"" jI/wc", = jwML

coefficient of coupling = y'[C1C~(C1+ Cm)(C2 + Cm)].

8. Impedance reflected into the primary circuit from the secondary by mutual coupling = w2M2/Zs.

Primary currentIp = V/(Zp + w2M2/Zs).

Voltage induced in secondary = - jwMlp.

Secondary current Is = - jwMlp/Zs = - jwMV/(ZpZs + w2 M~.

* E.g. See L. B. Arguimbau and R. B. Adler, Vacuum Tube Circuits and Transistors, Wiley, 1956, p. 249.


Zp =Rp + j(wLp -1/wCp) =Rp + jwLp(1-1/ci).

Zs = Rs + j(wLs - l/wCs) = Rs + jwLsCl - I/o?). Is = - jwMV/{RpRs - (1 - l/ci)2w2LpLs

+ w2M2 + j(1 - l/cx2)(wLpRs + wLsRp)}· Dividing the numerator and denominator by w2LpLs and noting that Qp =

wLp/Rp, ~ = wLs/Rs, M2 = k2LpLs and w = cxwn

Is = - jVk/cxwr.J(LpLs){k2 + I/Qp~ - (1 - 1/(12)2

+ j(1- l/cx2)(1/Qp + 1/~)} Is = - jwMV/[{Rp + j(wLp - l/wCp)}

{Rs + j(wLs - l/wCs)} + w2M2].

:.Is reaches a maximum value when the circuits are in resonance and w2M2

= RpRs·

For maximum Is, wM = .J(RpRs) = w.J(LpLs).J(Qp~). critical value of k = 1/.J(Qp~).

9. k = .J(300 X 300)60 = 0.2.

Primary current Ip = lO/(Z + w2M2/Z)

where

Z = j(wL - l/wC)

= j(2 X 106 X 300 X 10-6 -1012/2 X 106 X 103) = j 100.

Secondary current Is = - jwMlp/Z = - jO.273 A.

10. Input impedance = Zp + w2M2/Zs

where Zp = (j 21T X 106 X 200 X 10-6) il

Zs = (100 + j21T X 106 X 20 X 10-6) il

and M= [0.1.J(200 X 20 X 10-12)] H.

Zp = (6.1 + jI249.1) il.

11. Effective primary impedance (Zp)

= Rl + j(wL I - l/wC1) + w2M2/{R2 + j(w~ - l/w(2)}.

Substituting the given figuresZp is found to be (718 + jO) il,

Effective resistance = 718 il

SOLUTIONS 11-14 179

and Effective resistance = O.

Primary current = 100/718 A = 0.139 A.

Secondary current = wMX 0.139/R 2 = 1.306 A.

12. With the secondary open-circuited the impedance of the primary 1

winding is jWLl + -. - and the resonant frequency f= w/2rrV(L1Cl). JwCl

Cl = 1/(2rr X 500 X 103)2 X 1 X 10-3 F = 101 f.l.f.l.F.

When the secondary is short-circuited the impedance of the primary is [jwLl + l/jwCz + w2M2/jw~], and the resonant frequency =

1 . / t M2 1 where Cz is the new capacitance.

2rrA/ l(L l - ~)~ M

DE :. since the resonant frequencies are the same,

(Ll - M2/L2)C2 = Ll Cl

i.e. 10-3 X 101 X 10-12

10-\1 - 0.25) F = 135 f.l.f.l.F.

:. The change of capacitance = (135 - 101) f.l.f.l.F = 34 f.l.f.l.F.

13. The input impedance = Rl + jWLl + l/jwCl + w2M2/R 2. When this is purely resistive its value is

(2rr X 106)2 X (10 X 10-6)2 Rl + w2M2/R2 = 5 + = 202.4 il.

20

14. With the currents as shown the equations for the circuit are:

el = (Rl + jwLl)Il - jWMI2

and e2 = (R2 + j wL2)I2 - j wMIl

180

Now

and

PROBLEMS IN ELECTRONICS WITH SOLUTIONS

(R1 + jwL1) = (60 + j 1885 X 50 X 10-3) n

(R2 + jwL2) = (80 + j 1885 X 70 X 10-3) n

- jwM = - (j 1885 X 17.75 X 10-3) n

= 169.7 100y e1 V(2) L..:::..-

141.4 ° e2= V(2) 02.. y.

11 = (0.835 - jO.819) A = 1.168 / - 45.6° A

and 12 = (0.874 - jO.212) A = 0.903/ - 13.6° A.

The phasor diagram is as shown below:

o.-__ --~--~~~-----

IS. The two following equations apply:

e = (R1 + jwL1)I1 ± jwMI2

and e = (R2 + jwL2)I2 ± jwMI1.

Writing Z1 = R1 + jwLt. Z2 = R2 + jWL2 and Zm = ± jwM,

e = Z1I1 + ZmI2

e = z",I1 + Z2I2'

11 = e(Z2 - z",)/(Z1Z2 - z", ~

SOLUTIONS 1S-17 181

and 12 = e(Zl - Zm)/(ZlZ2 - ~ ~.

I = Ii + I?, = e(Zl + Z2 - 2Zm)/(ZlZ 2 - ~ 2).

Equivalent impedance

Zp = ell = (ZlZ2 - Zm2)/(Zl + Z2 - 2Zm).

16. When the coils are in series:

and

Ll + L2 + 2M = 2 (L + M) = 360 mH

Ll + L2 - 2M = 2 (L - M) = 40 mHo

L = 100 mH andM= 80 mHo

Using the result of the previous solution the equivalent inductance Le of the two coils in parallel is given by:

wLe = {wL1wL2 - (± wM)2}/{wLl + wL2 - (± 2wM}

i.e. Le = (L2 - M2)/2(L ±M) = (1002 - 802)/2(100 ± 80) mH

= 10 or 90 mHo

17. Using Thevenin's theorem the circuit to the left of points A and B can be replaced by a single e.m.f. acting in series with a single impedance. The e.m.f. (e) is the voltage between A andB when the network to the right of these points is disconnected. The impedance (Z) is equal to that which would be measured looking to the left at terminals A and B.

If the network is opened at A, B the current Ii flowing in mesh 1

= EI(Rl + jwL1).

e= -jwM1I 1

= - j X 2 X 106 X SO X 10-6 X 6/(40 + j200)

= - 2.94 / 110 19' V.

AlsoZ = jWL2 + w2M12/(Rl + jwL1) = GWL2 + 9.6 - j48) Q.

The original circuit therefore reduces to the following:


(9.6-j48)1l

- 2.94/11°19'V

A

B

}M2 ~

Mesh 3

L--c:=r---J Mesh 4

R4

Similarly, mesh 4 can be removed by adding an impedance w 2Ml/(R4 + jwL4) = (15.1 - j68) n in series with ~ as shown in the following figure:

For mesh 5

e = ft{9.6 - j48 + 15.1 - j68 + R2 + jw(L2 + L3 + Ls) + jWMJ2.

For mesh 3

0= j~MJl + IlR6 + jWL6). From these two equations 12 is found to be 0.00369 / - 640 6' A.

18. The total impedance of the primary loop

= {(3.4 + 5.1) + jw(55 + 725) X 10-6 - j/w7.6 X 10-9}n

= (8.5 -j174) n since w = 21T X 50 X 103•

The total impedance of the secondary loop

= {CO.5 + 120) + jw(106 + 450) X 10-6 - j/wC}n

where

l/C = 1/~ + 1/C3 = 106 [1/0.0421 + 1/0.0076]

i.e. total impedance of secondary loop = (120.5 - j320) n.

SOLUTIONS 18-20 183

The mutual impedance includes the impedance of the common branch and the mutual impedance resulting fromM. It is therefore j(wM + l/w~)

= j(21T X 50 X 103 X 268 X 10-6 + 1/21T X 50 X 103 X 0.0076 X 10-6) n

= j503 n.

The apparent impedance which the voltage source sees is

(8.5 - j 174) - 0503)2/(120.5 - j320) n = (271 + j522) n.

Primary current (/p) = 10/(271 + j522) = 10/589 /- 62.6°

= 0.017 /-62.6° A

where e is taken along the real axis. The current ratio/p/4 = (120.5 - j320)/j503 = 0.679 /20.6°.

4 = 0.017 / - 62.6% .679/20.6° = 0.0251 / - 83.2° A.

19. If two coils having inductances Ll and L2 respectively and a mutual inductance M are connected together their joint inductances are:

(a) In series aiding Ll + L2 + 2M. (b) In series opposing Ll + L2 - 2M. (e) In parallel aiding (LIL2 - M~/(Ll + L2 - 2M). (d) In parallel opposing (L1L 2 - M 2)/(L1 + L2 + 2M).

For (a) here joint inductance is therefore 450 pH. :. frequency range is

1/21TV(450 X 10-6 X 50 X 10-12) Hz

to 1/21TV(450 X 10-6 X 1000 X 1O-1~ Hz

which corresponds to a wavelength range of283 to 1265 m.

Similarly, the other ranges are found to be:

(b (b) 249 to 1115 m, (e) 122 to 546 m and (d) 107 to 481 m.

20. The impedance Z of the combination

= {(R + jwL)/jwC}/(R + jwL + l/jwC)

= (R + jwLX1- w2LC- jRwC)/((1- w2LC)2 + W 2C2R2}.

The effective resistance is therefore


Re = R/f(1 - w2LC)2 + W2C2R2 }

and the effective inductance is

Le = fL(1- w2LC) - R 2C}/{(1- w2LC)2 + W2C2R2}.

Substituting L = 5 X 10-3 H, R = 100 n, C = 5 X 10-12 F and W = 27T X 500 X 103, Re = 177 nand Le = 6.67 mHo

21. Let the inductance and self-capacitance of the coil be Land C respectively and let the original frequency be [hertz.

Then [= 1/27Ty[L(C+ Ct )] and 2[= 1/27Ty[L(C+ (2)]

where C1 = 250 WF and C2 = 55 WF.

C1 - 4C2 = 3C so that C = 10 p.p.F.

22. Apparent mutual inductance Ml is approximately*

M{1 + w 2(Lt C1 + L2C2)}

where Lt = 50 p.H, L2 = 200 p.H, Ct = 5 p.p.F, C2 = 7 p.p.F, W = 27T X 2 X 106

and M = 0.05Y(50 X 200)p.H.

23.

Mt = 6.3p.H.

For circuit (a) the impedance

Z = [_1 + ~:~: J a jWC 1+. L :--c JW t

JW t

For circuit (b) the impedance

t-1- + (jWL 2 + _1_)~ 1 Zb = ~ + _1_ + .~:C2 = [jWC2(1 ~ ~~~:~2L2C'~ .

. c' . C J 2 JW JW 2

* This is easily proved from first principles but readers may like to look at the following: F. K. Harris, Electrical Measurements, Wiley, 1952, pp. 671 and 672; S. Butterworth, 'Capacity and Eddy Current Effects in Inductometers,' Proc. Phys. Soc., 33, p.312, 1921; L. Hartshorn, 'The Properties of Mutual Inductance Standards at Telephonic Frequencies,' Proc. Phys. Soc., 38, p. 302, 1926.

SOLUTIONS 23-26

IfZa =Zb, ( C'

Cz(1- wZL1CI - wZL1C) 1 + Cz - wZLzC')

= C(1 - wZLzCz)(1- wZL1Cl).

Equating the w4 terms gives

C' = CC1/(C + Cl ) .

Equating the W Z terms gives

_ Ll fclcZ + C1C' + CCz + CC' - CC1] Lz - Cz L C - C' .

Substituting for C' from (1) gives

Lz = Ll(CI + ci/Cz

Equating the terms which do not contain w gives

Cz+C'=C

Substituting for C' from (1) gives

Cz = CZ/(C + Cl )

24. Impedance Z

= {(R + jwLXR + l/jwC)}/{(R + jwL) + (R + l/jwC)}

={R2 + L/C+ jR(wL -l/wC)}/{2R + j(wL -l/wC)}.

If L/C = R2 then Z = R.

2S.

r + jwl = RjwL/(R + jwL) = (jwLR 2 + w 2LZR)/(R2 + w2L~. Equating real and imaginary parts,

r = w2L2R/(R2 + w2L2) and 1 = LR2/(R2+ w2L2).

R = r + (w212/r) andL = 1 + (r2/w21).

26. Let the impedance of the source be Zs = Rs + jXs, the load impedance ZL = RL + jXL and the voltage of the source V. Then the load current h = V/[(Rs + RL) + j(Xs + XL)]'

the power in the load (W) = RL. V2/[(Rs + RL)2 + (Xs + XdZ].

185

(1)

(2)

(3)


If XL is variable, W is a maximum when XL = - Xs and W is then RL V2/

(Rs + RLi. This is a maximum when RL = Rs,

i.e. W is a maximum when ZL = RL + jXL = Rs - jXs'

Let the loudspeaker impedance be (R + jX) Q.

(R + jX)(- j5) Then the total load on the generator = .

R + jX-j5

Conjugate of source impedance = (3 - j4) Q.

(R + jX)(- j5) = 3 -J'4. R + jX-j5

Cross-multiplying and equating real and imaginary parts gives:

X-3R=-20 and 3X+R= 15

R=7.5Q and X=2.5Q.

27. For the series circuit

tan rp = l/wCp,

where rp is the phase angle between current and voltage.

Power factor, cos rp = p/";(p2 + 1/w2C~ ~ wCp.

F or the parallel circuit

tan I/J = wCr and cos I/J = 1/";(1 + C2w 2r) ~ 1/wCr

l/wCr = wCp or w 2C2pr = 1.

Here, cos rp = 0.001, w = 21(/, Cp = 25 X 10-10 so

/= 63.7 kHz.

28. The phasor diagram for the network is as shown. I is the current through Rand C. RI 1

SOLUTIONS 28- 29

(R2 + 1/C2w2)1!l = 5000

RwC= tan 300 = 1/V(3)

R = 2500 n and C= 0.037 fJF and Va lags behind V;.

29. Mesh analysis.

The equations for the three loops are:

187

(1)

(2)

E = (600 + j600 + 400)/1 - j6oo/2 - 40013 (1) 0= - j600/1 + (900 + j600 - j600)/2 - (- j600)/3 (2)

0= - 400/1 - (- j600)/2 + (600 + 400 - j600)/3 (3)

Currents Ito 12 and 13 can be found from equations (1), (2) and (3) using Cramer's Rule. They are:

and

and

11 = E(83.3 - j35.7) 10-5,

12 =E(23.8 + j23.8) 10-5,

13 = E(47.6) 10-5•

Pi. = E - 600/1 = E(0.499 + jO.214),

V2 = 400(/1 - 13) = E(0.143 - jO.143),

V3 = 600/3 = E(0.284).

Nodal analysis

The various admittances in the network are:

Yg = l/Rg = 1.67 X 10-3 S,

Y1 = 1/R1 = 1.1 X 10-3 S, Y2 = 1/R2 = 2.5 X 10-3 S, Y3 = 1/X3 = - j 1.67 X 10-3 S,

Y4 = 1fX4 = j 1.67 X 10-3 S,

11 = l/R, = 1.67 X 10-3 S.

The nodal equations are:

E/600 = Pi.(1.67 X 10-3 + 1.1 X 10-3 - j 1.67 X 10-3)

- Vz(- j 1.67 X 10-3) - fl(1.1 X 1O-~

0=- Pi.(- j 1.67 X 10-3)

+ Vz(2.5 X 10-3 - j 1.67 X 10-3 + j 1.67 X 10-3)

- flO 1.67 X 10-3) •

(4)

(5)


0=- Jil(1.1 X 10-3) - ViG 1.67 X 10-3)

+ ~(1.67 X 10-3) + 1.1 X 10-3 + j 1.67 X 10-3) (6)

Node voltages Jil, Vi and ~ can be found directly from equations (4), (5) and (6). They are:

Jil =E(0.499 + jO.214),

Vi = E(0.143 - jO.143),

and ~ = E(0.284)

30. The nodal equations are obtained by simply applying Kirchhoffs first law at nodes 1 and 2. Thus:

- El Y1 + V1(Y1 + Y3 + Y4 + Ys) - Vi(Y4 + Ys) = 0

and - E2 Y2 + V2(Y2 + Y4 + Ys + Y6) - Jil(Y4 + Ys) = 0

31. II = Yll Jil + Y12 V2 + Yi3 V3 + 12 = Y21 Jil + Y22 V2 + 123 V3 + 13= .

In = Yn1 Jil + Yn2 V2 + Yn3 V3 + Yu Y12 Y1k- 1I1 Y1k+1 •

:. Vk = Y21 Y22 Y2k- 1I2 Y2k+1 .

Ynl Yn2 Ynk-1In Ynk+1 •

.:l

Yll Y12 Y13 Yin where.:l = Y21 Y22 Y23 Y2n

+ Yin Vn + Yln Vn

+ Ynn Vn

Yin Y2n

Ynn

SOLUTIONS 31-33 189

where Ail' absorbs the algebraic sign associated with it. If only Ij is present

Vk = 1" ~ where ~ is the open-circuit transfer impedance. d --d

V, A Similarly, the open-circuit input impedance Zu = .2 =--.!!

~ d

32.

(i) X (ii)

33.

a (a) !j=O

X X With C3 I added I I I

f I

"iY ( b)

190

34. Xl

35.


Original

ir f

f

F or the part to the right of AA.

For the part to the right ofBB.

SOLUTIONS 35- 36 191

For whole circuit.

Xl Reactance//graph for (b)

~_+--7flC--+----J.,--+---=t o-t :, X_21--_rc--;--~~-T-~1 !:--R....:~'gnrelf~Pb 18

for inverse network Inverse

36. (a) tUXl + t12X2 + tlOXO = 0

. tlO t12 .. Xl = --xo-- X2

t11' t11

t2l Xl + t22X2 + t20XO = 0 . t20 t21 .. X2=--XO--Xl

t22 t22

Xl

Xo X2 Xo

Flow graph of equation (2)

Flow graph of - t 20

t22 equations (2) and (4)

(b) (i) Vi = aVo + dV2

V2=cVl +eV2

V3=bVl +JV2

. V3 -= act + ab (1 - e)

Vo l-e-cd

x2

networks

(1)

(2)

(3)

(4)

(1)

(2) (3)


(ii) G = ~ Gkt::..k t::..

= ab (1- e) + ael(1) l-cd-e l-ed-e

37. (a) The solution can be found elsewhere.*

(b) Using the general flow-graph equation

G = ~k Gkt::..k t::..

G1 = aed, t::..l = 1, t::.. = 1 - (ab + de + aed!) + abed

. J'6 (aed) 1 .. -=

Vi 1 - (ab + de + aed!) - abed

38. (i) Vi = aVo + eVi (1)

Vi = b Vi + dVi (2)

From (2) Vi = Vi (1-d)/b

Substituting in (1) Vi/Vo = ab/(1- d - be)

(ii) ~=ab,t::..= I-bc-d,t::..k = 1

. V-:/ _ (ab) 1 •• 2 Vo-

I-be-d

39.

I: ') -~' 3 Y, ' ~ 3

Paths

Y2 r, Y2 Yi

lC>2 4 (X) 2 -01 Y3 y. Y3 Y4

Residual networks

* See F. A. Benson and D. Harrison, Electric Circuit Theory, Arnold, 3rd Edition, 1975, pp. 280-281.

SOLUTIONS 39-40 193

Path values and co/actors are:

11 = Yi Y4 ~1 = Y2 + Y3

P2= Ys ~2= (Yi + Y4XY2 + Y3)

P3= Y2 Y3 ~3 = (Yi + Y4)

Network determinant ~

= YiY4 (Y2 + Y3)+ Ys (l'iY2 + YiY3 + Y4 Y2 + Y4 Y3) + Y2 Y3 (Yi+ Y4)

Ys ~ Y2 ~ Y2 Y3

Individual trees

40. Trees are: 2 • Y1

1; Y2

• 1 3

r, 2 Y2 3 •

7-1

4

Y1 2 --~---. 3

4

Network determinant ~ = Yi Y2 Y3 + Yi Y2 Y4 + Yi Y3 Y4

= Yi (Y2 Y3 + Y2 Y4 + Y3 Y4)

Tree value Yi Ya Y3

194

Yl


'Network for evaluating .0..

2 __ -~-.--~--~3

Path value = Yi Y2

Path cofactor = 1

A = (Y1 + Y3XY2 + Y4) + Y2 Y4

1,4 Yi Y2 :. Voltage ratio = ----.:.-:=----

(Yi + Y3XY2 + Y4) + Y2 Y4

41. Expansion in paths

3

"Or, ...

Residual network 1.

1··-..."Y,:-----<~-....,Y.,.,.2--3

/~ 4

Path 2 Residual network 2

Path values Co/actors

P1 =YiY3 A1 =Y2 +Y4

P2 = YiY2 Y4 A2 = 1

Network determinant A = Y1 Y3 (Y2 + Y4) + Yi Y2 Y4

SOLUTIONS 41-43 195

When transmission path is collapsed we get the following residual network.

42. Use 8 to replace jw.

The network determinant is:

.:l=

= 382

(- 28 X 10-4 - 13~)

(58 X 10-4 + 4 ~810)

4X 108 Dividing this expression by Au = 58 X 10-4 + gives the driving-

38 point impedance Z(8)

3384 X 10-8 + 1382 X lif + 1016

Z(8) = 8(1582 X 10-4 + 4 X 108)

Replacing 8 by jw and multiplying top and bottom by 108

1 {33W4 - 13w2 X 1012 + 1024 }

Z(w) =;; -15w2 X 104 + 4 X 1016

An alternative form is:

43.

Z(w) = jw 2.2 X 10-4 (w2 - 10.48 X 10 l OXw2 - 28.9 X lot~ w 2 (w2 - 26.67 X 101~


Substituting given values:

2S2 + 1 (s - j 1/y(2)(s + j 1/y(2) Z(s) = S(2S2 + 3) = s [s - jy(3/2)] [s + jy(3/2)]

Zeros occur at SI = j 1/y(2), S3 = - j 1/y(2)

Poles occur at S2 = 0, S4 = jy(3/2), S6 = - jy(3/2)

jW

,f372 s plane

-1/,/2

44.

The plot of poles and zeros is illustrated.

1.3 X 2.7 Z(2)=----

0.8 X 3.2 X 2.0

=0.68

(SL)(~) Input impedance Z(s) = 1 + R

Substituting numerical values

4s2+2s+1 Z(s) = 4s2 + 1

sL+-sC

s2RLC+sL +R

[s +t + jy(3/4)] [s +t - jy(3/4)] =

(s + j/2) (s - j/2)

The poles and zeros are shown in the figure.

SOLUTIONS 44 and 59

jW

0.5

9------ ./3/4

-14 I I I

6-- - --- -,[3/4 -0.5

5 plane

59. The following equation holds for the circuit:

di q L-+Ri+ -= V

dt C

197

where i is the current, V is the applied voltage and q is the charge on the capacitor.

Differentiating d2i Rdi i -+--+-=0. dt2 L dt LC

The solution of this equation depends on the relative magnitudes of the R2 1

constants R, Land C. In this case 4L2 < LC and the solution is:

i=Ae-~cos [j(L~-::2) t+BJ

where A and B are constants. Initially when t = 0, i = 0 and q = O.

A = - V/L J (L~ - ::i) and B = n/2.

V -& J( 1 R2) J(J.. _ R2) e 2L sin LC - 4L2 t amperes.

L LC 4L2

i=

Substituting the given Circuit values in this expression gives:

i = 40e-SOt sin 1000t amperes,


i.e. the current is oscillatory, of gradually decreasing amplitude and of

frequency 1000/21T Hz = 159 Hz.

40~------~--------~--------,

l' , , ,.. ..... 20 I-'-I-\+------+---~ ~ I' " \ ,

-40~------~~------~--------~

The current/time curve is plotted from the above expression. R2 1

When R = 10 n, 4L2 = LC and the solution of the above differential

equation is: Rt

i=(At+B)e-2i

where A and B are constants. When t = 0, i = 0 and q = O.

A = V/L andB = O.

V -~ i = L t e 2L amperes.

Substituting the given circuit values in this expression gives:

i = 40 OOOt e-1OOOt amperes.

The current/time curve is plotted from this expression.

16 ,,----....... / ... ,

-/ ... , I "", I ,

-/ " I "

..... 12 ~ ~ 8

t I ...... ...... ..::::--I -_ a 4

o I -

0.001 0.002 Time (5)

0.003 0.004

SOLUTION 59-60 199

R2 1 When R = 20 n, 4L2 > LC and the solution of the differential equation

is ,i = A 1em.t + B1e"'2t, whereA1 and B1 are constants and m1 and m2 are the R 1

two roots ofm2 +-,-m + -= o. L LC

With the same initial conditions as before

A1 = - B1 = V/L(m1 - m2)

and V

Using the given circuit values, m1 = - 3732 and m2 = - 268.

i = 11.54 (e-268t - e-3732~ amperes.

The current/time curve is plotted from this expression.

10

--8 < ':;'6 c f 4 L-

-/'--", I '-,I ..... -, .......... -

I --_

a 2 I --I --I ----

o 0.002 0.004 0.006 Time (5)

60. At any time t after closing the switch the current in the circuit (i) is given by:

di L - + Ri = 200 sin 628t = V sin wt.

dt

The solution of this equation is:

Rt

i = A e - L + V sin (wt - ())/V(R 2 + w 2L2)

where A is a constant and tan () = wL/R.

With the given initial conditions

A = V sin () /V(R2 + w 2L2). Rt Rt

:. the transient current i1 = A e - L = V sin () e - L /V(R2 + w2L~ and the

steady cyclic current i2 = V sin (wt - ())/V(R2 + w2L~.


Substituting the given circuit values in these expressions:

il = 15.52 e-100t amperes

and i2 = 15.71 sin (628t-81') amperes.

Thus il> ;2 and i can be plotted against t as shown. It is seen that after a

time corresponding to about three complete cycles of the supply voltage

il is small.

20

.u

f 0 L. :::J u

-10

-20

.. /';2 ,-\ ,-\ ' .. /, \ . r--/-\ /-\--, ...... \-11- I \ I , ..........

I \ .... _/- ... \ I ,

i -\-- --i--_' i \ 0.01 iTime(s)\ 0.02 I \ O.

I ,~' \~' , , ,I \ / \-I \ I ,

\ ,:/ '-' I I

03

,-,;= ;1+ /2

20r--/--\---+----------~---------1 I' "\ ," /' /,\ ~ 10'r-/-- \ 1-' /-,-

~ 1 \ 1 \ / \ ~ 0/ \ I \ i \ U ,0.01 .'T' () '., 0.02 ! , 0.03

\ lime s I -101----- ,I \_ 1 \_

-,I ,~

-20----------~----------~---------J

WhenR = 0, 0 = 90°, and with the same initial conditions

i = V(1 - cos wt)/wL,

i.e. i never becomes negative; its minimum value is O. It is also seen that the

voltage and current values are zero at the same instant.

61. The following equation holds for the circuit:

di q L-+Ri+-= Vcos(wt-O)

dt C

SOLUTION 61 201

where i is the current and q is the charge on the capacitor.

d2i R di i Vw dt2 + L dt + LC = Leos (wt - () + n/2).

L R C

C~---O=O J s V cos(wt-8) volts

R2 1 In this case 4L2 < LC and the solution is:

where tan ~ = ( I 2 and A, a: and () are constants determined by the 1 LC-w)

Rw/L

initial conditions that when t = 0, i = 0, q = 0 and the applied voltage is zero.

Using the given circuit values it is found that

i = 41.5 e-SOt cos (lOOOt + 173.5') + 41.3 cos (628t - 6°) amperes

The first term is the transient and it has a frequency of 159 Hz. It is

seen from the graph of this current against time that about 0.05 s after

closing S its amplitude is less than 10 per cent of its maximum amplitude

of 41.3 A.

202

50

~ 20 <{ ..... ..... c: 0 ~ .. ::J u -25

-50


-'\ i-, , \ 1 I \ ~

1\ , \ ; \ ! \ " ,.. ..~

, \ /0.01\ J \O.~2 \ja.O~/ -1>.04 ...... o. , \' V ... Time(s) ~ 1-'.1

05

50~----~------~----~----~----~

" __ /\ i2 J , /\ "', I 3 25 \-_1 \--, \--/1-\-_1 .u \ ','\ \ '\ I c: \ I \ !, ! \ I , ! ~ 0 L. \ ;0.01 \ 10.02 \ 10.03 \ ;0.04 \ 10.05 8 I ~\ I \ I \ I -25~\-i-~\-r-T\-'-~ \-l-~'-I-

..J ~nme(s)\'/~ \,J \J -50L------L----~~~~~----~----~

70~----~----~~----~----~----~

35 1-----1-4-

..... c: f OJ.-+----l-~_l_--~~.__-I-_4_+_____,~_\_--_l__J

.... ::J

U

,35

-50L-----~----~~----~ ____ ~ ____ ~

R2 1 62. In this case 4L2 < LC and the natural frequency of the circuit is

the same as the frequency of the applied voltage. Proceeding in the same way as in the previous solution it is found that

i = - 800 e-25t sin 1000t + 800 sin 1000t amperes

The ilf, iJt and i21t curves are plotted from this expression.

SOLUTIONS 62-63 203

~ 800r---_r----~--~~~_r----~--~----_,

~ 4001---+------1--i 0 .~~~~~~-~~~~~-4--~~_1 t: -400 :r U -800 '--__ ....... __ "--_--1. ____ ....... __ ......;.....:..... __ ---' ___ ......

R2 1 63. In this case 4L2 > LC and the solution of the differential equation

of Problem 61 is now of the form:

i = A e"',t + B e"'2t

+ :W cos (wt-O + ~-~) / j[(RW/L) 2 + (L~ --W2fJ. Substituting the given circuit values in this expression gives-

i = A e-1151t + B e-868t + 11.4 cos (314t - 0 + nl2 - 35.3°)

where A, B and 0 are constants depending on the initial conditions. If when t = 0, i = 0, q = ° and the applied voltage is at its maximum

positive value, i.e. 0 = 0, it is found that

i = - 131.6 e- l151t + 125 e-868t + 11.4 cos (314t + 54.7°) amperes.

The transient current i1 = 131.6 e- l151t + 125 e-868t amperes, the perma

nent current i2 = 11.4 cos (314t + 54.7°) amperes and i can therefore be

plotted against t as illustrated.


~ 10-"/\-. --1-----1 ~ I \ I, , \ ~ 5 , --,,-+------1 ~ I ,_ 5 0, -u 0.Op5 0i{)1

I Time (5) I -5L----~------~

,-..... 10~---+------+------'." --, -

/ " 5 , //----<>------1

~ \i2 / ~ 0 e---\ 0.005 0.01 1 0.015 ::J ,

U_ 5 "I 1// Time (5)

, /.1 -10~---' /..-+-------<>---------1

15

.... 5 c: f 3 0 U

-5

-10

1'--....

T\ 1-\ .... ~,-1 \ i // '. -~ / \ / \ /

\ 0.005 0.01/ 0.015 \ I 1/ Time (5)

'\ 1 ~--/.. ..... _"

64. The solution to this problem has been given elsewhere. *

65. The solution to this problem has been given elsewhere.t

66. The solution to this problem has been given elsewhere.:j:

67. The solution to this problem has been given elsewhere.§

* F. A. Benson and D. Harrison, Electric Circuit Theory, Arnold, 3rd Edition, 1975, pp.260-1. t Ibid., pp. 249-51. t Ibid., pp. 263-4. § Ibid., pp. 268.

SOLUTION 68 205

68. Suppose y(t) is a known function of t for values of t > O. Then the Laplace transform ji (s) ofy(t) is defmed as

ji(s) = [DO e-sty(t) dt (1) ·0

where s is a number sufficiently large to make the integral convergent. If a is any number, real or complex, then ji (s + a) is the Laplace trans

form of e-aty(t). Using (1) it is found that if y(t) = t"-l/(n - I)!, thenji(s) = l/sn. Thus, the transform of e-at,,-l/(n - I)! is l/(s + a)n. Similarly, using

(1) if y(t) = sin at, thenji(s) = a/(s2 + a2). Thus, the transform of e-bt sin at = a/{(s + b)2 + a2}.

I ...

R L

For the circuit illustrated,

l=dQ/dt

Also, L dl/dt + RI + Q/C= V

Itvl/!

The problem is to solve (2) and (3) with given initial values,

(2)

(3)

1=/0 , Q = Qo att = 0 (4)

Forming the subsidiary equations for (2), (3) and (4) in the usual way:

(Ls + R)I + Q/C= Llo + V (5)

Q=I/s + Qo/s (6)

From (5) and (6)

(Ls + R + l/Cs)1 = V + LIo - Qo/Cs (7)

If a constant voltage E is applied at t = 0 and 10 = Qo = 0, the subsidiary equation (7) becomes:

Thus,

(Ls + R + l/Cs)j=E/s

1 = E/L{(s + ay + ~2} (8)

(9)


The solution of (9) is I = Ete-<Y.t/L, when R = 2Y(L/C). Similarly, the

transient response of the circuit can be investigated using (7) no matter

what the form of the applied voltage.

69. (a) The subsidiary equations are:

Solving:

_ S4Xo + S2(3xO + 1) + 2 1 s(2xo - 1) s(2xo - 1) x= =-+ +~-==-~

S(S2 + 1)(s2 + 4) 2s 3(S2 + 1) 6(S2 + 4)

_ 2xo - 1 (2xo - 1) (1 1) Y = (S2 + 1)(S2 + 4) = 3 S2 + 1 - S2 + 4

x = t + !(2xo - 1) cos t + !(2xo - 1) cos 2t

and Y = !(2xo - 1) sin t - !(2xo - 1) sin 2t

(b) The subsidiary equations are:

Llsll + MSl2 = ii

MSII + (L2S + R2 + ~S) 12 = 0

-Msii i2 = ----::--::------

(LIL2 - ~)s2 + R 2Lls + L';C2

e.g. if v = a constant V, ii = Vis.

Then,

where

70. (a) Logarithmic decrement 0 = rrR/wL = R/2jL.

The frequency f corresponding to a wavelength of 300 m is 1 MHz.

SOLUTIONS 70-71, 77

/j = 10/(2 X 106 X 150 X 10-6) = 0.033.

(b) Let the number of oscillations be N.

207

Then the amplitude of the Nth oscillation, IN = 11 e-fjV-1)6, where /j is the logarithmic decrement.

100 = efjV-1)O.1 andN= 47.

71. The natural frequency of free oscillations is

1 J( 1 R2) - --- =478 kHz. 21T LC 4L2

To make the discharge non-oscillatory R must be at least equal to 2 J~, i.e. 12.1 .n.

77. For any waveform which is cyclic, repeating itself at intervals of21T,

f(O) = A + a1 sin 0 + ... + On sin nO + .. . + b1 cos 0 + ... + bn cos nO + ...

where 1 f211'

A = 2- f(O) dO 1T 0

1 f211' On = - f(O) sin nO dO

1T 0

and 1 f211'

bn = - f(O) cos nO dO. 1T 0

In the case of a half-wave rectifier f(O) = E sin 0 from 0 to 1T andf(O) = 0 from 1T to 21T.

Esin e cu /\/ ~,' \ ~ I \ > , \ ,

/-, I ,

I \ , \ , \ I , ,

21T Angle

1 J,1I' A = - E sin 0 dO = E/1T. 21T 0


1 f1r lln = - E sin 0 sin nO dO which is zero except for n = 1 in which case 1T 0

al =E/2. 1 1r

bn = - f. E sin 0 cos nO dO. 1T 0

When n is odd, bn = 0; when n is even bn = - 2E/1T(n2 - 1). The Fourier expansion is therefore

[ 1 1 2" cos nO J E -+-sinO-- L. 2 • 1T 2 1T n=2,4,6 (n - 1)

The full-wave circuit consists essentially of two half-wave circuits, one circuit operates during one half-cycle and the second operates during the next half-cycle. Also analysing each half-wave separately using the above result it is seen that for the full-wave circuit the component fundamentals cancel out, the negative even cosine harmonics are coincident, and are therefore present with twice the amplitude, and it is evident that the value of the constant term is twice the value for the previous case.

The Fourier expansion for the full-wave case is therefore

E [~_i L c~snol. 1T 1T n=2,4,6 . •• (n - 1)J

The r.m.s. value = E.../[4/1T2 + 16/21T2{(1/3)2 + (1/15)2 + ... )]

=E/.../2.

78. The method of solution is the same as that already given for Problem 77 and is therefore not given again here.

79. The method of solution for the first part of the problem is the same as that already given for Problem 77 and is therefore not given again here.

The solution to the second part of the problem can be found elsewhere. *


* W. H. Middendorf, Analysis of Electric Circuits, Wiley, 1956, Chapter 17.

SOLUTIONS 81-84


82. The solution to this problem has been given elsewhere.*

83. The formulat for t(t) is:

t(t) = f~oo g(w)~ dw

The relationship between the transform pairs and the Fourier series is discussed in many textbooks.:!:

i.e.

I f+oo . g(w) =- t(t)e-jWt dt

21T -00

= - e-(O<+.JW)t dt I i 00 •

21T 0

I r e-(o<+jw)t 1 00

= 21T t-(a+ jw)J 0

g(w) = 1/21T(a + jw)

209

For amplifier of complex gain G(w) spectrum is G(w)g(w) and output is:

or

F(t) = !!o f~: G(w)g(w)ejwt dw

I roo G(w) . F(t) = .2 - ejWt dw

0<-+021T • -00 (a + jw)

84. The Fourier transform of t(t) is

g(w) = f:oo t(t) e-jwt dt

* F. A. Benson and D. Harrison, Electric Circuit Theory, Arnold, 3rd Edition, 1975, pp.202-204 t See Footnote after the Question. :j: See, for example, R. E. Scott, Linear Circuits (complete), Addison Wesley, 1960,

Chapter 20.

210

For (a):

For (b):


g(w) = f~T(1+t/1)e-jwtdt

+ f: (1 - t/1)e-jwt dt

= T{Si:~~/2r

eoh2n2 91. 'n=--2

mnZe

8.855 X 10-12 X (6.624 X 1O-34i X 1010n2 .

= 1r X 9.107 X 10-31 X (1.602 X 10 1~2Z angstrom umts

= 0.529 n2/Z angstrom units

me4 Z2

Wn = - 8eo2h2n2 J

9.107 X 10-31 X (1.60 X 1O-1~4 Z2 J

8 X (8.855 X 1O-12i X (6.62 X 1O-34i n2

21.8 X 10-19 Z2 J

n2

21.8 X 10-19 Z2 eV n2 X 1.602 X 10-19

= - 13.6z2/n2 eV

92. (a) The permissible energy levels Wn may be expressed as*:

Wn = - (13.6z2/n2) eV (see Question 91)

where Z is the atomic number of the atom. For a hydrogen atom Z = 1. The lowest energy state (n = 1) for hydrogen is therefore - 13.6 eV.

* See, for example, D. J. Harris and P. N. Robson, The Physical Basis of Electronics, Pergamon, 1974.

SOLUTIONS 92-94

When n = 2, W2 = - (13.6/4) eV = - 3.4 eV

When n = 3, W3 = - (13.6/9) eV = - 1.51 eV

When n = 4, W4 = - (13.6/16) eV = - 0.85 eV

(b) Energy released = (13.6 - 3.4) eV = 10.2 eV

Frequency of radiation if) is given by:

93.

hf= (10.2 X 1.602 X 1O-1~ J

f= 10.2 X 1.602 X 10-19 Hz 6.624 X 10-34

= 2.465 X lOIS Hz

(see Problem 91)

For Z = 1, n = 1, € = 11.7

13.6 Wn = -( )2 = 0.099 eV

11.7

(see Problem 91)

11.7 X 8.855 X 10-12 X (6.62 X 10-34)2

'IT X 9.107 X 10 31 X (1.602 X 10 1~2 m

= 6.16 A

i.e. the smallest orbit passes outside the nearest neighbour atoms so is essentially in the bulk crystal.

94. Bohr postulate:

h For n = 1 mvr =-

, 2'IT

nh Angular momentum * = -

2'IT

0.529n2 (see Problem 91) r = -Z-- angstrom units

211

* See D. J. Harris and P. N. Robson, The Physical Basis of Electronics, Pergamon, 2nd Edition, 1974, p. 25.


= 5.29 X 10-11 m

6.624 X 10-34 _ v= ms 1

21T X 5.29 X 10-11 X 9.107 X 10-31

= 2.19 X 106 ms-1

v 2.19 X 106 Ratio-= = 7.3 X 10-3

C 3 X 108

95. The conductivity U of a semiconductor is given by*

u = e(p/lh + nile)

where n = electron density

p = hole density

Ile = electron mobility

Ilh = hole mobility

For an intrinsic semiconductor p = n = rlj, where nj is the density of holes and electrons. Therefore the intrinsic conductivity Uj is given by:

Uj = nje(lle + Ilh)

n. = 1 m-3 I 0.47 X 1.602 X 10-19 (0.36 + 0.17)

= 2.5 X 1019 m-3

96. Density of donor impurities = (4.4 X 1028) X 1O-6 m-3

= 4.4 X 1022 m-3

The intrinsic density = 2.5 X 1019 m-3

(2.5 X 101~2 hole density = ----- = 1.42 X 1016 m-3

4.4 X 1022

Electron density = 4.4 X 1022 m-3

1 Resistivity = 4.4 X 1022 X 1.602 X 10-19 X 0.36

=4 X 10-4 Urn

* See, for example, D. J. Harris and P. N. Robson, The Physical Basis of Electronics, Pergamon, 1974, Section 3.4.1.

SOLUTIONS 97-98 213

97. The Einstein relation between mobility Il. and diffusion constant D is D = kTIl/e where k is Boltzmann's constant and T is the absolute temperature,

1.38 X 10-23 X 300 X 0.36 __ For electrons, Dn = 1.602 X 10-19 = 9.3 X 10 3m2S I

For holes, D = 1.38 X 10-23 X 300 X 0.17 = 4.4 X 1O-3m2s-1 p 1.602 X 10-19

98. The conductivity is:

a = nelle + pellh

and np=n/

a n/ -=nll +-Ilh e e n

This is a minimum when d(a/e) = 0 dn

i.e. where

jllh n=nj -Ile

or

d2 (aje) , 'ti' th t' , t' " ~ IS POSI ve so e urnmg porn IS a mlmmum.

n/ jlle p = -;; = nj P;;

Under intrinsic conditions:

(J = nje(}.l.e + Ilh)

= 2.5 X 1019 X 1.6 X 10-19 (0.57) Sm-I

= 2.28 Sm-1

Under the minimum conductivity condition:

a = njeY(llhJJ.e) + njeY(lleJJ.h)

= 2njeY(llhlle)

= 5 X 1019 X 1.6 X 1O-19Y(0.38 X 0,19) Sm-I

= 2.12 Sm-I


The conductivity is equal to the intrinsic one when:

n·2

neJ.l.e + _I eJ.l.h = ni e (J.l.e + J.l.h) n

or

i.e.

or

But

Also

n2J.1.e - nni (J.l.e + J.l.h) + ni2 J.l.h = 0

ni (J.l.e + J.l.h) ± y'[n/ (J.l.e + J.l.h)2 - 4J.1.e n/ J.l.h] n=

2J.1.e

n = ..!!L (0.57 ± y'[(0.57? - 4 (0.38) (0.19)] 0.76

n· n =.2 (3 ± 1)

4

n =1= ni' so n = n;/2 = 1.25 X 1019 m-3

p = n/ln = 5 X 1019 m-3

99. For the p-type material op = ppeJ.l.h where op is the conductivity of the p material and J.l.h the mobility of holes.

104

so Pp = 0.17 X 1.602 X 10-19 3.68 X 1023 m-3

For the n-type material

_ 100 _ 21 -3

nn - 0.36 X 1.602 X 10-19 - 1.75 X 10 m

Now Pn = n/lnn = (2.5 X 101~211.75 X 1021

=3.57X 1017 m-3

Contact potential

V; = kT In (!!E.) / e Pn

= 1.38 X 10-23 X 300 n (3.68 X 1023 )

1.602 X 10-19 1 3.57 X 1017

= 0.35 V

100. Total saturation current density is given by*

* See D. I. Harris and P. N. Robson, The Physical Basis of Electronics, Pergamon 1974, Appendix 3B.

SOLUTIONS 100-102 215

J. = DpePn + DneTZp s Lp 4z

where Dp and Dn are the diffusion coefficients and Lp and Ln the diffusion lengths.

From the previous solution

Pn = 3.57 X 1017 m-3

and n = n.21p, = 1 7 X lOIS m-3 P I I, P •

Also kT kT

- Dp = -; Ilh and Dn = -; Ile

so

J. = (1.38 X 10-23) 300 {3.57 X 1017 X 0.17 + 1.7 X lOIS X 0.36} sIX 10-3

= 0.25 Am-2

Hole saturation current = IlhPn since L = L Electron saturation current Ilenp p n

= 100

101. The solution to this problem can be found elsewhere.*

102. J = Js (exp (~~) - I} exp (e V) _ 1 = lOS = 4 X lOS

kT 0.25

eV = 12.9 kT

V= 12.9 X 1.38 X 10-23 X 300 1.602 X 10-19

= 0.33 V

* See, for example, M. V. Joyce and K. K. Clarke, Transistor Circuit Analysis, Addison-Wesley, 1961, p. 9, or D. J. Harris and P. N. Robson, Vacuum and Solid State Electronics, Pergamon, 1963, pp. 107 and 243.


103. Mean free path

where r is the radius of the spherical scattering centre and N is the density.

In this case A = (1/1020 X 7r X 25 X 104 X 10-2'1 m

= 1.27 X 10-6 m

Mean time between collisions is T = Ale where e is the mean electron speed.

Also ! m* «(3)2 = ~ kT

Mobility

( m*)t T= A 3kT

{ 1.57 X 9.107 X 10-31 }

= 1.27 X 10-6 3 X 1.38 X 10-23 X 300

= 1.36 X 10-11 s

eT 1.602 X 10-19 X 1.36 X 10-11

J1=- = m* 1.57 X 9.107 X 10-31

104. The conductivity a = e(nJ1e + PJ1h)

For intrinsic material n = P = nj

aj = nje(J1e + J1h)

or

At 310 K: 3.56 _ 4 X 1018 -3

nj = 1.602 X 10-19 (0.36 + 0.17) - 2 m

At 273 K: 0.42 18 3 n· = = 4.7 X 10 m-

I 1.602 X 10-19 (0.36 + 0.17)

In doped material (n + p)p = n? where n is the ionized donor density = 1021 m-3

n»nj»p

SOLUTIONS 104-106

At 310 K: n/ (42 X 1018)2

P = -;;= 1021

a = 1.602 X 10-19 (1021 X 0.36 + 176 X 1016 X 0.17)

= 57.6 n-1 m-1

At 273 K: n/ (4.7 X 1018)2

P = -;;= 1021

a = 1.602 X 10-19 (1021 X 0.36 + 2.2 X 1016 X 0.17)

= 57.6 n-1 m-1

lOS. The conductivity a of an intrinsic material is given by

= nje(lle + Iln)

217

where nj is the intrinsic density and 11 the mobility. If the mobilities remain constant, nj is the only term which varies with temperature, T, as follows.

where N is some constant. Hence the resistivity and thus the resistance R varies as nj-l or

R = Ce-Egf2kT

where C is another constant. At 290 K,R = 500 n so

1.08 X 1.06 X 10-19

500 = Ce 2 x 1.38 X 10-23 X 290 = Ce21.6

Similarly at 325 K

Dividing:

which gives

R2 = Ce19•3

500 = e(21.6-19.3)

R2

R 2 = 50 n

106. The derivation of the formula can be found elsewhere. * The formula is:

* See, for example, W. Shockley, Electrons and Holes in Semiconductors, D. van Nostrand, 1956, p. 217.


Hall coefficient R = (Pllp 2 - nlln 2)/e(Pllp + nllni

(a) For intrinsic material, p = n = nj

R = nj(llp2 -lln2) en/(Ilp + Iln)2

i.e. R = (Ilp -lln)/enj(llp + Iln)

(b) For th~ highly-droped n-type material, n > p

1 (-nil 2) R=- __ n_ =-I/ne e n21ln2 ---

107. The conductivity of the semiconductor is

I u=-=peIlH

RA

where 1 is its length, A its area of cross section, R is its resistance and p the hole concentration, assuming it to be p-type.

Also the Hall coefficient is given by

1 VH.d RH =-=-

pe IB

where VH is the Hall voltage and d the sample thickness

1 5 X 10-3 X 1 X 10-3

p--'X'I-.6-X-IO--=19 = 10 X 10-3 X 0.5

which gives p = 6.25 X 1021 •

Substituting this value in the fIrst equation gives:

I 30 X 10-3

IlH = RApe = 500 X 6 X 10-6 X 6.25 X 1021 X 1.6 X 10-19

= 10-2 m2 V-I S-1

Notice that for constant dimensions, field and current the Hall voltage is proportional to the reciprocal of the carrier density. Hence the Hall voltage with a copper sample, VHc is:

_ (~) _ -3 6.25 X 1021 _

VHc - VH(I)-5X 10 X 8.5X 1028 V-0.37nV -p

SOLUTIONS 108-109

108. In equilibrium, charge neutrality exists and

n+Na=p+Nd

whereNa andNd are the acceptor and donor concentrations and the impurities are all assumed to be ionized.

(a) The net impurity concentration = Na - Nd = P - n

219

= 1020 - 2 X 1019 = 8 X 1019 m-3

(b) (Na - Nd ) is positive so acceptors are in a majority and the compen

sated material is p-type.

or

(c) The intrinsic density, nj, is obtained from:

nl' = rlp = 1020 X 2 X 1019

109. The conductivity a of the intrinsic material is

1 a = - = enj(}l.h + l1e)

p

where the intrinsic density, nt, varies with temperature as exp (- Eg /2k1). Hence

or

or

1 E 3 -0: exp (_....:::L) Tz P 2kT

~ 3 P = Ce 2hT T· where C is a constant

In (pr-t) = In C + ~ 2kT

3 Hence, if the assumptions are correct a graph of In (pT-.) versus 1 /T should yield a straight line of slope Eg /2k. The data for the graph is given in the question and when drawn the graph is a straight line of slope 4670.

The gap energy is then obtained from:

~=4670 2k

or 1.38 X 10-23

Eg= 2 X 4670 X 1.6 X 10-19 = 0.8 eV


110. The incremental junction capacitance of an abrupt junction, q, is given by*:

where Jj is the junction voltage and Na and Nd are the doping levels at either side of the junction. Hence for a particular diode,

k Cj = (V+ Vo)t

where k is a constant, V is the reverse bias voltage and Vo is the contact potential.

When V= 2 V, Cj = 200 J1.J1.F, so

k = 200 X 10-12 X (2 + 0.85)t= 3.38 X 10-10

Then, when Cj = 100 J1.J1.F, the reverse bias voltage necessary is found from

3.38 X 10-10 100 X 10-12

(V+ 0.85)t

or V= 10.6 V

111. The conductivity, a, of the channel region is given by

a= J1.eeNd

where Nd is the doping concentration and J1.e is the electronic mobility. Hence

N. -~- 20.9 _ 21 -3 d - J1.e e - 0.13 X 1.6 X 10-19 - 10 m

The pinch-off voltage, lj" is given byt

eNd a2 lj,=--

2e,to

where a is the half width of the channel when the gate voltage, Vg, is zero, so

V. _ 1.6 X 10-19 X 1021 X (2.5 X 10-6)2 _ P - 12 X 8.85 X 10-12 - 4.71 V

The drain-source resistance, R, can be deduced fromt

* See J. Allison, Electronic Engineering Materials and Devices, McGraw-Hill, 1971, p.205. t See J. Allison, Electronic Engineering Materials and Devices, McGraw-Hill, 1971,

p.269.


~ R= I 1- (Vg/Yp)2"

where ~ is the resistance when Vg = o. When R = 250 n, Vg is given by:

250 = 50 I

1 - (Vg/4.71)2

which leads to Vg = 3.01 V (negative with respect to the source).

112. The ratio of hole to electron current at a p-n junction is approximately equal to the ratio of the conductivities of the p- and n-type materials which constitute the junction for all bias conditions. Hence

Jh =!l? = eP.nNa Je an eP.eNd

where Na is the acceptor concentration and Nd the donor concentration. So, in this case

I n 1022 X 0.2 -=

1024 X 0.4 1:200

113. The rectifier equation which both diodes obey is

I = Is [exp (~V/k1) - 1]

where Is is the saturation current. So for the germanium diode:

or

{ 1.6 X 10-19 V } 100 X 10-3 = 10-6 exp 1.38 X 10-23 X 293

V=288mV

A similar calculation for the silicon diode using Is = 10-8 A gives V = 407mV. -

114. The diode current is given by the rectifier equation

1= 10 {exp (:~)-1) where 10 is the saturation current and V the bias voltage. Hence, since 0.1 V is dropped across the diode:

I ( 1.6 X 10-19 X 0.1 ) 1=3 X 10-6 \exp 1.38 X 10-23 X 293 - 1 = 154 p.A


the series resistance, R, is then

V 0.1 R=-= =65012

I 154 X 10-6 --

115. It can be shown* that the transconductance,gm' can be estimated from:

- J.LeCg V, gm- (2 d

where Cg is the gate capacitance, 1 its length, J.Le the electron mobility in the channel and V d the drain-source voltage at saturation. Now

= €A = 3.7 X 8.854 X 10-12 (0.84 X 10-3 X 5 X 10-6) = 0 917 F Cg d 150XI0-9 . J.LJ.L

So = 0.02 X 0.917 X 10-12 X 10 = 34 X 10-3 S &n (5 X 10-6)2 _7. ___ _

116. The concentration of acceptors at a depth x and time t,N(x,t) for this limited source diffusion is given byt

N(x,t) = Y(~Bt) exp {~:: }

where Q is initial surface concentration of boron atoms. At the junction (x = 1 J.Lm), N(x,t) = background concentration = 1022•

Hence

Q (1O-12 } 1022 = y(n 10-16 3600) exp 4 X 10-16 X 3600

144. Let the voltage across the diode be V and the current through the diode ImA,

V = 200 - 20 (:0 + I) ,i.e. V = 150 - lSI

which is the equation of the load line.

* See J. Allison, Electronic Engineering Materials and Devices, McGraw-Hill, 1971, p.278. t J. Allison, Electronic Integrated Circuits, their Technology and Design, McGraw

Hill, 1975, p. 44.

SOLUTIONS 144-145

7----~--~--~----~--~--~--~

61----I--__+_

51---+--~---'-~-~--~

,...... 41---+----I---+-« E '-i 3

2

o 50 v (V)

100 150

223

At the point of intersection of the I/V curve and the load line 1= 3 rnA.

145. The static characteristic has been plotted from the given figures.

V; = Va + R,la (1)

where R; = 2500 U.

If (1) is plotted on the same sheet as the static curve a straight line (the load line) results.

A typical load line (for V; = 40 V) is shown. The point of intersection of the load line with the static curve, P, indicates the current flowing. Thus the dynamic curve can be plotted because the current is that corresponding to P when the input voltage is 40 V and so the first point on the dynamic curve isP'.

Iv; o-----------------~

Hence, when the supply voltage is 50 V the load current is 14.5 rnA. The

voltage across the load = 2500 X (14.5 X 10-3) = 36.25 V.


From the static curve and the 50-V load line the voltage across the diode is 13.75 V. The voltage across the load is therefore (50 - 13.75) V = 36.25 V which agrees with the result already obtained.

....... « E ~

...,

60

~4O L L :::I o

o

/ '" Static

/' characteristic , /

/ /

/ ~

Load line for Vi / / Dynamic characteristic

P/~~ _________ P'~-~1-~ ~ ----~

" -- I .,.-fIII"''' , .... __ ...... -- I ---10 20 30 40 50

Anode voltage (V)

146. The control characteristics are as shown. From these, when Va = 400 V, the change in critical grid voltage required is seen to be about 2.8 V.

~--,:----. 600

500

...... 400 ~

., 300 g' ...,

g 2eO .:l

o I: «

1---i-ir-lk-\--i 'OO

2.8V

-10 -5 0 Grid voltage (V)

When the temperature is 40°C and Vg = - 4 V, Va = 320 V. But Va = 350 sin 8, therefore 8 = sin-1 (320/350) = 66° 5'.

SOLUTIONS 146-149

When the temperature is 70°C and Vg = -4 V, Va = 150 V and

() = sin-1 (150/350) = 25° 22'.

225

147. The effective mutual conductance is the sum of the individual mutual conductances since the anode currents add directly, i.e. (2 + 5 + 3) mS = 10 mS.

The equivalent anode resistance is obtained by adding the individual anode resistances as one adds resistances in parallel, i.e.

1/(1/5000 + 1/4000 + 1/10000) Q = 1818 Q.

The equivalent amplification factor

= 1818 X 10 X 10-3 = 18.18.

148. Thermionic emission of electrons is in accordance with the expression:

1= AT2 e-b/T Am-2 ,

where Tis the absolute temperature. For the thoriated-tungsten filament:

85 X 10-3 = 3 X 104 X (1900)2 e-30 SOO/I900 X area (1)

For the pure-tungsten filament:

i X 10-3 = 602 X 103 X (2500)2 e-S2400/2S00 X area (2)

Dividing (2) by (1):

i 60.2 X (2500)2 e-S2400/2S00 . 85 = 3 X (1900)2 e-30S00/1900 and I = 21.8 rnA.

149. The Child-Langmuir equation for plane-parallel electrodes gives* J = 2.34 X 10-6 Va 3/2/d2 , where J is the current density (Am-2), Va is the anode voltage in volts and d is the anode-cathode distance.

Here Va = 200 V, d = 2 X 10-3 m so J = 1.65 X 103 Am-2.

* This equation is proved in many textbooks. For example, see S. Seely, Electron· tube Circuits, McGraw-Hill, 2nd Edition, 1958, p. 12, or P. Parker, Electronics, Arnold, 1950, pp. 93-6; or K. R. Spangenberg, Vacuum Tubes, McGraw-Hill, 1948, pp.170-1.


150. The current is given by the following expression: * 1= 1.47 X 10-5 Va 3/2 l/ra(32 amperes

where Va is the anode voltage, 1 is the active length of the valve, ra is the anode radius and (32 is a quantity that is determined from the ratio of anode radius to cathode radius (r,).t

For the first valve ra = 2 mm, 1 = 20 mm, r, = 0.05 mm and Va = 25 V. Thus ra/r, = 40 and (32 = 1.0946,

1= 17mA.

For the second valve r,= 0.75 mm, ra/rt= 2.67 and (32 ~ 0.45,

1=41 rnA.

151. The solution to this problem can be found in many textbooks.:/:

152. The current I under the condition of an accelerating field of E

(Vm-1) at the cathode surface is§ II e+O·44E1/2/T, where II is the zero-field thermionic current and T is the absolute temperature of the cathode.

10g10 (1//1) = 0.4343 X 0.44 X (106)1/2/2600 = 0.07345.

thus 1/11 = 1.184, which shows that the Schottky theory predicts an increase of 18.4 per cent over the zero-field emission current.

153. (a) The amplification factor J1 = - 2rraJag In (2 sin rrrw/ag) where a2 is the grid-anode spacing, ag is the grid-wire spacing and rw is the grid-wire radius.1I

* For example, see S. Seely, Electron-tube Circuits, McGraw-Hill, 2nd Edition, 1958, p. 12, or P. Parker, Electronics, Arnold, 1950, pp. 98-9, or K. R. Spangenberg, Vacuum Tubes, McGraw-Hill, 1948, p. 176.

t where c< = lo'6e (ralr,).

Values of {32 corresponding to various values of the ratio (ralr,) have been plotted in Parker's book, Fig. 82, and tabulated in Appendix II of that book.

:j: E.g. see K. R. Spangenberg, Vacuum Tubes, McGraw-Hill, 1948, pp. 181-2; or P. Parker, Electronics, Arnold, 1950, pp. 99-100.

§ For the proof of this expression see J. Millman and S. Seely, Electronics, McGrawHill, 1951, Section 5-19, pp. 151-6.

II For the proof of this see, for example, K. R. Spangenberg, Vacuum Tubes, McGraw-Hill, 1948, pp. 125-8.


Since ag is large compared with rw, p, ~ 2rraJag In (ag/2rrr w). Now il2 = 1.9 X 10-3 m, ag = 1.27 X 10-3 m and rw = 6.4 X 10-5 m.

p,~8.

(b) The amplification factor p, ~ 2rrrg In (ra/rg)/ag In (ag/2rrr w) where ra is the anode radius, rg is the radius of the grid-wire circle, rw is the radius of the grid wire and ag is the linear distance between the grid-wire centres at radius rg .*

If N = l/ag, p, ~ 2rrNrg In (ra/rg)/ln (1/2rrNrw). Here, p, = 20, ra = 1.05 X 10-2 m, rg = 5 X 10-3 m and rw = 4 X 10-4

m,soN~3.

Total number of grid wires = 2rrNrg ~ 10.

(c) The expressions are:

(i) For plane-electrode triode,

p, = {2rril2/ag -In cosh (2rrrw/ag)}/{ln (coth 2rrrw/ag)}

where il2 is the grid-anode distance, ag is the grid-wire spacing and r w is the grid-wire radius.

(ii) For cylindrical triode,

p, = {2rrNrgln (ra/rg) -In (cosh 2rrNrw)}/{In coth (2rrNrw))

where the symbols having the same meaning as in the solution to the previous problem.

The derivations of the expressions can be found elsewhere.t

154. The load line passes through the points A (0, 4 rnA) and B (6 V, 0). The quiescent working point is at Q. When an input signal fo 40 p,A peak current is applied, the peak-to-peak input signal will be 80 p,A; the base current will vary between 0 and 80 p,A. The extremes of the working range are given by points X and Y.

Peak-to-peak collector-emitter voltage excursion is X' y' ~ 4.7 V.

At Q, collector current ~ 2 rnA. Power supplied by battery = (2 X 6) mW = 12 mW.

* See, for example, K. R. Spangenberg, Vacuum Tubes, McGraw-Hill, 1948, pp. 135 -7. t See F. B. Vodges and F. R. Elder, 'Formulas for the Amplification Constant for

Three-element Tubes,' Phys. Rev., 24, pp. 683-9,1924. W. G. Dow, Fundamentals of Engineering Electronics, Wiley, 2nd Edition, 1952, Chapter 4. K. R. Spangenberg, Vacuum Tubes, McGraw-Hill, 1948, pp. 142-52.


(6/1500) A = 4rnA

X/1 2 3 4 5 6 7 Collector voltage (V)

Power dissipated as heat in 1500-il load = (22 X 10-6 X 1500) W = 6mW.

power dissipated in the transistor itself = (12 - 6) mW = 6 mW.

ISS. Consider first the Ie/Veb characteristics. With the collector-base voltage constant at - 4 V a change in Ie from

1 rnA to 5 rnA gives a change in collector current from - 1.03 rnA to -4.95 rnA.

Thus, a = - {- (4.95 - 1.03)} = 0.98 (5-1) -

Consider now the Ie/Vee curves and a constant value of collector-emitter voltage of - 4 V. A change of Ib from - 20 IlA to - 80 IlA gives a change of collector current from - 1.1 rnA to - 4.5 rnA.

Thus,

But

so

i.e.

- (4.5 - 1.1).10-3

{j = - (80 - 20) . 10-6 ~ 57

{j = (f>ic/f>ib)

f>ib = - (f>ie + f>ic)

(j = - f>ic/(f>ie + f>ic)

= - (f>ic/f>ie)/(f>ie/f>ie + f>ic/f>ie)

(j = a/(l-a)

From this expression for {j it is seen that:

a = (j/(1 + a)


This equation can also be obtained directly from the definition of Q,

substituting - (8ib + 8ie) for 8ie and dividing each term in the numerator and denominator by 8ib•

156. Current gain {3 = OIe/8h When Vc = - 5 VandIb = -70 IlA,Ie = 2.46 rnA

When Vc = - 5 V andIb = - 50 IlA,Ie = 1.72 rnA

. (2.46 - 1.72) 10-3

gam = (70 _ 50)10--6 = 37

Load line

-3 -6 -9 Collector voltage Vc (V)

The load line is as shown. It passes through the points Vc = - 9, Ie = 0 and 9 X 103

Vc = O,le = 1800 rnA (i.e. Ie = 5 rnA)

For Vc = - 4 V, operating point is Q where Ib ~ - 82 IlA.

157. The equivalent circuit is as shown.

A

looon

c 3000n

f'\,I 2

loooon

(-j 15920)n


Reactance of capacitance

1 106

= 2rr[C = 2rr X 2000 X 0.005 = 15920 il.

Let currents II and Iz in milliamps circulate as shown.

For the Iz mesh:

(10 + 1 + 3 - jI5.92)lz - 3/1 = 0

For the II mesh:

(8+ I +3)/1 -3Iz+20Vg=0.

Also, Vg = 1 + Iz

From (1), (2) and (3)/z = (- 0.1556 - jO.1357) rnA. The capacitor blocks the d.c. and the meter reads the product of its

resistance and the a.c. through it, i.e. Iz.

.. meter reads 10 [0.1556z + 0.1357zF/z = 2.06 V.

158. The equivalent circuit is as shown. Let the currents 110 Iz and 13 circulate as shown.

If

For the 1\ mesh:

el::::; 1 + jO

ez = 2 (cos 30° + j sin 30)

::::; 1.73 + j1.

(ral + RL + r ~ - jXc)/l

(1)

(2)

(3)

- (RL + ra)lz - (- jXc)/3+ III V-g -Ilz V-g = 0 (I) I I 2


For the 11, mesh:

(RL + ra2 + R1,)I1, - (RL + ra)/1 - R1,13 + 111, Vg: = 0 (2)

For the 13 mesh:

Also,

and

(R l - jXc + R1,)/3 - (- jXc)/1 - R1,I1, = 0

Vg. = el + Rl/3 = 1 + Rl/3

Vg: = ~ + R1,(I1, - 13) = 1.73 + jl + R1,(I1, - 13)

:. II> 11, and 13 can be found.

159. For a triode, a change {j/a in the anode current la can be written

( ~) (~) 1 - oVa + - oVg = -oVa + gm oVg. oVa Vgconst. oVg Vaconst. ra

Current generator

9mo~

Anode

RL (load resistor)

(3)

(4)

(5)

The current-source equivalent circuit shown follows from this expression.

160. The equivalent circuit of the arrangement is as shown.

r---------1----------.0'

500011 10000.n. 2000011

-:t 2~ rv ~ 80V 3 80V ~

~--------~--------~O


Millman's Theorem* states that

, VOl Yl + VOl Y2 + V 03 Y3 Voo = ~~---=~---=~ Yl +Y2 +Y3

where V 00' is the voltage drop from 0 to 0'

" " " 0 to 1, etc.

In this case VOl = 80, VOl = 80, V03 = 0, Yi = 1/5000, Y2 = 1/10000, Y3 = 1/20000.

Voo' = 68.6 V.

Let the currents in the two meshes be x and y rnA.

For the x mesh: 5x + lO(x - y) - 80 + 80 = O.

" "y 10CY - x) + 20y + 80 = O.

From these equations x = - 2.3 mA,y = - 3.4 rnA.

The valve currents are - x = 2.3 rnA and x - y = 1.1 rnA.

161. Consider the common-base transistor connection. The emitter and collector voltages, Ve and Vc, measured with respect to the base, are functions of the independent variables Ie and Ie, the emitter and collector currents.

i.e.

and

Ve = fi (Ie ,Ie)

Vc = !2(Ie,le)

For small-signal variations the voltage variations are given by:

and (8Vc) (8Vc) 8Vc = -- Me + - Ole Me Ie Me Ie

If 8Ve, 8Vc, Me and Me are written as Ve, Ve, ie and ie respectively, these equations may be written as:

Ve = ruie + r12ic

Ve = r21ie + r22ic

(1)

(2)

(3)

(4)

(5)

(6)

* See J. Millman, 'A Useful Network Theorem,' Proc. I.R.E., 28, pp. 413-17,1940, and F. A. Benson, Electric Circuit Problems with Solutions, Chapman and Hall, 2nd Edition, 1975, pp. 183-4.

SOLUTION 161

where the coefficients 1'u, 1'12, 1'21 and 1'22 are defined as:

and

((H~) 1'11 = Me Ic

( c5 Ve) 1'12 = c5L I

c e

( c5 Vc) Me Ic

( c5 Vc) Mc Ie

233

(7)

(8)

(9)

(10)

It is possible to draw several equivalent circuits which satisfy equations (5) and (6). These four-terminal networks are active, not passive, so four independent parameters are needed to specify their performances. In some equivalent circuits the four parameters used are 1'e, 1'b, 1'c and 1'm (or a). By comparing the mesh equations for the various networks it is easily shown that*:

1'11 = 1'e + 1'b (11)

1'12 = 1'b (12)

1'21 = 1'b + 1'm (13)

1'22 = 1'b + 1'c (14)

and a = 1'2J1'22 (15)

Equations (5) and (6) can be re-arranged to give the voltage ve and current ic in terms ofie and vc' The h parameters (or hybrid parameters) are then defined by these equations as follows:

(16)

ic = hfbie + hobVc (17)

Similar parameters may be defined for common-emitter and commoncollector arrangements. The relationships between the hand l' parameters can easily be determined as follows:

From equation (5), 1'11 = velie with ic = O. Under this condition:

ve = IZjbie + hrbVc

* See, for example, S. Seely, Electronic Engineering, McGraw-Hill, 1956, Chapter 16 or L. M. Krugman, Fundamentals a/Transistors, Rider and Chapman and Hall, 2nd Edition, 1959, or F. A. Benson and D. Harrison, Electric-Circuit Theory, Edward Arnold, 3rd Edition, 1975, pp. 342-3.


and

Thus,

Similarly,

Then,

Also,

Then,

and

so

Finally,

Then,

i.e.

Also,

It follows that:

and

o = hfbie + hobVe

'11 = velie = (h;bhob - h,.bhfb)lhob

'21 = velie when ic = 0

0= hfbie + hobve

'21 = - htJhob

'12 = velie with ie = 0

ve = hrb

ic = hobVe

"12 = h,.JhOb

'22 = velie with ie = O.

ic = hObVe

'22 = Ilhob

a = '2J'22 = - hfb .

'e = '11 - '12 = h;b - hrb (1 + hfb)lhob

'b = '12 = h,.Jhob

'e = '22 - '12 = (1 - h,.b)lhob ~ Ilhob

'in = '21 - '12 = - (hfb + h,.b)/hob •

In the example given:

'11 = (35 X 1 X 10-6 + 7 X 10--4 X 0.976)/(1 X 10--6) = 718.2 il

'12 = (7 X 10--4)/(1 X 10-6) = 700 il

'21 = {0.976/(l X lO--6)}il = 976 kil

'22 = 1/(1 X 1O-6)il = 1 Mil

a= -0.976

'e = (718.2 -700) = 18.2 il

'b = 700 il

'e = '22~ 1 Mil

'm = (976000 - 700)il = 975.3 kil

(18)

(19)

(20)

(21)

(22)

(23)

(24)

(25)

(26)



163. Consider the two arrangements shown at (a) and (b).

(0) (b)

For circuit (a) For circuit (b)

(Z1 + Z2)/1 + Z2/2 = V (Z1 + Z2)/1 + Z2/2 = 0, (Z2 + Zm)/1 + (Z2 + Z3)/2 = 0 (Z2 + z",)/1 + (Z2 + Z3)/2 = V

1 (Z1 + Z2) V 1

. (Z2 + Zm) 0 .. 12 = 1 (Z1 + Z2) Z2 I·

(Z2 + Zm) (Z2 + Z3)

In general, 11 =1= 12 , so the original circuit does not satisfy the reciprocity

condition. For the reciprocity condition to be satisfied Zm must be zero.

164. It is foundt that 0: varies with frequency according to the following expressions:

0: = 0:0 {I + j~/fa)} where 0:0 is the low-frequency value of 0: and k, called the alpha cut-off frequency, is that frequency where 0: = 0:0/Y(2).

Thus, if 0:0 = 0.96,/a = 5 MHz andf= 10 MHz,

0: = 0.96/Y(1 + (10/5)2)

* See, for example, S. Seely, Electronic Engineering, McGraw-Hill, 1956, Chapter 16. See also, L. M. Krugman, Fundamentals of Transistors, Rider and Chapman & Hall, 2nd Edition, 1959, and F. A. Benson and D. Harrison, Electric-Circuit Theory, Edward Arnold, 3rd Edition, 1975, Chapter 15. t R. L. Pritchard, Frequency Variations of Current-Amplification Factor for Junction

Transistors, Proc. I.R.E., 40, p. 1476, 1952, and D. E. Thomas, Transistor Amplifier Cut-off Frequency, ibid., 40, p. 1481, 1952.


=0.43

If ao = 0.96, a = 0.6 and/= 5 MHz,

1=1.../[(ao/a)2-1]

= (5.../(0.96/0.6)2 - 1) MHz

= 6.25 MHz.

165. a' = a/(1 - a)

0:0/ {1 + j (ff fo,) } ao

1 - ao/ (1 + j(//lrJ}- (1 - 0:0) + j (f/lrJ The cut-off frequency is defined as that for which the gain falls to

1/.../2 of its original value. This occurs for the common-emitter circuit when the frequency isfo,' such thatfo,'/fo, = 1 - ao

Le. la' = la(1 - 0:0)

166. The drain current ID is a function of the gate voltage VGS and drain voltage VDS, Le.

ID = I (VGS, VDS)

OlD = (aID/a VGS)\!' a VGS + (aID/a VDS)v a VDS DS OS

or id = gm Vgs + (l/rd) Vds

where OlD, aVGS and aVDS are written as id , Vgs and Vds respectively.

gm = (aJD/a VGs)v is the mutual conductance or transconductance DS

and l/rd = (aID/a VDS)v and rd is the drain resistance. GS

An equivalent circuit which satisfies the above expression for id is shown below.

----------~----+--------os

177. (a) Mean load current

1 rll' 300..;2 sin e (Id •c) = 21T Jo (150 + 1000) de = 117 rnA.

SOLUTIONS 177-179

(b) Alternating load current

(Irm.s) = [~ {7r (300y'2 sin 0)2 dol 1/2 = 184 rnA. • 21T Jo 1150 J

(c) D. C. power supplied to the load = (Id•c)2 X 1000 = 13.8 W.

(d) Power supplied to the circuit = (lr.m.sY X 1150 = 39.1 W.

(e) Rectification efficiency

d.c. output power 0 Of Of -----=---=----- X 1 010 = 35.3{0 power supplied to the circuit --

(f) Ripple factor = [(Ir.m.s}Id.cY _1]1/2 = 1.21.

178. (a) The d.c. load voltage

(Ed.c) = ~ f2 (300y'2 sin 0 - 10) dO 21T o.

where 01 and O2 are the angles at the striking and extinction points. Since 300y'2 ~ 10,01 may be taken as zero and O2 as 1T.

Ed.c. = 130 V.

(b) D.C. power supplied to load = (Ed•cY/I000 = 16.9 W.

(c) Input power to circuit

= 211T f: 300y'2 sin 0 eOOy'~~~~O -10) dO = 43.7 W.

16.9 (d) Rectification efficiency = - X 100 per cent = 38.7 per cent.

43.7

(e) It can be shown* that the ripple factor is

approximately 1.21 ~ + 0.5 X 30~~i] = 1.225.

237

179. Mean load voltage Ed •c. = Mean load current Id•c. X R i , where R; is 230y'2

the resistance of the load and Id c = 5 0 R ). ··1T(0+1

* J. Millman and S. Seely, Electronics, McGraW-Hill, 2nd Edition, 1951, p. 350.


Ed.c. = 230";2/rr - sOO/d.c.

Ed.c. changes from 103.5 V when Id.c. = 0 to 63.5 V when Id.c. = 80 rnA, Le. regulation is (103.5 - 63.5) V = 40 V.

(/dcYRt 40.6 Efficiency = (l )2(·R + 500) X 100 per cent = -sao per cent.

r.m.s I 1 +_ R,

:. efficiency decreases from 40.6 per cent when I d•c. = 0 to 24.9 per cent

when I d•c• = 80 rnA.

Maximum output power is obtained when R, = 500 n and the efficiency is then 20.3 per cent.

,. the current at which maximum power is obtained is given by 20.3 = 40.6[1 - sOO/d.c}103.5].

Id.c. = 103.5 rnA.

2 ( 300";2 ) 180. (a) Mean load current (/d•c) =; 500 + 2000 = 108 rnA.

1 ( 300";2 ) (b) Alternating load current (Ir.m.s) = ";2 500 + 2000 = 120 rnA.

(e) D.C. output power = (Id.cY X 2000 = 23.3 W.

(d) Input power = (/r.m.sY(sOO + 2000) = 36 W.

( 23.3 e) Rectification efficiency = 36 X 100 per cent = 64.8 per cent.

if) Ripple factor = [(J;..m~.//d.cl- I pf2 = 0.482.

2 (g) D.C. output voltage Ed .c. = - X 300";2 - Id •c. 500.

rr

:. Ed.c. changes from 270 V when Id.c. = 0 to 216 V when Id.c. = 108 rnA, Le. regulation is 54 V.

[ 1 r" J1 /2 181. R.m.s. current = 5 A = 2rr Jo 1m2 sin2 (J d(J = Im/2

where 1m is the maximum value of the current.

SOLUTIONS 181-182

1 Moving-coil ammeter reads -

21T

For full-wave rectification:

Im= 10 A.

i 7T L 1m sin 0 dO =..!!! = 3.18 A.

o 1T -----

A.C. ammeter reads r.m.s. value = Im/../2 = 7.07 A.

2 Moving-coil ammeter read!> mean value = - 1m == 6.3 7 A.

182.

12

« S E

'-4

o

,"

1T -----

I I 1_1 .... ~t--:c-:-:--I v : SOi

v

0.4 O.S .1.2 V (volts)

239

The following table can be drawn up using the rectifier characteristic:

Current i (rnA) 2 4 6 8 10 12 14

80i volts 0.16 0.32 0.48 0.64 0.80 0.96 1.12

v volts. 0.60 0.76 0.82 0.88 0.93 0.98 1.03

V = (v + 80i) volts. 0.76 1.08 1.30 1.52 1.73 1.94 2.15

The i/V characteristic is drawn. The current wave corresponding to the positive half-cycle of voltage can then be obtained as shown. The current during the negative half-cycle is so small that it can be neglected. The moving-coil ammeter reads the mean current taken over the whole cycle. This is found to be 3.32 rnA.


3 /-'. • \ i

/ ';1 , '(

12

/ \ • V' Mean height

/ ...... ~ \/ of this curve i /// ........ " \ = 6.64 rnA . , , , 8 ...... 2

I/)

<l ..... E "0

~

J' '\ l t\ I' \ \ 1/ . \

I • \ \ 1/ • , . ,~

~

4

o 30 60 90 120 150 180 Angle (degrees)

183. Current flows in each cycle for an angle e, where cos e/2 = VR/V, The current through R is VR/R which must equal the mean current through the rectifier,

I 1 f fJ/2 V(cos t/> - cos e/2) dt/> VR'R=-

21T -fJ/2 10

But 8 is given as 21T/6, so R = 585 n.

/(i~ -= = 1-=-1- 10.n. I I I I \ V V

QI I I I I \ R 01 I I I I , c I I \ ...,

I ;g ~8'" \ Angle I \ I \ I , I ,_/

Component of fundamental frequency in the a.c. supply is

1 LfJ/2 11 = -10 V(cos t/> - cos 8/2) cos t/> dt/> = 0.0287 X V/I0 1T -fJ/2

[v.

SOLUTIONS 183-185

A.C. power input =! VI. = t V2 X 0.0287/10

D.C. power output = (VR )2/R = (V cos O/2)2/R = V2 eJ;)js85. Efficiency of rectification

d.c. power output -.....:....---. ""':""-X 100 per cent = 89.4 per cent. a.c. power mput

241

184. It is evident from the diagram that, since 200y2~ 10, conduction may be assumed to continue until the end of each positive half-cycle.

(a) R.m.s.load current

Angle

2~ I

/ I

=j [~ f 1r (200Y2 sin 0 - 10) 2 d~ = 0.63 A. 21T 1r/3 200 J

(b) R.m.s. value of voltage across valve

= JU1T [fo1r'3 (2ooy2 sin 0)2 dO + f...:3102 dO + f:1r (2ooy2 sin8i dO ])

= 155 V.

(c) Power

1 J,1r (200Y2 sin 0 -10) = -2 (2ooy2 sin 0) dO = 77 W. 1T 1r/3 200 --

185. The filter is shown in the diagram. Any losses in the rectifying elements, transformer and choke will be neglected.


Suppose e is given by the first two terms of the Fourier-series represen

tation of the rectifier output voltage, i.e. ~m [I - ~ cos 2wtJ where Em

is the maximum value of the transformer voltage to the centre-tap .

. 1 Em 2 R.m.s. value of e(l!r.m.s.) = y2 -. -.

1T 3 E 2 1

a.c. through the circuit is approximately y2 ;. 3" . XL = Ir.rn.s.

where XL = 21T(2f)L andfis the supply frequency.

:. ripple voltage across load is approximately Ir.m.s. Xc = Er.m.s., where Xc = 1/21T(2f)C

Ripple factor = Er.m.s.l(2Em/1T) = y2Xc /3XL = y2/3(41Tf)2LC = 10/300 in this case.

:. whenf= 50 HZ,LC= 35.86 X 10-6, and whenf= 60 HZ,LC= 24.87 X 10-6.

If the rectifier is to pass current throughout the whole cycle the peak current delivered must be less than the direct current, i.e. 4Em/31TXL ~ 2Em /1TR,.

The limiting condition for this is when L = R,/61Tf. In this caseR, = 300/0.12 = 2500 n.

:. when (a)f= 50 HZ,L = 2.65H, and when (b),f= 60 HZ,L = 2.2IH.

The above expressions give the minimum values of Land LC that may be used to obtain the required results. Since the minimum value of L =

2.65H for case (a) and 2.21H for case (b) choose a IO-H choke, in both

instances. This is a readily available item and its size must be such as to

carry the necessary current.

If L = 10 H andf= 50 Hz, C= 35.86 X 10-6/10 = 3.586 f.1F. If L = 10 H andf= 60 Hz, C= 24.87 X 10-6/10 = 2.487 f.1F.

:. in both cases choose a 4-f.1F capacitor which is also readily available.

SOLUTION 186

186. (a) Simple inductor filter.

2Em 4Em cos (2wt - 1/» Load current = --- -;----=---~

1TR, 31T V(R,2 + 4w2L2

243

where the symbols Em' R, and L have the same meanings as in the previous solution, and tan I/> = 2wL/R,.

3:~2 V(R? + 4w2L ~ the ripple factor = --'--------

2Em/1TR,

If w = 1001T,L = 20 H andR, = 2000 n. the ripple factor = 0.074.

If w = 1201T, the ripple factor = 0.062.

(b) Simple inductor filter.

WhenL = 40 H, the ripple factor is 0.037 forf= 50 Hz and 0.031 for

f= 60 Hz.

(c) Simple capacitor filter.

cL-~~-~--~-----f ""df -~.,... \ --9' \ ~ \,--1 " \ I \ I \

I \ ,/ \,' \ Ed.c. Em I \ \ I \

, \ " \ I \ , \ \ I \

The diagram shows the voltage curves.

The r .m.s. value of the ripple voltage er.m.s. = Ed /2V3. Assume the capacitor discharge continues for the fuil half-cycle at a

constant rate equal to the average value of the load current Id.c.* In the time for half a cycle (1/21) the capacitor wiillose an amount of

charge Id•cj2f coulombs.

* This gives a ripple factor for a given C which is too large. See J. Millman and S. Seely, Electronics, McGraw-Hill, 2nd Edition, 1951, p. 393, and S. Seely, Radio Electronics, McGraw-Hill, 1956, pp. 71-3.


Idc /(2fC.2v'3) The ripple factor = er.m.s)Ed •c. = .. 1 R

d.c. I

In this case C = 16 I1F and RI = 2000 n. :. whenf= 50 Hz, the ripple factor = 0.090, and whenf= 60 Hz, the

ripple factor = 0.075.

(d) Simple capacitor filter.

When C= 3211F the ripple factor is 0.045 for f= 50 Hz, and 0.0375

for f= 60 Hz.

(e) Single L-type filter.

In the previous solution it has been shown that the ripple factor for a single L-type fIlter is v'2/3(4rrfiLC.

Substituting the values off, Land C it is found that the ripple factor is 0.0037 for f= 50 Hz, and 0.0025 for f= 60 Hz.

if) Single L-type filter.

When L = 40 H and C = 32 I1F the ripple factor is 0.0009 for f = 50 Hz,

and 0.0006 for f= 60 Hz.

(g) Double L-type filter.

The reactances of the chokes are much larger than the reactances of the capacitors. Assume reactance of C small compared with RI •

impedance between P2 and Q2 is approximately Xc = 1 2rr(2nC

" " " Xc

P Q XL = 2rr(2nL

AI · I . . 1 v'2 2Em 1 ternatmg current lIS approXImate y -. - . - . 3 rr XL


Alternating voltage across PI QI is fIXC'

Also

y2 2Em (Xc) 2 Alternating voltages across P2Q2 = f2XC = 3' ~ XL

y2 (X. )2 ripple factor = - S . 3 XL

When L = 20 Hand C = 16 fJ.F the ripple factor is found to be 2.95 X 10-5 for f= 50 Hz, and 1.42 X 10-5 for f= 60 Hz. --

187. An upper limit to the ripple can be found by assuming that cutout takes place for the entire half-cycle.* The triangular ripple waveform shown in the solution to Question 186 becomes a triangular wave with vertical sides.

The Fourier analysis of such a waveform gives

Ed ( sin 4wt sin 6wt ) Ed -- sin2wt----+-- - ... ,c'n: 2 3

where Ed = fd.c)2fC as in Question 186. Harmonics greater than the second will be neglected. R.m.s. second-harmonic voltage E2 = fd.c)2n:fCY2 and this is impressed on an L-section filter.

The output ripple is therefore approximately E2 . XC/XLI

where XCI = 1/2n:(2!)CI and XLI = 2n:(2!)LI'

Ripple factor = E2 . XCJXLI . Ed•c. = y2/LICICR,(2n: .2/)3. If C and CI are in microfarads and f = 50 Hz, ripple factor = 5700/

CCILIR,. In this case R, = 250 X 1000/50 = 5000 Q and the ripple factor =

0.01/100.

A value for LI is usually chosen to be that of a readily available item.

* J. Millman and S. Seely, Electronics, McGraw-Hill, 2nd Edition, 1951, p. 402.


For example if Ll = 20 H, C= 23.9I1F.

Alternatively, if Ll = 40 H, C= 16.9I1F.

The capacitors chosen for these two values of Ll would need to be not less than the corresponding figures quoted.

Having chosen a suitable choke its d.c. resistance will be known and therefore the d.c. voltage drop across it can be calculated. This gives the voltage drop across the first capacitor from which the peak transformer voltage to the centre-tap can be evaluated.

If C and C1 are in microfarads andf= 60 Hz, ripple factor = 3300/ CC1L1R,. In this case, if C1 = C, C2Ll = 6600.

If now Ll = 20 H, C= 18.2I1F. Alternatively, if Ll = 40 H, C= 12.8I1F.

188. A 1T-section filter with a resistor replacing the inductor may be analysed as in the previous solution.

:. the ripple factor is now

y2Xc . Xc/R, . R instead of y2Xc . Xc/R, . XL,

i.e. for the same ripple factor R = XL, = 41TjLl , = 12568 n for 50 Hz, and 15082 n

for 60 Hz. When output current = 100 rnA, power dissipated is

(a) (0.1)2 X 12568 W = 125.7 W.

(b) (0.1)2 X 15082 W = 150.8 W.

When output current = 10 rnA, power dissipated is (a) 1.257 W, (b) 1.508 W.

189. (a) To obtain the minimum value of R the Zener current Iz must be at an optimum value. This current must satisfy the regulation require-ment

Iz = 8 + (65 -15) = 8 + 50 = 58 rnA

If the Zener current is 58 rnA it can decrease to 8 rnA to allow the load current to go from 15 to 65 rnA. If the load current is at the maximum 65 rnA the Zener current can take the excess current if the load decreases to 15 rnA, for (65 - 15) + 58 = 108 rnA which is less than 120 rnA.

R = (65 - 33)/1

1 = 15 + 58 = 73 rnA

SOLUTIONS 189-191

so R = 32/73 X 10-3 = 440 n.

(b) From (a)/z = 58 rnA.

(c) Vrnax = (58 + 55) 10-3 X 440 + 33 = 82.7 V

Vrnax = (8 + 55) 10-3 X 440 + 33 = 60.6 V.

190. The solution to this problem can be found elsewhere. *

191. A num.ber of solutions are possible. The following one possibly represents the simplest approach with a minimum of extra components.

SelectR1 to provide 3 rnA in the Zener diode (operational amplifier input circuits neglected)

Rl = {(30 - 6.5)/3 X 1O-3}n = 7.8 kn.

247

Assuming the operational amplifier has high gain and low output voltage

R2/R3 = 6.5/{15 - 6.5) = 6.5/8.5

In the absence of further information let a current of 1 rnA flow in R2 andR3

(un stabilised ) 30V 1

6.SV

R2 = 6.5 kn

R3= 8.5 kn

11SV (stabilised)

A, R2 and R3 effectively form a negative feedback amplifier for which

R the output resistance Ro' is given by Ro' = 1 + ;Av where (3 = R2/(R2 + R3)

* See R. J. Maddock, Intermediate Electronics, Book 2, Butterworths, 1970, pp.236-9.


and Ro is the open-loop output resistance of A

200 Ro'= il 15.4mil

( 6.5 ~ 1 + 15 X 3 X 104,


193. Let resistance of series resistor be R ohms.

Current through R = (20 + 30) rnA = 50 rnA.

Voltage across R = (400 - 200) V = 200 V.

R = 200/50 X 10-3 = 4 kil.

Since the load current = 20 rnA, and the tube current may vary from 10 to 50 rnA, the current through R varies from 30 rnA to 70 rnA.

the voltage across R varies from 120 V to 280 V,

i.e. the input voltage varies from 320 V to ~80 V.

Load current can vary from zero to 40 rnA (when tube current is at its minimum value of 10 rnA),

i.e. load resistance varies from 5 kil to 00, since voltage across load is 200 V.

194. The solution to this problem has been given elsewhere by the author.t

195. The solution to this problem has been given elsewhere by the author.:j:

196. The voltage-current .curve is as shown.

* See F. A. Benson, 'Voltage Stabilizers,' Electronic Engineering Monograph, 1950, pp. 30-2. See also E. W. Titterton, 'Some Characteristics of Glow-discharge Voltage Regulator Tubes,' 1. Sci. Instrum., 26, p. 33, 1949. t F. A. Benson, 'Voltage Stabilizers,' Electronic Engineering Monograph, 1950,

Chapter 4. * F. A. Benson and G. V. G. Lusher, 'Voltage Stabilizers for Microwave Oscillators,' Electronic Engineering, 26, p. 106, 1954.

SOLUTIONS 196 and 205-206

0.6

_----T ... T Barretter

~ 0.4 characteristic « '" .... c ~ L :::I

u 0.2

80 100 120 140 160 Voltage .(V )

249

If V is the voltage across the barretter and I the current through it, then for the 200-V input

200 = v + 100l

This is the equation of straight line A which cuts the barretter characteristic at 0.5 A.

For 180-V input line B is obtained which cuts the barretter characteristic at 0.5 A.

For 220-V input line C is obtained which cuts the barretter characteristic at 0.504 A.

Current variation if input voltage changes by ± 10 per cent is 0.004 A.

20S. Supply voltage = {250 + (10 X 103 X 9 X 1O-3)} V = 340 V.

Resistance ofload = (430 - 250)/9 X 10-3 Q = -20 kQ.

206. Resistance = 8/(9 X 10-3) = 889 Q.

The capacitor should have low reactance compared with 889 Q. The greater the capacitance the more effective is the capacitor in taking most of the alternating component of the anode current.

Suppose the reactance of the capacitor is chosen to be 1/10 of the resistance. Then at 1000 Hz, C = 106/2rr X 1000 X 88.9 J1F,


i.e. C= 1.79 IlF, say 2 IlF.

At 100 Hz,C= 17.9 J1F,say 201lF.

207. Power input = 1/600 W.

Power output = 1012, where I is the load current.

10 10glO (1012 X 600) = 60, so I = 12.9 A

60 dB = 60 X 0.1151 nepers = 6.9 nepers.


0.8 H

1= 100 A a 8000 + 1000 + j(21T X 300 X 0.8)

= (10.81 - jI.81) rnA.

The output voltage

Vo = - (10.81 - j 1.81X1000 + j 1508)10-3 V

= - (13.54 + j 14.49) = 19.61/- 133°.

19.61/- 133° ° The gain A = 5 = 3.92 /- 133 .

The phasor diagram is therefore as illustrated. When the frequency is 2000 Hz:

100 fa = 8000 + 1000 + j(21T X 2000 X 0.8) A = (4.94 - j5.52) rnA.

SOLUTIONS 208-210

The output voltage

Va = - (4.94 - j5.52)(I000 + j 10060) X 10-3 V

= 74.87 / - 143.8°.

the gain A = 14.97 / - 143.8°.

251

The gain at 2000 Hz is greater than the gain at 300 Hz, i.e. frequency

distortion is present. The results also show that phase-shift distortion exists

in the amplifier.


210. The equivalent circuit of the arrangement is as shown.

ega

pV-y;,-V-Y3 Using the Millman Theorem, Va = y. g y, y"g y; where Ya = lIra,

a+ ,+ 2+ 3 12 = jwCac, 1'3 = jwCga and 11 = liZ,.

the gain = - Va = 13 - &n =A say. Vg y;,+1I+12+Y3

In this case, since w = 21T X 10000, Y2 = j2.26 X 1O-7 S, and Y3 = j1.88 X 10-7 S. Also 11 = 1.11 X 10-5 S, Ya = 2.5 X 10-5 S and &n = 1.5 X 10-3 S.

. -1.5 X 10-3 + j 1.88 X 10-7

gam = 3.61 X 10-5 + j4.14 X 10-7 •

Thus the j terms which come from Y2 and Y3 are negligible. Neglecting these j terms the gain is - 41.6.

Since A is real the input impedance consists of a capacitance of value q=Cgc +(1 +A)Cga

* See, for example, J. D. Ryder, Engineering Electronics, McGraw-Hill, 1957.


= (3.0 + 42.6 X 3.0)J.LJ.LF = 130.8 JJp.F.

For a two-stage amplifier the input impedance of the second stage acts as a shunt for the load of the first stage. Thus q, along with the Cae of the first tube, shunts the load. It should also be remembered that every 1 IlIlF of stray capacitance between the leads to the anode and grid of the second stage adds effectively 42.6 IlIlF across the load resistor of the first stage. It is reasonable to assume therefore that the 90 OOO-Q load of the first stage is shunted by a capacitance of 200 IlIlF (a conservative figure).

Yi = 1.11 X 10-5 + j 1.26 X 10-5 s. . Y3-gm -gm

the gam = -----'::...--.;:;.:.:.:..--Ya + Yi+ Y2 + Y3 Ya + Yi+ Y2 + Y3

= - 36.79 + j13.24 = 39.1 /160.2°.

211. The equivalent circuit of the arrangement is shown.

"a = 5000011

From the equivalent circuit:

ia = IlVg/(R, + ra + Ze)

From (1) and (2):

Now

From (3) and (4),

(1)

(2)

(3)

(4)

Va/V; = -1lR,/{ra + R, + Ze(1 + Il)} (5)

2000/jwC ButR, =100000Q,Ze= / ,ra=50000Qandll=80

2000 + 1 jWC


vofv; = 8000/ {150 + 162 } . 1 + 2000jwC

:. vo/v; is a maximum when w ~ 00 and is 53.3, and vo/v; is a minimum when

w ~ 0 and is 25.6.

vo/v; is equal to 0.707 of its maximum value when the frequency (w/21f) = 121.2 Hz.

212. The equivalent circuit of one stage is as shown. Applying the Millman Theorem between points 0 and B:

TT _ VoAYC "-B-

o Yc + YRg + YCg

A c B

o

where Yc = jwC, YRg = l/Rg and YCg = jwCg.

Applying the Millman Theorem between points 0 and A.:

_ pVgYa + VoBYC _ _ VoA - Y. + y, + y- ' where Ya - l/ra and Yi - l/R,.

a , C

:. gain A = - VoB/Yg -PYa Yc =--------=--=-------

(Yc + YRg + YCgXYa + Yi) + YdYRg + YCg)'

At intennediate frequencies where Yc is large and YCg is small,

A = Ao = - PYa/(Ya + Yi + YRg).

At low frequencies the effect of Cg is negligible and

A = Al = - plqYc/{YdYa + Yi + YRg) + YRg(Ya + Yi)}.

AI_ 1 where!! = YRg(Ya + Yi) Ao I - j!tI! 21fC(Ya + Yi + YRg)'


A 1 If the load is a pure resistance It is real and A: = v[1 + ([1/f)2]

i.e. It represents that frequency at which the gain falls to l/v2 of its intermediate-frequency value.

At high frequencies YRg and YCg can be neglected in comparison with Yc and A = A2 = - JIYa/(Ya + Y, + Y Rg + YCg)'

A2 1 - = ·fi'/.'I" , where f2 = (Ya + Y, + YRg)/21rCg • Ao I+J J2

A 1 If the load is a pure resistance f2 is real and A: = v(1 + ([/12)2]' i.e. f2

represents that frequency (at the high-frequency end) where the gain falls to l/v2 of its intermediate-frequency value.

In this case,

and

Ya = 1.3 X 10"'" S, Y, = 0.2 X 10"'" S,

YRg = 0.02 X 10"'" S.

Ao = - 17.1'[1 = 31 Hz andf2 = 121000 Hz.

Ao= -17.1 so when A = 14

A 14 the minimum gain ratio - = - = 0.8187.

Ao 17.1

If r is the low frequency where the gain drops to 14 and f" is the high frequency where the gain drops to 14,

1 1 v[1 + ([1/f')~ = 0.8187 = v[1 + if'lfJ2]

I' = 44 Hz and I" = 84960 Hz.

213. Gain per stage = v(6000) = 78.

Using the symbols introduced in the solution to Problem 212:

I ~: I = [1 + :11/2)2] n(2

where n is the number of stages (in this case 2)

0.95 = [i + (~//2)2] 2/2


i.e. fl12 = 0.223

Since f= 100 kHz,12 = 450 kHz.

Now 12 = 1/21TCgR" where Cg = 20.5 JlJlF (given)

R, = 1/(21T X 20.5 X 10-12 X 450 X 103) = 17200 n

The gain per stage at mid-frequency is

- gmR, = - 5.2 X 10-3 X 17 200 = - 89.4

The overall gain = 89.42 = 7992.

214. It is shown in Solution 212 that:

Ao = - JlYc,/(Yc, + II + YRg)

With pentodes it is possible to assume that ra ~ R,. Further, it may be assumed that Rg ~ R, because R, must be made small to raise 12, whereas Rg must be large to lower fl. With these assumptions Ao = - gmR, and the high-frequency gain becomes:

A2 = - gmR,/{l + jwCgR,)

The gain ratio for a pentode is then

I ~: I = [1 + :fl12)2] t

where 12 = 1/21TCgR, = bandwidth in hertz sincefl will be low.

Figure of merit (gain X bandwidth) = gmR,/21TCgR, = gm/21TCg.

This is constant for a given type of valve. If Cg = Cae + Cgc (the irreducible minimum value of Cg for pentodes) = 9.2 JlJlF in this case and 1Jm = 5.7 mS:

Figure of merit = 5.7 X 10-3/(21T X 9.2 X 10-12)

=98.6.

215. If an amplifier is made up of n cascaded RC-coupled stages, not necessarily identical, the overall high-frequency gain ratio is:

I~ I ~ ~ + ~/f,'),r [I + ~h")r If the stages are identical


I Azi [ I ] n/2 Ao = I + (1/12)z

If Is designates the 3 dB point for the overall amplifier of n stages:

l/y2 = [i + (~/12)Z] n/2

Similarly, for the three non-identical stages

l/y2 = L + (~/12')Z]! [I + (~/12")2] t [I + (~/lz"')Z] t 1/2 = [ I + (~250)Z] [1 + (f~350)ZJ [i + (~550)Z]

This expression gives 1 at the high-frequency end in kHz.

A similar calculation gives the low-frequency limit of the overall bandwidth:

I ~ 1= l/y2 = L + (I~o/Iir L + (180/f)2r 8 + :50/f)Z] t


217. The solution to this problem can be found elsewhere.t

218. Consider first the low-frequency equivalent circuit (a) of the amplifier. The secondary of the transformer feeds the grid circuit of the next valve which is assumed to have infinite impedance.

Ip = JlVg/(ra + Rp + jwLp) (1)

Voltage across primary = jwLplp (2)

= JlVg/ {I - j(ra + Rp)/wLp} (3)

* See J. D. Ryder, Engineering Electronics, McGraw-Hill, 1957, p. 108. t See, for example, V. C. Rideout, Active Networks, Constable, 1954, pp. 224-7;

or S. Seely, Electron·tube Circuits, McGraw-Hill, 2nd Edition, 1958, pp. 319-21.

SOLUTION 218 257

(0)

Voltage across secondary is n times as large as this.

gain = ± nll/{l- j(ra + Rp)/wLp} (4)

The magnitude of the gain (A) = nll/v[1 + {era + Rp)/wLp}2] . (5)

When wLp > (ra + Rp), A = nil = 30 in this case. The gain drops off at low frequencies because wLp is not large compared with (ra + Rp). When wLp = (ra + Rp) the gain is only 70.7 per cent of its value nil approached at higher frequencies. Thus the gain drops to 70.7 per cent of 30, i.e. 21.2 when!= (8000 + 3500)/(21T X 70) = 26 Hz.

Now consider circuit (b) which is the high-frequency equivalent circuit.

L = O.SH

c= lOO°tlpF

An analysis of this circuit shows the gain to be

± jnll (wlc) R+j (WL- wIC)

The magnitude of the gain A = J r I J LR2 + (WL - wc)

nil (Wi c)

(6)

(7)

At low frequencies, wL is small and l/wC is large, therefore the gain approaches nil as already stated. At very high frequencies I/wC is small and wL is large and the gain falls rapidly to zero. The gain passes through a


maximum between these two extremes which is found by putting dA/dw = 0 to occur when

l/wC= {2(wL)2 + R2}/2wL.

Since wL is usually much greater than R, the maximum occurs when wL = l/wC. This is the condition for series resonance.

In this case the frequency for maximum gain

1 = = 7117 Hz.

2trv'(0.5 X 1000 X 1O-1~

An analysis of the circuit shows that the corresponding gain is

;J(~) = 153~00 )[1000°/10-12] = 44.7.

A gain-frequency curve can be plotted by finding other values of gain at the low-frequency end using equation (5) and at the high-frequency end using equation (7). Both expressions give a gain of nJ1. = 30 for the midband frequencies. A sketch of the resulting response curve is shown in diagram (c).

44.7 ------- ----11\

I: " 30 ,,' I ------,.._._. I

c / I '0 2).:.2 __ _ l : t!): I

I I I I I 7117

26

10 100 1000 10000 Frequency (Hz) (c)

219. Let the input voltage be sinusoidal and of the form Vg = Yg coswt. The anode current ia is of the form

fa + Bo + Bl cos wt + B2 cos 2wt

Bo, Bf, etc., can be found from the characteristic curves of the valve. From the figure when:

wt = 0, ia = fmax

(1)

SOLUTION 219

wt = n/2, ~ = Ia

wt = n, ~ = Imin.

Substituting in (1):

Imax = Ia + Bo + BI + B2

fa =Ia +Bo-B2

Imin =Ia +Bo-BI +B2

From (3), Bo=B2 •

From (2) and (4), BI = (Imax - Imin)/2

From (2) and (6), B2 = (Imax + Imin - 2fa)/4

I I , I

pi ___ L 1a

I I I I L I 1min ---I---r I I

------~--~~--;-~~~~ I I r ....... I

I t=o '~ I ~ I ,-'

-+:' I ,''Ir/2 ,/' I , +1T '... I ........ ~

I

259

(2)

(3)

(4)

(5)

(6)

(7)

In this case maximum current corresponds to a grid voltage of (- 8 + 6) V = - 2 V and minimum current corresponds to a grid voltage of (- 8 -6)V=-14 V.

From the characteristic curves and the load line,

Imax = 21.6 mA,Ia = 13.2 rnA andImin = 6.3 rnA.

B2 = Bo = (21.6 + 6.3 - 26.4)/4 = 0.375 rnA.


Total steady current = (13.2 + 0.375) rnA = 13.575 rnA.

Peak fundamental current is

Bl = (21.6 - 6.3)/2 rnA = 7.65 rnA.

7.65 Peak fundamental output voltage = 1000 X 8000 = 61.2 V.

Fundamental gain = 61.2/6 = 10.2.

Percentage of second-harmonic distortion

= 100SJB1 = 100 X 0.375/7.65 = 4.9 per cent.

220. The quiescent point Q is determined by drawing the load line through the point fa = 0, Va = 300 V with a slope fixed by the resistance Rl of the choke. Since Rl is generally small the static load line is almost vertical. Since anode dissipation is 25 W, anode current permissible is 25/300 = 83 rnA. This corresponds to a grid bias of about - 20 V.

.... c

160

120

t 80 ::J o 41

" o c

<{

40

o

I I I I I I ,

I I I

83mA I I Q ---t---j--, I I I I I I I

1'Ig=-21V I I I I I I

, I Vy= -32V I

I ! Static I I I I I I

I ~-T-T-Ioad line I

, I

I I I I

I I I !

100

I I ,

I I

I I

0"

200 300 400 500 Anode voltage (V)


Permissible grid swing is about 20 V peak; distortion occurs from the nonlinear parts of the valve characteristics. Thus a minimum anode current of about 20 rnA is set where characteristics begin to curve. Therefore, dynamic load line is as shown (Q' QQ"). Voltage swing is (425 -175) V = 250 V. Corresponding current change is (148 - 20) rnA = 128 rnA. Load resistance = 250/(128 X 10-3) il = 1.95 kil.

Output power = (Vmax - VrninXlmax - Irnin)/8 = (250 X 0.128)/8 = 4 W.

Efficiency = (4/25) X 100 per cent = 16 per cent.

221. (a) The analysis required can be found in many textbooks.* (b) The method of determining the composite characteristic con

sists of inverting the characteristic of valve (2) with the quiescent point P2

immediately under PI and then adding algebraically the corresponding ordinates of the two characteristics.

~ E ..., c G) L. L. ::I 0 G)

" 0 c <l:

80

60

40

20

Composite ''''''--+-characteristic

100 400 Anode voltage (V) O~--~--~~-+~~r-~----,O

Anode voltage (V) 400 100 <l:

-f---t---i 20.5-..., c G)

C --+----!----\ 40 ~.

G)

" o ---1-----\ 60 .'i

* See, for example, P. Parker, Electronics, Arnold, 1950, Section 73; or Cruft Electronics Staff, Electronic Circuits and Tubes, McGraw-Hill, 1947, Chapter 13.


At the points P, slopes of characteristics

= 1/2000 S, so ra = 2000 Q.

At Q, ra = 1000 Q.

222. (a) The solution to this problem can be found in many standard textbooks. *

(b) Power output = ! Vi~ where Vi and ~ are respectively the peak values of the voltage across, and the current through, one section of the output transformer primary. Now ~ = /max and Vi = V - Vmin where V = 500.

power outputPa.c. = !Imai500 - Vmin).

But Vmin = 1000/max' soPa.c. = !/mai500 -lOOO/max). This is a maximum when/max =!A whenPa.c. = 31.25 W.

Power drawn from h.t. supply,Pd.c. = 2/a V where /a is the mean anode current of either valve (valves assumed identical). The pulses of anode current may reasonably be taken as half sine waves so /a = /max/1T and therefore Pd•c. = 2 V /max/1T.

The efficiency = Pa.c)Pd.c. = 1T(l - Vmin/V)/4.

In this case the efficiency = 1T(l- 250/500)/4 = 0.393 = 39.3 per cent.

(c) The efficiency of a class-C amplifier can be shown to bet

(l - Vmin/V)(O - sin 20/2)/2 (sin 0 - 0 cos 0)

where 20 is the angle of flow, V min is the minimum anode voltage and V is the supply voltage. In the ideal case, Vmin = 0 and if 20 is then 1200 the efficiency is found to be 89.6 per cent.

1 90 223. Mean value of anode current = 2" X 2.5 X 360 = 0.31 A.

Power supplied by h.t. source is 0.31 X 2500 = 775 W.

Output power = 0.8 X 0.8 X 750 = 480 W

480 efficiency = 775 X 100 per cent = 61.9 per cent.

* For example, see P. Parker, Electronics, Arnold, 1950, p. 340. t For the proof of this expression see, for example, P. Parker, Electronics, Arnold,

1950, pp. 343-4; or A. T. Starr, Electronics, Pitman, 2nd Edition, 1959, p. 272.


At the point where anode current commences to flow, the instantaneous angle voltage is:·

Va = Vh.t. - Va.c. cos 8 (28 is the angle of flow)

= 2500 - 2000 cos 450 = 1086 V.

224. Let Z be the impedance of the parallel resonant circuit shown.

Then

Let

C R

1 1 1 -=-+-+jwC ZjwLR .

w02Le = 1 and Q = R/woL.

1 R (1 - W2/WO~ + jwR/wr/2 =--'-----'-~-"---'--=

Z RjwR/wr/2

Z=R/[1 + jQ(w/wo-wolw»).

When w is near to Wo let w = Wo + ilw.

Then Z = R/[l + 2jQilw/wo).

For the anode load of the valve the voltage across L is

vL = - J,J.Zvg/(Z + 'a)·

the voltage across ~ is .VL . jwM = vLM/L = vo. jwL

Vo = - J,J.ZvgM/L(Z + 'a)

=-50Mv L 1 +--I[ { 30000 g 15000

ilw 500 Now-=-- V = 1 V M= ImH

w 20000' g' ,

* See, for example, J. D. Ryder, Electronic Fundamentals and Applications, Pitman, 3rd Edition, 1964, Chapter 12.


L = R/woQ = 15 000/(21T X 20000 X 45).

Ivol = 3.485 V.

225. Referring to the diagram:

Vo=AVg (1)

and Vg = V; + f3Va (2)

Va A V; = Af = 1 - {3A (3)

In this case, gain with feedbackAf = 20 'I + -.!.... 20) = 6.67. 1\ 10 -

Vi Vg Amplifier

V. normal gain A o r--

'--- Feedback ~ (Wo network

226. Original input voltage = 60/120 = 0.5 V

" distortion" = 10 X 60/100 = 6 V.

The distortion voltage has to be reduced 10 times, therefore:

1 - f3 X 120 = 10, i.e. f3 = - 0.075 where f3 is the feedback factor.

The gain will also be reduced by a factor of 1O.Af = 120/10 = 12. The

added distortionless gain needed ahead of the feedback amplifier is 120/12

= 10, and this amplifier must supply a signal voltage of 60/12 = 5 V.

227. With feedback and normal supply voltage the gain

Afl = 24000(1 + 24000/1000) = 960.

With feedback and a 25 per cent reduction of supply voltage from its normal value the gain Af2 = 16000(1 + 16000/1000) = 941.

SOLUTIONS 228-232

228. The gain of each stage!:::! - gmR, {Rg +Rf/jwc }

!:::! -jwCRggmR,

1 + jw CRg

{- jw C R gmRl)3 loop gain of amplifier = i3 1 + jW~Rg

_. {wCRggmRl) 3

- Ji3 1 + jwCRg

265

According to Nyquist's criterion, instability will occur if the loop gain> 1 when there is no phase-shift around the loop, i.e. when tan (wCRg) = 30° or when wCRg = 1 /Y3.

Under this condition the loop gain is i3(g",R,i/8. Thus, for stability, i3(gmR,)3/8:l> 1

i.e.

229. The cathode follower has been analysed by many authors. * The output impedance is nearly equal to l/gm. In this case the output impedance = 103/4 = 250 n.


231. (a) For the solution to this problem see the book by Seely.:\: (b) For the solution to this and a similar problem see the book by

Rideout.§

232. The input resistance of a common-base amplifier is given byil:

R j = re + rb -- rb(rb + rm)/(rb + rc + R,)

= r11 - r12r21/(r22 + R,),

* E.g. see S. Seely, Electron-tube Circuits, McGraw-Hill, 2nd Edition, 1958, Chapter 5; or A. T. Starr, Telecommunications, Pitman, 1954, pp. 232 and 233. t See S. Seely, Electron-tube Circuits, McGraw-Hill, 2nd Edition, 1958, pp. 176-8. :j: S. Seely, Electron-tube Circuits, McGraw-Hill, 2nd Edition, 1958, pp. 164-6. § V. C. Rideout,Active Networks, Constable, 1954, pp. 175-8.

II See, for example, S. Seely, Electronic Engineering, McGraw-Hill, 1956, Chapter 16; L. M. Krugman, Fundamentals of Transistors, Rider and Chapman & Hall, 2nd Edition, 1959; J. D. Ryder, Engineering Electronics, McGraw-Hill, 1957, Chapter 12; R. B. Hurley, Junction Transistor Electronics, Wiley, 1958.


where 'b, 'e, 'e, 'm, '11, '12, '21 and '22 are the usual transistor parameters and R/ is the load resistance.

In this case, Ri = {550 - 500 X 1.9 X 106/(2 X 106 + R/)} n. Thus, Ri

varies from 75 n to 550 n as R/ changes from 0 to 00.

The output resistance of the arrangement is given by·:

Ro = 'e + 'b - 'b('m + 'b)/('b + 'e + Rg)

= '22 - '12'21/(Rg + '11),

where Rg is the resistance of the source (zero in this case). Thus,

Ro = (2 X 106 - 500 X 1.9 X 106/550) n = 2.72 X 105 n The maximum possible voltage gain·

= '21/'11 = 1.9 X 106/550 = 3454.



235.

Voltage gain (Av)

'e = '11 - '12 = 20 n Tb = '12 = 800 n 'e = '22 - '12 ~ 2 Mn

'm ='21 - '12 ~ 1.98 Mn

* See, for example, S. Seely, Electronic Engineering, McGraw-Hill, 1956, Chapter. 16; L. M. Krugman, Fundamentals of Transistors, Rider and Chapman & Hall, 2nd Edition, 1959; J. D. Ryder, Engineering Electronics, McGraw-Hill, 1957, Chapter 12; R. B. Hurley, Junction Transistor Electronics, Wiley, 1958. t See, for example, S. Seely, Electronic Engineering, McGraw-Hill, 1956, Chapter

16; L. M. Krugman, Fundamentals of Transistors, Rider and Chapter & Hall, 2nd Edition, 1958; J. D. Ryder, Engineering Electronics, McGraw-Hill, 1957, Chapter 12; R. B. Hurley, Junction Transistor Electronics, Wiley, 1958; F. A. Benson and D. Harrison, Electric-Circuit Theory, Arnold, 3rd Edition, 1975, Chapter 15.

SOLUTIONS 235-236

430 ~--------------------------------800 - {(820)(2 + 106 + 430)/(1.98 X 106 - 20)}

~-15.2.

Current gain (A;)

1 =-------------1 + (rc + R1)/(re - rm)

1 ~--------~----------~----

1 - (2 X 106 + 430)/(1.98 X 106 - 20)

~99

Input resistance (R;)

re(rc + R1) = rb + ---=....:.-"-----"'---

(rc + re - rm + R1)

20(2 X 106 + 430) ~ 800 + (2 X 106 + 20 - 1.98 X 106 + 430)

~2755 n

267

236. The solution to the first part of the Question can be found elsewhere.*

. -hie + 11.5 Current gamAie = 1 + hoeRl = 1 + 12.5 X 10-6 X 16000 = 9.6

-h R Voltage gain Ave = '[e I

h;e(1 + hoeR/- IzrehfeR/)

11.5 X 16000 =--------------:----------------:-----------500 (I + 12.5 X 10-6 X 16000 + 2.5 X 10--4 X 11.5 X 16000)

= 285.

IzrehfeRI Input resistance R; = h;e 538 n

1 +hoeRI

Output resistance Roe is found to be

* See, for example, F. A. Benson and D. Harrison, Electric Circuit Theory, Arnold, 3rd Edition, 1975, pp. 366-367.


where Rg = 460 n Roe = 64.5 kn

237. (a) Low frequencies

First neglect C to calculate the mid-frequency gain. Current summation gives:

-- il = - v, + - + -( a) [1 1 1] 1 -a c rc(1-a) Rl Rie

But Vc = i2Rie

~= - arC R I =A·· i1 [R1 + rc(1-a)]r.Rie + Rl rc(1-a) J 1/

L Rl + rc(1-a)

For low frequencies include C

Now C ~Jil = - Vc ~c(11-a) + ~1 + Rie -1j/WC]

and Vc = i2 (Rie - j/wC)

= Ail

Ail -=---------- 1 =---

1 - jftlf

SOLUTION 237 269

where

High frequencies i'

1 + jf/12

(b) 11 hie 12

hoe R,

R+ Intermediate-frequency circuit

11 hie 12

hoe R, R.[ V, Cee

High-frequency circuit


The intennediate-frequency voltage gainAvej = Vo/Vs

= _ g, ( Rie' ) ( RpRie ) m Rs + Rie' Rp + Rie

where gm = hfelhre

1 1 -=h +Rp oe R,

The high-frequency voltage gain Aveh

Greatest value of Ave; is when Rs = 0

Then

Define

__ RpRie Avei - &n R + R-

p Ie

12 = Rie + Rp 2rrCceRreRp

Avei X 12 = ....&.. = gain-bandwidth product. 2rrCce

238. The expression for the stability factor has been derived elsewhere. *

s= Rb+Rc Rc + (l-a)Rb

110 -----=9.2 10 + 0.02 X 100

239. The expression for the stability factor has been derived elsewhere.*

*See, for example, H. Henderson, Transistors, B.B.C. Engineering Training Department, Supplement No. 12, 1958. Methods of obtaining bias and stabilizing the operating point are described in many books, e.g., 1. D. Ryder, Engineering Electronics, McGrawHill, 1957. 1. D. Ryder, Electronic Fundamentals and Applications, Pitman, 3rd Edition, 1964. A. H. Seidman and S. L. Marshall, Semiconductor Fundamentals, Wiley, 1963. L. M. Krugman, Fundamentals of Transistors, Rider and Chapman and Hall, 2nd Revised Edition, 1958. Mullard Reference Manual on Transistor Circuits, 1 st Edition, 1960.

SOLUTIONS 239-241

Here,

so

S = ( where Rb = R1R2 /(R 1 + R2) Re +Rb I-a)

Rb = (50 X 20)/70 = 14.3 kll

S = 14.3 + 2.5 = 6. 2.5 + 14.3(0.02) -

271

240. Using a constant-voltage source equivalent circuit for the F.E.T. the original circuit can be replaced by the one illustrated here.

Let the loop currents be il and ;" rnA as shown.

20kfl 60kfl

The voltage between G and S

VGS' = vol' - 20i1

Also tJ. = g",rd = 5 X 10-3 X 30 X 103 = 150

For the illoop: V = i1(20 + 60 + 30) - ;"30 - tJ.vGS'

From equations (1) and (2) and substituting for tJ.

15lvs= 311Oi1 - 30;"

Now Vo = 15;"

and for the il loop

From equations (3), (4) and (5)

vo/vs = 2.7

241. The equivalent circuit is shown below.

(1)

(2)

(3)

(4)

(5)


200n lfJF i o---e::=::J--I 1

r~ ~1 o--------------~

lkO

1 Impedance of I-JlF capacitor = lOOn X 10--1; il = 3200 il

Impedance of 100-JlF capacitor = 32 il (negligibly small compared to say RL )

Also l/hoe = 106/40 = 25 kil.

RL is made up of the load resistance RL in parallel with 25 kil. i.e. RL = 25 X 10

35 = 7.15 kil.

Vi = h;eii

Vo = - hteiiRL

Vo= -h{eRL = -120 X 7.15 = _ 860 Vi h;e 1

~ = 1000 = 0 295 Va V[(1200)2 + (3200)2] .

Total gain = - 860 X 0.295 = - 254

Output voltage = - 254 X 3 X 10-3 = 0.76 V

. 1200 ° Phase angle between Vi and va is given by tan cf> = 3200 = 0.3 75 so cf> = 20.6 .

Phase from input to output = 180 + 20.6 = 200.6°.

242. The solution is similar to that for Problem 236 so will not be repeated here.

243. The solution is similar to that for Problem 236 so will not be repeated here.

244. The solution can be found elsewhere. * * See, for example, F. A. Benson and D. Harrison, Electric·Circuit Theory, Arnold,

3rd Edition, 1975, pp. 367-368.

SOLUTIONS 245-246

245. The solution can be found elsewhere. *

246.

hre h[eR, __ 2400 _ (2.5 X 10-4)(- 50)(25 X 10) R~ = h;e

1 + hoeR, 1 + 0.1125

=2680 il.

273

12 is divided between 20 kil, 5 kil and R; in parallel. Let the equivalent 2

resistance of these three be Re(kil)

1 1 1 1 -=-+-+- Le:Re = 1.6 kil Re 20 5 2.68

Voltage drop across each parallel branch = 1600/2 and 13 = 1600/2 /2680 = 0.597/2

13/12 = 0.597

Re , through which 12 flows, is the load R'I on the first stage

A - ~ - - hIe _ 24 _ i l - II - 1 + hoeR'1 - 1 + (20 X 10-6)(1.6 X 103) - 23.2

R- = h. _ hreh[eR'1 = 800 _ (8 X 10-4)(- 24X1.6 X 103)

'I Ie 1 + hoeR'1 1 + (20 X 10-6)(1.6 X 10)

= 830 il.

The 2.5-kil bias resistor is paralleled by Ril so not all the generator current enters the transistor.

* See P. K. Yu and C. Y. Suen, 'Analysis of the Darlington Configurations', Electronic Engrg., 40, 38, January, 1968.


2500lg • II = 2500 + 830 .. Idlg = 0.752

2500Rj R-= ! = 623 U

I 2500+Ri --I

. I, I, 13 12 II Overall current gam = - = - . - . - . -

Ig 13 12 II Ig

= 44.4 X 0.597 X 23.2 X 0.752

=463


( 30)2 1 248. Maximum power in load = y2 X 8= 56 W

2 Vee Now Pomax occurs when Ip =--1fRL

and Pc =Pdc-Pac

_ ?:}p _ Ip 2 RL Pe - 1f Vee 2

_ 4 Vce2 _ 2 Vce2 _ 2 Vce2

Pemax - 1f2 RL 1f2 RL - 1f2 RL

Hence if RL is reduced to 4U and Pemax = 30 W

30 = 2 Ve/ 1f2 X 4

so Vcc= 24.4 V


* J. D. Ryder, Electronic Fundamentals and Applications, Pitman, 4th Edition, 1970, Chapter II.

Readers will probably fmd the following reference useful: J. K. Skilling, Pulse and Frequency Response, The General Radio Experimenter, 42, Nos. 11, 12,3, Nov.Dec. 1968. t S. Seely, Electron·tube Circuits, McGraw-Hill, 2nd Edition, 1958, Chapter 8.


285. This circuit and modifications of it have been analysed by many authors so the solution can be found elsewhere. *

286-288. For the solution see the books by Seely.t

289-290. The solutions to these problems can be found elsewhere.:!:

293. The condition for maintenance of oscillation in a tuned-anode oscillator is M = (L + raRC)/ Jl, where M is the mutual inductance between the two coils.

With the higher anode voltage

M = (175 + 220 X 10-6 X 18 X 9000)/9 JlH = 23.4 JlH.

The co efficiency of coupling

= 23.4/V(175 X 60) = 0.228.

With the lower anode voltage

M = (175 + 220 X 10-6 X 18 X 11 000)/9 JlH = 24.3 JlH.

The coefficient of coupling

24.3 = v(175 X 60) = 0.237.

294. The frequency of oscillation of a tuned-anode oscillator is given by

I Jrra +R) 21T \ raLC '

where R, Land C are the usual constants of the anode coil.

= ~ J( 1800 + 11 ) . = X-6 25 21T 1800 X 0.6 XC·· C 67.9 10 F.

* See for example, S. Seely, Electronic Circuits, Holt, Rinehart and Winston, 1968, Chapter 5. J. F. Pierce, Transistor Circuit Theory and Design, Merrill Books Inc., 1963, Section 6.8. F. C. Fitchen, Transistor Circuit Analysis and Design, Van Nostrand, 1960, p. 213. R. J. Maddock,lntermediate Electronics, Book 1, Butterworths, 1969, pp. 158-161. t S. Seely, Electron·tube Circuits, McGraw-Hill, 2nd Edition, 1958, Chapter 8, and

S. Seely, Electronic Circuits, Holt, Rinehart and Winston, 1968, Chapter 5. :j: S. Seely, Electron-tube Circuits, McGraw-Hill, 2nd Edition, 1958, Chapter 8.


The condition for maintenance of oscillation is given in the previous solution.

M= (0.6 + 67.9 X 10-6 X 11 X 1800)/5 = 0.389 H.

M must have at least this value for oscillation. The maximum mutual inductance available is

0.32 X 0.6 = 0.192 H

:. the circuit will not oscillate.

From the maintenance condition the value of C corresponding to the mutual inductance 0.192 H is found to be 18.2 pF.

This capacitance gives a frequency of oscillation of 48.3 Hz.

295. The expressions for the frequency of oscillation and the condition for maintenance of oscillation are identical with those for the tuned-anode oscillator given in the previous two solutions.

296. The tuned-collector oscillator has been analysed by many authors so the solution can be found elsewhere. *

297. The condition for maintenance of oscillation in a tuned-grid circuit is:

pM {l- M} =Rr C pL a

where M is the mutual inductance between the grid and anode coils.

M must be as large as ~ - [(~) 2 - RraLC 1

~ [9 \180 J ((9 X 2180)' - 26 X 11 000 X 180 X 0.0012)] jill

= 39 pH.

But the maximum available M is 0.3 y(I80 X 50) = 28.5 pH. :. the circuit will not oscillate.

* See for example, J. R. Abrahams and G. J. Pridham, Semiconductor Circuits: Worked Examples, Pergamon, 1966, pp. 163-164.

SOLUTIONS 298-299

298. The theorem is well known and will not be proved here. The equivalent circuit of the tuned-anode oscillator is shown.

r--__ ..... Anode

c

'-----""---4 Cathode

. 1 (R + jwL)-. -

Z= jWC 1

R+jwL+-jWC

h = Va/(R + jwL).

Vg = - jwM. Va/(R + jwL).

N= - jwM/(R + jwL).

277

Substitution of the expressions for Z and N in Z + ra/(I + p.N) = 0, equating real and imaginary parts and rearranging gives:

(a) the frequency of oscillationf= ~ = .!..J( ra L+cR ), 271" 271" ra

and (b) the maintenance conditionM= (L + raRC)/IJ..

299. The frequency of oscillation* = 1/271"V[(C(L1 + ~ + 2M)] = f. IfM= 0,/= 1/271"V(0.1 X 10-6 X 40 X 10-3) = 2517 Hz.

In the second case, 2000 = 1/271"V[0.1 X 10-6(40 + 2M)1O-3], where Mis in mHo

M= 11.67 mHo

The coefficient of coupling = 11.67/20 = 0.584.

* See solution to Question 300.



c

The voltage across the bottom coil is

Vg = (R2 + jw~)ic + jwM(ic - ia)

Applying Kirchhoffs second law to the ia and ic loops:

(1)

ia(ra + Rl + jwLl) = ic (R l + jw(Ll + M)} - J.lVg (2)

and ia(Rl + jwLl) = ic(Rl + jWLl + jwM + l/jwC) + Vg (3)

Substituting (1) in (2) and (3) and then dividing (2) and (3)

ra + Rl + jWLl - J.ljwM Rl + jWLl + jwM

Rl - J.LR2 + jw(Ll + M) - jw(~ + M)J.l

Rl + R2 + jWLl + jw~ + 2jwM + l/jwC

Simplifying, and equating the imaginary terms, gives:

2 1 + Rl/ra ( 2 w = = 2rrf) C [(Ll + L2 + 2M) + (Rl~ + ~:L1Xl + J.l)]

. where f is the frequency of oscillation.

Equating the real terms gives:

(ra + R1XRl + R2) - w3(Ll - JJM)(L I + ~ + 2M) + (Ll - JJM)/C

= Rl(Rl - J.LR2) - w2(Ll + M)(L I + M - J.LLz - JJM).


Vg = ic/jwC2

(4)

(1)

SOLUTIONS 301-304

Applying Kirchhoffs second law to the ia and ic loops:

ia/jwC1 = ic(R + jwL + 1/jwC1 + l/jw(2)

and ia(ra + 1/jwC1) = ic(1/jwC1 - p,/jw(2)

279

(2)

(3)

Dividing (2) and (3), simplifying, and equating imaginary and real terms, as in the previous solution gives:

where f is the frequency of oscillation.

(1 + P,)/W2C1C2 = 'aR + L/C1 which is the condition for maintenance of

oscillations.


303. The method of solution is the same as that for Problem 302 so is not given here.

304. The equivalent circuit is shown.

r-------., ega I I I I I C2 I I ra I C I

L - - - - -~ Equivalent circuit of crystal

* See J. D. Ryder, Engineering Electronics, McGraw-Hill, 1957, Chapter 12.


Vg = - i1GwL + l/jwC)

Applying Kirchhoffs second law to the three meshes in turn:

i1GwL + l/jwC + l/jwC1) = ~/jWCl

(1)

(2)

i1GwLIl + ll/jwC- l/jwC1) + ~(1/jWCl + l/jwCga + Ta) = i3Ta (3)

and i1GwLIl + ll/jwC)

(4)

Eliminating i3 from (3) and (4), dividing by (2), cross-multiplying and equating the imaginary terms gives:

W z = _1 [1 + (1 + C';Cga + C/Cga)/Il] = 21T z LC 1 + (1 + CdCga)/1l ( 1)

where f is the frequency of oscillation.

305. The feedback network is shown in the diagram.

outputj. voltage

v

v = E. ZZ/(ZI + Zz)

ZI = Rl + l/jwC1

Input voltage V

Zz = Rz/(l + jRzw(2)

RzV v= ------~--------

(Rz + Rl + Rz(2) + j (R1R2WCZ __ 1_) C1 wC1

:. v will be in phase with Vat a frequency given by:


The system will then oscillate at this frequency if the associated amplifier gain is greater than 1 + Rt/R2 + ~/Cl'

IfC1 = ~ = O.OOIIlF andRl = R2 = 120 kil,

f= 1326 Hz.

306. The phase-shift network is as shown.

C C C

v[) !R) !:~I""""-'-Z!R i' Let currents x,y and z circulate as shown:

Then (R + -. 1_) x - Ry = V jWC

(2R+-.1-)y-RX-RZ=0. jWC

(1)

(2)

and ( 2R + -. 1_) z - Ry = 0 jWC

(3)

i.e.

From (3) Y = (2R + -. 1 ) z/R jWC

(4)

V + (2R + -. 1_) z jWC

From (1) and (4) x = 1 (5) R+-

jWC

Substituting (4) and (5) in (2) gives:

z (R 3 + 6R2 _~ __ 1_) = VR 2 • (6) jWC W2C2 jw3C3

6R2 1 There is no j term when wC = W3C3' i.e. when w 2 = 1/6R2C2,

whenf= 2rrRCv6'


At this frequency z(- 29 R 3) = VR 2 :. v = Rz = - V/29, Le. the attenu

ation ratio of the network is 29, and the total phase shift is 1800 when f = 1/2rrRCv'6.

In this case,f= 1/2rr X 105 X 0.0005 X 1O-6v'6 = 1300 Hz.

307. The phase-shift network is as shown.

R R R

Proceeding with the analysis as in the previous solution it is found that the frequency f at which the network produces 1800 phase shift is v'6/2rrCR.

In this case f = v'6/2rr X 105 X 0.0005 X 10-6 = 7800 Hz.

The attenuation ratio of the network is again 29.

308. Let currents x, y, z and p circulate as shown.

Then

and

From (4)

C C C C

?--II ~ .. ~ II ~ II ~

vIlflflf/f (R +~)X-RY= V

jWC

( 2R + -. 1_) Y - Rx - Rz = 0 jWC

( 2R +-. 1_)Z-RY-RP=O jWC

( 2R + -. 1_) P - Rz = 0 jWC

Z = (2R +_.1_) p/R JWC

(1)

(2)

(3)

(4)

(5)


From (3) and (5)

y = (3R2 + 4R/jwC - 1/w2C~p/R2 (6)

From (2), (5) and (6),

x = (4R3 + 10R2/jwC - 6R/W2C2 - 1/jw3C~p/R3 (7)

From (1), (6) and (7),

p [R4+ 10R3 _15R2 -~ + _1_] =ER3.. (8) jwC W2C2 jw3C3 W4~

There is no imaginary term when w 2 = O.7/R2C2.

At this frequency,pR = v = - V/18.39,

i.e. the attenuation ratio of the network is 18.39.

In this case

f= V(O.7)/21rRC = V(O.7)/21r X 105 X 0.0005 X 10-6 = 2663 Hz.

309. When S is closed, the voltage across C rises exponentially as shown until it reaches Vs. C is then suddenly discharged until the voltage across it falls to Ye. ------------T

v

The voltage across C(Vc) at any time t after closing S is given by

Vc = V(1- e-tfCR)

:. at points A and B

and

Ye = V(1 - e-TdCR)

Vs = V(1- e-TJCR.)


:. period of oscillation

T = T2 - Tl = CR In {(V - Ve)/(V - Va)}·

310. Using the solution to Question 309 and the same symbols,

T= 0.01 X 10-6 X 500 X 103 In {230/(250 - Va)}.

Also Va - Ve = 100 V, so that

Va = 120 V and T= 2.83 X 10-3 s.

Since control ratio is 30, Vg = (- 120/30) V = - 4 V.

311. An expression for the period of oscillation of a multivibrator has been developed by Seely.* For a symmetrical multivibrator the expression can be reduced to:

[V, - V.] T = 2CRg In 1 Vg s

where here.Rg = 50 X 103 n, C= 0.005 J.lF, Va = 250 V, Vi = 110 V and Vg = -20 V.

f= l/T= 1027 Hz.

312. The frequency of a 300-m signal is 1 MHz.

beat frequency = (1.3 - 1) MHz = 300 kHz.

The frequency of a 400-m signal is 0.75 MHz.

new oscillator frequency = (750 + 300) kHz = 1050 kHz.

313. (a) For feedback type of oscillator,

Cmax = (10000)2 = 4 X 104. Grun 50

(b) For beat-frequency oscillator,

Gnax= ( 100+ 10)2 == 12 Grun 100 + 0.05 _ ..

* See S. Seely, Electron·tube Circuits, McGraw-Hill, 2nd Edition, 1958, p. 429.

SOLUTIONS 314 and 324

or

314.

Z;=R + jwL + r/jwC r+ l/jwC

= (R + 1 + ~2C2"" ) + jw (L - 1 + ~C72 ) The real term is zero when

w=±~ J[- (1+~)] Since r is negative

fR=_1 J(~-I) 2rrCr R

The quadrature component is zero when

Cr2 L- =0

or

with values given

1 + w 2C2r2

fx= 2rrJ(LC) j(l- ~2) fR = 504 MHz

fx= 186 MHz

285

fx - impedance is negative resistance and reactance is zero - corresponds to natural frequency of oscillation of circuit.

fR - at frequencies <fR input impedance has a negative resistance component. This is the upper frequency limit for which the circuit is capable of oscillating with additional series capacitance.

324. The thermal agitation noise voltage is given by


where E is the r.m.s. noise voltage in volts. T" " absolute temperature. R" " resistance in ohms. [" " frequency in hertz. [1 and f2 are the limits of the frequency band.

When the integration is carried out over the band,

E = V[4kTR(j2 - [1)] volts.

In this case

E=V[4X 1.38 X 10-23 X 290 X l000(I0,)] V = 12.66JN.

The effect of thermal-agitation noise may be expressed either as an e.m.f. in series with the resistor considered noiseless, or as a constantcurrent generator in parallel with the resistor considered noiseless, as shown.

Noisefree resistorR

¥ Infinite-impedance constant - current generator providing

I r.m.s .

The output current of the generator is obtained by dividing the expression for the r.m.s. noise voltage, given in the previous solution, by R.

fr.rn.s. = j ekT~ - [1») amperes.

In this case fr.ms. = 12.66 X 10-9 A.

325. The r.m.s. value of the noise-current components in a bandwidth Ui - fi) hertz is given by

(ir.m.sl = 2 efUi - [1) amperes2

where f is the average current in amperes.

ir.m.s. = V{2 X 1.602 X 10-19 X 10-3 X 2 X 104) = 2.53 X 10-9 A.

326. The mean square of the fluctuation components of the current depends only on the magnitude of the emission current 10 and the frequency bandwidth.

SOLUTIONS 326-330

Ir.m.s. = V{2 e1o(f2 -II)} amperes

= V(2 X 1.59 X 10-19 X 10 X 10-3 X 20 X 103) A

= 7.98 X 10-9 A.

327. As in the solution to Question 324 the thermal-agitation noise voltage is given by

The impedance of the parallel combination is

(RI - jRI2WC)/(l + RI2W2C2)

R = R1/(l + W2C2R12)

E2 = 4kT tOO Rl dl/(l + 41T2f2C2R12) = kT/C

E = V(kT/C).

287

328. The value of the noisy resistor is approximately 2.5/gm * ohms = 2.5/2.6 X 10-3 = 961 n.

2.5 ( la ) ( 84 ) 329. The equivalent resistor is - -- 1 + - *t ohms gm la + 4 g",

2.5 ( 10 ) ( 8 X 2.5 X 10-3 ) = 9 X 10-3 12.5 1 + 9 X 10-3 ohms = 716 n.

330. The equivalent resistance * = (20Ri + 4 X 104Ia/g",3)/ohms where Rg is the shunt resistance of the grid circuit in ohms, and 1 is the control-grid current in amperes.

* See W. A. Harris, 'Space Charge Limited Current Fluctuations in Vacuum Tube Amplifiers and Input Systems,' R.C.A. Review,S, 50S, 1941 and 6, 114, 1941. K. R. Spangenberg, Vacuum Tubes, McGraw-Hill, lst Edition, 1948, Chapter 12. L. B. Arguimbau and R. B. Adler, Vacuum Tube Circuits and Transistors, Wiley, 1956, Chapter 15. t D. O. North, 'Fluctuations in SpaceCharge Limited Currents in Multi-collectors,'

R.C.A. Review,S, 244, 1940. :j: See K. R. Spangenberg, Vacuum Tubes, McGraw-Hill, 1st Edition, 1948, Chapter 12.


the resistance

= {20 X 1010 + 4 X 104 X IO-J/(5 X IO-J)J}0.01 X 10-6 n = 2003.2 n.

331. Current throughR due to VI with Vz short-circuited

Rz { VI } = R + Rz Rl + RRz/(R + Rz)

= Rz V1/(RR 1 + RzR 1 + RRz)

R

Similarly, current through R due to Vz with VI short-circuited

= Rl VZ/(RR 1 + RzR 1 + RRz)

The Johnson formula for thermal noise generated in a resistor R J is:

Vn z = 4kTRJdf

therefore total mean-square noise current through R per unit bandwidth

4k(T1R 1Rzz + TzRzR 1Z)

(RRI + RzR 1 + RRz)z

4kR1Rz(T1Rz + TzRI)

(RR 1 + RzR 1 + RRz)z

N . 4kR1Rz(T1Rz + TzRl)R Olse power P = z

{(Rl + Rz)R + R 1Rz}

This is a maximum when R = RIRz/(Rl + Rz)

P. = 4kR1Rz(T1Rz + TzR 1)R1Rz max. (Rl + Rz){R1Rz + RIRz}z

= k(T1Rz + TzR 1)/(Rl + Rz)

SOLUTIONS 332-333

332. (a) For two resistors Rl and R" in series, total resistance

=R1+R"

:. per unit bandwidth

4kT1Rl + 4kT"R" = Teff. 4k(Rl + R,,)

where Teff. is the effective noise temperature.

T1Rl + T"R" T. = - ~-=----=:........=. eu. - (Rl + R,,)

. _ T1Rl + T"R" = T1Rl + T"R" USlOg the statement, Teff. - (Rl + R,,) (Rl + R,,) (Rl + R,,)

(b) When the resistors are in parallel

Using the statement, Teff. Tl(~) T2(~) __ ..:.1_+ 2

I I I I -+- -+R" Rl R2 Rl

T1R2 + T"R1 T. - ~-=----=:........=. eff. - (Rl + R,,)

Thus, the statement has been verified for the two cases.

Using the statement:

289

Teff. = Tl {l - exp (- 2al)} + T" exp (- 2aJ) (see diagram)

/,-exP<-2aO

o

---[]eXP(-2ao

I--333. Open-circuit voltage due to Rl with sources of e.m.f. in R2 and R3

branches short-circuited


(RlR2 + RlR3 + R2R3)2

By repeating this calculation for the other two resistors and adding the results the total voltage

4kRlR2R3(TlR2R3 + T~lR2 + T2R~l) (RlR2 + RlR3 + R2R3i

For single resistor at temperature T this voltage squared would be

4kTRlR2R3

(RlR2 + RlR3 + R2R3)

because for the three resistors in parallel

When the resistors are in series:

and 4kTlRl + 4kT2R2 + 4kT3R3= 4kT(Rl + R2 + R3)

T= TlRl + T2R2 + T~3 (Rl +R2 + R3)

334. To receive maximum signal power from antenna the input impedance of the circuit is made equal to the radiation resistance of the antenna.

In bandwidth dfpower radiated from antenna = dV,//4R/

thermal radiation power picked up = 4kTR/df/4R/ = kTdfwatts

If P is the noise power generated in the receiver and G is the power gain

2(Gk 300 + P) = Gk 900 + P

P= 300 Gk


Noise figure = 10 log10 X

{ Output noise power from actual receiver at room temperature }

Output noise power from a perfect receiver that introduced no noise

{300 Gk +P} noise figure = 10 log10 300 Gk = 10 log10 2 ~ 3 dB.

335. Let calibration of signal generator read a power of P watts. Then voltage generated in signal generator = Vwhere V2/(4 X 500) =

P or V2 = 2 X 103 P. Signal on grid of triode = (VX 1000)/1500 = V/1.S. Signal output from receiver IX (V/1.S)2. Noise output power produced by a SOO-n resistor in parallel with a

- 1000 loo0-Sl resistor IX Vn2 , i.e. 4kT -3- . Llfsince 1/500 + 1/1000 = 3/1000.

4kT 1000 Llf= 4 X 1.38 X 10-23 X 1000 X 300 X 104 3 3

= 5.52 X 10-14

(V/l.5)2 = 5.52 X 10-14

V2 = 5.52 X 10-14 X 1.52 = 2 X 103 P

soP=6.2X 1O-17 W

336. The mean-square deflection 82 is given by:

!d2 =kT 2 2

where C is the specific couple of the suspension

0-2 kT 1.38 X 10-23 X 300 0 12 = C = 10-10 = 41.4 X 1 -

i.e. 9" = V(41.4 X 1O-1~ = 6.44 X 10-6 radian.

Thus, r.m.s. deflection = 2 X (optical arm length) X V02

= 2 X 1000 X 10-6 X 6.44 mm = 0.0129 mm.

0.0129 Minimum detectable current = 75 X 1000 mpA = 0.172 mJ.lA.


337.

v c v =---

c 1 + jwCR

where w = 2rr[.

1 fileR dw Mean square output voltage = 4kTRG2 -

2rr w=o 1 + W 2C2R2

Put wCR=x

4kTRG2 II dx Mean square output voltage = 2 --2

rrCR ol+x

kTG2 =--

2C

343. R + jwL = 10.4 + j5000 X 3.67 X 10-3 = 10.4 + j 18.35

= 21.08 /60° 27' Q

G + jwC= 0.8 X 10-6 + j5000 X 0.00835 X 10-6

= (0.8 + j41.75)10-6 = 41.76 /88° 55' X 10-6 S.

Characteristic impedance

J(R + jWL) ° , Zo= G+jwC =711/-14 14 = (689-jI75)Q.

Propagation constantP = V[(R + jwL)(G + jwC)] = a + j{3, where a is the attenuation constant and {3 is the wavelength or phase constant.

P = 0.0297 /74° 41' = 0.00785 + jO.0287

a = 0.00785 neper/km-1

and {3 = 0.0287 rad km-1

The wavelength A = 2rr/{3 = 219 km.

Velocity of propagation

v = fA = w/{3 = 5000/0.0287 = 174300 km S-I.


344. Since the line is terminated by Zo the input impedance is also Zoo Input current when connected to generator (i1)

=2/(600+689-j175)A=0.001539/7°44' A.

Current at receiving end (~)

= i1e-P1 where 1 is 300 km = 0.0001458 /- 485° 16' A.

Voltage across load = i2Zo = 0.1036 /- 499° 30' V.

345. In this case

R + jwL = 10.4 + j18.35 + (7.3 + j5000 X 0.246)/7.88

= 11.32 + j174.35 = 174.35 /86° 18' n G + jwC= 41.76/88° 55' X 10--6 S.

Zo =J(R + ~WL) = 2038/-1° 19' n G+JwC

P = 0.0036 + jO.0850 = a + jJ3

A = 2rr/J3 = 74 km

v = w/J3 = 58800 km S-I.

346. Zo = 689 - j175 = 711 /- 14° 14' n

a = 0.00785 :. al = 0.785 and el:il = 2.192

13 = 0.0287 :.131 = 2.87 radians = 164° 20'.

Load impedance = 500 /45° = (353.5 + j353.5) n = Zr.

Assume the initial voltage at the sending end is 1 L!2 V = Es. Then the

receiving-end voltage due to this is Er = (1/2.192) /- 164° 20'

= 0.456/-164° 20' = (- 0.440 - jO.123) V.

Reflected-wave voltage at receiving-end is

E/ = Er(Zr - ZO)/(Zr + Zo)

E/ = 0.270 / - 51 ° 36' = (0.168 - jO.211) V.


Voltage at receiving end due to E/ is

" 0.270 / - 51 ° 36' ° , . Er = 2.192 /164° 20' = 0.123 / - 215 56 = (- 0.10 + JO.072) V.

Let Is be the component of the sending-end current due to Es' then

Is = 1/(711 / - 14° 14') = 1.407 X 10-3 / 14° 14'

= (1.364 + j0.347)1O-3 A.

If Is" is the component of the sending-end current due to Er", then

0123 / - 215° 56' 1"= . X 1/180°=0.173 X 10-3 /-21°42' s 711 / - 14° 14' -

= (0.161 - jO.064)10-3 A.

Sending-end voltage is

Es + Er" = 1 - 0.10 + jO.072 = (0.9 + jO.072)

= 0.9/4° 35' V.

Sending-end current is

Is + Is" = (1.364 + j0.347)1O-3 + (0.161 - jO.064)10-3

= (1.525 + jO.283)1O-3 = 1.55/10° 30' A.

Receiving-end voltage is

Er + E/ = (- 0.440 - jO.123) + (0.168 - jO.211)

= (- 0.272 - j0.334) = 0.431 /230° 52' V.

Input impedance of line

Z. = 0.9 L.±.l2. Q = 580 / - 5° 55' Q I 1.55 / 10° 30' -

= (576 - j60) Q.

Actual sending-end voltage = 16~: :~i I = 0.986 V.

Since the original assumption that Es = 1/Jl gave a value of sending-end voltage of 0.9 V,

IEr + E/I = 0.431 X 0.986 0.472 V = receiving-end voltage 0.9

SOLUTIONS 346-348

and 111= IEr+E/I= (0.472) A=944 A. r IZrl 500 m

347. 70.n. ,..--c::::::r--+--...,-...,

tv 10V f=lOOOHz

(0 )

c= 0.001 ... F

L= o.002H

lOon

z

(b)

295

0.002 H

100.n.

The actual circuit (a) can be replaced, using Thevnin's theorem by the one at (b).

E= r lOX ~ 1 -j1592 l (10 + 70 + j~C) = 80 - j159.2

Z= .1CX 80 ~(-. 1_+ 80) =-80G159.2)/(80-j159.2) JW '/ I JWC

load current = E/(Z + 100 + jwL) = 0.71 rnA.

av ai --=Ri+L-ax at 348.

ai --=Gv ax

a2v ai a2i - ax2 = R ax + L axat

_ a2i = G av axat at a2v av

--=-RGv-LG-ax2 at Assume V = Vo sin (rrx/l)e-'Yt

a2v rr2 av - =--- V;- =-rV ax2 [2 at


so

i.e.

1r2/1 2 = - RG + -yLG

-y = R/L + 1r2/LG/ 2

349. The line must be half a wavelength long. In free-space the wavelength corresponding to 20 MHz is 15m.

if € is the permittivity of the dielectric 5 = 7.5/V€,

€ = 2.25.

For a short-circuited line of length I the input impedance

ZI = Zo tanhPI.

For an open-circuited line of length I the input impedance

Z2 = Zo coth PI

ZIZz=Z02.

In this case Zo = V(4.61 X 1390) = 80 il.

I 2· J(4.61) Also, ZI Z2 = tanh PI, .. tanh PI = 1390 = 0.05758.

The attenuation constant = 0.05758/5 nepers m-I = 0.1 dBm-l .

Velocity of propagation = velocity of light/V € = 2 X 108 ms -I.

350. If a line with a characteristic impedance Zo is terminated by an impedance ZL the voltage reflection coefficient pei° is (ZL - ZO)/(ZL + Zo).

Let the voltage of the wave travelling towards the load be V cos w(t + x/v) at a distance x from the load. Then the voltage at the load due to this wave is V cos wt. Thus, at a distance x from the load, the reflected wave is pV cos {w(t - x/v) + O}. The total voltage at distance x = Vx = V cos

w (t +;)+ pV cos {w(t - x/v) + O}. The amplitude of Yx = I Yxl =

Vv[1 + p2 + 2p cos (0 - 2wx/v)]. Voltage standing-wave ratio (as a quantity> 1)

I Yxlmax 1 + P =r=--=--IYxlmin 1 - p.


Also, position of first voltage maximum is given by 8 - 2wx/v = 0, i.e. when 8 = 4rrx/A.

In this case r=2 :.p=(r-1)/(r+ 1)= 1/3

x = A/12 :.8 = rr/3.

(ZL - Zo)/(ZL + Zo) = pei° eirr/3

1 + pei° 1 +3" 8 + j3"'3 ZdZo = 1 - pei° = ei"/3 = 7

1--3

ZL = 70(8 + j3"'3)/7 = 80 + j52 ll.

351. The solution to this problem can be found elsewhere.·

352. Normalized terminating impedance

= (37.5 + j52.5)/75 = 0.5 + jO.7.

This point (A) can be located on either the Cartesian or Smith Charts (not shown).

On the Cartesian Chart, at A the values of u and v are:

Uo = 0.325, Vo = 0.11 A.

At the input B: UI = 0.325 since there is no loss and

VI = O.llA + 0.30A = OAlA.

B corresponds to z = 0.42 - j0.55.

Hence, input impedance = 75(0.42 - j0.55) = (31.5 - j41.2) ll.

On the Smith Chart the value of U need not be found since the U circles are all centred at the origin of the chart. Hence, movement of point A on the circle centred at the origin through a distance O.3A towards the generator locates point B. Again B corresponds to z = 0042 - j0.55 and so the input impedance is (31.5 - j41.2) ll.

The Cartesian diagram will now be used to find the input impedance when the loss in the line is 1.15 dB.

* See W. Jackson, High Frequency Transmission Lines, Methuen, 3rd Edition, 1958, Ch!lpter 6.


1.15 dB = 1.15/8.686 = 0.132 neper.

Hence, input impedance is located at a point H' for which VI' = Vo + 0.30X = 0.41X and Ul' = Uo + 0.132 = 0.457. Point H' gives z = (0.56-j0.48), i.e. input impedance = (42 - j36) n.

In using the Smith Chart to solve this part of the problem the value of Uo can be read off the pre-calibrated cursor to be 2.85 dB. Then ut' = Uo + 1.15 dB = 4.0 dB. Point H' is then located at the intersection of the 4 dB circle and the 0.41A line.*

The voltage standing-wave ratio existing in the line is equal to the intercept of the U circle on the resistive axis, i.e. 0.315.

353. The input impedance (Zl) of a fmite short-circuited line oflength I is Zo tanh PI, where P = Q: + j~ is the propagation constant.

If 1= nX/4 where n is an odd integer and X is the wavelength, Zl = Zol tanh al which is approximately Zolal.

Now Q: is approximately R/2Zo.

Zl = 8Zo2/RnX

R = 41.6V![1/a + I/b ]10-9 nm- l

Zo = 138 10giO (b/a) n

:. if n = I, in this case, Zl = 248600 n.

354. The selectivity of a parallel tuned circuit may be expressed in terms of the 'Q' of the coil; and, by analogy, an expression for the 'Q' of a resonant line may be obtained.

R c

L

* Further information on transmission-line charts can be found in: W. Jackson and 1. G. H. Huxley, 'The Solution of Transmission Line Problems by Use of the Circle Diagram of Impedance,' J.I.E.E., 91, Part III, 105, 1944, and P. H. Smith, 'Transmission Line Calculator,' Electronics, 12,29, January, 1939.


Consider the parallel tuned circuit shown. Its impedance

Z= R +jwL 1 - w 2LC + jwCR

(1)

At resonance WR 2LC = 1 and Q = wJ!-/R is large compared with 1, hence

ZR is approximately L/CR (2)

Now the impedance at WR + liw near resonance is

Z = jL(WR + liW)/U(WR + liw)CR - 2wRliwLC1 (3)

Z = ZR(1- j2Qliw/WR) (4)

The impedance of a short-circuited "A/4 line at resonance is ZR = ZoIotI, as mentioned in the previous solution, and its impedance in general is Z = Zo tanh (a + j/3)/, where /3 = 21Tf/e = WR/C and e is the velocity oflight.

When operation is at wR + liw,

/3 = (WR + liw)/e= 21T/"A + liw/e

/31 = 1T/2 + I. liw/e.

Z = Zo tanh (otl + jliw. I/e + j1T/2)

Since otI and liw. I/e are small:

Z = ZR [1 - j2(wR/2ea)liw/wR1

(5)

(6)

Now equations (4) and (6) are of the same form so the Q of the resonant line is

wR/2ea = 1Tf/ac = 21TfZo/Re . (7)

Substituting for Zo and R and putting in the values of the constants gives:

Q= 1468.

355. For a short-circuited line Zl = Zo tanh (otl + j/3l), where a is the attenuation constant and /3 is the phase constant.

Expanding and manipulating:

Zl = Zo{sinh 2ot1 + j sin 2/3l}/ {cosh 2ot1 + cos 2/3l}

But, as otI is small, cosh 2ot1 !:::! 1 and sinh 2 otI !:::! 2ot1

Zl = Zo{otI/cos2 /31 + j tan /3l}.


Now a ~ R/2Zo so,

Zl = Rc(I/X)/2f cos2 2rr(I/X) + jZo tan (2rrl/X).

In the case of ordinary reactances, the change of reactance ~X produced by a fractional change of frequency ~f/f is ~X = X . ~f/f

In the case of the above line, X = Zo tan (2rrl/X) = Zo tan (2rrfl/c)

dX = 2rr(I/X)Zo(df/f)/cos2 (2rr/jA).

Multiplying both numerator and denominator by tan (2rrl/X) and reducing gives

dX = 4rr(I/X)(df/nX/sin (4rrl/X)

Selectivity factor of line reactance 4rrl/X

Selectivity factor of lumped reactance sin (4rrl/X)"

When I/X = 0.2 this ratio is 4.28.

356. The attenuation in a coaxial line is given by Jackson* as:

a = 9.95 X 1O-6v (fe/eo) l/avUa + l/bvob /loglo(b/a)

+ 9.10 X lO-sv(e/eon tan [) dB m-1

where f is the frequency in hertz, e/ eo is the ratio of the permittivity of the line dielectric to that of air, a and b are the radii of the inner and outer conductors respectively in metres, 0a and 0b are the conductivities of the inner and outer conductors in siemens per metre cube and tan li is the power factor of the dielectric.

Here Ua = 5.62 X 107 siemens per metre cube, 0b = 1.54 X 107 siemens per metre cube, a = 5.65 X 10-4 m,b=3.97X 1O-3 m,e/eo= 1, and tan li =0.

The characteristic impedance of a coaxial line is also given by Jackson t as:

Zo = 138v(eo/e) loglo (b/a) ohms = 117 n.

* See W. Jackson, High Frequency Transmission Lines, Methuen, 3rd Edition, 1958, p.50.

tibid, p. 46.


357. Using the expression quoted in the previous solution and remembering that e/fo = 2.25 and tan [) = 0.0004 it is found that,

0: = 0.977 dB m-1 and Zo = 78 n.

358. The formulae for calculating the characteristic impedance Zo and the attenuation 0: are given by Blackband and Brown. *

0: = 8.686 tanh-1y(gmin/gmax) dB.

From the circle (given with the problem),

gmin = 0.489,gmax = 1.202, so 0: = 6.55 dB.

Zo = Zo' y(gmin . gmax), where Zo' is the characteristic impedance of the measuring line.

Zo = 75/Y(0.489 X 1.202) = 97.8 n.

359. The zero-susceptance points are seen from the circle (given with the problem) to be g'min = 0.492 and g'max = 1.198.

0: and Zo are now calculated as in the previous solution.

0: = 6.60 dB.

Zo=97.7 n.

360. A quarter-wavelength section of line of characteristic impedance Zo = y(150 X 75) n, i.e. 106 n provides the desired transformation,

eliminating a reflected wave on the 75-n line.

361. The admittance of the load = 1/(100 + j100) S. = (1 - j)/200 S.

Susceptance of load is thus (- j/200) s, so that of the stub must be (+ j/200) S. The conductance of the load is then (1/200) S, i.e. the load resistance = 200 n. To match this load to a line of characteric impedance 500LQo n, the quarter-wave line must have an impedance ofy(500 X 200)n

= 316 n. * W. T. Blackband and D. R. Brown, 'The Two-point Method of Measuring Charac

teristic Impedance and Attenuation of Cables at 3000 Mc/s,' J.LE.E., 93, Part IlIA, p. 1383, 1946.


The susceptance of a short-circuited line of characteristic impedance Zo and length I is

- j/Zo tan (31 = - j/Zo tan (2rrl/'A).

Thus, - j/Zo tan (2rrl/'A) = j/200.

tan (2rrfl/c) = - 200/Zo = - 200/316 = 0.6328

2rrfl/c = 2.578 radians.

Now f= 100 X 106 Hz, c = 3 X 108 ms-1, so the minimum length ofline I is 2.578 X 3 X 108/2rr X 108 = 1.23 m.

362. For the explanation asked for see the book by Kraus.*

363. The mesh equations are:

Vt = 60/1 - 50/3

90/3 = SOIl + 20/4

80/4 = 20/3 + 10/2

f2= 10(/4 - 12)

If 13 and 14 are eliminated two equations remain which can be put in the form:

Vt = A f2 + BI2 and 11 = Cf2 + DI2

where A = 20.8, B = 179 n, C = 0.68 S and D = 5.9.

(1)

(2)

(3)

(4)

364. Let the network be represented by the T -section shown. If the image impedance Zj is connected across terminals 3 and 4 the impedance measured between 1 and 2 is also Zj.

2o-______ ~------~4

* J. D. Kraus, Electromagnetics, McGraw-Hill, 1953, pp. 439-441.

SOLUTION 364 303

(1)

If the current flowing into 1,2 is il and the current flowing out of 3,4 is ~, the image transfer constant is given by

e+6 = il/~ (2)

Voltage across terminals 3 and 4 = i2Zt

Voltage across Zb

From (2) and (3),

:. from (1) and (4),

Z? = Za2 {I + 2 (Zj/Za + 1)/(e8 - I)}

200Jl 146.8Jl 146.8Jl

I 10V 1153Jl lOOOJl

In this case 8 = 0.5 and Zt = 600 12 :. Za = 146.812.

Also Zb = Za(Zj/Za + 1)/(e8 - 1) = 115312.

R 146.8Jl

\ 100011 If

L

Using Thevenin's Theorem, the network can be replaced by the one shown where

E = 10 X 1153/(1153 + 200 + 146.8) = 7.68 V

and R = 1153 X 346.8/(1153 + 346.8) = 266 12.

h = 7.68/(266 + 146.8 + 1000) = 5.44 rnA.

(4)

(5)


365. The conditions for zero output are:*

w2 = I/R3el~(Rl + R2) and w2 = (e1 + ~)/(el~R1R2)' In this casef= w/2rr = 1240 Hz.

366. (a) With the output short-circuited,

and

" "

Zsc = Rl + R1R2/(R1 + R2)

" open-circuited, Zoe = Rl + R2

Z/ = ZoeZsc = R12 + 2R1R2

tanh2 8 = Zsc/Zoe = (R12 + 2R1R2)/(R1 + R2)2.

It follows that R2 = Z;!sinh 8 and Rl = Zj tanh (8/2).

If N is the voltage ratio = efJ,

Rl = Zj(N - 1)/(N + 1) and R2 = 2NZ;!(N2 - 1).

When the loss is 10 dB,N= 3.162; alsoZj = 600 n Rl = 311.8 n andR2 = 421.6 n.

When the loss is 20 dB,N= 10; alsoZj = 600 n Rl = 491 n andR2 = 121 n.

(b) Let the elements of the attenuator have resistances (Rl/2) and R2 as illustrated.

Then,

and a: = cosh-1(1 + RdR2)

From (1),600 = 480(480 + R2)

R2 = 135 n

* A. T. Starr, Electronics, Pitman, 2nd Edition, 1959, pp. 155-6.

(1)

(2)

SOLUTIONS 366-369

From (2) a = cosh-1 (1 + 480/135) nepers = 19.1 dB.

367. With terminals 3 and 4 open-circuited:

Za + Zc = (250 + jl00) n With terminals 3 and 4 short-circuited:

Za + ZbZc/(Zb + Zc) = (400 + j300) n With terminals 1 and 2 open-circuited:

305

(1)

(2)

Zb + Zc = 200 n (3)

From (1), (2) and (3): Za = (150 + j300) n,zb = (100 + j200) n,zc= (100 - j200) n.

368. The solution to this problem has been given elsewhere. *

369. The equation relating the input voltage Vi and current 11 with the output voltage J-2 and current 12 are:

For circuit (a),

and

Vi = A J-2 + BI2

11 = CJ-2 + DI2

Vi = J-2 + ZI2

11 = 12

Comparing equations (1) and (3), also (2) and (4),

A = 1, B = Z ohms, C = 0, D = 1.

The transfer matrix [AJ is there [~ ~J.

For circuit (b),

and

Vi=J-2·

11 = YJ-2 + 12 Comparing equations (1) and (5), also (2) and (6),

A = 1, B = 0, C= Ysiemens, D = 1.

(1)

(2)

(3)

(4)

(5)

(6)

* F. A. Benson and D. Harrison, Electric Circuit Theory, Arnold, 2nd Edition, 1963, p.128.


.. [1 0] The transfer matnx [A] IS therefore y 1 .

The transfer matrix of the network and load in cascade is

[A]=[~ ~l[I:Z1 ~l=[~:~i ~]. Thus, the input impedance

and the voltage gain

=---

For the common-base transistor:

so

where

[Z] = ['11 '12] '21 '22

r IZJ [A]= '~1 '21 '22

The input resistance

'21 '21

'11RI + '11'22 - '12'21 '11RI/'21 + IZI!'21

RI/'21 + '22/'21

= '11 - '12'2d('22 + RI)

370. Circuit (a) consists of a series impedance and a shunt admittance in cascade. Therefore, using the results of the previous solution, the transfer matrix [A] is:

~ ~J . [~ ~] = [0 +yZY) ~J

SOLUTIONS 370-371

Circuit (b) consists of a shunt admittance and a series impedance in cascade. Thus,

[AJ= [~ ~1· [~ ~] = [~ (1fzY)]·

307

Circuit (c) can be considered as a series impedance, a shunt admittance and a series impedance in cascade. Therefore:

371. The transfer matrix for the first network excluding the load is:

[AJ = P jWL]. [.1 01. p jWL] . [.1 0] LO 1 ]wC IJ LO 1 ]wC 1

= r(1 - 3w2LC + w4L 2C2) (2jwL - jw3L 2C)] L (2jwC - jw3LC~ (1 - w2LC) .

The transfer matrix of a network which has general parameters A, B, C and D, and a load Z, in cascade is:

fA +B/Z, Bl LC+D/Z, DJ.

Thus, ~/Vi = I/(A + B/Z,).

Here

and

A = 1-3w2LC+ w4L2C2,

B = 2jwL - jw3L2C,

Z, = V(L/C) + jwL/2.


The transfer matrix for the second network excluding the load is:

[A] = [(1- w 2LC) (2jwL - jw3L2c)1 OwC) (1 -w 2LC) J

Rl ~/Vi= 2 3 2 (1- W LC)Rl + (2jwL - jw L C)

372. (a)

For network 1:

For network 2:

Network 1

Network 2

t t [AJ = [AI Bl]

C, 0, [AJ = [~ ~ J

Vi =A~ +B12

II = C~+DI2

~ = Al Ji3 + BI/3

12 = CI V3 + DI/3

Vi = (AAI + BCI) V3 + (ABI + BDDI2

II = (CAl + DCI) V3 + (CBI + DDI)/3

(b) When the two networks are connected in order 1, 2 the transfer matrix of the combination is the product of the original transfer matrices, i.e.

[A ] = [A ] . [A ] = P·50 11 J . [1.66 4J = p3.5 39.0] 1,2 I 2 LO.25 2.5 1 3 L 2.92 8.5.

The Y parameters of the networks are as follows:

SOLUTIONS 372-373

Network 1 Network 2

Yll = DIB 2.5/11 3/4

Y12 =-IAI/B ~1/11 -1/4

Y2l = liB 1/11 1/4

Y22 = -AlB -1.5/11 -1.66/4

The admittance matrix of the networks in parallel is

[Yi,2] = [Yi] + [1'2]

= [2.5/11 - 1/11] [3/4 - 1/4 J 1/11 - 1.S/11 + 1/4 - 1.66/4

r43/44 - lS/44] ltS/44 - 73/132 .

373. The mesh equations for the circuit are:

Z2(Il - z) + Zl(Il - i + 12') = Vi

~(i - 12') - Zl (II - i + 12') = J.-2

309

(1)

(2)

Zli + Z2(i - 12') = Zl(Il - i + 12') + Z2(J1 - i) (3)

From (3), i = (II + 12')/2 (4)

Substituting (4) in (1) and (2) there results:

~(Il - 12') + Zl(Il + 12') = 2Vi .

Z2(Il - 12') - Zl (II + 12') = 2 J.-2 •

Thus, the [Z] matrix is seen to be

rZl + Z2)/2 (Z2 - Zl)/21 L(h - Zl)/2 (Zl + Z2)/2J .

(S)

(6)

Equations (S) and (6) can be re-arranged to give the A, B, C and D parameters of the [A] matrix. Alternatively,

A = ZlJZ12 = Z2JZ12 = D = (Zl + Z2)/(Z2 - Zl),

B = (ZllZ22 - Z122)/Z12 = 2Z1Z2/(Z2 - Zl),

310

and


c = l/Z12 = 2/(~ - Zt).

[A J = f(Zt + ~)/(Z2 - Zt) L 2/(~ -Zt)

2Zt~/(Z2 - Zt) J (Zt + Z2)/(Z2 - Zt) .

374. The transfer matrices are:

ForZh [~

For Y, [~

Zt] 1 ,(see Solution 369)

0] . 1 ,(see SolutIOn 369)

r1 Z2] For Z2, Lo 1 ,(see Solution 369)

For the transformer, [~ l~n]' (for a transformer Vi = nV2 and It =

I21n)

:. Transfer matrix for whole arrangement is:

[~ ~tJ. [~ ~] . [~ l~n]' [~ Z~] = rn (1 + Zt Y ) ~(1 + Zt Y ) + Zt/n 1

nY nZ2 Y + lin J

375. For the 1( network

For the lattice network

For the T network

:. for the whole network

[AJ= [4~R ~] [AJ = [2~R 4:]

[AJ = [3/2 5R/21 1/2R 3/2J

[A] = [4~R ~]. [2~R 4R] r 3/2 5R/2] 3 . 1l/2R 3/2

= [57/2 119R/2] 79/2R 165/2

SOLUTIONS 375-377

. A + BIZ, AZ, + B Input Impedance Zu = C + DIZ, CZ, + D

57R, + 119R =

79R,fR + 165

Vout -=---2R,

Vin A + BIZ,

376. I, ...... f----- ~ 12

where

~ 00-------------00 ~

v,! 00-------------00

Vi = J.1 cosh 11 + I2Zo sinh 11

It = J.1 Yo sinh 11 + 12 cosh 11

11 = 0 and Yo = I/Zo.

[A] = r cosh 0 Zo sinh OJ lro sinh 0 cosh 0 J

For the T network

[A] = P ZlJ. rIO] . [1 Zl] Lo 1 It/~ 1 0 1

= [1 + Zt/Z2) Zl(ZtZ 2Z2)]

I/Z2 (1 + Zt/Z2)

Equating the two [A] matrices it is found that

ZI = Zo(cosh 0 - 1)/sinh 0

and Z2 = Zo/sinh 0

377.

(a) ::I::[?" :

311

Xc = l/jwC

G= I/R


The A element only is required

this is 1 + 6XcG + 5 Xc2 G2 + Xc3 G3

5 6 1 or 1 - -- + -- - ----:,...-:---:

W 2C2R2 jwCR jw3C3R 3

Le. 1 1

w = y6CR or f= 2rry6CR

Under this condition A = - 29 so the required gain of the amplifier is >29.

(b)

1 Z=R+ jwC

1 • C Y Y=7i+/ w

Transfer function = 1/(1 + ZY)

3+j (WCR-_1 ) wCR

The j term is zero when w = I/RC Le.f= 1/2rrRC.

(i)

Gain of amplifier required is then~.

378.

1 - n2 [1 + jB]

1 + n2 [1 + jB]

SOLUTIONS 378 and 391-395

I-n2 -jn2 B _(I-j) . = 1 + n2 + jn2 B - 3 + j (given)

:. Equating real and imaginary parts

Check:

(li)

n =y2 B = 0.5

1 - [1/n2 + jB] S22 = 1 + [1/n2 + jB]

(n2 -1)-jn2 B I-j . - --(Olven) - (n2 + 1) + jn2 B-3 + j 0-

n=y2 B=O.5

Vi- + 1 1 S12 = - when Vi = 0 = - = -f2+ n y2

391. The solution to this problem can be found in certain standard textbooks. *


393. Critical wavelength Ac = 2 X 0.0762 = 0.1524 m.

Guide wavelength Ag is given by I/Ag2 = 1/0.12 - 1/0.15242, since a wavelength of 0.1 m corresponds to the frequency of 3000 MHz

Ax = 0.133 m.

394. The solution to this problem can be found in certain standard textbooks.t

313

395. It can be shown that the attenuation in nepers per unit length of a guide carrying an eyanescent mode is given by::j:

* E.g., see H. R. L. Lamont, Wave Guides, Methuen, 3rd Edition, 1950, Chapter 1. t E.g., see H. R. L. Lamont, Wave Guides, Methuen, 3rd Edition, 1950, Chapter 2. :j: See, for example, H. M. Barlow and A. L. Cullen, Microwave Measurements,

Constable, 1950, p. 245; or L. G. H. Huxley, A Survey of the Principles and Practice of Waveguides, Cambridge U.P., 1947, p. 57.


Here ~ = 2.61a, where a is the cylinder radius· and A = 0.3 m. Also a = 1000 dB m- I = 1000 X 0.115 nepers m- I .

a= 0.0206 m.

396. A general formula for calculating the attenuation in a waveguide caused by losses in the wall metal has been given by Kuhn.t For the case of the HOI mode in a rectangular waveguide the attenuation a is given by the following expression:

[c PI (€) 1/2] 1/2 1 1 (1) 1/2 a= _7I.-'; J; b3/2·[Ae(I_ Ae:)11/2 "2

L~r ~r J [t: + ~J (1)

where a and b are the short and long internal dimensions of the guide respectively, Ae is the wavelength in the unbounded dielectric, Acr is the critical wavelength of the guide, a is the conductivity of the wall metal, PI is the permeability of the wall metal, € and P are the dielectric constant and relative permeability of the dielectric respectively, and c is the velocity of electromagnetic waves. In the case of an air-filled copper guide with

resistivity 1.7 X 10-8 ilm the factor [~. :1 .(~f/2r/2 = 0.2065, if a is

measured in dB m-I , the guide dimensions and wavelengths being in centimetres.

Expression (1) gives a in a form in which the ratio Ae/~r is the only parameter involving wavelength. An alternative expression is:

a = [~.;;. (; y/2] 1/2 b C:j3/2 [~: + ~] . where Ag is the guide wavelength and

1 1 0=0-) 2

g e ''Cr

* See Problem No. 392.

(2)

(3)

t S. Kuhn, 'Calculation of Attenuation in Waveguides,' J. Ins tn. Electrical Engrs., 3, Part IlIA, 663-78, 1946.


Consider the case where Ae = 3.1 cm,

b = 2.54 cm, b/a = 2, her = 2b = 5.08 cm

1 1 1 Ag2 = 3.12 - 5.082' i.e. Ag = 3.92 cm.

From (2),

_ 0.2065 X 3.92 X [(3.1)2 + 11 dBm-1 = 0 0801 dBm-1 a - 2.54 (3.1)3/2 li5.08)2 J . .

Similarly if Ae = 3.2 em, a is found to be 0.0814 dB m -I.

397. The losses due to the wall metal can be calculated from equation (2) of the solution to Question 396. It should be remembered, however,

that the constant 0.2065 previously used for [~. :1 (~) 1/2] 1/2 should

now be multiplied by (2.55)1/4 since € = 2.55. In this case Ae = 1O/V2.55 = 6.318 em, her = 2b = 9.6 cm, so Ag =

8.391 cm. Further b/a = 3,

a = 0.05543 dB m-I •

A formula for calculating the loss in the dielectric of a waveguide has been given by Kuhn.* The attenuation constant ad in dB m-I is given by:

2726 I [ ( Ae ) 2] 1/2 ott = x:- [tanli] 1- her .

In this case, Ae = 6.318 em, tan li = 0.0006 and her = 9.6 cm.

ott = 0.3438 dB m-I .

The total value of the attenuation in this guide is

(0.05543 + 0.3438) dB m-I = 0.399 dB m-I •

398. The guide wavelength Ag = 2(5.731 - 3.749) X 10-2 m = 0.03964m.

·S. Kuhn, 'Calculation of Attenuation in Waveguides,' J. [nstn. Electrical Engrs., 93, Part IlIA, 663-78,1946.


The voltage standing-wave ratio is*t

r = [K2 - cos2 (1TW/Ag)]~/sin (1TWIXg) = 15.25

[For high voltage standing-wave ratios r ~y'(~ -1)Xg/1TW, i.e. r ~ 15.17, i.e. less than 0.5 per cent error.]

The loss in the component

0: = 10 10glO [(r + 1 )/(r - I)} dB = 0.581 dB.

402. The cut-offfrequency fc:j: = l/'rry'(LC) = 1000 Hz

L L

Now load resistance = y'(LIC) = 50 n . From (1) and (2), C= 6.37 J.lF andL = 15.92 mHo

Attenuation constant 0: (per section)

= cosh-1 (- 1)(1 + Zt/2Z2)

where Zl = jwL, and Z2 = l/jwC,

= cosh-1 (- 1)(1 - 2f2lfc 2)

0:= cosh-1 (-1) {1-2 (~fc2) / fc2 } = 1.928.

(1)

(2)

* See S. Roberts and A. von Hippell, 'A New Method of Measuring Dielectric Constant and Loss in the Range of Centimetric Waves,' 1. App, Phys., 17, 610, 1946; or H. M. Barlow and A. L. Cullen, Microwave Measurements, Constable, 1950, Chapter 5; or F. A. Benson, 'Waveguide Attenuation and its Correlation with Surface Roughness, Proc. IE.E., 100, Part III, 85,1953. t Here the voltage standing-wave ratio is measured as a quantity greater than unity. :j: See, for example, L. C. Jackson, Wave Filters, Methuen, 3rd Edition, 1960,

Chapter 2.


m = v(l - W//Woo~

When

= VI - 10002/12002

= 0.55

Lim = mL = 0.55 X 15.92 = 8.8 mH

~= (1-m2)L/4m = 5 mH

~ = mC= 0.55 X 6.37 = 3.5211F --

403. (a) The frequencies I! and 12 at the ends of the pass band are given by:*

and

11 =fc [J(ci + 1) -J(ci)] 12=fc [J(ci + 1) + J(ci)]

where fc is the resonant frequency of both ar~s = 1000 Hz, C1 is the capacitance in the series arm and C:2 is the capacitance in the shunt arm.

and

11 = 1000 [V1.01 -VO.01] = 905 Hz

12 = 1000[V1.01 + VO.01] = 1105 Hz

bandwidth = (J2 - fi) = 200 Hz.

(b) The T-section is illustrated in the figure.

* See, for example, L. C. Jackson, Wave Filters, Methuen, 3rd Edition, 1960, Chapter 2.


The angular frequencies W1 and W2, corresponding to frequencies!1 and 12 at the ends of the pass band are, from the previous solution:

and

where

W1=WC [J(~+I)- J(~)]

W2=Wc [J(~+ 1)+J(~)J wc2 = I/L1C1 = 1/L,.~

The iterative impedance*

where

From these equations and assuming Zo = R:

L1 = 2R/(~ - W1) = 2 X 600/3000 X 21T = 63.6 X 10-3 H

each series inductance = 31.8 mH

~ = Lt/R2 = (63.6 X 10-3 X 106)/6002 pF = 0.177 pF

L,. = (~- w1)R/2~Wl

= (3000 X 21T X 600 X 106/2 X 21T X 120 X

21T X 123 X 106)pH

=9.7 pH

C1 = L,./R2 = (9.7 X 10-6 X 1012/6002) ppF = 26.95 ppF

each series capacitance = 53.9 ppF.

404. The cut-off frequency fc * = 1 /4rr<./LC = 2500 Hz

Now load resistance = v(L/C) = 600 n L = 19.1 mH and C= 0.053 pF.

* See, for example, L. C. Jackson, Wave Filters, Methuen, 3rd Edition, 1960, Chapter 2.

(1)

(2)

SOLUTIONS 404-410

L



407. The solution to this problem can be found elsewhere.:j:

408. The solution to this problem can be found elsewhere.§

409. Cut-off frequency II fc = 1/1TV(LC) = 796 Hz.

Terminating impedance II = v(L/C) = 600 n.

319

It follows that L = 240 mH and C = 0.666 JlF. A T-section would therefore be constructed from two 120-mH (L/2) series inductors and a 0.666-JlF capacitor in the shunt arm.

For a 1T-section there would be a 240-mH series inductor with two 0.333 JlF (C/2) capacitors as shunt elements.

410. The series-derived T-section is shown in diagram (a) and the shuntderived 1T-section in diagram (b).

Here Zl = wL and Z2 = l/wC

:. for (a) each series inductance

= mL/2 = 0.6 X 120 mH = 72 mHo

Also, inductance in shunt arm

= (1 - m2)240/2.4 mH = 64 mHo ---* See, for example, L. C. Jackson, Wave Filters, Methuen, 3rd Edition, 1960,

pp. 12 and 13. t ibid., pp. 27 to 30. See, also, the Solution to Problem 410. :j: ibid., p. 32. § ibid., p. 43. " ibid., pp. 15-17.


mZ,

mZ,12 mZ,12

4mZ2/(1-m2) 2Z/m 2Z,Jm

(a) (b)

Capacitance in shunt ann

= 0.666 X 0.6p.F = OAp.F.

Similarly, for 7T-section, inductance in series arm

= 0.6 X 240 mH = 144 mHo

Also, capacitance in series arm

= 0.666 X (1- m2)/4m p.F = 0.178p.F.

Capacitance of each shunt arm

= 0.333 X 0.6p.F = 0.2p.F.

411. The limits of the pass band are given by:

Z 4~ = 0 or - 1 where Zt = l/jweC and Z2 = jWeL

i.e. - 1/4we2 LC= 0 or-I

we = 00 or 1/2v'CLC)

Ie = 00 or 1/47Tv'CLC)

Zo=v'[Zt~(1 +Ztl4~)]= Jl~ 1-4)LC1 = J[% (1-~)] o w

a

o -7TI---J

w

SOLUTIONS 412- 413

412. (i) section has/oo = 1.05 fc (ii) section has /00 = 2 fc

(iii) section has /oa = 00

(iv) section has/oo = 1.25 fc

413. Magnitude function is:*

. 1 INOw)1 = [1 + e2 en 2 (w)]t

The tolerance band edge is unity

1. IN(jI)1 = (1 + e2)t

For allowable 0.5 dB tolerance N (1) = 0.945

e = 0.347

Parameter n (no. of poles) given from

1 IN (2)1 = (1 + e2 cosh2 n cosh-1 2)t = 0.125

n = 2.94

n must be an integer so take n to be 3. Knowing nand e

321

* Modern filter theory is discussed in the books F. R. Connor, Networks, Arnold, 1972, Chapter 6 and J. D. Ryder, Introduction to Circuit Analysis, Prentice-Hall, 1973, Section 10.8.


1T X = - (2k + 1) which gives location of poles in left-hand plane

2n

Poles are found to be s, = -0.628

S2= -0.314 + j 1.022

s3=-0.314 -j(1.022)

K N(s) - -------

- S3 + 1.256s2 + 1.549s + 0.72

where K is a constant.

416. The radiation resistance of a small loop antenna* is

R = 31171 (A/A?)2 ohms

where A is the area of the loop (m2) and X is the wavelength (m).

Here A = 1T(X/41Ti/4 and R = 0.77 12.

417. The radiation resistance R of a short dipole ist 801T2(f/X)2 ohms where f/X is the line length in wavelengths.

In this case, R = 801T2 X (1/12)212 = 5.5 12.

The directivity is defined as the ratio of the maximum radiation intensity to the average radiation intensity. For a short dipole this is * 1.5. ---

418. It is often useful to consider that a receiving antenna possesses an aperture, or equivalent area, over which it extracts energy from a radio wave.

The aperture A is given by§ DX2/41T where D is the directivity and X is the wavelength.

In this case A = 90 X 4/41T m2 = 28.6 m2•

* See 1. D. Kraus, Electromagnetics, McGraw-Hill, 1953, pp. 486-9; or S. Ramo and 1. R. Whinnery, Fields and Waves in Modern Radio, Wiley, 2nd Edition, 1953, 189-190 and 457-8. t See 1. D. Kraus, Electromagnetics, McGraw-Hill, 1953, 500-I. :j: ibid., 502-4. § ibid., 504-7.


419. The field strength due to a distant transmitting station is, neglecting absorption, given by*:

where

E = 377(hI/Ad) Vm-I

h = effective height of transmitting antenna in metres,

d = distance in metres,

X = wavelength in metres

and I = antenna current in amperes.

The power radiated from an antenna * W = 1.58h2f2 /X 2 kW.

E = 300v'(W)d Vm-I •

Here W = 100, d = 100 X 103 m, so E = 0.03 Vm-I •

420. Using the expression for W in the previous solution and noting that h = 100 m,I= 450 A and X = 7.5 X 103 m, Wis found to be 56.9 kW.

Radiation resistance* = 1580 X h2/X2 ohms

= 15.8/7.52 n Efficiency = 15.8/(7.52 X 1.12) = 0.251 = 25.1 per cent.

421. In an antenna array with finite spacing, the total field in a direction at an angle (J with the normal to the array is*

E = EI sin (Na/2)/sin (a/2),

where EI is the field due to one antenna,

N is the number of vertical antennas,

and a is the phase difference between the radiations of consecutive antennas in the given direction.

Also a = (21ra/X) sin (J ± cp,

where a is the antenna spacing,

X is the wavelength,

and cp is the phase difference between the currents in adjacent antennas.

* See, for example, Admiralty Handbook of Wireless Telegraphy, Vol. II, 1938, Section R.


E is zero when Na./2 = 1T, 21T, etc. If the antenna currents are in phase, as in the present problem, cp = 0

and the first zero occurs when sin 8 = A/Na. In this case,

sin 8 = A/(lOA/2) = 1/5.

The angular width of the broadside beam = 28 = 23 0 4' .

422. The voltage received in a frame antenna in the plane of propagation of the wave is* 21TEAN/A volts, where A is the frame area in m2, N is the number of turns, E is the field strength in Vm -1 and A is the wavelength in m.

Here E = 0.01, A = 1, N = 12 and A = 300, so the voltage received = 25.14 X 10-4 V.


424. The radiation pattern, E(sin 8), of a finite aperture distribution E (y) is given by:j:

ra/2 E(sin 8) = E(y) exp Gky sin8) dy

"-0/2

For regions in the vicinity of boresight, sin 8 ~ 8 and for uniform illumination, E(y) = 1.

E(8) = I a/2 exp (jky8) dy = [expo (jky8)] a/2 -0/2 Jk8 -0/2

=a sin (1T8a/A)

(1T8a/A)

The nulls occur whenE(8) = 0, i.e.1T8najA = n1T, n = 1,2,3 ...

nA 8n =-or

a

Whenf= 10GHz, A = 3 cm and

81 = 1.91 0 82 = 3.820 83 = 5.730

* See, for example, Admiralty Handbook of Wireless Telegraphy, Vol. II, 1938, Section T. t F. R. Connor, Antennas,. Arnold, 1972, Section 2.1. :j: L. Thourel, The Antenna, Chapman and Hall, 1960, 222.


The sidelobe maxima occur at the maxima and minima of E(8), i.e. when rr8a/X = nrr/2, n = 3, 5, 7 ... The fIrst sidelobe maximum relative to the main beam amplitude is

_I a sin (3rr/2)/ sin (0) 1_ 2 El - (3rr/2) a (0) - 3rr

El = - 13.46 dB

425. IfE2 andEI are the received fIeld strengths with and without the presence of the blockage, then at any point in the radiation pattern

sin (rr8a/X) E 1 = a --'---'--.:.

(rr8a/X) and* E = E - b sin (rr8b/X)

2 1 (rr8bjA)

(a) 8 = 0° El = 0.9 E2 = 0.9 - 0.1 = 0.8

Change = 20 10g10 E2 = 20 10g10 (~) = - 1.038 dB. El 9

(b) 8 = 2.865° :.rr8a/X = 3rr/2, rr8b/X = 0.1667rr

0.9 X 2 El = -~= -19.1 X 10-2

E = (-19.1- 0.1 X 0.5) X 10-2 = - 28.67 X 10-2

2 0.523

E2 (28.67) Change = 20 10g10 - = 20 IOg10 -- = + 3.52 dB. El 19.1

426. Input noise power at 300 K

= 1.38 X 10-23 X 300 X 5 X 106 = 2.07 X 10-14 W

Receiver noise level = 10 10g10 (2.07 X 10-14) + 10 ~ - 133 dBW

Wavelength X = 3 X 106/1.2 X 1010 = 0.025 m

{ 4rr X 4 X 107} Path loss = 20 10g10 0.025 = - 206 dB

100 kw effective radiated power = +50 dBW

* S. Silver,Microwave Antenna Theory and Design, McGraw-Hill, 1949, 190.


receiver signal level = - 206 + 50 = - 156 dBW

If receiver signal-to-noise level is to be +20 dB the antenna must supply x dB.

so

-156 - (-133) + x = 20

x=46 dB

431. The carrier frequency

1 1 fc = -271'-.J--'-(L-C-) = 271'V(50 X 10-6 X 0.001 X 10-6) = 712 kHz.

:. sidebands are offrequencies 712 ± 10 kHz,

i.e. frequency range occupied is 702 to 722 kHz.

432. Let the amplitude of carrier current = I, then sidebands each have amplitude mI12.

Power in carrier a: f2 = k/2 say.

( m2I2) Power in sidebands = k 4 X 2.

total power radiated = (carrier power) (1 + ~2),

i.e. 8.932 = 82 (1 + ~). m = 0.7 and percentage modulation = 70 per cent.

Let new antenna current be II when m = 0.8.

2 _ 2 ( 0.82 ). _ Then II - 8 1 + 2 ,I.e.II - 9.19 A.

433. As in the previous solution, total power radiated

= (carrier power) (1 + m; ). 10.125 = 9 (1 + ~), i.e. m = 0.5.


Radiated power = 9 (1 + (0;)2 + (0;4)2) = 10.845 kW.

434. Let the values of m for the several frequencies be m" m2, m3, etc. Then ml + m2 = m3 +, etc., must not exceed unity otherwise overmodulation will occur.

Total power of all sidebands

( m12 ml ml ) = Carrier power 2" + 2" + ""2 +, etc. .

If ml + m2 + m3 + ... , etc., does not exceed unity then ml2 + ml + ml +, etc., is less than unity.

total power of all sidebands <! X carrier power.

435. Letl=A +aV+bV2 •

Let V = Es sin wst + Ee sin wet.

I = A + a(Es sin wst + Ee sin Wet)

+ b (Es sin wst + Ee sin wet)2

b = A + '2 (E/ + E/) + aEs sin wst + aEe sin wet

b --E 2 cos 2w t 2 e e

b 2 -'2 Es COS 2wst + bEsEe COS (We - Ws)t

- bEsEe COS (We + ws)t.

Substituting the given values:

la = 10.2725 + 3 sin wst + 10 sin wet

- 0.25 cos 2wet - 0.0225 cos 2wst

+ 0.15 cos (we - ws)t - 0.15 cos (We + ws)t rnA.

where W s = WOOt and we = 4 X 106t.


The carrier has an amplitude of 10 rnA and the sidebands have amplitudes of 0.15 rnA.

m = (2 X 0.15)/10 = 0.03.

436. If 1m is the highest modulating frequency the highest sideband frequency is fc + 1m, where fc is the carrier frequency, and the lowest sideband frequency is fc - 1m.

:. bandwidth of transmission is (fc + 1m) - (fc - 1m)

= 21m = 2 X 3.4 kHz = 6.8 kHz

The upper sideband will extend from (104 kHz + 300 Hz) to (104 kHz + 3.4 kHz), i.e. from 104.3 kHz to 107.4 kHz.

The lower sideband will extend from (104 kHz - 300 Hz) to (104 kHz - 3.4 kHz), i.e. from 103.7 kHz to 100.6 kHz.

These frequencies will be present in the transmitted wave in addition to

the carrier frequency of 104 kHz.

437. Total sideband power is given by*

Now

Psb = Pc [2/(t) + j"2(t)]

I(t) =...!.. f.Tm I(t)dt = 0 Tm 0

f'(t)~ ~ L~ GJ' dt~ r:' m~:m ~0.333 2 2

Total sideband power = 0.333 Pc

The modulation index for the fundamental component of I(t) is

Similarly

E1 2/tr m1= -=- = 0.637

Ec 1

2 ~=-=0.318

2tr

* A. Bruce Carlson, Communication Systems, McGraw-Hill, 1968, 171.

SOLUTIONS 437-439

Now

2 m3= -=0.212

311"

2 1'1'l4=-=0.159

411"

ml2 m22

PSb, = Pe "2 = 0.203 Pe Ps~ = Pe ""2 = 0.051 Pe

m32 mi PSb3 =PeT= 0.0225Pe PSb4 =PeT= 0.0126Pe

329

438. Let the carrier voltage be Ee sin wet and the audio-frequency voltage be Ea sin wat.

At the input to the first non-linear element there is a voltage (Ea sin wat + Ee sin wet).

At the input to the second element there is a voltage (Ee sin wet - Ea sin wat).

II = I + a(Ea sin wat + Ee sin wet) + bl(Ea sin wat + Ee sin wet)2

and

12 = I + a(Ee sin wet - Ea sin wat) + b2(Ee sin wet - Ea sin wat)2.

In the output II - 12 results

i.e. . b1 -b2

2aEa sm wat + -2- (Ee2 + E/)

Therefore the carrier frequency is suppressed.

439. The circuit arrangement of the anode-modulated Class-C amplifier is shown below.

(1)

The amplitude of the r.f. current is directly proportional to the anodesupply voltage. This proportionality may be expressed as

It = kea where k is a constant . (2)


From (1) and (2)

where

1---. Eo It nT2

Modulation

It = kEa(1 + m sin wit)

m =E,/Ea .

The instantaneous r.f. current

it = It sin wet = kEa(1 + m sin wit) sin wet

which is of the usual form.

(3)

(4)

(5)

The r.f. voltage across the tank circuit is also proportional to the total anode-supply voltage, or

Et = k' Ea(1 + m sin w,t) where k' is a constant (6)

Voltage va is the algebraic sum of Ea, E, sin wit and et. The maximum value of et is k'(Ea + E,) as seen from equation (6). In practice k' is about 0.9. With m = 1 therefore the voltage Va can reach the value 3.8 Ea.

With no modulation the power developed in the tank circuit is the carrier power

Pe = (Et)2/2R, since Et is the peak value (7)

With modulation the power delivered is

P = Pe (1 + ~) = (k' Ea)2 (1 + ~2 ) / 2R, . (8)

Assume that the anode-supply current ia = k" ea where k" is a constant.

ia = k"(Ea + E, sin Wit) (9)

Average power from d.c. source is

(10)


Average power delivered by modulating transformer

= tE,(k"E,) = tk"m2E/ (11)

Total average power supplied by anode-supply source is

Pa = k"Ea2 + k"m2Ea2/2 (12)

Anode-circuit efficiency = P/Pa = (k')2/2k"R, (13)

In the problem under consideration: (a) m =E,/Ea = 1400/2000 = 0.7.

(b) Maximum value of va = (Ea + E,)(1 + k') = 3400(1.9) = 6460 V.

(c) Power delivered by d.c. supply

= Eala = 2000 X 200/1000 = 400 W.

(d) Power delivered by modulation transformer = Elk"/2. From equation (10) k"E/ = 400 W, so k" = 400/(2000)2.

power delivered by transformer = 98 W.

(e) R.f. output power without modulation = 0.8 X 400 = 320 W.

(f) R.f. output power with modulation = 320(1 + m2/2) = 398 W.

(g) The modulating voltage E, causes the anode-supply current to have a component of amplitude k"E, in phase with E,. Thus the load on the modulation transformer is effectively a resistor whose resistance is E,/k"E" i.e. Ea/k"Ea = Eaii;. In this case the resistance = 2000/0.2 = 10 000 il.


441. The solution to the previous problem gives*

i = l[Jo(M) sin ct + J1(M) {sin (c + a)t - sin (c - a)t}

+ J2(M){- sin (c + 2a)t + sin (c - 2a)t}

+ J3(M){sin (c + 3a)t - sin (c - 3a)t}

+ J4(M) { - sin (c + 4a)t + sin (c - 4a)t}

+ ... ]. * See, for example, L. B. Arguimbau and R. B. Adler, Vacuum Tube Circuits and

Transistors, Wiley, 1956, Chapter 12, Section 11.


In this caseM= 50/5 = 10 so that, disregarding signs

Jo(M) ~ 0.24, J1(M) ~ 0.05, J2(M) ~ 0.26, J3(M) ~ 0.05, etc.,

as are readily found from graphs of these functions. * :. carrier amplitude = 240 X 10'; Vm-1 and sideband amplitudes are 50 X

10'; Vm-1, 260 X 10-6 Vm-1, 50 X 10'; Vm-1, etc.

442. C=K1V-t andf,.= 1/2rr..J(LC)

so 1. 50 X 106 6

/'r = K2 V 4 and K2 = 1 = 25 X 10 164

1 _~ 25 X 106 X 0.1 of,. = ;JK2 V 40V= 63 = 78125 Hz

4X 1 4

= 78.125 kHz.

of,. 78.125 M= 1m = 1O~7.8

f,. = K2 V t = K2 {16 + 0.1 sin 2rr X 104 d t

1 (0.1 )-t = K2 . 164 1 + 16 sin 2rr X 104 t

= 2K2 (I + ! ~; sin 2rr X 104 t

+!~i) (0.1 sin~~x 104tf + ... }

6 [0.1 4 3 (0.1) 2 1 = 2 X 25 X 10 1 + 64 sin 2rr X 10 t - 32 16 '1

(1 - cos 4rr X 104 t) + ... ]

~ (°.1)2 1 32 16 '1

Second-harmonic distortion = 0.117 per cent.

(~:)

* See, for example, L. B. Arguimbau and R. B. Adler, Vacuum Tube Circuits and Transistors, Wiley, 1956, Chapter 12, Section 11.

SOLUTIONS 443-444

443. When the carrier is at its first zero the modulation index, mf = 2.405.

The modulator conversion factor is defined as

_ mrfm _ 2.405 X 3 X 103 _ 085 H V-I kf - - - 2 z Em 3.46

For the modulating signal 2.11 sin 4000 TTt

m = kfEm 2085 X 2.11 = 2.2 'f 1m 2000

From tables of Bessel functions the significant sideband amplitudes, IN(mf)' are found and the spectrum is drawn as follows:

0;6000

0.3951

0.1623

444. - = 20 dB = 100 = -(S) ~ N A.M. 2'T/Br

333

where Pc is the transmitter power, Br is the receiver bandwidth and 'T/ is the noise power spectral density.

( ~) = 12 dB= Pc = 16. N F.M. 'T/Bt

200Br = 16Bt

( 200 X 3) Bt = 16 kHz = 37.5 kHz

But Bt = 61m + 2.13 lim

( 37.5 - (6 X 3)} lim = kHz = 9.1 kHz

2.13


(So) (frm)2 Pc (9.1)2 " Now - = 3 = --= 3 - 100 = 2750 = 34.4 dB No F.M. Br 21/ B, 3

Improvement = 34.4 - 18 = 16.4 dB

445. The modulation index, mp = kpEm = 3 rad. Therefore, * the number of significant sideband pairs, Sp = 6 and the maximum value of 1m is given by:

/, = Bandwidth = 40 = 3.33 kHz m 2Sp 12

For suppression of the carrier component, Jo(mp) = 0

Le. mp = 2.405, 5.520,8.654, 11.792 ... rad

Le.

m Em = i: = 2.405, 5.520, 8.654, 11.792 ... V

F or suppression of the third sideband pair component, J3(mp) = 0

mp = 6.380,9.761,13.015,16.223 ... rad

m Em =:.::2.= 6.380,9.761,13.015,16.223 ... V

kp

446. The solution to this problem can be found in certain standard textbooks.t

447. Problem 80 gives the answer to the first part of this question and the method of solution is the same as that already given for Problem 77 so will not be dealt with here. The final expression for the P.A.M. train follows directly from the other expressions given in the Question.

448. The maximum permissible value for the time constant of a diode detector RC circuit is given by:t

RC ~ v(1 - m2)/(2trnm.

* Goodyear, C.C. Signals and In/ormation, Butterworths, 1971, 142. t For example, see K. R. Studey, Radio Receiver Design, Part 2, Chapman & Hall,

1954, 250 and F. R. Connor,Modulation, Arnold, 1973,34 and 94. :j: J. D. Ryer, Electronic Fundamentals and Applications, Pitman, 3rd Edition, 1964,

493-4.


In this case, 220 X 103 X 100 X 10-12 <; y(l - m2)/(21T X 6000) m.

i.e. m<;O.77.

449. The solution to this problem can be found elsewhere* in the case of a valve as the device but the Ia/Vg relationship for a valve having two control grids is stated without any discussion of why the relationship should hold. The reader can find such a discussion, however, in the book Thermionic Valve Circuits, by E. Williams.t Readers may also wish to read about dual-gate MOS field-effect transistors.:j:

450. The oscillator frequency = 700 + 465 = 1165 kHz. Let 1= ao + al V + a2 V2 + a3 V 3 and assume V = (Es cos wst - Eh cos

Wht) where s refers to the signal and h to the oscillator. The expansion of the a3 Vg3 terms shows that there are eight frequencies in the output, namely Is, 31s,fh, 3fh, 21s + th, 21s - fh, 2fh +Is, 2fh - Is-

Undesired-signal frequencies of351, 816,1867 and 2797 kHz produce 2 kHz whistles because:

1165 -2 X 351 = 463 kHz

2 X 816 - 1165 = 467 kHz

2 X 1165 -1867 = 463 kHz

2797 - 2 X 1165 = 467 kHz.

451. (a) The oscillator circuit is as shown, where L is the tuning inductance, Co the coil self capacitance and Cp the padding capacitance.

L

--, I I • .,.. Co I I

--",

C

* K. R. Studey, Radio Receiver Design, Part I, Chapman & Hall, 3rd Edition, 1965, 430. t E. Williams, Thermionic Valve Circuits, Pitman, 4th Edition, 1961,270-2. :j: See H. M. Kleinman, Application of Dual·Gate MOS Field-Effect Transistors in

Practical Radio Receivers, R.C.A. Publication ST-3486, reprinted from IEEE Transactions on Broadcast and T.V. Receivers, July 1967.


The oscillator frequency

(1)

This equation can be satisfied simultaneously for any two values !hI and f~ by a suitable choice of Land Cpo Suppose that for these two frequencies C has values C1 and ~.

Then

and

1 (-1j 2 = L {Co + CpC./(Cp + C1)}

21T hI)

1 (21Tf~2 = L {Co + Cp~/(Cp + ~)}

(2)

(3)

From these two equations Land Cp can be found. Let !hI andf~ correspond to signal frequencies Is and Is . The most suitable values of Is and Is

I 2 I 2

are those giving the least error over the frequency band and to find them the shape of the error/frequency curve must be known. Assume the error/ frequency curve is parabolic and that there are equal errors at the ends and centre of the range as illustrated. Let frequency be represented by x and let x = - 1 when f = fa the lowest frequency of the range and let x = + 1 when f = fb the highest frequency of the range. The maximum error is d kHz.

X= -1 x=+l

The general equation of the parabola isy = ax2 + bx + c, but dy/dx = o when x = O,y = d when x = ± 1 andy = - d when x = O.

the equation is y = d(2x2 - 1).

Thus the frequencies for zero error are given by x = ± 1/";2.

Is = fc - 0.707(fc - fa) I

1s2 = fc + 0.707(fb - fa)·

In this case,fc = 1025 kHz,fa = 550 kHz andfb = 1500 kHz.

Is = 689 kHz and Is = 1361 kHz. I 2

SOLUTION 451 337

The capacitance required to tune the signal coil at these two frequencies can easily be calculated since the inductance is given as 156 J.lH. Not all this tuning capacitance is found in the tuning capacitor itself. It is reasonable to assume that about 40 J.lJ.lF is due to stray capacitance (of range switch, wiring, self-capacitance of coil, trimmer). Hence the actual value of tuning capacitance is found by subtracting 40 J.lJ.lF from the calculated figures. Similarly assume the oscillator tuning circuit has 20 J.lJ.lF stray capacitance (there is no trimmer here).

The value of C at a signal frequency of 689 kHz is therefore the calculated value 342 J.lJ.lF -40 J.lJ.lF + 20 J.lJ.lF = 322 J.lJ.lF = C1•

Similarly

Assume

c,. = 67.6 J.lJ.lF.

Co= 10 J.lJ.lF.

Also!~ = (Is, + 465) kHz andJh2 = (fs2 + 465) kHz.

From equations (2) and (3),

Cp = 288 J.lJ.lF and L = 117.3 J.lH.

(b) The oscillator circuit is as shown, where Cp is the trimmer capacitance. In the equations below Ct represents the total capacitance across L, including self and stray capacitances. The method of solution is similar to that of (a).

L

c

Zero error can now be obtained at three oscillator frequenciesJh,Jh and , 2

Jh corresponding to signal frequencies/ .. ,Is and Is . Suppose that for these 3 -, 2 3

frequencies Chas values Ch c,. and C3 respectively.

Then _1_ =L [Ct+~] (1) (27rJhi Cp + C1

(27r~l = L [Ct + c~pf~ ] (2)

and _1 -2 = L [Ct + C'pGJ ] (3) (27r!h) Cp + C3


From these equations L, Cp and Ct can be found. Proceeding as before we can now assume the error/frequency curve is cubic of the form illustrated

t x=-l

Frequency ,.

The general equation of the curve is

y = ax3 + bX2 + ex + d

~

x='tl

(4)

When X = O,y = 0 and when x = + 1,y = ± e so the equation reduces to

y=ax3 +ex

and (a +e)=-e

At frequency fi where x = Xl> dy/dx = O.

Xl = +v(-e/3a)

From (7), (5) and (6), sincey = e at X = Xl

e= -3a/4

From (5) and (8), y = ax(X2 - 3/4)

for zero error X = 0 or ± V(3/4)

i.e.

and

1s2 = fc Is, = fc -v(iXfc - fa)

Is, = fc + V(i)(fb - fc)·

(5)

(6)

(7)

(8)

In this case,fa = 550 kHz,fb = 1500 kHz andfc = 1025 kHz, sols, = 614 kHz,1s2 = 1025 kHz andis, = 1436 kHz.

Making the same assumptions as in (a)

C1 = 410.5 WF, C2 = 134.58I1P·F and C3 = 58.74 J.1J.1F.

Now fh = (Is + 465) kHz,fh = (Is + 465) kHz " 2 2

SOLUTIONS 451 and 457-458 339

and fit =. (fa + 465) kHz. 3 3

:. from equations (1) to (3),

Cp = 601 JlJlF, Ct = 36.5 JlJlF andL = 77.4 JlH.

457. Number of possible combinations for a given increment = 64 X 16 = 210 i.e. 10 bits per increment.

i.e.

There are 5 X 105 X 100 increments per second i.e. 5 X 107•

Number of bits per second = 5 X 107 X 10 = 5 X 108

The Hartley-Shannon law is:

C=Blog2 (1 +~) §. = 30 dB = 103 N

B= C = 5X108

log2103 3.3210g10 103

B = 5 X 108/9.96 ~ 50 MHz.

458. H(w) = exp (- aw2 - jbw)

--

Impulse response h (t) = f": exp (- aw2 - j bw) exp Owt) dt.

where w = 2rrt.

1 00

so h(t) = 2rr f -00 exp {jw (t -.b)} exp (- aw2) dw

h(t)=_l- exp {_ ~-b)2} 2y(rra) 4a

Thus resulting output signal reaches its maximum at t = b.

The signal falls to lIe of the maximum value when (t - b)2 = 4a

i.e. when t - b = ± 2ya

or t=b ± 2ya

The width = 2 (2ya) = 4Va.


459. Entropy = ! log2 4 + ! log2 j

= 0.812

Group Probability Bits

AA 9/16 1 9/16

AB 3/16 2 6/16

BA 3/16 3 9/16

BB 1/16 3 3/16

27/16

:. 27/32 bits per symbol on average = 0.845

Efficiency = 0.812/0.845 = 96 per cent.

460. The solution to the first part of the Question can be found in many books. *

Entropy = - (0.5 log 0.50 + 0.15 log 0.15 + ... )

= 2.21 bits per symbol.

Symbol P N NP

A 0.5 1 0.5

B 0.15 001 0.45

C 0.12 011 0.36

D 0.10 010 0.30

E 0.04 00011 0.20

F 0.04 00010 0.20

G 0.03 00001 0.15

H 0.02 00000 0.10

~ 2.26

* See, for example, F. R. Connor, Signals, Arnold, 1972, Section 6.1.

SOLUTIONS 460 and 467-468 341

Efficiency = 2.21/2.26 = 97.7 per cent.

Code A 000 B 001 C 010 D 011 E 100 Efficiency = 2.21/3 = 73.5 per cent.

F 101 G 110 H 111

461. C = channel capacity B = bandwidth S = signal power on reception N = noise power on reception.

[Readers may like to consult books by F. R. Connor and C. C. Goodyear. * Connor shows that the relationship is plausible.]

SIN is increased by a factor of 2

:. Cis increased by B log22 = B = 107 bits per second.

i.e.

i.e.

467. ! mil2 =! kT

Thus, when T= 273 K,

When T= 373 K,

-2 3 X 1.38 X 10-23 X 273 v =

2 X 14 X 1.67 X 10-27

iI = 492 ms-1

il2 = 3 X 1.38 X 10-23 X 373 2 X 14 X 1.67 X 10-27

ji = 575 ms-1•

468. The fractional number of particles having velocities in the range v to v + dv is given by:

* F. R. Connor, Signals, Arnold, 1972, Section 6.2, C. C. Goodyear, Signals and Information, Butterworths, 1971, Section 9.6.


dN. ( m )3/2 mv2

_V = 41T -- V2 e -2kT dv N 21TkT

The most probable velocity vp = )(2'::)

When v = vp:

dN. (m) 3/2 (2kT) _V = 41T -- -- e-1 dv N 21TkT m

469.

so

( m )3/2 '!!Yi:. 00 ~ aN =N -- e -2kT dv f e -2kT dv

x 21TkT x _00 y f 00 mvi

e -2kT dvz "-00

Now f _~ e~p.r ds = V(1T/{3)

and

( m )3/2 ~ dNx = N -- e - 2kT dvx (2kT1T/m)

21TkT

( m )t ~ =N -- e- 2kT dv 21TkT x

Now 1

Vp = (2kT/m)2

SOLUTIONS 469-471

( m)t •• dN = N -- e-vx /vp dv x 21TkT x

Let

Then

N(l) '/' =- - e-vx vp dvx V1T vp

W = vx/vp

dw =dvx/vp

dNx = Ne-w' dwlv1T

343

470. First find the number of molecules with velocity v perpendicular to the wall of the box that strike the wall in unit time.

Let the wall be perpendicular to vX • In unit time distance moved is vX •

number striking unit area of wall = dNx Vx

so number with velocity greater than v is

fOO dNx Vx = _~ foo e-w2 dw vp w v V1T v/Vp

= Nvp [e-w'-] 00 = Nvp e-(v/vp)' V1T 2 v/Vp 2V1T

=Nvp e-w'/2V1T

471. Number striking unit area of wall per second = Vx dNx. Partial pressure = (2mvx) Vx dNx

total pressure p = 2mN (~) t f 00 e - r;:¥ v 2 dv 21TkT 0 x x

Put mvx2/2kT= S2, i.e. Vx = sv(2kT/m)

Now

I

( m ) 2 2kT J 2kT f 00 p = 2mN -- -- . - (e-SJ) S2 ds 21TkT m m 0

4NkT roo 2 =-- L (e- S )s2ds

V1T ·0

f_: e-(lY2 ds = v("/(3)


f 00 S2 e-13s2 ds = Y(7T)(!) -00 N{3

When (3 = 1, fooo s2e-s2 ds = Y7T/4

p=NkT.

472. The electrons which escape in unit time are those with a velocity Vx such that tmvx 2 > <p, i.e. Vx 2 > 2<p/m

where

where

473.

Nvp 2 Nvp 2/ 2 number = e-w = e-vx vp 2Y7T 2Y7T

= Nvp e-2r/J/mvp2 2Y7T

vp2 = 2kT/m

number = ~J(2kT) e-2mr/J/m2kT 2Y7T m

= ~JPk) veT) e-r/J/kT 2Y7T \ m

=A veT) e-r/J/kT

A=~J(2k) 2Y7T m

dN. = 27TN - E 1/2 e-E /kT dE ( 1 ) 3/2 E 7TkT

( 1 )3/2 foo Mean energy = 27T -- E3/2 e-E /kT dE

7TkT 0

Put E/kT = x 2, then dE/kT = 2x dx and E3/2 = (kT)3/2 x 3

( 1 3/2 f 00

mean energy = 27T -) (kT)3/2 x 3 e-X22xdx(kn 7TkT 0

( 1 )3/2 f 00 2

= 47T -:; kT 0 x4 e-X dx

SOLUTIONS 473-474

But foOO x4 e-x2 dx = 3yrr/8

mean energy = 3kT/2

E I !2e-E /kT is a maximum for most probable energy.

Write y = E I /2 e-E / kT•

Then :. = EI /2 e-E /kT (- k~) + !E-I !2 e-E /kT = 0 for maximum

E/kT=-1/2

474. (i)

(li)

most probable energy = kT/2

3kT/2 = 1.602 X 10-19

1.602 X 10-19 X 2 T= 3 X 1.38 X 10-23 = 7740 K

v

v=OO fV kV3 N = f dNv = K v2 dv = -

v=0 0 3

K= 3N/V 3

Iooo v dNv K Ie:' v3 dv 3 r (iii) Average speed v = = V 2 =-3

JooodNv K Io v dv 4V

v= 3V/4

Mean square speed

345


K J.v V4 dv 3 V S = 0 =_

K fer v2 dv 5V3

r.m.s. speed y(ji2) = y(3) V/y5

Most probable speed = ~

475. (a) The kinetic energy !mv2 = Ve.

! X 9.107 X 10-31 X v2 = 11.6 X 1.602 X 10-19

where v is in m S-I.

(b) Excess energy

= ! X 9.107 X 10-31 X (2.02 X 106)2 - 10.4 X 1.602 X 10-19

= 9.107 X 1O-3I V2•

(c) Let the field strength be x Ym-I . Then the acceleration imparted to the electron is xe/m. If v is the electron velocity acquired in a distance s m,

v2 = 2xes/m

But v2 = 21.5 X 2 X elm.

x= 21.5/s= 21.5/7.9 X 10-4= 2.72 X 104 Ym-l •

(d) If an electron falls from an energy level WI to another level W2, then WI - W2 = hfwhere h is Planck's constant andfis the frequency of the emitted radiation. The wavelength of the emitted radiation is in A given by 12400/(WI - W2).

In this case the two wavelengths Al and A2 are given by:

Al = 12400/(7.93 - 6.71) and A2 = 12400/6.71

i.e. Al = 10 160 A and A2 = 1848 A.

476. Suppose no electrons per second are released from the cathode by external means. Consider a plane distant x from the cathode.

Number of electrons crossing this plane per second (n) = no + number produced in x.

SOLUTION 476 347

The n electrons produce in a further distance dx an additional number of electrons dn = Q ndx, where Q is Townsend's first ionization coefficienct. Hence,

jn dn = Q i X dx 110 n 0

i.e. n = noeooc•

Thus, the value of the current i for a given point on the current/x characteristic = ioeooc, where io is the current produced at the cathode by external means.

Experimentally it is found that a log plot of i against x does not give a straight line but increases faster than exponentially with x. This is attributed nowadays to secondary emission from the cathode. The theory above works well, however, for small currents.

Now let lla = the number of electrons to the anode per second; n+ = the number of electrons released from the cathode per

second by positive-ion bombardment;

and r = secondary emission coefficient at the cathode, i.e. the number of electrons from the cathode per incident ion.

Thus, lla = (no + n+)ead

where d is the anode-cathode spacing. Also, n+ = r{lla - (no + ~)}.

na = noead/ (I - r(ead - I)},

or ia = ioead/ {l- r(ead - I)}.

Here, from a plot ofln! against x, for 0 <x < 0.8 em, the slope Q = 3 cm-t •

Also, In io = 0, so io = lWA. If d = 1.6 em, from the above expression for ia:

200 = e4 •8/ {I - r(e4 •8 - l)}.

r= 0.0033.

From the above expression for current ia it is seen that ia -+ 00 if r(ead - I) = l, or if ead = (r + l)/r. Usually r ~ 1, so for breakdown read ~ 1. Here 0.0033 e3d ~ 1, so d ~ 1.91 cm.


477. At breakdown, 'Y eM = 1

so 0.02 eo.SQ = 1

i.e. 0.5 a = loge 50

a= 7.82 cm-1

At 200 Y, with d = 2.5 mm, a is the same since E is unaltered

e 7.82XO.2S

multiplication = 1 _ 0.02(e 7.82XO.2S _ 1) = 8.06.

478. The theory of probes and details of probe measurements can be found in many textbooks. *

The electron current Ie, corresponding to a probe current Ip, is given by:

Ie = Ip + I; = Ip + 0.08,

where I; is the saturation value of the positive-ion current. The slope of the 10& lelVp graph is found to be 0.65 y-1 and this equals

elkTe, where k is Boltzmann's constant and Te is the electron temperature. Thus, in this case,

Te = 1.602 X 10-19/0.65 X 1.38 X 10-23 = 17800 K

The random electron current density

Je = lelA = 36.4/0.033 = 1100 mA/cm-2 = 1.1 X 104 A/m-2•

Therefore, 1.1 X 104 = eNe V(kTe /21Tm),

where Ne is the electron concentration and m is the electron mass, i.e. Ll X 104

= Ne . 1.602 X 1O-19V(1.38 X 10-23 X 17 800/21T X 9.107 X 10-31)

Ne = 3.3 X 1017/m-3•

Plasma potential = - 11 Y, soE = (11 - 5)/12 = 0.5 y/cm-1, whereE

is the voltage gradient. The drift current density = 1.1/6 A cm-2 = 0.18 A cm-2•

'" For example, see J. Millman and S. Seely, Electronics, McGraw-Hill, 2nd Edition, 1951,286-9; P. Parker, Electronics, Arnold, 1950,637-44; and F. G. Spreadbury, Electronics, Pitman, 1947,93-8.

SOLUTIONS 478-480 and 485

Mobility of electrons = 0.18/NeeE = 0.18/3.3 X 1011 X 1.602 X 10-19 X 0.5 = 6.8 X 106 cm2 V-I S-l.

479. The current I at any point x can be represented by

1= loe-Ax = 10(1 - Ax ... ),

349

where 10 is the current at x = 0 and A = QN. N is the number of particles per unit volume and Q is the total collision cross-section. Here, Ax = 1/10 and x = 20 cm, so A = 1/200. N = 2.7 X 1019 (p/760), where p is the gas pressure in torr.

1/200 = 10-16 X 2.7 X 1019(p/760) so p = 1.405 X 10-3•

480. 10= J(;:) Hz.

At 8 mm,/o= 3.75 X 101o Hz,son= 1.75X 1013 cm-3. At 3 cm,to = 1010 Hz, so n = 1.24 X 1012 cm-3•

If concentration is no at t = 0, then at time t:

and

n = nolO + nocx t)

1.75 X 1013 = nolO + no cx 5.7 X 10-6)

1.24 X 1012 = nolO + no cx 81 X 10-6)

cx~ 10-8

485. Choose axes as shown.

y IE =lo4 vm-'

Vy -- --I V

I I

t 300 !

B = mWb m2J1-_--...I __ ..... X

z


If the initial velocity of the electron was directed along the fields the magnetic field would exert no force on the electron. The electron would then move in a direction parallel to the fields with constant acceleration. If the initial velocity has a component perpendicular to the magnetic field, as in the present case, this component together with the magnetic field, will give rise to circular motion. Because of the field E the velocity along the fields changes with time so the resultant path of the electron is helical.

The velocity v can be resolved into two components Vx and vy.

Vx = 1.19 X 107 sin 30° = 5.95 X 106 ms-1

Vy = 1.19 X 107 cos 30° = 1.03 X 107 ms-1•

The acceleration along the

- Y direction = a = eE/m = 1.759 X 1015 ms-2•

The projection of the electron path on the XZ plane is a circle of radius

r = mvx/eB = 5.95 X 106/1.759 X 1011 X 5 X 10-3 = 0.00677 m.

The velocity along the fields is not constant but is given by

vy' = 1.03 X 107 - 1.759 X 1015t ms-1

andy, the distance moved = 1.03 X 107t-!X 1.759 X 1015t2•

The electron begins to move in the + Y direction but because the acceleration is along the - Y direction, it will gradually come to rest and will then reverse its motion in the Y direction. This reversal will occur after a time t' for which vy' = o. i.e. t' = 1.03 X 107/1.759 X 1015 = 5.86 X 10-9 s.

The distance travelled in the + Y direction in this time t' is

y' = 1.03 X 107 X 5.86 X 10-9

-! X 1.759 X 1015 X (5.86 X 1O~2 = 0.03 m.

After the reversal of motion the electron continues moving in the - Y direction and does not reverse again. There is, of course, no reversal of the direction in which the electron traverses the circular component of its path.

The angular velocity is constant.

= w = Be/m = 1.759 X 1011 X 5 X 10-3 = 8.8 X 108 rad S-l.

The periodic time T= 21T/W = 7.14 X 10-9 s.

SOLUTION 486 351

486. The force on an electron due to E is directed along the + X axis. Any force due to the field B is always at right angles to B. Thus there is no component of force along the Y direction.

The two following equations therefore hold:

dvx m-=eE-evB

dt z

dvz m- =evxB.

dt

d2vx eB dvz dvz 2 From (1)- =-- -=-w-=-w v from (2) .

dt2 m dt dt x

where w=Bejm.

The solution of equation (3) is of the form

Vx = P cos wt + Q sin wt

(1)

(2)

(3)

(4)

where P and Q are constants determined by the initial conditions that Vx = Vz = 0 when t = 0

From (1) when t = 0

and from (4) when t = 0

p=o. dvx eE -=-dt m

dvx -=Qw dt

eE E Q=-=-=u say.

mw B

Thus, Vx = u sin wt

Using this in equation (1),

I dvx v = u - - - = u - u cos wt z w dt

u From (5) x = fVx dt = fu sin wt dt = -- cos wt + constant.

w

Since x = 0 when t = 0, constant = ~. w

From (6) u .

z = fvz dt = f(u - u cos wt) dt = ut - - sin wt + constant. w

(5)

(6)


Since z = 0 when t = 0, constant = O.

Thus u u

x = - (1 - cos wt) and z = ut - - sin wt. w w

z

1Ir""'"-----,~X

.-Rolling circle

The electron path is a common cycloid, the path generated by a point

on the circumference of a circle of radius u/w which rolls along a straight

line, the Z axis, as shown.

487. The axes are chosen as in Question 486. As shown in the previous solution the electron path is cycloidal in the

XZ plane. The electron travels with constant velocity along the Yaxis. Referring to the previous solution:

w = eB/m = 1.759 X 109 rad S-I

u = E/B = 106 ms-I •

The time t for which the electron is in the region between the plates = 0.02/106 = 2 X 10-8 s.

Angle () turned through in this time t

= wt = (1.759 X 1O~(2 X 10-8) = 35.18 rad

= (31.42 + 3.76) radians = (101T + 3.76) rad.

i.e. the electron enters upon its sixth revolution before leaving the plates. Referring to the previous solution and noting 3.76 rad = 2150 :

u x = -(1 - cos wt) = 0.0569 X 10-2(1 - cos 215)

w

= 1.035 X 10-3 m

SOLUTIONS 487-488

z = ( ut - : sin wt) = 0.0569 X 102(35.18 - sin 215)

= 2.034 X 10-2 m.

353

:. distance from axis when electron leaves the region between the plates is V(x2 + Z2) = 2.04 X 10-2 m.

488. The electric field E has two components Ex = E sin 20° and Ey = -E cos 20°.

It follows that the equations of motion are those given as (1) and (2) in the solution to Question 486 where E is replaced by - E sin 20° but the force in the Y direction is no longer zero but is e(E cos 20°). The equations for x and z are then those given at the end of the solution to Question 486 with E replaced by - E sin 20° and the expression for y is VOyt + 1 e(E cos 20)t2

- , where VOy is the initial velocity in the Y direction. 2 m

-z

-x +x --~---

+z

Thus, again, the projection of the path in the XZ plane is a common cycloid as shown. The equation of the cycloid is given by:

and

u -x = - (1- cos wt)

w

u -z =-(wt-sinwt)

w

E sin 20° 5 X 103 X sin 20° 6 -1 U = B = 103 = 1.71 X 10 ms

w = eB/m = 1.759 X 108 rad S-I.


489. The deflection at the screen* = Vl(I/2 + L)/2Ed = 0, where 1 = 1.5 em, d = 0.3 em, L = 20 em, E = 1500 volts and V is the voltage applied to the deflector plates.

V/o = 2 X 1500 X 0.3/1.5 X 20.75 = 28.9 Vern-I.

490. The deflection at the screent = Hlv'(e/2mE)[I/2 + L] = 0, where 1= 0.015 m, L = 0.2 m, E = 1500 V and H is the magnetic field produced by the poles.

:. H/o = 1/0.015 X 0.2075v'(1.759 X 1011/3000)=0.042 Wbm-2permetre.

492. The Schering bridge circuit is shown in the diagram. At balance:

1/jwC1 1/(1/14 + jwC4) =

P2 + l/jw~ Equating real and imaginary parts of this equation gives:

and

The loss angle 02 is given by

~=14Ct!R3

P2= C~3/Cl

tan02= WP2~

(1)

(2)

(3)

(4)

In this case, ~ = €€oA/t farads, where € is the dielectric constant, €o = 8.855 X 10-12, A is the area of the plates in sq. metres and t is the distance between the plates in metres.

* See F. A. Benson, Electrical Engineering Problems witk Solutions, Spon, 1954, 219-20. t ibid., 220-1.


~ = 0.000213 /IF.

From (2) therefore 14= 4260 ll.

From (4) tan 9' = (21T X 50)Pl X 0.000213 X 10-6

i.e. Pl = 39200 ll.

:. from (3) C4 = 0.00196I1F.

493. At balance the following equations are obtained:

il (R + jwL) = ~ Q + i3 r (1)

il P = i3/jwC . (2)

and i3r + MjwC = (~ - i3)S (3)

From (1), (2) and (3), eliminating it. ~ and i3:

{(R+jwL)/jwCP-r}S=Q(r+S+ l/jwC) (4)

Equating real and imaginary parts of (4):

R = PQ/S and L = CP(rQ + SQ + rS)/S.

In this case,R = 500 X 1000/1000 = 500 II

and L = 2 X 10-6 X 500(200 + 1000 + 200)103/1000 = 1.4 H.

494. At balance the following equation holds:

(Rl + jwL1)/(R2 + jw£,) = (14 - j/W~)/(R3 - j/wC1) (1)

Equating the real and imaginary parts of (1):

RIR3 + L./C1 = Rl14 + £,/~ (2)

and (3)

From (3) it follows that Ll CIR3~ must equal £,~CI14.

i.e. LIR3 = £'14 (4)

Also from (3)Rl~ must equal R1C1

From (2) it follows that if RIR3 = Rl14, L./C1 must equal Ll/~.


495. The network is shown in the diagram.

At balance: Q/S = (RI + PI + 1/jwC1)/(R2 + P2 + l/jw~) (1)

Equating real and imaginary parts of (1):

Q/S = (RI + PI)/(R2 + P2) = ~/CI

Here RI = 11.4 n, P2 = 0,R2 = 10 n, ~ = 0.023 J.1F

:. from (2) PI = 1.1 nand C1 = 0.0184 J.1F.

496. Let the currents and voltages be as shown on the diagram.

(2)

SOLUTIONS 496-497

.... ----'-- ~ .... --"---

... ___ 2-__ _

At balance: (Rl + jwL)il =R2~

R3il = (~+ l/jwC)~ .

Dividing (1) and (2) and equating real and imaginary parts gives two equations in terms of Land Rl from which:

L = R2R3C/(1 + W2~2C2)

and

The phasor diagram for the network is as shown.

497. With S open the balance condition is:

Cl = CzC3/(Cz + C3) •

357

(1)

(2)

(1)


WithS closed the balance condition is:

From (1) and (2),

Cx = Cl(e,. - C/)/(C/e,. + e,.'C3 ~ C2C3).

With S open and with C1 adjusted then at balance

C1 = 1000 X 50/1050 = 47.6 J.LJ.LF.

WithSclosed,47.6=e,.'X 51/(e,.'+51) :. C2'=714WF.

Thus e,. - e,.' = (1000 -714) WF = 286 J.LJ.LF.

(2)

There are two readings, one at the maximum setting of e,. and the other at e,. = 286 J.LJ.LF, so the reading error is ± 10 X 100/286 per cent = ± 3.5 per cent.

498. Let Vx = l/1 sin (wt + (1) and Vy = l/2 sin (wt + (2).

Vx = l/l (sin wt cos 81 + cos wt sin (1)

and Vy = l/2 (sin wt cos 82 + cos wt sin ( 2)

From (1) and (2) eliminating wt gives:

Vx2 ~ 2 Vx 1), _ . 2 2 + 2 - cos (01 - ( 2) - sm (01 - (J2). l/l ~ l/ll/2

(1)

(2)

This is the equation of an ellipse whose major and minor axes coincide with the x and y axes respectively when (81 - ( 2) = n/2. In general, the trace gives an ellipse the orientation of which depends on the phase difference between the two voltage waves.

499. The solution to this problem can be found e1sewhere.*



* F. A. Benson and A. O. Carter, 'A Critical Survey of Some Phase-angle Measurements Using a Cathode-ray Tube,' Electronic Engineering, 22, 238-42, 1950.


502. y y

~ ~

~Jt ~ ,

~ ~ q, = 0 ,

1/

x

... '" / :,'"

'" // ~'" /

'" ",' / "'~ q,= 30° I

It '" '" ~'

x

y y

'" ... " '" '" q,= 60°

/ " x

,,~ .... .... .- , / a, = 90°, I

x I /

/~ \ " .. ....

\ // , , ,'" .... -503. y

... - ... , ... 4>=0 "'~ "- h" " , ,,"

" " , " '" , ...

' ... - ... '"

y

x

1', -...... q,=300' ' ....... ... , ;1' ,.. ...

, ... I-- " " J if" ... ' --

x

y y ---1"- .... ,

cb = 90° ............ .... ...

) x

, " ... --

n=2

360

504.


y ---~ cP = 0 -.::: ... "

"'~ ... " , , .... -...... -- --

y

..... ~ .......... "" -- .. - .... " cP=60o , ... ~ ...... .... -... ' ... ' , , ... ' ... ~<-

" -.. _-t..,_ -"J

y

'> ... -, , ......... ~ ><<11= 0 X

I '" 4' \ 1_' '" ... , I

\ '" " ''Y.~ '<. " .... '" , , ..... k.'" , - ... -y

,-;, .... ---" ~ d>=60o ", , , ~" I'. , ....

, ". '" " '" , ""i '.<' "," ."': k " ,,_ ...... ..... ......... ' y

x

x

)(

x

y --.. --~ ... ~ ....... ... ~ ~" f--__ "'-=--f-,-r=--I X

y Ito- --"I ..... -' d>=90~-:' Ir .... -....... -... ~ ... ~", x , '" ...... ... '"

... '-'" .... ' ~ ...... ... - -... ...

y

y , .... lid>=90o

.., , \

~, , " '" ,

'" " 10-'" ,.,.;,," ' ... ........

x

SOLUTIONS 505-507 and 510 361

505. One method of finding the frequency ratio is as follows: draw a line parallel to the x-axis, which does not pass through any intersections of different parts of the curve, and count the number of points where the line intersects the curve; repeat this procedure with a line drawn parallel to the y-axis. It should be noted that the lines must not be part of the boundary rectangle. The ratio of the two numbers so found will be the required frequency ratio. The number of intersections on the line parallel to the x-axis is proportional to the frequency of the y-variation, and vice versa. *

506. For equal sensitivities

R = l/wC= 1/2rr X 50 X 2 X 10-6 n = 1592 n.

For differing sensitivities

R = 1592 X 0.55/0.45 = 1946 n.

507. Voltage range = (5 X 30 - 20) = 130 V.

Length of time base = 130 X 0.8 mm = 0.104 m.

If T is the time of the sweep,

0.01 X 10-6 X 130 = 1.5 X TX 10-3•

Frequency f= l/T= 1154 Hz.

510. Tabulate all values of ABCD as below.

ABCD a) b) c) f

0000 0 0 0 0

0001 0 0 0 0

0010 0 0 0 0

0011 0 0 0 0

0100 0 0 0 0

0101 0 0 0 0

0110 0 0 0 0

0111 0 0 1

* For further information see G. Parr and O. H. Davie, The Cathode-Ray Tube and its Applications, Chapman & Hall, 3rd Edition, 1959, Chapter 6. See also, F. E. Terman, Radio Engineer's Handbook, McGraw-Hill, 1943,955.


:( [1

511.

ABCD

1000

1001

1010

1011

1100

1101

1110

1111

r-;-'

1 1

11 1

1 "",-,

Inputs

A B

0 0

0 0

0

0 1 1 0 1 0

1

P

0

1

0

0 1

0

1

S = ABP + ASP + ASP + ABP

= P (AB + AS) + P (AS + AB)

= P (AB + AS) + P (AB + AS)

= PS' + PS'

where S' is sum of half adder

a) b) c) f

0 0 0 0

0 0 1

0 0 0 0

0 0 1 1

0 0 1

0 0 1

1 0 0

0 0 1

f= AB + AD + BCD

A B 1

A D f 1

B C D

Outputs

S C

0 0

0 P = previous carry

1 0

0 S = sum

1 0 C = carry

0 1

0 1

1 1

SOLUTIONS 511-514

C = ABP + ABP + ASP + ABP

= P (AB + AS) + AB (P + P)

=PS' + AB

=PS' + C'

where C' is carry of half adder.

512.

Clock

o ---Lo' ___ ....

513. A A B A ~

~f1TOl B~ ~ c(~

C B

A ~~

~~B C

514. (a) Clock

, , I

L

!"L...Jr--------, OiJ 3 ----.... , ..... _-,-

03 IL..J L

363


(b) Shift register.

515. G4 G3 G2 G1 D C B A

0 0 0 0 0 0 0 0

0 0 0 0 0 0 1

0 0 0 0 1 0

0 0 1 0 0 0 1 1

0 1 1 0 0 0 0

0 1 1 0 0 1

0 1 0 1 0 1 1 0

0 0 0 1 0 0 0

1 1 0 0 1 0 0 1

1 1 0 0 0

By Karnaugh Map method using 'don't care' states.

G2 G, ~

7 ., 1 1

X X X X X X X X ..... '-' ~ ....

1 X X X X 1

1 1 ..... ....

X X X X X X X X

,..... X X 1 X X 1

1 (1 1J 1 .....


516. Count J1 Kl Ql J2 K2 Q2 J3 K3 Q3

0 1 X 0 0 X 0 0 X 0 1 X 1 1 X 0 0 X 0 2 1 X 0 X 0 0 X 0 3 X 1 1 X 1 X 0

4 0 X 0 0 X 0 X 1 1

0 1 X 0 0 X 0 0 X 0

The equations below will satisfy the above table.

J1= 03 J2= Ql J3 = QIQ2

Kl = 1 K2= Ql Ka= 1

517. Decimal column is Ql Q2 Q3 in binary. J1 is the exclusive or function of Ql and Q2.

Decimal Clock J1 Kl QdJ2 OdK2 Q2/J3 02/K3 Q3

0 0 0 0 1 0 1 0 terminates

1 0 0 1 0 0 0 1 0 1 0 0 0

terminates

2 0 1 0 0 1 1 0 0 5 1 1 0 0 0 1 1 6 2 0 1 0 0 0 3 3 0 0 1 0 1 5 4 0 0 0 1 1

continuous 5 ~ 6 ~ 3 ~ 5

4 0 1 0 0 0 0 6 1 0 1 1 0 1 0 0 3 2 1 0 0 1 1 0 1

5 3 1 0 1 0 0 1 1 6 4 0 1 1 0 0 0



Decimal

7

3 5

6

3

Clock

0

2

3

4

J1

0

1

0 1

Kl QdJ2 Ql/K2 1 0

0 0 1

0 1 0 1 1 0

0 0 1


Entry at 0 and 1 terminates in 0 All other entries lead to loop sequence 3 ~ 5 ~ 6 ~ 3.

518. C(

o 1

Q2/K3 0

0 1

0

0

Q, L-___ ..... ! Q,/J4

(3,

Q 21 J,IK4

Q2 /K ,

(32

Q31h -.-J Q31 K2 ----,

(33

Q4 1 J3 -1 Q4 /K 3 -,

~4

521. The skin depth 8 = l/.../C1TlJfa)

where a is the conductivity, IJ the permeability andfthe frequency.*

Q3 1

1

0

* See, for example, I. C. Slater,Microwave Transmission, McGraw-Hill, 1942, 114; or R. A. Bailey, 'A Resonant-cavity Torque-operated Wattmeter for Microwave Power,' Proc. I.E.E., Part C, 103,60,1956.


For copper,

8 = l/y(rr X 47T X 10-7 X fX 5.88 X 107) m

When f= 300 MHz, 8 = 3.79 X 10-6 m.

When f= 10 000 MHz, 8 = 6.56 X 10-7 m.

522. The inductance of a straight piece of wire at very high frequencies is:*

L = 0.002/(2.303 10giO 411d - 1 + d121) J.LH,

where 1 is the length of the wire in centimetres and d is the wire diameter in centimetres.

In this case, 1 = 2.54 cm and d = 0.0254 cm so L = 0.014 J.LH.

At 500 MHz the reactance

= 27T X 500 X 106 X 0.014 X 10-6 Q = 44 Q.

523. The intrinsic impedance of the"plate Zo = y(ZI Z2),

where Zl = intrinsic impedance of air

and Z2 = intrinsic impedance of dielectric medium.

Now Zl = yOlo/eo) = 377 Q and Z2 = 377/y4 Q = 188 Q

Zo = y(377 X 188) Q = 266 Q.

The relative permittivity required = (377/266? = 2.

524. The output voltage is as shown in the diagram.

* See K. R. Spangenberg, Vacuum Tubes, McGraw-Hill, 1948,477.


i.e.

or

i.e.

de Rate of rise of waveform = - = wEm cos wt ~ wEm.

dt Time taken to rise from - E to + E is oT~ 2E/wEm,

oT= 2 X 2/C21T X 105 X 200) s = 0.032 JlS.

525. If V is independent of z:

Let V = fCr) cos 40

I 0

r or

Suppose f = rn

I 0 (OV) I 02V -;: or \rTr + ~ 082 = 0

( of) 16f r - cos 40 - - cos 40 = 0 or r

o ( Of) r- r- =16f or or

df = nrn-1 and r df = nrn dr dr

r~ (r df ) = n2rn = n2f dr dr

n2 = 16 or n = ±4

fCr) = r4 or r-4

The solution is r-4 since V ~ 0 as r ~ 00

so

i.e.

oV E=--

r fir

V = Ar-4 cos 40

Er = -4Ar-s cos 40

E1:2ro) = (2rofS = T 5

~ro) (rofS

E(2ro) = Eo/32.

526. For the first layer: capacitance = Aeof(/d

For the second layer: capacitance = AeoK/d

conductance = A a/d.


d d d Au . jwAeoK

Total impedance =. A~ II' + d d JW "0"". _ + __ _

Au jwAeoK

= jW~eoK {I + 1 + u/~weoK }

For the composite dielectric impedance = 2d/jwAeoK'

;, = k {I + 1 + U/~WeoK }

or K' = 2K(K - ju/weo)/(2K - ju/weo).

527. The electric field at an atom site is the sum of that due to the external field and that due to all the induced electric dipoles in the material. The latter will be proportional to the electric polarization.

Here E; =E+P/3eo

The relative permittivity €r is related to the susceptibility X by:

€r = I + X = 1 + P/eoE

But P=NOt.Ei

(1)

(2)

(3)

where N is the atomic number density and Ot. is the atomic polarizability.

or

or

EliminatingE; from (1) and (3) gives:

From (2) therefore,

P=NOt.E +NOt.P/3 eo

P/eoE = NOt./eo (1- NOt./3eo)

€ _ I + NOt./eo r - (1 - NOt./3eo)

NOt. €r-1 -=--3eo €r + 2

which is the Clausius-Mosotti relationship.

NOt. 5 X 1028 X 2 X 10--40 =-------

eo 8.855 X 10-12 (a)

370

(b)


NOt./fo €r = 1 + = 2.81

(1 - NOt./3fo) -

E; = 1 + ~ = 1 + 1.81 = 1 6 E 3eoE 3 _._.

528. The Clausius-Mosotti relationship* is:

€r-1 NOt. --=-€r + 2 3fo

To calculate the polarizability Ot. of a sphere:

Dipole moment = 41Tfo,-3 E

where E is the electric field and r the sphere radius.

Thus,

Ot. = 41Tfo,-3

10-9

= 41T X - X (10-3)3 361T

= 10-18/9

N= (_1_)3 = 109/64. 4/1000

€r - 1 109 10-18 361T €r + 2 = 64 . -9- . 3 X 10-9

€r = 1.21.

529. The Clausius-Mosotti relation is:

€r-1 NOt. --=-€r + 2 3fo

NOt. 1.25 -=-3€o 4.25

For the expanded polythene,

NOt. (1.25) 5 -= - -=0.0148 3fo 4.25 100

* See, for example, B. I. Bleaney and B. Bleaney, Electricity and Magnetism, Oxford, 1957,493, also the previous Solution.

SOLUTIONS 529-531

i.e.

€ -1 _r-=0.0148 e,.+2

€r = 1.058.

371

530. (a) Light is considered as composed of particles of energy hI so that electron emission is possible only if the frequency of the impinging light is greater than the 'threshold' value fc = eifJ/h; where ifJ is the work function in electron volts, e is the charge on the electron and h is Planck's constant.

ch 12400 Corresponding threshold wavelength, Xc = - = --A.

eifJ cp

In this case ifJ must be less than 12400/8000 = 1.55 eV.

For caesium, Xc = (12400/1.8) A = 6890 A.

(b) The electron-volt equivalent of the energy of the incident photons is 12400/2537, i.e. 4.89 eV.

According to Einstein* the maximum energy of the emitted electrons is 4.89 - 4.3 = 0.59 eV.

The maximum velocity

v = J (2; E) = 5.93 X 10s../Ems-1

= 5.93 X IOSY0.59 ms-1 = 4.56 X 105 ms-1•

531. If the charge per unit length of electrode is Q, then b

E = Q/21rr€o and V = f E dr = Q In (b/a)/21r€o, a

where a and b are the anode and cathode radii respectively and r is any radius between a and b.

E = Vir In (b/a)

Let the number of ion pairs be n at any point. Then,

dn/n = Ape-Bp In (b/a)r/V dr

* A. Einstein, 'tiber einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt,' Ann. Physik., 17, 32, 1905.


Here A = 13.6, B = 17.3 X 13.6 ~ 235, b/a = 20,

so In (b/a) "'" 3 and p = 1

dn/n = 13.6 e-235X3r/V dr,

b

and In (n/no) = 13.6 f e-70sr/V dr a

i.e.

Write 705r/V = s, then 705 dr/V = ds, so

f70S /V In (n/no) = 13.6 e-S (V/705) ds

70S/20V

= (13.6 V/705){e-7os/2ov - e-70SfV }

~ (V/52)e-3s.3 fV.


533. The solution to this problem can be found elsewhere.f

534. The solution to this problem can be found elsewhere.:j:

535. Emax = 3 X 104 V cm-1

Hmax = Emax/1201T '1 1

P =! f 0 fa (Jt2 /Zo) dx dy = Jt2 /2Zo

P=9X 108/2 X 1201T= 1.19X 106 Wcm-2 •

* See, for example, P. Parker, Electronics, Arnold, 1950,50-3 and 980-2; or R. Latham, A. H. King, L. Rushforth, The Magnetron, Chapman & Hall, 1952, Section 6.4.2. t See, for example, J. Millman and S. Seely, Electronics, McGraw-Hill, 1951,80-2.

See also, P. Parker, Electronics, Arnold, 1950, Section 107. :j: See W. W. Harman, Fundamentals of Electronic Motion, McGraw-Hill, 1953,

207-15.

INDEX

Page numbers are given first, followed by problem numbers set in bold and in parentheses.

A.c. bridges, 154 (492-494), 155 (495-497), 159 (508)

Adder, 160 (511) Additive frequency changing, 139 (449) Admittance matrix, 116 (372), 123 (390) Alpha cut-off frequency, 50 (164, 165) Amplifier,

359),112 (364), 119 (380, 381), 121 (385),126 (402), 128 (411) in waveguide, 124 (395-397), 125

(399,400) Attenuator, 112-133 (366), 124 (395) Avalanche breakdown, 41 (134)

difference, 90 (284, 285), 91 (286), Balanced modulator, 134 (438) 92 (289, 290), 93 (292) Band-pass mter, 126 (403)

response to step-voltage input, 30 Bandwidth,4 (6), 18 (48),66 (214,215), (83) 88, (278), 126 (403), 131 (426),132

Amplifiers, 64-89 (205-283) (430) 134 (436), 136 (444), 142 Amplifier, (457), 144 (464), 145 (465)

stability of, 88 (280) Barretter, 61 (196) summing, 91 (287,288),93 (291) Barrier, potential, 40 (130) thermal noise in input impedance of, Beat frequency, 99 (312)

105 (337) Beat-frequency oscillator, 99 (313) transistor, 114 (369) Biasing circuit for transistor, 74 (238,

signal flow graph for, 14 (37) 239),89 (282, 283) with feedback, 102 (323) Blackband and Brown's two-point method,

Amplitude modulation, 133 (431), 134 110 (359) (436,437), 136 (444), 139 (450, Bohr atom, 33 (91) 453),140 (454) Boltzmann distribution, 147 (473)

Anderson's bridge, 154 (493) Boolean expression, 164 (520) AND gate, 160 (510) Breakdown, 148 (476,477), 168 (535) Anode fall, 148 (478) avalanche,41 (134)

modulated class-C amplifier, 134 (439) Bridge-T network, 21 (57) Antenna array, 122 (388), 130 (421), Bridge network, 15 (39)

132 (428) Bridges, a.c., 154 (492-494), 155 (495-Antennas, 130-132 (416-430), 133 (432) 497), 159 (508) Aperture, effective, of antenna, 130 (418), Buncher, 167 (533)

131 (424,425) Array, antenna, 130 (421), 132 (428) Artificial dielectric, 166 (528) Attenuation constant, 107 (343), 108

(349),109 (356,357),110 (358,

Cable, coaxial, 110 (358), 120 (381), 121 (386)

Capacitance, measurement of, 155 (497)

373

374 INDEX

self, of coil, 8 (20,21),9 (22) Capacitor,

losses in, 10 (27) power factor of, 10 (27) voltage-controlled, 36 (110),41

(137) with composite dielectric, 166 (526),

170 (541) Carrier, 133 (431-435), 134 (436-438),

136 (443,445) 139 (452,453), 140 (454),141 (456)

Cascode type of difference amplifier, 91 (286)

Cathode follower, 70 (229) Cathode-ray tube, 149 (479), 151 (487),

152 (489,490),156 (498,499), 157 (500,502,503), 158 (504,505), 159 (507)

Channel capacity, 142 (457) Characteristic impedance, 107 (343,344),

108 (349,351,352),109 (353, 355-357) 110 (358-360), 111 (361,362), 118 (379), 120 (382, 384), 121 (387),125 (401)

Characteristics of valves and transistors, 44 (144-146)

Chebyshev filter, 129 (415) Choke-capacitor coupling in amplifier,

68 (220) Circle diagram, transmission line, 108

(351,352) Circuit analysis, 3-21 (1-58) Circuits, identical at all frequencies, 9

(23) Circularly-polarized plane wave, 169

(537) Circular time base, 158 (506) Class-A amplifier, 67 (217),79 (254),

82 (260) Class-B amplifier, 69 (222) Class-C amplifier, 69 (222,223)

anode-modulated, 134 (439), 135 (441)

Clausius-Mosotti formula, 161 (527-529) Clipping circuit, double-diode, 165 (524) Coaxial line (see Concentric line) Code converter, 161 (515)

Coding efficiency, 143 (459,460) Coefficient of coupling, 4 (7), 5 (8, 9),

18 (48-50),94 (293) 96 (299) Coils, loading, on transmission line, 107

(345) Collision cross-section, 149 (479) Colpitts oscillator, 96 (301, 302) Communication and information theory,

142-145 (457-468) Complex frequency plane, 16 (43) Composite characteristic, 69 (221), 79

(254) filter, 128 (412)

Computing circuits, 90-93 (284-292) Concentric line, 109 (353,356,357),

120 (383, 384), 121 (385) Conduction in and kinetic theory of

gases, 146-150 (467-484) Constant-k filter, 126 (402-404), 127

(409), 128 (411,412) Contact potential, 35 (101), 38 (120) Control ratio of thyratron, 159 (507) Counter, 162 (516) Coupled circuits, 4 (7),5 (8-11),6 (12-

14),7 (15-17),8 (18),9 (22), 18 (48-50),69 (224) multi-mesh, 7 (17)

Critical wavelength in waveguide, 124 (391,393,394)

Crystal oscillator, 97 (304), 101 (321)

Darlington connection, 76 (245), 83 (266)

Decibel. 81 (258) Depth of modulation, 133 (432,433,

435), 135 (439),138 (448), 139 (452)

Detection, modulation and frequency changing, 133-141 (431-456)

Detector, diode, 138 (448) Determinant, network, 15 (39), 16

(40,41) Dielectric, artificial, 166 (528) Difference amplifier, 90 (284,285),91

(286),92 (289,290), 93 (292) Diffusion, 37 (116),42 (140),43 (143)

INDEX 375

coefficient, 34 (97), 37 (116), 38 (118)

length, 38 (118) Diode,44 (144, 145),45 (150-152),52

(170),55 (171, 178, 180),61 (197) Diode detector, 138 (448)

junction, varlable-capacitance, 135 (442)

saturated, in time-base circuit, 159 (509)

temperature-limited, noise in, 103 (326),106 (340)

Zener, 58 (189),59 (191),62 (200, 201),63 (203)

Dipole, 130 (417), 132 (428) Dipole moment, 170 (544) Directivity, 130 (417,418) Discharge gap, 148 (471) Discharge tube, 148 (476,478), 149

(480), 150 (483) Discharge tube relaxation oscillator, 98

(309),102 (322) Distortion,64 (208), 67 (219),70 (226),

77 (247),82 (260) Distribution, Boltzmann, 147 (473) Distribution function, 146 (469), 147

(474),149 (481,482) Dividing circuit, 92 (290) Double-diode clipping circuit, 165 (524) Double-stub tuner, 111 (362) Drift velocity, 39 (122) Driving-point impedance, 16 (42)

Ebers-Moll model, 50 (162) Efficiency, antenna, 130 (420), 90-93

(284-292) Electric and magnetic fields, motion of

electrons in, 151-153 (485-491) Electron,

concentration, 148 (478), 150 (484) emission, 146 (472) temperature, 148 (478), 150 (484)

Electronic computing circuits, 90~93 (284-292)

Emission, photoelectric, 167 (530) Emitter-follower amplifier, 76 (244),78

(248),86 (274) Energy level, 33 (92), 147 (475)

Entropy, 143 (460), 144 (463) Equal-ripple approximation, 128 (413) Equipartition law, 106 (341) Equivalent circuits, 18 (52),48 (157,

158),49 (159, 160, 162),52 (173), 70 (230) 73 (237),90 (284),91 (288),99 (314), 118 (378)

Espley method for determining harmonic components, 79 (253)

Excitation, 147 (475) Expansion in paths, 16 (41)

Feedback, 70 (225-228, 230), 71 (231), 80 (256), 102 (323)

Feedback summing amplifier, 91 (287) F.E.T., 36 (111),41 (136),51 (166),74

(240),84 (267,268),85 (269-271), 86 (272,273),93 (292), 94 (295), 123 (390)

Field components for E and H waves in waveguide, 124 (393) patterns for HOI mode in waveguide,

124 (392) Figure of merit (gain Xbandwidth), 66

(214), 74 (237) Filter half section, 127 (408)

pi-section, 126 (405), 127 (406,409), 128 (410)

rectifier, 57 (185, 186),58 (187, 188) Filters, 126-129 (402-415), 144 (464),

145 (465) Filter T section, 126 (405), 127 (406,

407,409), 128 (410-412), 129 (414) Flip-flop, J-K, 160 (512), 161 (514),162

(516) Fourier,

series, 28 (71-79), 29 (80-82), 30 (83), 31 (85,86), 134 (437), 137 (447), 140 (453)

transform, 30 (83, 84) 31 (87,88),32 (89,90)

Four-point probe method, 42 (141, 142) Four-terminal network, 111 (363), 112

(364), 113 (367-369), 114 (370), 115 (371,372), 122 (389)

Frame antenna, 130 (422) Frequency,

changing,

376 INDEX

additive, 139 (449) modulation and detection, 133-141

(431-456) multiplicative, 139 (449)

deviation, 135 (441), 136 (444), 140 (455)

modulation, 135 (440-442),136 (443, 444), 140 (455)

multiplication, 140 (455) stability of oscillator, 96-97 (302)

Full-wave rectifier, 55 (180), 57 (185, 186),58 (187,188),61 (197) waveform of, 28 (77)

Gain of antenna, 131 (423,426),132 (429)

Galvanometer, 105 (336), 106 (342) Gap energy, 36 (109) Gases, kinetic theory of and conduction

in, 146-150 (467-484) Gas,

pressure exerted by, 146 (471) valve containing, noise in, 104 (330)

Gate, AND, 160 (510) OR, 160 (510)

Glow discharge, 59 (192), 60 (193), 148 (478), 150 (484)

Graphs, reactance-frequency, 11 (32), 12 (33,

34),13(35) signal flow, 13 (36), 14 (37), 15 (38)

Gray code, 161 (SIS) Guide wavelength, 124 (393), 125 (398,

401)

Half-adder, 160 (511) Half section of filter, 127 (408), 128

(411,412) Half-wave rectifier, 55 (177-179),56

(181, 183),62 (199) waveform of, 28 (77)

Hall effect, 35 (106), 36 (107), 39 (125, 126),40 (127,133),41 (135)

Hartley oscillator, 96 (299, 300), 97 (303), 100 (318)

Hartley-Shannon law, 143 (461)

Hay's bridge, 155 (496) Heaviside's bridge, 159 (508) High-pass filter, 126 (404) Huffman binary code, 143 (460) Hybrid parameters of transistor, 49 (161),

52 (171, 172),53 (174, 175), 73 (237),75 (241-243),76 (244,245) 80 (257)

Identical circuits at all frequencies, 9 (23) Image impedance, 112 (364), 127 (408) Image parameters, 113 (368) Impedance, driving point, 16 (42) Impedance matrix, 116 (373)

open-circuit input, 11 (31) transfer, 11 (31)

Impurity concentration, 36 (108), 37 (116), 38 (119),42 (141),43 (143)

Inductance of straight wire, 165 (522) Information and communication theory,

142-145 (457-468) Interference whistles, 139 (450) Intermediate frequency, 139 (450,451) Inverse network, 11 (32), 12 (34), 13

(35), 19 (54) Ionization, 147 (475) Ionization coefficient, 148 (476,477) Iterative impedance, 112-133 (366),

126 (405), 128 (411), 129 (414, 415)

J-K flip-flop, 160 (512), 161 (514), 162 (516)

Johnson noise formula, 104 (332,333) Junction, microwave, equivalent circuit

for, 118 (378)

Karnaugh map, 160 (510), 161 (513), 164 (520)

Kinetic theory of and conduction in gases, 146-150 (467-484)

Klystron, 167 (533), 168 (534), 171 (546)

Ladder phase-shift oscillator, 98 (306-3<*) Langmuir probe, 148 (478), 150 (484) Laplace equation, 166 (525)

transform, 25 (68,69), 26 (74, 75)

INDEX 377

Loading .coils on transmission line, 107 (345)

Load line, 46 (154), 82 (259) Logarithmic decrement, 26-27 (70,71) Logical switching circuits, 160-164

(510-520) Loschmidt's number, 149 (479) Loss angle, 170 (542) Losses in a capacitor, 10 (27) Loudspeaker,9 (26) Low-pass f"llter, 126 (402), 127 (407,

409),128 (410,412,413),129 (414, 415), 145 (465)

L-type f"llter, 57 (185, 186)

Magnetic and electric fields, motion of electrons in, 151-153 (485-491)

Magnetron, 167 (532) Matching, 111 (361,362) Matrices, 114 (369,370),115-116

(372), 116 (373, 374), 117 (375-377), 118 (378), 123 (390)

Maximum power transfer from source to load, 9 (26)

Maxwell's equations, 124 (391, 392) 169 (539)

m-derived f"llter, 126 (402), 127 (406-408),128 (410,411)

Mean free path, 147 (475) Measurements, 154-159 (492-509) Mesh analysis, 10 (29) Microwave antenna, 131 (424)

data-link, 144 (461) junction, equivalent circuit for, 118

(378) Millman's theorem, 71 (230),91 (288) Mobility, 34 (95,98,99),35 (10H06)

36 (107, 109),37 (111, 112),38 (119, 120), 39 (121, 124, 125),40 (130, 132),41 (133, 135, 136),42 (139-142)

Mobility of electrons in plasma, 148 (478)

Modulation, amplitude, 133 (431), 134 (436,437),

136 (444), 139 (450,452,453), 140 (454)

detection and frequency changing, 133-141 (431-456)

frequency, 135 (440442), 136 (443, 444), 140 (455)

Modulation index, 135 (442) phase, 136 (445) pulse-amplitude, 137 (447) pulse-width, 141 (456)

MOST, 37 (lIS) Motion of electrons in electric and mag

netic fields, 151-153 (485-491), 167 (532)

m-phase rectifier, waveform of, 29 (82) Multi-mesh coupled circuit, 7 (17) Multiplication in gas discharge, 148

(477), 150 (483) Multiplicative frequency changing, 139

(449) Multiplying circuit, 92 (290) Multivibrator, 99 (311) Mutual inductance,S (8, 9, 11), 6 (l2-

14), 7 (15, 17), 8 (18, 19),9 (22), 18 (48), 24 (67),25 (69),69 (224)

Neper, 64 (207) Network determinant, 15 (39), 16 (40,

41) function, 128 (413)

Nodal analysis, 10 (29), 11 (30,31), 19 (53)

Noise, 103-106 (324-342), 131 (426) 132 (429,430), 142 (457), 144 (464), 145 (465,466)

Noise figure of receiver, 132 (429,430) Nyquist,'s criterion and diagram, 70 (228),

71 (231),102 (323)

Open-circuit impedance, input, 11 (31) transfer, 11 (31)

OR gate, 160 (510,511) Oscillators, 94-102 (293-323), 117 (377),

133 (431), 135 (442),139 (451), 140 (455)

Oscillator circuit, 25 (70)

378 INDEX

Parallel circuits,4 (4-6),9 (24), 18 (47) coupled, 7 (15)

Parameters, hybrid of transistor, 49 (161) Path, 15 (39), 16 (41) Path cofactor, 15 (39)

values, 15 (39) Pentode, noise in, 103 (329) Percentage modulation, 133 (432) Permittivity, 108 (349), 120 (381, 383),

121 (384),122 (387), 165 (523), 166 (526-529),169 (537,539), 170 (540-544), 171 (545)

Phase-angle measurement, 156 (499), 157 (500, SOl)

Phase, constant, 107 (343), 112 (364),

118-119 (379), 125 (401),128 (411)

modulation, 136 (445) -shifting network, 10 (28) velocity, 119 (380), 125 (401)

Phasor diagram, 6 (14),64 (208) Photoelectric cell, 167 (531)

emission, 167 (530) Pinch-offvoltage, 37 (111),41 (136) Pi-section filter, 126 (405), 127 (406,

409),128 (410) Plane wave, 121 (387),165 (523), 168

(535), 169 (537), 170 (540) Plane wave guided between parallel strips,

121 (387) Plasma, 148 (478), 149 (480), 150 (484) Plasma frequency, 168 (534) Polarizability, 166 (527, 529), 170 (543,

544),171 (545) fularization, 166 (527) Poles, 16 (43), 17 (44),21 (58) Positive column, 148 (478) Potential barrier, 40 (130) Power factor of capacitor, 10 (27) Power, radiated, from antenna, 130

(420) Pressure exerted by gas, 146 (471) Probability, 142 (459), 143 (460),

144 (462), 145 (466) Probe, 170 (540)

Langmuir, 148 (478), 150 (484)

Propagation constant, 168 (534) Pulse modulation,

amplitude, 137 (447) width, 141 (456)

Push-pull, amplifier, 68 (221),79 (254),82 (260) signals, circuit for producing, 90 (284)

Q factor, 3 (1,2) 4 (4,5),5 (8),17 (45), 18 (47-50), 69 (224),80 (255)

Q factor for crystal, 321 (102) transmission line, 109 (354)

Quarter-wavelength plate, 165 (523)

Radar, 132 (429) Radar equation, 131 (423) Radiated power from antenna, 130 (420),

131 (423,426),132 (428), 133 (433) Radiation, 147 (475) Radiation pattern of antenna, 131 (424)

resistance, 130 (416,417) RC-coupled amplifi~r, 65 (209,212,

213),66 (214-216),70 (228),73 (237), 77 (246, 247)

Reactance-frequency graphs, 11 (32), 12 (33,34), 13 (35), 19 (54),20 (55)

Reactance-valve circuit, 137 (446) Receiver noise figure, 132 (429)

noise measurements on, 105 (334, 335)

superheterodyne, 139 (450) Reciprocity, 50 (163) Recombination,

coefficient, 169 (538) time, 38 (118)

Rectification, 55 (177-180),56(181-183),57 (184-186),58 (187, 188), 61 (197),62 (198, 199)

Rectifier filter, 57 (185, 186),58 (187, 188) full-wave, waveform of, 28 (77) half-wave, waveform of, 28 (77) metal, 56 (182) m-phase, waveform of, 29 (82)

Reflection coefficient, 169 (538)

INDEX 379

Relaxation oscillator, 98 (309),99 (310, 311),102 (322)

Residual network, 15 (39) Resonance, 17 (45,46) Resonant frequency, 3 (1), 4 (5), 5 (8),

6 (12), 18 (47) Rhombic antenna, 132 (427) Ripple factor, 55 (177, 178), 56 (180),

57 (186)

Satellite, 131 (426) Saturated diode in time-base circuit,

159 (509) Scattering matrix, 118 (378) Schering bridge, 154 (492) Schottky barrier diode, 37 (114) Secondary emission coefficient, 148

(476) Selectivity, 109 (355) Self-capacitance of coil, 8 (20, 21), 9

(22) Semiconductor devices, 3343 (91-143) Series circuit, 3 (1-3), 17 (45,46),27

(76) Series-resistance bridge network, 155

(495) Shift register, 161 (514),162 (517) Sidebands, 133 (431,434),134 (437),

135 (440,441),136 (445), 139 (452, 453),140 (454)

Sidelobe, 13 (424) Signal flow graphs, 13 (36), 14 (37), 15

(38), 21 (56) Skin depth, 165 (521),169 (536) Smith chart, 108 (351, 352), 111 (362),

120 (383) Solid-state electronics, 3343 (91-143) Spectrum, 134 (438), 135 (440),136

(443) Square-root circuit, 92 (289) Stability of amplifier, 80 (280) Stabilization circuit to give transistor

biasing, 61 (196),62 (200, 201), 63 (202-204),74 (238,239),89 (282, 283)

Stabilized power supplies, 58 (189),59 (190-192),60 (193-195)

Standing-wave, measurements, 125 (398) ratio, 108 (350,352), 110 (358), 120

(382, 383), 125 (398) Stub matching, 111 (361,362) Summing amplifier, 91 (287,288),93

(291) Superheterodyne receiver, 139 (450) Susceptibility, 170 (543) Switching circuits, logical, 160-164 (510-

520)

Telemetry system of satellite, 144 (463) Temperature-limited diode, noise in, 103

(326),106 (340) Thermionic emission, 45 (148-152),51

(169) Thermistor, 35 (105) Threshold wavelength, 167 (530) Thyratron,44 (146),51 (167,168),57

(184),159 (507,509) Thyratron relaxation oscillator, 99 (310) Time base, 158 (506), 159 (507,509) Topology, 15 (39),16 (40,41),21 (57) Transfer,

function, 142 (458), 144 (464) matrix, 114 (369, 370), 115 (372),

116 (373,374),117 (375-377) Transformer,6 (12), 18 (48), 118 (378) Transformer-coupled amplifier, 67 (218),

79 (251,252) Transformer, quarter-wavelength, 110

(360), 122 (388) Transient,22-25 (59-69),26 (72-75) Transient response of amplifier, 77 (247) Transistor,

amplifier,71 (232),72 (233, 234), 73 (235-237), 114 (369) signal flow graph for, 14 (37)

biasing circuit, 74 (238,239) characteristics and equivalent circuits,

46 (154),47 (155, 156),49 (161, 162),50 (163-165),51 (166),52 (171, 172),53 (174-176) 73 (237)

oscillator, 94 (295),95 (296),96 (302),97 (303), 100 (316), 101 (320), 117 (377)

380 INDEX

Transmission coefficients, 169 (538) Transmission line, effective temperature

of length of, 104 (332), 132 (430), 144 (463)

Transmission lines and networks, 107-123 (343-390)

Transmission line, T network as equivalent circuit for section of, 117 (376)

Transmitter, 133 (432,433), 139 (452), 140 (455)

Trees, 15 (39),16 (40) Triode,45 (147),46 (153),49 (159, 160),

64 (205,208),67 (217,219),68 (220, 221),69 (224), 78 (249),79 (251-254),94 (293, 294), 95 (297), 100 (315,317), 101 (319) noise in, 103 (328)

Truth table, 160 (511) T-section attenuator, 112 (366)

fIlter, 126 (405), 127 (406,407,409), 128 (410412),129 (414)

Tuned amplifier, 80 (255) Tuned oscillator,

anode, 94 (293, 294), 95 (298), 100 (315)

collector, 95, (296),100 (316) drain, 94 (295) grid, 95 (297), 100 (317)

Tunnel-diode equivalent circuit, 99 (314) Twin-T network, 112 (365)

Valve, characteristics and equivalent circuits,

44 (144-146),45 (147-152),46 (153),48 (157, 158),51 (167, 16~ 52 (173)

containing gas, noise in, 104 (330) oscillators,94 (293, 294), 95 (297,

298),96 (300, 301),97 (304), 10( (315,317,318),101 (319)

Variometer,7 (16) Velocity modulation, 168 (533) Velocity of propagation, 107 (343), 108

(349) Vodges-Elder expressions for amplifi

cation factor, 46 (153)

Waveform analysis, 28-32 (77-90) Waveguides, 124-125 (391-401) Wavelength, 107 (343)

threshold, 167 (530) Wavemeter,8 (19) Wien-bridge oscillator, 98 (305) Work function, 51 (169), 167 (530)

Zener diode, 58 (189),59 (191), 62 (200, 201), 63 (203)

Zeros, 16 (43), 17 (44), 21 (58)

Section 2 - Springer978-94-009-5770-1/1.pdf · winding is jWLl + -. - and the resonant frequency f=...

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Transcript of Section 2 - Springer978-94-009-5770-1/1.pdf · winding is jWLl + -. - and the resonant frequency f=...