CHAPTER 13

Exercises

E13.1 The emitter current is given by the Shockley equation:

−

= 1exp

T

BEESE V

vIi

For operation with 1exp have we , >>

>>

T

BEESE V

vIi , and we can write

≅

T

BEESE V

vIi exp

Solving for BEv , we have

mV 4.7181010ln26ln 14

2

=

=

≅ −

−

ES

ETBE I

iVv

V 2816.457184.0 −=−=−= CEBEBC vvv

9804.05150

1==

+=ββα

mA 804.9== EC ii α

A 1.196 µβ=C=B

ii

E13.2 α

αβ−

=1

α β 0.9 9 0.99 99 0.999 999

E13.3 mA 5.0=−= CEB iii 95.0/ == EC iiα 19/ == BC iiβ

E13.4 The base current is given by Equation 13.8:

−

×=

−

−= − 1

026.0exp10961.11exp)1( 16 BE

T

BEESB

vVvIi α

which can be plotted to obtain the input characteristic shown in Figure 13.6a. For the output characteristic, we have BC ii β= provided that

431

V. 0.2 elyapproximat ≥CEv For V, 0.2 CCE iv ≤ falls rapidly to zero at .0=CEv The output characteristics are shown in Figure 13.6b.

V 0.8- and8.0in =v E13.5 The load lines for V are shown:

As shown on the output load line, we find

V. 0.1 and V, 5 V, 9 minmax ≅≅≅ CECEQCE VVV

432

E13.6 The load lines for the new values are shown:

As shown on the output load line, we have

V. 0.3 and V, 7 V, 8.9 minmax ≅≅≅ CECEQCE VVV

433

E13.7 Refer to the characteristics shown in Figure 13.7 in the book. Select a point in the active region of the output characteristics. For example, we could choose the point defined by mA 5.2 andV 6 =−= CCE iv at which we find A. 50 µ=Bi Then we have .50/ == BC iiβ (For many transistors the value found for depends slightly on the point selected.) β

E13.8 (a) Writing a KVL equation around the input loop we have the equation for the input load lines: 08000)(8.0 in =+−− BEB vitv The load lines are shown:

Then we write a KCL equation for the output circuit: CEC vi =3000+9 The resulting load line is:

From these load lines we find

434

,A 5 A, 24 A, 48 minmax µµµ ≅≅≅ BBQB III V 3.8 ,V 3.5 V, 8.1 minmax −≅−≅−≅ CECEQCE VVV (b) Inspecting the load lines, we see that the maximum of vin corresponds to IBmin which in turn corresponds to VCEmin. Because the maximum of vin corresponds to minimum VCE, the amplifier is inverting. This may be a little confusing because VCE takes on negative values, so the minimum value has the largest magnitude.

E13.9 (a) Cutoff because we have V 5.4 andV 5.0 −=−=< CEBEBCBE VVVV which is

less than 0.5 V. (b) Saturation because we have .BC II β< (c) Active because we have V. 2.0 and 0 >> CEB VI

E13.10 (a) In this case ( the BJT operates in the active region. Thus the equivalent circuit is shown in Figure 13.18d. We have

)50=β

A 5.717.0µ=

−=

B

CCB R

VI mA 575.3== BC II β

=−= CCCCCE IRVV 11.43 V Because we have ,2.0>CEV we are justified in assuming that the

transistor operates in the active region. (b) In this case ( ,the BJT operates in the saturation region.

Thus the equivalent circuit is shown in Figure 13.18c. We have )250=β

V 2.0=CEV A 5.717.0µ=

−=

B

CCB R

VI mA 8.142.0=

−=

C

CCC R

VI

Because we have ,CB II >β we are justified in assuming that the

transistor operates in the active region.

E13.11 For the operating point to be in the middle of the load line, we want

V 102/ == CCCE VV and mA 2=−

=C

CECCC R

VVI . Then we have

(a) A 20/ µβ == CB II Ω=−

= k 9657.0B

CCB I

VR

(b) A 667.6/ µβ == CB II Ω=−

= M 985.27.0B

CCB I

VR

435

E13.12 Notice that a pnp BJT appears in this circuit. (a) For ,50=β it turns out that the BJT operates in the active region.

A 3.197.020 µ=−

=B

B RI mA 965.0== BC II β

V 35.1020 −=−= CCCE IRV (b) For it turns out that the BJT operates in the saturation region.

,250=β

V 2.0−=CEV A 3.197.020 µ=−

=B

B RI mA 98.12.020

=−

=C

C RI

Because we have ,CB II >β we are assured that the transistor operates in the active region.

E13.13 V 521

2 =+

=RR

RVV CCB EB

BEBB RR

VVI)1( ++

−=

β

BC II β= )( BCECCCCCE IIRIRVV +−−= β BI (µA) CI (mA) CEV (V) 100 32.01 3.201 8.566 300 12.86 3.858 7.271 For the larger values of 21 and RR used in this Exercise, the ratio of the collector currents for the two values of β is 1.205, whereas for the smaller values of 21 and RR used in Example 13.7, the ratio of the collector currents for the two values of β is 1.0213. In general in the four-resistor bias network smaller values for 21 and RR lead to more nearly constant collector currents with changes in β.

E13.14 Ω=+

= k 333.311

121 RR

RB V 521

2 =+

=RR

RVV CCB

( ) A 13.141

µβ

=++

−=

EB

BEBBQ RR

VVI mA 239.4== BQCQ II β

( )Ω=== 1840

mA 238.4mV 26300

CQ

T

IVr β

π

436

Ω 7.66611

1=

+=′

CLL RR

R 7.108−=′

−=π

βrRA L

v

0.163oc −==π

βr

RA Lv Ω=

++= 1186

1111

21 πrRRZin

43.64−==L

invi R

ZAA 7004== ivAAG

Ω== k 1Co RZ

)sin(46.76in

inin t

RZZvAvAv

ssvvo ω−=

+==

E13.15 First, we determine the bias point:

Ω=+

k 00.5011

121 RR

=RB V 1021

2 =+

=RR

RVV CCB

( ) A 26.141

µβ

=++

−=

EB

BEBBQ RR

VVI mA 279.4== BQCQ II β

Now we can compute πr and the ac performance.

( )Ω=== 1823

mA 279.4mV 26300

CQ

T

IVr β

π Ω 7.66611

1=

+=′

ELL RR

R

( )( ) 9910.0

11

=′++

+′=

L

Lv Rr

RAββ

π

( )( ) 9970.0

11

oc =+++

=E

Ev Rr

RAββ

π

( ) Ωk 10.40]1[11

1=

′+++=

LBin RrR

Zβπ

74.39==L

invi R

ZAA

38.39== ivAAG Ωk 333.811

1=

+=′

sBs RR

R

( ) 18.33111

Ω=+

+′+

=

Es

o

RrR

Z

π

β

Problems

P13.1

437

P13.2 The sketch should resemble Figure 13.1a in the book. In normal operation, current flows into the base, into the collector, and out of the emitter.

P13.3 To forward bias a pn junction, the p-side of the junction should be

connected to the positive voltage. In normal operation of a BJT, the emitter-base junction is forward biased and the collector-base junction is reverse biased.

P13.4 The emitter current is given by:

−

= 1exp

T

BEESE V

vIi

P13.5 For a BJT, the parameters are defined as:

E

C

ii

=α

α

αβ−

==1B

C

ii

It is assumed that the base-emitter junction is forward biased and that the collector-base junction is reverse biased.

P13.6* BCE iii +=

mA 3.93.09 =+=

9677.03.9

9===

E

C

iiα

303.0

9===

B

C

iiβ

P13.7* The emitter current is given by the Shockley equation:

−

= 1exp

T

BEESE V

vIi

For operation with 1exp have we , >>

>>

T

BEESE V

vIi , and we can write:

≅

T

BEESE V

vIi exp

Solving for BEv , we have

438

mV 5.6581010ln26ln 13

2

=

=

≅ −

−

ES

ETBE I

iVv

V 341.9106585.0 −=−=−= CEBEBC vvv

9901.0101100

1==

+=ββα

mA 901.9== EC ii α

A 01.99 µβ== C

Bii

P13.8 From Equation 13.9, we have

α

α−1

β =

Solving for α , we have

9950.0201200

1==

+=ββα

P13.9 ( ) mA 2A 10200 =×== µβ BC ii

mA 01.2=+= BCE iii P13.10 mA 5.0=−= CEB iii

9524.05.10

10===

E

C

ii

α

205.0

10==

B

C

i=

iβ

P13.11 Solving the Shockley equation, we have

−

=

1expT

BE

EES

VviI

Also, we have qkTVT /= Substituting values for 300 K, we obtain mV 875.25=TV A 10500.8 13−×=ESI At 310 K, we have mV 74.26=TV A 10794.3 12−×=ESI

439

For the 10 K increase in temperature, we find that IES has increased by a factor of 4.464.

P13.12 From the circuit, we can write:

mA 8.0k 10

715=

Ω−

=Ci A 03.21k 680

7.015 µ=Ω

−=Bi

Then, we have 04.38==B

C

iiβ

P13.13 We have

−

= 1exp

T

BEESE V

vIi

Solving for the saturation current:

143

10030.21

026.07.0exp

1010

1exp

−−

×=−

×

=−

=

T

BE

EES

VviI

For mA 1=Ei , we have

V 6401.010030.2

10ln026.0ln 14

3

=

×

=

≅ −

−

ES

ETBE I

iVv

For A 1 µ=Ei , we have

V 4605.010030.2

10ln026.0ln 14

6

=

×

=

≅ −

−

ES

ETBE I

iVv

P13.14 Writing a current equation at the collector of 1Q , we have:

mA 1211 =++ BBC iii However, 1112 and BCBB iiii β== , so we have mA 1111 =++ BBB iiiβ A 804.91 µ=Bi A 804.92 µ=Bi mA 9804.0121 === BCC iii β mA 9902.01121 =+== CBEE iiii As in the solution to Problem P13.7, we have:

V 6583.010

109902.0ln026.0ln 14

3

21 =

×=

≅= −

−

ES

ETBEBE I

iVvv

440

P13.15 The Shockley equation states:

−

= 1exp

T

BEESE V

vIi

Thus, ESE Ii to alproportion is , and we can write: 12 10 EE ii = Because are the same for both transistors, we also have: βα and 1212 10 and 10 CCBB iiii == Writing a current equation at the collector of 1Q , we have mA 1211 =++ BBC iii mA 110 111 =++ BBB iiiβ A 009.91 µ=Bi A 09.9010 12 µ== BB ii mA 9009.011 == BC ii β mA 009.922 == BC ii β mA 9099.0111 =+= CBE iii

mA 099.910 == ii 12 EE As in the solution to Problem 13.13, we have

V 6561.010

109099.0ln026.0ln 14

3

1

112 =

×=

≅= −

−

ES

ETBEBE I

iVvv

P13.16* We have:

−

= 1exp11

T

BEESE V

vIi

−

= 1exp22

T

BEESE V

vIi

Adding the respective sides of these equations, we have:

( )

−

+=+= 1exp2121

T

BEESESEEEeq V

vIIiii

Thus, we have: 13

21 102 −×=+= ESESESeq III A

Also, we have ,21 BB ii = and we can write:

10010010021

21

21

21 =++

=++

==BB

BB

BB

CC

Beq

Ceqeq ii

iiiiii

ii

β

441

P13.17 We have 11211121121 )1( BBEBCCCeq iiiiiii ++=+=+= βββββ . Then we can write

)1()1(121

1

11211

1

21 ++=++

=+

== ββββββ

βB

BB

B

CC

Beq

Ceqeq i

iii

iiii

P13.18* At 180 mA 1.0 and =BiCo , the base-to-emitter voltage is approximately:

( ) V 4.030180002.07.0 =−−=BEv

P13.19* We select a point on the output characteristics in the active region and compute BC ii /=β . For example on the curve for A, 20 µ=Bi we have

mA 8=Ci in the active region. Thus, 400 A) =mA)/(20 8(= µβ . Then, we have α .9975.0=)1/( +ββ=

P13.20 In the active region (which is for V 2.0>CEv ), we have BC ii β= . In saturation, the current falls to zero. The sketches are:

442

P13.21

P13.22 Distortion occurs in BJT amplifiers because the input characteristic of

the BJT is curved (rather than straight) and because the output characteristic curves are not uniformly spaced.

P13.23 The slope of the load line does not change as VCC changes.

443

P13.24* Following the approach of Example 13.2, we construct the load lines shown. We estimate that V 12 and V, 6.15 V, 4.18 minmax === CECEQCE VVV . Thus, the voltage gain magnitude is ( ) 164.0124.18 =−=vA

P13.25 Following the approach of Example 13.2, we construct the load lines

shown. From the input load line, we have 0maxmin ≅≅≅ BBQB III . Thus, we have 20minmax ≅≅ CECE VV . Thus, the voltage gain magnitude is:

444

( ) 04.02020 =−=vA

P13.26 The input load line is the same as in Problem P13.24. Thus, we again

have A 10 and A 5.5 A, 2 maxmin µµµ === BBQB III . The output load line is shown below. Because minCECEQ VV = , we know that the output waveform is

severely distorted. It is not appropriate to compute voltage gain for such a severely distorted output waveform.

445

P13.27

P13.28*

446

P13.29* (a) and (b) The characteristics and the load line are:

(c) We are given )2000sin(510)()( ttiti sB π+== . From which we

determine that A. 15 µ and A, 10 A, 5 maxmin µµ === BBQB III Then at the intersections of the load line with the respective characteristics, we determine that mA 5.1max = and mA, 0.1 mA, 5.0min == CCQC III and that

V. 5 V, 10 max V, 51min −≅−≅ CE−=CE CEQ VVV (d) The sketches of vCE(t) are:

(e) We are given )2000sin(520)()( ttiti sB π+== . From which we determine that A. 25min µ and A, 20 A, 15 maxµµ === BBQB III Then at the intersections of the load line with the respective characteristics, we determine that mA 0.2min and mA, 0.2 mA, 5.1 max === CCQC III and that

V. 2.0V, 0.2 max V, 5min −≅− CE≅−= CEQCE VVV A sketch of vCE(t) is shown above.

447

P13.30 995.0/ == EC iiα .199)/(/ =−== CECBC iiiiiβ

P13.31 The magnitude of BEv is decreased by about 2 mV for each C increase in temperature. Thus at 180 o C, we have:

o

( ) V 4.030180002.07.0 −=−+−=BEv

P13.32 We can write

)1()1()1(21

1

112

1

12

1

2 +=+

=+

=== βββββ

βB

B

B

C

B

E

Beq

Ceqeq i

ii

iii

ii

P13.33 See Figure 13.16 in the text. P13.34 See Figure 13.16 in the text.

P13.35

P13.36* In the active region, the base-collector junction is reverse biased and

the base-emitter junction is forward biased. In the saturation region, both junctions are forward biased. In the cutoff region, both junctions are reverse biased. (Actually, cutoff applies for slight forward bias of the base-emitter junction as well, provided that the base current is negligible.)

448

P13.37 (a) Active region. (b) Cutoff region. (c) Cutoff region (because V 4.0 for 0 =≅ BEB VI at room temperature). (d) Saturation region (because BC II β than less is ).

P13.38 (a) Cutoff region (because V 3.0 for 0 −=≅ BEB VI at room

temperature). (b) Saturation region (because BC II β than less is ). (c) Active region.

P13.39 (a) mA 5.0=−= CEB III

The transistor is in saturation because we have BC II β< Thus, we have V. 2.0 and mA, 5.1 mA, 1 === CEEC VII (b) The transistor is in the active region because we have 2.0>CEV

and .0>BI Thus, we have V. 5 and mA, 1.10 mA, 10 ===β= B CEEC VIII

(c) This pnp transistor is in the active region because we have 2.0−<CEV and .0>BI Thus, we have

V. 6.4 and mA, 02.2 2 mA, −===β= BC CEE VIII (d) This pnp transistor is in cutoff because both junctions are reverse

biased. Thus, we have V. 3 and 0, −==== CEEBC VIII P13.40 For the Darlington pair, we have V. 2.121eq =+= BEBEBE VVV

For the Sziklai pair, we have V. 6.01eq == BEBE VV

P13.41* 1. Assume operation in saturation, cutoff, or active region.

2. Use the corresponding equivalent circuit to solve for currents and voltages. 3. Check to see if the results are consistent with the assumption made in step 1. If so, the circuit is solved. If not, repeat with a different assumption.

449

P13.42 The fixed base bias circuit is:

It is unsuitable for mass production because the value of varies by typically a factor of 3:1 between BJTs of the same type. Consequently, the bias point varies too much from one circuit to another.

β

P13.43 See Figure 13.22a in the text.

P13.44* The results are given in the table:

Circuit

β

Region of operation

CI (mA)

CEV (volts)

(a) 100 active 1.93 10.9 (a) 300 saturation 4.21 0.2 (b) 100 active 1.47 5.00 (b) 300 saturation 2.18 0.2 (c) 100 cutoff 0 15 (c) 300 cutoff 0 15 (d) 100 active 6.5 8.5 (d) 300 saturation 14.8 0.2

P13.45 The results are given in the table:

Circuit

β

Region of operation

I (mA)

V (volts)

(a) 100 active 2.38 5.25 (a) 300 saturation 4.45 9.80 (b) 100 cutoff 0 10 (b) 300 cutoff 0 10 (c) 100 active 4.26 -10.74 (c) 300 active 4.29 -10.71 (d) 100 active1Q

active2Q 9.53 9.53

(d) 300 active1Q saturated 2Q

14.8 14.8

450

P13.46 V 521

2 =+

=RR

RVV CCB Ωk 67.66/1/1

121

=+

=RR

RB

A 071.2)1(

µβ

=++

−=

EB

BEBB RR

VVI

mA 4141.0== BC II β V 697.6)( =+−−= BCECCCCCE IIRIRVV P13.47* The BJT operates in the active region. We can write the voltage

equation: EEBB IRIR ++= 7.05

However, we can substitute using the relations:

ββ

β CBCE

IIII =+

= and 1

Thus, we have:

CEC

B IRIRβ

ββ

13.4 ++=

mA. 4 require we ,100 For == CIβ Furthermore, for require we ,300=βmA. 5=CI Thus, we have two equations:

( ) ( )33 1004.41004.03.4 −− ×+×= EB RR ( ) ( )33 10017.51001667.03.4 −− ×+×= EB RR Solving, we find: Ω=Ω= 753 and k 5.31 EB RR

P13.48 It turns out that for the values given the BJT operates in the saturation

region and the equivalent circuit is:

451

Enclosing the transistor in a closed surface (supernode) and writing a KCL equation, we have

07.0157.0152.0

21

=+

+−+

+−+

+R

VR

VR

VRV EE

C

E

E

E

Substituting the resistance values and solving we obtain 533.7=EV V.

Then we have mA 7267.0)2.0(15=

+−=

C

EC R

VI

mA. 0266.0

and

=−= CEB III As a check we note that we have CB II >β as required for operation in the saturation region.

P13.49* From Equations 13.20 through 13.23, we have:

( ) EB

BEBBC RR

VVII1++

−==

βββ

21 11

1RR

RB +=

21

2

RRRVV CCB +

=

Maximum CI occurs for k 7.495.0 ,200 minmax =Ω×==== EE RRββ

=Ω×==Ω=Ω×==Ω k 4705.1 and ,k 95k 10095.0 , 4465 max22min11 RRRR .k 35.49 Ω With these values, we calculate:

Ω= k 48.32BR V 128.5=BV mA 952.0max =CI Minimum CI occurs for k 7.405.1 ,50 maxmin =Ω×==== EE RRββ

==Ω=Ω×==Ω min22max11 and ,k 105k 10005.1 ,k 935.4 RRRR .k 44.65k 4795.0 Ω=Ω× With these values, we calculate:

Ω= k 33.31BR V 475.4=BV mA 6667.0min =CI

P13.50 A 101043.1

156

11 µ=

×−

= BEB

VI mA 111 =β= BC II

)101(107.0)10(1010)(1015 23

234

23

2214

BBEBEBC IIIVII +++=+++= − A 74.382 µ=BI mA 874.322 =β= BC II mA 913.3)1( 22 =+β= BE II V 613.410 22

31 =+= BEECE VIV

V 213.7101015 23

23

2 =−−= ECCE IIV

452

P13.51 The circuit is:

We have:

A 157k 100

7.015 µ=Ω

+2 =I A 20 µ

β== C

BII

A 17721 µ=+= BIII

Ω=×−

=−

= − k 3.2410177

7.056

11 I

VVR BECE

mA 177.21 =+= III C

Ω=−

= k 59.415

IVR CE

C

P13.52

mA 104.0

k 1507.015

2 =Ω

+=I

( ) ( ) 7.01047470015 23

2 ++×+++= BBC IIIII Substituting BC II β= and solving for BI , we obtain:

453

( )( ) A 96.8

5170014700517007.015 2 µ

β=

++−−

=IIB

mA 79.1== BC II β ( ) V 06.6470015 2 =++−= BCCE IIIV

P13.53 See Figure 13.26 in the text.

P13.54 CQ

T

IVr β

π =

P13.55 CQ

T

IVr β

π =

CQI πr

A 1 µ ΩM 6.2 mA 1.0 Ωk 26

mA 1 Ωk 6.2

P13.56* We use the same approach as in Section 13.7. We can write ( )(10) 25 tvti BE

−=B 25 )]([10)( tvVtiI beBEQbBQ +=+ −

)(10)(10210)( 25525 tvtvVVtiI bebeBEQBEQbBQ−−− +×+=+ (1)

However for the Q-point, we have 2510 BEQBQ VI −= . Therefore, the first term on each side of Equation (1) can be canceled. Furthermore, the last term on the right hand side of Equation (1) is negligible for small signals [i.e., )(tvbe much less than VBEQ at all times]. Thus, Equation (1) becomes

( )(102) 5 tvVti beBEQ−×≅b

We have defined

)()(

titvr =π

b

be

and we have

CQCQBQBEQ IIIV

1581100/10

10510

105102

15

4

5

4

5 =×

=×

=× −πr =

For =CQI 1 mA, we obtain =πr 50 kΩ.

454

P13.57 The equivalent circuit is:

)1()(

1211

111211

eq

eqeq β

βππ

πππ ++=

++== rr

iiirir

iv

rb

bbb

b

be

P13.58 The equivalent circuit is:

11

11

eq

eqeq π

ππ r

iir

iv

rb

b

b

be ===

P13.59 Coupling capacitors are used to provide a path for ac signals while

maintaining an open circuit for dc. In connecting the signal source or load, coupling capacitors prevent the source or load from affecting the bias point of the amplifier. When it is desired to amplify dc signals, coupling capacitors should not be used.

P13.60 The common-emitter amplifier is inverting. Both the voltage gain and current gain magnitudes are potenitally greater than unity.

455

P13.61 See Figure 13.27a in the text.

P13.62 The dc circuit is:

The bias point calculations are:

Ω=+

k 197.311

121 RR

=RB V 796.421

2 =+

=RR

RVV CCB

( ) A 3.391

µβ

=++

−=

EB

BEBBQ RR

VVI mA 93.3== BQCQ II β

( )

Ω=== 662mA 93.3mV 26100

CQ

T

IVr β

π

Then, we compute the amplifier performance:

Ω 50011

1=

+=′

CLL RR

R 5.75−=′

−=π

βrRA L

v

0.151oc −=β

=πr

RA Cv Ω 548

1111

21

=++

=πrRR

Zin

4.41−==L

invi R

ZAA 3124== iv AAG

Ω== k 1Co RZ

P13.63* The solution is similar to that of Problem P13.62. The results are:

456

High impedance amplifier (Problem 13.63)

Low impedance amplifier (Problem 13.62)

ICQ 0.0393 mA 3.93 mA πr 66.2 kΩ 662 Ω vA -75.5 -75.5 ocvA -151 -151 inZ 54.8 kΩ 548 Ω

iA -41.4 -41.4 G 3124 3124

oZ 100 kΩ 1 kΩ

P13.64 (a) The small-signal equivalent circuit is:

in which we have defined

LCL RR

R/1/1

1+

=′

(b) From the equivalent circuit we can write: bEbin iRirv )1( ++= βπ (1) bLo iRv ′−= β Then we have

E

L

bEb

bL

in

ov Rr

RiRir

iRvvA

)1()1( ++′−

=++′

==ββ

ββ

ππ

(c) From the equivalent circuit we can write:

bB

inin i

Rvi +=

Then solving Equation (1) for ib and substituting, we have

E

in

B

inin Rr

vRvi

)1( +++=

βπ

[ ]EBin

inin RrRi

vR)1(/1/1

1+++

==βπ

457

(d) Because the coupling capacitors are open circuits for dc, we can ignore the signal source and the load in the dc analysis. To attain ICQ = 1 mA, we must have IBQ = (1 mA)/β = 50 µA. Writing a voltage equation we have

EBQBEQBQBCC RIVIRV )1( +++= β Substituting values and solving, we obtain RB = 375.9 kΩ. (e) First, we have Ω 520005.0/)026.0100(/ =×== CQt IVr βπ . Then using the equations from parts (a) and (b) we determine that Av = -9.416 and Rin = 10.33 kΩ.

P13.65 See Figure 13.30a in the text.

P13.66 The voltage gain of an emitter follower is positive and less than unity in magnitude. The current gain and power gains are potentially greater than unity.

P13.67* The dc circuit is:

The bias point calculations are:

Ω=+

= k 511

121 RR

RB

V 5.721

2 =+

=RR

RVV CCB

( ) A 1.641

µβ

=++

−=

EB

BEBBQ RR

VVI

mA 41.6== BQCQ II β Now we can compute πr and the ac performance.

( )Ω=== 405

mA 41.6mV 26100

CQ

T

IVr β

π

458

Ω 33311

1=

+=′

ELL RR

R

( )( )

98.011

=′++

+′=

L

Lv

RrRAβ

β

π

( )( ) 996.0

11

oc =+β++β

=π E

Ev Rr

RA

( )[ ] Ωk 36.4111

1=

′+++=

LBin RrR

Zβπ

61.8==L

invi R

ZAA 51.8== iv AAG

( ) Ω 83311

1 which in 111

=+

=′+

+′+

=sB

s

Es

o RRR

RrR

Z

π

β

Ω= 1.12oZ P13.68 The solution is similar to that of Problem P13.67. The results are:

High impedance amplifier (Problem 13.68)

Low impedance amplifier (Problem 13.67)

ICQ 64.1 µA 6.41 mA πr 40.5 kΩ 405 Ω vA 0.988 0.988 ocvA 0.996 0.996 inZ 436 kΩ 4.36 kΩ

iA 8.61 8.61 G 8.51 8.51

oZ 1210 Ω 12.1 Ω

459