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A :

A

A / C

A (C )

, .

, :

,

, .

.

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(.) A

A B

, ,

B ( )

, ,

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The BJT is biased in the active region by dc voltage

source VBE. Q-point is set at (IC, VCE) = (1.5 mA, 5 V)

withIB = 15 A (F = 100)

Total base-emitter voltage is: vBE= VBE+ vbe

Collector-emitter voltage is: vCE= 10 iCRC This

is the load line equation.

Assuming:

VBE=0.7 V,IB=15A

(peak) = 8 mV Causes 5A change in vBE

varies between [0.70.008 and 0.7 + 0.008] = [0.692 and 0.708]iB varies between [15A 5A and 15A + 5A] = [10A and 20A]

iC= iB iC varies between [1 mA and 2mA]

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If changes in operating currents and voltages

are small enough, then iCand vCEwaveforms

are undistorted replicas of the input signal.

A small voltage change at the base causes a

large voltage change at collector.

is 180o out of phase of

Assuming:

VBE=0.7 V,IB=15A

(peak) = 8 mV Causes 5A change in vBE

varies between [0.70.008 and 0.7 + 0.008] = [0.692 and 0.708]iB varies between [15A 5A and 15A + 5A] = [10A and 20A]

iC= iB iC varies between [1 mA and 2mA]

vCE varies between [3.4 V and 6.7V] (peak) =

= 1.65V

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Av =Vce

Vbe=

1.65180o

0.0080o= 206180o = 206

180

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C

.

1 2

.

3

,

(

)

.

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A

, , , 3 .

C

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==

==

k100k22

k300k100

3

21

RRL

R

RRR

C

B

Capacitors are replaced by short circuits

C

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:

rd=1

gd

:

dDDi

dDD

iI

vVv

+=

+=

:

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= 0

. A ,

:

gd = IS

VT and rd =

VT

IS

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iD =IS exp vDVT

1

ID+id=IS exp

VD+vdVT

1

ID+i

d=I

Sexp

VDVT

1

+ISexp

vD

VT

vd

VT

+12

vd

VT

2

+16

vd

VT

3

+...

SubtractingID from both sides of the equation,

id= (I

D+I

S)

vd

VT

+12

vd

VT

2

+16

vd

VT

3

+...

For idto be a linear function of signal voltage vd, vd

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( )

The bipolar transistor is assumed to be operating in the Forward-Active Region:

Using a two-porty-parameter network:

The port variables can represent either time-varying part of total voltages and

currents or small changes in them away from Q-point values.

ic = gmvbe + govce

ib = gvbe + grvce

iC IS exp vBE

VT

1+

vCE

VA

and iB

iC

F VCE VBE( )

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&

o= g

mr= F

1IC

1

F

F

iC

Qpo int

IM is the collector current at which is maximum.

o >Ffor iC

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B

Transconductance:

Input resistance:

Output resistance:

gm=IC

VT

40IC

r=oVT

IC=ogm or

o=

gmr

ro=VA+

VCEIC

VAIC

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,

B B

.

vbe = ibr

gmvbe =gmibr =oib

ic =oib +vce

rooib

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For linearity, ic should be proportional to vbe with vbe

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.

A .

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DefineRiB as the input resistance looking into

the base of the transistor:

The input resistance presented to vi is:

The signal source voltage gain is:

(o +1)RE

RiB

=v

b

ib= r

Rin=R

I+R

B R

iB=R

I+R

B r

Av

CE=v

o

vi

=v

o

vb

vb

vi

=Avt

CE R

B r

RI+RB r

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AvCE=Avt

CE RB r

RI +RB r

AvtCE Avt

CE= gmRL RL = ro RC R3

Typically: ro >>RC and R3 >>RC AvCE gmRC = 40ICRC

ICRCrepresents the voltage dropped across collector resistor RC

A typical design point is ICRC =VCC

3

AvCE 40

VCC

3

=13.3VCC

To help account for all the approximations and have a number that is easy to

remember, our "rule-of-thumb" estimate for the voltage gain becomes

AvCE

10VCC

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C ,

= 100, = 75 , (0.245 A, 3.39 )

,

=. . .

gm = 40IC = 40 0.245mA( ) = 9.80 mS r =ogm =

100

9.8mS=10.2 k

ro =VA +VCE

IC=

75V+3.39V

0.245mA= 320 k RB =R1 R2 =160k 300k=104 k

RL = ro RC RL = 320k 22k 100k=17.1 k RB r =104k 10.2k= 9.29 k

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(.)

Av = gmRLRB r

RI +RB r

= 9.8mS 17.1k( )

9.29k

1k+ 9.29k

= 168 0.903( ) = 151

Rin =RI +RB r =10.3 k

vbe = viRB r

RI +RB r

vbe 0.005V vi 5mV

10.3k

9.29k

= 5.54 mV

Check the rule-of-thumb estimate: AvCE

10 12( ) = 120 (ballpark estimate)What is the maximum amplitude of the output signal: vo 5.54mV 151( ) = 0.837 V

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(.)

5, 10 .

,

.

.

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,

= 65, = 50

A , = 0.7 , .

Analysis: To find the Q-point, the

dc equivalent circuit is constructed.

IB=3.71 A

IC=65I

B=241 A

IE=66I

B=245 A

105IB+VBE+(F+1)IB(1.6104)=5

5104ICV

CE(1.6104)I

E(5)=0

VCE

=3.67 V

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.

Rin=R

Br=6.31 k

Rout =RC ro = 9.57 k

AvCE=

vo

v i= gm Rout R3( )

Rin

RI +Rin

= 84.0

Gain Estimate: AvCE

10 VCC+VEE( ) = 100

6.74 k

()

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.

(a) Total power dissipated in the C-B and E-B

junctions is:

PD = VCEIC+ VBEIB where VCE = VCB + VBE

Total power supplied is:

PS = VCCIC + VEEIE

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vCE =VCE Vm sin t where Vm is the

output signal. Active region operationrequires vCE vBE So: Vm VCE VBE

Also: vRc t( ) =ICRCVm sint 0

Vm

min ICRC, VCE

VBE( )