Chapter 6 Exercise Problems EX6.1

(a) ( ) 2 0.7 2

650BB BE

BQB

V V onI A

Rμ

− −= = ⇒

Then ( )( )

( )( )100 2 0.20

5 0.2 15 2 CQ BQ

CEQ CC CQ C

I I A mAV V I R V

β μ= = == − = − =

(b) Now 0.20 7.69 /

0.026CQ

mT

Ig mA V

V= = =

and ( )( )100 0.026

13 0.20

T

CQ

Vr kIπβ= = = Ω

(c) We find

( )( ) 137.69 15 2.2613 650

v m CB

rA g R

r Rπ

π

⎛ ⎞= − ⎜ ⎟+⎝ ⎠

⎛ ⎞= − = −⎜ ⎟+⎝ ⎠

EX6.2 ( ) 0.92 0.7

100BB BE

BQB

V V onI

R− −= =

or 0.0022 BQI mA=

and ( )( )150 0.0022 0.33 CQ BQI I mAβ= = = (a)

( )( )

0.33 12.7 /0.026

150 0.02611.8

0.33

200 606 0.33

CQm

T

T

CQ

Ao

CQ

Ig mA V

V

Vr kI

Vr kI

πβ

= = =

= = = Ω

= = = Ω

(b)

( ) and o m o C sB

rv g v r R v v

r Rπ

π ππ

⎛ ⎞= − = ⋅⎜ ⎟+⎝ ⎠

so

( )

( ) ( )11.812.7 606 1511.8 100

which yields 19.6

ov m o C

B

v

v rA g r R

v r R

A

π

π π

⎛ ⎞= = − ⎜ ⎟+⎝ ⎠

⎛ ⎞= − ⎜ ⎟+⎝ ⎠= −

EX6.3 (a)

( ) 1.145 0.750

BB EBBQ

B

V V onI

R− −= =

or 0.0089 BQI mA=

( )( )Then 90 0.0089 0.801 CQ BQI I mAβ= = =

Now

( )( )

0.801 30.8 /0.026

90 0.0262.92

0.801

120 150 0.801

CQm

T

T

CQ

Ao

CQ

Ig mA V

V

Vr kIVr kI

πβ

= = =

= = = Ω

= = = Ω

(b) We have ( )o m o CV g V r Rπ=

and sB

rV V

r Rπ

ππ

⎛ ⎞= −⎜ ⎟+⎝ ⎠

so

( )

( ) ( )2.9230.8 150 2.52.92 50

which yields 4.18

ov m o C

s B

v

V rA g r R

V r R

A

π

π

⎛ ⎞= = − ⎜ ⎟+⎝ ⎠

⎛ ⎞= − ⎜ ⎟+⎝ ⎠= −

EX6.4 Using Figure 6.23 (a) For 0.2 ,CQI mA= 7.8 15 , 60 125,ie feh k h< < Ω < < 4 46.2 10 50 10 ,reh− −× < < × 5 13 oeh mhosμ< < (b) For 5 ,CQI mA= 0.7 1.1 , 140 210,ie feh k h< < Ω < < 4 41.05 10 1.6 10 ,reh− −× < < × 22 35 oeh mhosμ< < EX6.5

( )

1 2

2

1 2

35.2 || 5.83 5

5.83 55.83 35.2

TH

TH CC

R R R k

RV VR R

= = = Ω

⎛ ⎞ ⎛ ⎞= ⋅ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

or 0.7105 THV V=

Then ( ) 0.7105 0.7

5TH BE

BQTH

V V onI

R− −= =

or 2.1 BQI Aμ=

and ( )( )

( )( )100 2.1 0.21

5 0.21 10CQ BQ

CEQ CC CQ C

I I A mAV V I R

β μ= = == − = −

and 2.9 CEQV V=

Now 0.21 8.08

0.026100 476 0.21

CQm

T

Ao

CQ

Ig mA

VVr kI

= = =

= = = Ω

And ( ) ( ) ( )8.08 476 10v m o CA g r R= − = −

so 79.1vA = −

EX6.6

( ) ( )

1 2

2

1 2

250 75 57.7

75 575 250

TH

TH CC

R R R k

RV VR R

= = = Ω

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

or

( )( ) ( )( )

1.154

1.154 0.71 57.7 121 0.6

TH

TH BEBQ

TH E

V VV V on

IR Rβ

=− −= =

+ + +

or

( )( )3.48

120 3.38 0.418 BQ

CQ BQ

I AI I A mA

μβ μ

== = =

(a) Now

( )( )

0 418 16 080 026

120 0 0267 46

0 418

CQm

T

T

CQ

I .g . mA / VV .

.Vr . kI .π

= = =

β= = = Ω

We have o m CV g V Rπ= −

We find ( ) ( )( )1 7.46 121 0.6ib ER r Rπ β= + + = +

or 80.1 ibR k= Ω

Also 1 2

1 2

250 75 57.7 57.7 80.1 33.54 ib

R R kR R R k

= = Ω= = Ω

We find

1 2

1 2

33.5433.54 0.5

ibs s s

ib S

R R RV V V

R R R R⎛ ⎞ ⎛ ⎞′ = ⋅ = ⋅⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

or ( )0.985s sV V′ =

Now

( )1 1211 1 0.67.46s EV V R V

rπ ππ

β⎡ ⎤⎛ ⎞+ ⎡ ⎤⎛ ⎞′ = + = +⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥ ⎣ ⎦⎝ ⎠⎣ ⎦

or ( ) ( )( )0.0932 0.0932 0.985s sV V Vπ ′= =

So

( )( )( )( )16.08 0.0932 0.985 5.6ov

s

VA

V= = −

or 8.27vA = −

EX6.7

( )

1 2

2

1 2

15 85 12.75

85 1215 85

TH

TH CC

R R R k

RV VR R

= = = Ω

⎛ ⎞ ⎛ ⎞= ⋅ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

or 10.2 THV V=

Now ( ) ( )1CC BQ E EB BQ TH THV I R V on I R Vβ= + + + +

so ( ) ( ) ( )12 101 0.5 0.7 12.75 10.2BQ BQI I= + + +

which yields 0.0174 BQI mA=

and ( )( )

( )( )

( )( ) ( )( )

100 0.0174 1.74 101 0.0174 1.76

12 1.76 0.5 1.74 4

CQ BQ

EQ

ECQ CC EQ E CQ C

I I mAI mA

V V I R I R

β= = == == − −= − −

or 4.16 ECQV V=

Now ( )( )100 0.026

1.49 1.74

T

CQ

Vr kIπβ= = = Ω

We have ( )π

ππ π

π

=

⎛ ⎞= − − +⎜ ⎟

⎝ ⎠

o m C L

s m E

V g V R || R

VV V g V Rr

Solving for Vπ and noting that ,mg rπβ = we find

( )( )

( )( )( )( )

1

100 4 21.49 101 0.5

C Lov

s E

R RVAV r Rπ

ββ

−= =

+ +

−=

+

or 2.56vA = − EX6.8 Dc analysis

( )( )10 0.7 0.00439

100 101 200.439 , 0.443

BQ

CQ EQ

I mA

I mA I mA

−= =+

= =

Now ( )( )100 0.026

5.92 0.439

0.439 16.88 /0.026100 228

0.439

m

o

r k

g mA V

r k

π = = Ω

= =

= = Ω

(a) Now

( )

100 5.92 5.59

o m o C

Bs

B S

B

V g V r R

R rV V

R r RR r k

π

ππ

π

π

= −

⎛ ⎞= ⋅⎜ ⎟+⎝ ⎠

= = Ω

( )5.59Then 0.9185.59 0.5 s sV V Vπ

⎛ ⎞= ⋅ =⎜ ⎟+⎝ ⎠

( ) ( ) ( )Then 16.88 0.918 228 10ov

s

VA

V= = −

or 148vA = − (b) 0.5 100 5.92 6.09 in S BR R R r kπ= + = + = Ω

and 10 228 9.58 o C oR R r k= = = Ω

EX6.9 (a)

( ) ( )( )

0.25 9.615 /0.026

100 400 0.25

9.615 400 100 769

CQm

T

Ao

CQ

v m o c

Ig mA V

VVr kI

A g r r

= = =

= = = Ω

= − = − = −

(b) ( ) ( )( )9.615 400 100 100

427v m o c L

v

A g r r rA

= − = −= −

EX6.10

( )( )

( )( ) ( )( )

5 0.7 0.00672 10 126 50.84 , 0.847 10 0.84 2.3 0.847 5

BQ

CQ EQ

CEQ

I mA

I mA I mAV

−= =+

= == − −

or 3.83 CEQV V=

dc load line ( ) ( )CE C C EV V V I R R+ −≅ − − +

or ( )10 7.3CE CV I= −

ac load line (neglecting ro) ( ) ( ) ( )2.3 5 1.58ce c C L c cv i R R i i= − = − = −

1.37

0.84

3.83 10

IC(mA)

AC load line

DC load line

Q-point

VCE(V) EX6.11 (a) dc load line

( ) ( )12 4.5EC CC C C E CV V I R R I≅ − + = − ac load line

( )ec c E C Lv i R R R≅ − + or ( ) ( )0.5 4 2 1.83ec c cv i i= − + = − Q-point values

( )( )12 0.7 10.2 0.0150

12.75 121 0.51.80 , 1.82

3.89

BQ

CQ EQ

ECQ

I mA

I mA I mAV V

− −= =+

= ==

2.67

1.80

3.89 12

AC load line

DC load line

IC(mA)

Q-point

VEC(V) (b)

( )( )( )

1.80 1.8 1.83 3.29

min 3.89 3.29 0.6

C CQ

EC

EC

i I mAv V

v V

Δ = =Δ = =

= − =

So maximum symmetrical swing 2 3.29 6.58 V= × = peak-to-peak EX6.12

(a)

( )( ) ( )( )( )

2

1 2 1

1

0.1 1 0.1 121 1

TH CC TH CC

TH E

RV V R VR R R

R Rβ

⎛ ⎞= ⋅ = ⋅ ⋅⎜ ⎟+⎝ ⎠= + =

so

( ) ( )1

112.1 , 12.1 12= Ω =TH THR k VR

We can write ( ) ( )1CC BQ E EB BQ TH THV I R V on I R Vβ= + + + +

We have 1.61.6 , 0.0133 120CQ BQI mA I mA= = =

Then

( )( )

( )( )1

112 0.7 12.1 120.0133

12.1 121 1BQR

I− −

= =+

which yields 1 15.24 R k= Ω

Since 1 2 12.1 ,THR R R k= = Ω we find 2 58.7 R k= Ω Also ( )( ) ( )( )12 1.6 4 1.61 1 3.99 ECQV V= − − =

(b) ac load line ( )ec c C Lv i R R= − Want 0.1 1.6 0.1 1.5 c CQi I mAΔ = − = − = Also 3.99 0.5 3.49 ecv VΔ = − =

Now 3.49 2.327 1.5

ecC L

c

v k R Ri

Δ= = Ω =

Δ

So 4 2.327 LR k= Ω which yields 5.56 LR k= Ω EX6.13

( )

1 2

2

1 2

25 50 16.7

50 525 50

TH

TH CC

R R R k

RV VR R

= = = Ω

⎛ ⎞ ⎛ ⎞= ⋅ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

or

( )( ) ( )( )

3.33

3.33 0.71 16.7 121 1

TH

TH BEBQ

TH E

V VV V on

IR Rβ

=− −= =

+ + +

( )( )

or0.0191

Also120 0.0191 2.29

BQ

CQ

I mA

I mA

=

= =

Now

( )( )

2.29 88.1 /0.026

120 0.0261.36

2.29

100 43.7 2.29

CQm

T

T

CQ

Ao

CQ

Ig mA V

V

Vr kI

Vr kI

πβ

= = =

= = = Ω

= = = Ω

(a)

( ) ( ) ( ) ( )

1 2

1 2

1 1.36 121 1 43.7

ibs s

ib S

ib E o

R R RV V

R R R RR r R rπ β

⎛ ⎞′ = ⋅⎜ ⎟+⎝ ⎠= + + = +

or 1 2120 and 16.7 ibR k R R k= Ω = Ω

Then 1 2 16.7 120 14.7 ibR R R k= = Ω

Now

( )14.7 0.96714.7 0.5s s sV V V⎛ ⎞′ = ⋅ =⎜ ⎟+⎝ ⎠

and

( ) 1o m E o E o

VV g V R r V R r

r rπ

π ππ π

β⎛ ⎞ ⎛ ⎞+= + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

We have s oV V Vπ′ = +

then ( )0.967

1 11 1

ss

E o E o

VVV

R r R rr r

π

π π

β β′

= =⎛ ⎞ ⎛ ⎞+ ++ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( )

( ) ( )( )

We then obtain

10.967

1

1

0.967 11

E oo

vs

E o

E o

E o

R rrV

AV

R rr

R rr R r

π

π

π

β

β

ββ

⎛ ⎞+⎜ ⎟⎝ ⎠= =

⎛ ⎞++ ⎜ ⎟⎝ ⎠

+=

+ +

Now 1 43.7 0.978 E oR r k= = Ω

Then ( )( )( )

( )( )0.967 121 0.978

0.9561.36 121 0.978vA = =

+

(b) ( )( )1ib E oR r R rπ β= + +

or ( )( )1.36 121 0.978 120 ibR k= + = Ω

EX6.14 For RS = 0, then

1o E or

R R r π

β=

+

Using the parameters from Exercise 4.13, we obtain 1.361 43.7121oR =

or 11.1 oR = Ω

EX6.15 (a) For 1.25 and 100, we find

1.26 and 0.0125 CQ

EQ BQ

I mAI mA I mA

β= =

= =

Now 10CEQ EQ EV I R= − Or

( )4 10 1.26 ER= − which yields

4.76 ER k= Ω Then

( )( ) ( )( )( )0.1 1 0.1 101 4.76TH ER Rβ= + = or

48.1 THR k= Ω We have

( ) ( )2

1 2 1

110 5 10 5TH THR

V RR R R

⎛ ⎞= − = ⋅ −⎜ ⎟+⎝ ⎠

or

( )1

1 481 5THVR

= −

We can write ( )

( )0.7 51

THBQ

TH E

VI

R Rβ− − −

=+ +

Or

( )

( )( )1

1 481 5 0.7 50.0125

48.1 101 4.76R

− − +=

+

which yields 1 65.8 R k= Ω

Since 1 2 48.1 ,R R k= Ω we obtain

2 178.8 R k= Ω (b)

( )( )100 0.0262.08

1.25

125 100 1.25

T

CQ

Ao

CQ

Vr kI

Vr kI

πβ= = = Ω

= = = Ω

We may note that ( )m m b bg V g I r Iπ π β= =

Also ( ) ( )

( ) ( )1

2.08 101 4.76 1 100π β= + +

= +ib E L oR r R R r

or 84.9 ibR k= Ω

Now

( )1E oo b

E o L

R rI I

R r Rβ

⎛ ⎞= +⎜ ⎟⎜ ⎟+⎝ ⎠

where

1 2

1 2b s

ib

R RI I

R R R⎛ ⎞

= ⋅⎜ ⎟⎜ ⎟+⎝ ⎠

We can then write

( ) 1 2

1 2

1E ooI

s E o L ib

R r R RIA

I R r R R R Rβ

⎛ ⎞ ⎛ ⎞= = +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

We have 4.76 100 4.54 E oR r k= = Ω

so

( )4.54 48.11014.54 1 48.1 84.9IA ⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

or AI = 29.9

(c) 2.084.76 1001 101o E o

rR R r π

β= =

+

or 20.5 oR = Ω EX6.16 (a)

( ) ( )

1 2

2

1 2

70 6 5.53

610 5 10 570 6

TH

TH

R R R k

RVR R

= = = Ω

⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

or 4.2105 THV V= −

We find ( )

( )( )1

4.2105 0.7 52.91

5.53 126 0.2BQI Aμ− − − −

= ⇒+

and ( )( )

( )1 1

1 1

125 2.91 0.364

1 0.368 CQ BQ

EQ BQ

I I A mA

I I mA

β μβ

= = =

= + =

At the collector of Q1, ( )

( )( )11

11 2

0.7 551

CCCQ

C E

VVI

R Rβ− − −−

= ++

or

( )( )( )11 0.7 55

0.3645 126 1.5

CC VV − − −−= +

which yields 1 2.99 CV V=

also ( )( )1 1 1 5 0.368 0.2 5E EQ EV I R= − = −

or 1 4 93 EV . V= −

Then ( )1 1 1 2 99 4 93 7 92 CEQ C EV V V . . . V= − = − − =

We find ( )1

2

0 7 54 86

1 5C

EQ

V .I . mA

.− − −

= =

and

( )2 1125 4 86 4 82

1 126CQ EQI I . . mAββ

⎛ ⎞ ⎛ ⎞= ⋅ = =⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠

We find 2 1 0 7 2 99 0 7 2 29 E CV V . . . . V= − = − =

and 2 25 5 2 29 2 71 CEQ EV V . . V= − = − =

(b) The small-signal transistor parameters are:

( )( )

( )( )

11

11

22

22

125 0.0268.93

0.364

0.364 14.0 /0.026125 0.026

0.674 4.82

4.82 185 /0.026

T

CQ

CQm

T

T

CQ

CQm

T

Vr kI

Ig mA V

V

Vr kII

g mA VV

π

π

β

β

= = = Ω

= = =

= = = Ω

= = =

We find ( ) ( )( )1 1 11 8 93 126 0 2ib ER r R . .π β= + + = + Or

1 34 1 ibR . k= Ω and

( )( )( )( )

2 2 210 674 126 1 5 10 165

ib E LR r R R. . k

π β= + += + = Ω

The small-signal equivalent circuit is:

Vs����

R1��R2

RE1 RE2 RL

RC1

�

�

�

�

Vo

V�1 r�1 gm1V�1 gm2V�2V�2 r�2

We can write

( ) ( )2 21o b E LV I R Rβ= + where

( )12 1 1

1 2

1 11

π

π π

⎛ ⎞= −⎜ ⎟+⎝ ⎠

= ⋅

Cb m

C ib

s

ib

RI g VR R

VV rR

Then

( ) ( ) 1 1 12

1 2 1

1 πβ

=

⎛ ⎞⎛ ⎞−= + ⎜ ⎟⎜ ⎟+⎝ ⎠⎝ ⎠

ov

s

C mE L

C ib ib

VAV

R g rR RR R R

so

( ) ( ) 5 125126 1 5 105 165 34 1

⎛ ⎞ ⎛ ⎞= − ⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠vA .

.

or 17 7= −vA .

(c) 1 2 1 70 6 34 1 4 76 i ibR R R R . . k= = = Ω

and 2 12

0 676 51 51 126

Co E

r R .R R .π

β+⎛ ⎞ +⎛ ⎞= =⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠

or 43 7 oR .= Ω

Test Your Understanding Exercises TYU6.1

( )( )

0.25 9.62 /0.026120 0.026

12.5 0.25

150 600 0.25

CQm

T

T

CQ

Ao

CQ

Ig mA V

V

Vr kI

Vr kI

πβ

= = =

= = = Ω

= = = Ω

TYU6.2 75

200 A A

o CQCQ o

V Vr II r kΩ

= ⇒ = =

or ICQ = 0.375 mA

TYU6.3

As a first approximation, Cv

E

RA

R≅ −

Resulting gain is always smaller than this value. The effect of RS is very small. Set

10C

E

RR

=

Now ( )5 C C E CEQI R R V≅ + +

or ( )( )5 0 5 2 5C E. R R .= + +

which yields 5 C ER R k+ = Ω

So 10 RE + RE = 5 or

0 454 ER . k= Ω and 4 54 CR . k= Ω We have

0 5 0 005 100

CQBQ

I .I . mAβ

= = =

and ( )( ) ( )( )( )0 1 1 0 1 101 0 454TH ER . R . .β= + =

or 4 59 THR . k= Ω

also

2

1 2 1

1TH CC TH CC

RV V R V

R R R⎛ ⎞

= ⋅ = ⋅ ⋅⎜ ⎟+⎝ ⎠

or

( )( )1 1

1 234 59 5THV .R R

= =

We can write ( ) ( )1TH BQ TH BE BQ EV I R V on I Rβ= + + +

or

( )( ) ( )( )( )1

23 0 005 4 59 0 7 101 0 005 0 454. . . . .R

= + +

which yields 1 24 1 R . k= Ω

and since 1 2 4 59 R R . k= Ω we find 2 5 67 R . k= Ω TYU6.4 dc analysis

( )

1 2

2

1 2

15 85 12 75

85 1215 85

TH

TH CC

R R R . k

RV VR R

= = = Ω

⎛ ⎞ ⎛ ⎞= ⋅ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

or VTH = 10.2 V We can write

( ) ( )( )12 0 7 12 0 7 10 2

1 12 75 101 0 5TH

BQTH E

. V . .IR R . .β

− − − −= =+ + +

or IBQ = 0.0174 mA

and 1 74 CQ BQI I . mAβ= =

ac analysis ( )o fe b C LV h I R R=

and

( )1s

bie fe E

VI

h h R−

=+ +

so ( )( )1

fe C Lov

s ie fe E

h R RVA

V h h R

−= =

+ +

For ICQ = 1.74 mA, we find ( ) ( )( ) ( )max 110, min 70max 2 , min 1.1

fe fe

ie ie

h hh k h k

= == Ω = Ω

We obtain

( ) ( )( )( )

110 4 2max 2 59

1 1 111 0 5vA .. .−

= = −+

and

( ) ( )( )( )

70 4 2min 2 49

2 71 0 5vA ..

−= = −

+

TYU6.5

First approximation, Cv

E

RA

R≅ −

This predicts a low value, so set 9C

E

RR

=

Now ( )CC CQ C E ECQV I R R V≅ + +

or ( )( )7.5 0.6 9 3.75E ER R= + +

which yields 0.625 ER k= Ω and 5.62 CR k= Ω

We have ( )( ) ( )( )( )0.1 1 0.1 101 0.625TH ER Rβ= + =

or 6.31 THR k= Ω

Also

( )( )1 1

1 1 6.31 7.5TH TH CCV R VR R

= ⋅ ⋅ =

We have 0.6 0.006 100

CQBQ

II mA

β= = =

The KVL equation around the E-B loop gives ( ) ( )1CC BQ E EB BQ TH THV I R V on I R Vβ= + + + +

or

( )( )( ) ( )( ) ( )( )1

17.5 101 0.006 0.625 0.7 0.006 6.31 6.31 7.5R

= + + +

which yields 1 7.41 R k= Ω

Since 1 2 6.31 ,THR R R k= = Ω then 2 42.5 R k= Ω TYU6.6

We have ( ) ( )0.951

C Cv

E E

R RA

r R Rπ

ββ

⎛ ⎞−= = − ⎜ ⎟+ + ⎝ ⎠

or ( ) 20.95 4.750.4vA ⎛ ⎞= − = −⎜ ⎟

⎝ ⎠

Assume 1.2 r kπ = Ω from Example 4.6. Then

( )( )( )

24.75

1.2 1 0.4β

β−

= −+ +

or 76β =

TYU6.7 dc analysis: By symmetry, VTH = 0

1 2 20 20 10 THR R R k= = = Ω We can write

( )( )( )

( )( )

0 0.7 5 0.00672

10 126 5125 0.00672 0.84

BQ

CQ BQ

I mA

I I mAβ

− − −= =

+= = =

Small-signal transistor parameters: ( )( )125 0.026

3.87 0.84

0.84 32.3 /0.026

200 238 0.84

T

CQ

CQm

T

Ao

CQ

Vr kII

g mA VV

Vr kI

πβ

= = = Ω

= = =

= = = Ω

(a) We have

( )o m o C LV g V r R Rπ= − and sV Vπ = so

( )

( )( )32.3 238 2.3 5

ov m o C L

s

VA g r R RV

= = −

= −

or 50.5vA = −

(b) 238 2.3 2.28 o o CR r R k= = = Ω

TYU6.8 We find 0.418 ,CQI mA= then

( )( ) ( )( )1215 0.418 5.6 0.418 0.6120CEQV ⎛ ⎞= − − ⎜ ⎟⎝ ⎠

or VCEQ = 2.41 V So

( )2.41 0.5 2CEvΔ = − × or

3.82 CEv VΔ = peak-to-peak TYU6.9 dc load line

( ) ( )10 10CE CQ C EV I R R≅ + − + or

( )20 10CE C EV I R= − + ac load line

( )10ce c C cv i R i= − = − Now

( ) ( )0.7 10 10CE CEQ C CQv V i IΔ = − = Δ = so

( )0.7 10CEQ CQV I− = We have that

( )20 10CEQ CQ EV I R= − + then

( ) ( )10 0.7 20 10CQ CQ EI I R+ = − + or

( )20 19.3CQ EI R+ = (1) The base current is found from

( )10 0.7

100 101BQE

IR

−=+

so ( )( )

( )100 9.3

100 101CQE

IR

=+

Substituting into Equation (1), ( )( )

( ) ( )100 9.320 19.3

100 101 EE

RR

⎡ ⎤+ =⎢ ⎥+⎢ ⎥⎣ ⎦

which yields 16.35 ER k= Ω

Then ( )( )

( )( )100 9.3

0.531 100 101 16.35CQI mA= =

+

and ( )( )20 0.531 10 16.35 6.0 CEQV V≅ − + =

Now 0.7 6 0.7 5.3 CE CEQv V VΔ = − = − =

or, maximum symmetrical swing 2 5.3 10.6 V= × = peak-to-peak TYU6.10 We can write

( )( )( )

0 0.7 106.60

100 131 10BQI Aμ− − −

= ⇒+

and ( )( )130 6.60 0.857 CQI A mAμ= =

Assume nominal small-signal parameters of: 4 , 134

10, 12 83.3

ie fe

re oeoe

h k h

h h S kh

μ

= Ω =

= = ⇒ = Ω

We find

( )( ) ( )

11

4 135 10 ||10 || 83.3 641

ib ie fe E Loe

R h h R Rh

k

⎛ ⎞= + + ⎜ ⎟

⎝ ⎠= + = Ω

To find the voltage gain:

100 641100 641 10

B ibs s s

B ib S

R RV V V

R R R′ = ⋅ = ⋅

+ +

or ( )0.896s sV V′ =

Also ( )

( )1

1feo

s ie fe

h RVV h h R

′+=

′ ′+ +

where 1 10 10 83.3 4.72 E Loe

R R R kh

′ = = = Ω

Then ( )( )( )

( ) ( )0.896 135 4.72

0.8914 135 4.72

ov

s

VA

V= = =

+

To find the current gain:

( )

( )

1

11

10 83.3 10013510 83.3 10 100 641

⎛ ⎞⎜ ⎟

⎛ ⎞⎜ ⎟= = + ⎜ ⎟⎜ ⎟ +⎝ ⎠+⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠

Eoeo B

i fei B ib

E Loe

RhI RA h

I R RR R

h

or 8.59=iA

To find the output resistance: 1

1

4 10 10010 83.3 96.0 135

ie S Bo E

oe fe

h R RR R

h h+

=+

+= ⇒ Ω

TYU6.11 We find

( )

1 2

2

1 2

50 50 25

1 5 2.5 2

TH

TH CC

R R R k

RV V VR R

= = = Ω

⎛ ⎞ ⎛ ⎞= ⋅ = =⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠

Now ( )

( )

( )( )

( )( )

15 0.7 2.5 0.00793

25 101 2and

100 0.00793 0.793

CC EB THBQ

TH E

CQ

V V on VI

R R

mA

I mA

β− −

=+ +

− −= =+

= =

The small-signal transistor parameters:

( )( )

0.793 30.5 /0.026100 0.026

3.28 0.793

125 158 0.793

CQm

T

T

CQ

Ao

CQ

Ig mA V

V

Vr kIVr kI

πβ

= = =

= = = Ω

= = = Ω

(a)

Define 2 0.5 158 0.40 E L oR R R r k′ = = ≅ Ω Now

( )( )

( )( )( )( )

1 101 0.41 3.28 101 0.4v

RA

r Rπ

ββ

′+= =

′+ + +

or Av = 0.925 (b)

( ) ( )( )1 3.28 101 0.4ibR r Rπ β ′= + + = + or

43.7 ibR k= Ω

Also 3.282 1581 101o E o

rR R r π

β= =

+

or 32.0 oR = Ω TYU6.12

For VECQ = 2.5, 5 2.5 5 0.5

CC ECQEQ

E

V VI mA

R− −= = =

then

( )75 5 4.93 76

5 0.0658 76

CQ

BQ

I mA

I mA

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

Small-signal transistor parmaters: ( )( )75 0.026

0.396 4.93

75 15.2 4.93

T

CQ

Ao

CQ

Vr kI

Vr kI

πβ= = = Ω

= = = Ω

Define the small-signal base current into the base, then m bg V Iπ β= −

Now,

( )1E oo b

E o L

R rI I

R r Rβ

⎛ ⎞= +⎜ ⎟+⎝ ⎠

and

1 2

1 2b i

ib

R RI I

R R R⎛ ⎞

= ⋅⎜ ⎟+⎝ ⎠

The current gain is

( ) 1 2

1 2

1o E oI

i E o L ib

I R r R RA

I R r R R R Rβ

⎛ ⎞ ⎛ ⎞= = +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

We have

( )( )( )( )

0.5 1

0.396 76 0.5 0.5 15.2 19.1

E L

ib E L o

R R kR r R R r

kπ β

= = Ω= + += + = Ω

and 0.5 15.2 0.484 E oR r k= = Ω

Then

( ) 1 2

1 2

0.48410 760.484 0.5 19.1I

R RA

R R⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

which yields 1 2 6.975 R R k= Ω

Now

2

1 2 1

1TH CC TH CC

RV V R V

R R R⎛ ⎞

= ⋅ = ⋅ ⋅⎜ ⎟+⎝ ⎠

or

( )( )1

1 6.975 5THVR

=

We can write ( )

( )1CC EB TH

BQTH E

V V on VI

R Rβ− −

=+ +

or

( )( )5 0.7

0.06586.975 76 0.5

THV− −=

+

which yields

( )( )1

11.34 6.975 5THVR

= =

or 1 26.0 R k= Ω and 2 9.53 R k= Ω

TYU6.13 (a) dc analysis:

( )

( )

10 0.7 0.93 10

100 0.93 0.921 1 101

EE EBEQ

E

CQ EQ

V V onI mA

R

I I mAββ

− −= = =

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠

( )( )( ) ( ) ( ) ( )10 0.93 10 0.921 5 10

ECQ EE EQ E CQ C CCV V I R I R V= − − − −= − − − −

or 6.1 ECQV V=

(b) Small-signal transistor parameters:

( )( )100 0.0262.82

0.921

0.921 35.42 /0.026

T

CQ

CQm

T

Vr kI

Ig mA V

V

πβ= = = Ω

= = =

Small-signal current gain: o mI g Vπ= and sV Vπ =

also 1s

i m s mE E

VI g V V g

R r R rππ π

⎛ ⎞= + = +⎜ ⎟⎜ ⎟

⎝ ⎠

Then ( )

( )

( )( )( )( )

11

35.42 10 2.821 35.42 10 2.82

m Eo mI

i m Em

E

g R rI g VA

I g R rV g

R r

ππ

ππ

π

= = =+⎛ ⎞

+⎜ ⎟⎝ ⎠

=+

or 0.987IA =

(c) Small-signal voltage gain:

o m C m s CV g V R g V Rπ= = or

( )( )35.42 5ov m C

s

VA g R

V= = =

or 177vA =

TYU6.14 (a)

( )( ) ( )( )

10 0.71 100 101 10

EE BEBQ

B E

V V onI

R Rβ− −= =

+ + +

or 8.38 BQI Aμ= and 0.838 CQI mA=

( )( )100 0.0263.10

0.838

0.838 32.23 /0.026

0.838

T

CQ

CQm

T

Ao

CQ

Vr kI

Ig mA V

VVrI

πβ

= = = Ω

= = =

∞= = = ∞

(b) Summing currents, we have

sm

E S

V V V Vg V

r R Rπ π π

ππ

⎛ ⎞⎛ ⎞− − −+ = + ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

or 1 1 1 s

E S S

VV

r R R Rππ

β⎡ ⎤⎛ ⎞+ + + = −⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

We can then write

1s

E SS

V rV R R

Rπ

π β⎡ ⎤⎛ ⎞

= − ⎢ ⎥⎜ ⎟+⎝ ⎠⎣ ⎦

Now ( )o m C LV g V R Rπ= −

So ( )

( )( )( )

1

32.23 10 1 3.10 10 11 101

C Lov m E S

s S

R RV rA g R R

V Rπ

β⎡ ⎤⎛ ⎞

= = ⎢ ⎥⎜ ⎟+⎝ ⎠⎣ ⎦

⎡ ⎤= ⎢ ⎥⎣ ⎦

or 0.870vA =

Now

( )

( )

10 0.76310 3.10

1001 101

10 0.90910 1

Ee i i i

E

c e e

Co c c c

C L

RI I I IR r

I I I

RI I I IR R

π

ββ

⎛ ⎞ ⎛ ⎞= ⋅ = ⋅ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠⎛ ⎞ ⎛ ⎞= ⋅ = ⋅⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠⎛ ⎞ ⎛ ⎞= ⋅ = ⋅ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

So we have

( ) ( )1000.909 0.763101

oI

i

IA

I⎛ ⎞= = ⎜ ⎟⎝ ⎠

or 0.687IA =

(c) We have

3.10101 101i E

rR R π

β= =

+

or 30.6 iR = Ω

Also 10 o CR R k= = Ω

TYU6.15 dc analysis: We can write ( )5 BQ B BE EQ EI R V on I R= + + or

( ) ( )( )( )

( )

5 0.7 4.3101 101

100 4.3101

BQB E B E

CQB E

IR R R R

IR R

−= =+ +

=+

Also 5 5CQ C CEQ EQ EI R V I R= + + − or

10110100CEQ CQ C EV I R R⎡ ⎤⎛ ⎞= − + ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

ac analysis: ( )o m C LV g V R Rπ= −

and

1 Bs B

V RV V R V

r rπ

π ππ π

⎛ ⎞= − − ⋅ = − +⎜ ⎟

⎝ ⎠

or

sB

rV V

r Rπ

ππ

⎛ ⎞= − ⋅⎜ ⎟+⎝ ⎠

Then

( )ov C L

s B

VA R R

V r Rπ

β= =+

where mg rπβ =

For 1 ,CQI mA= ( )( )100 0.026

2.6 1

T

CQ

Vr kIπβ= = = Ω

Now ( )( )100 2 2

202.6v

B

AR

= =+

which yields 2.4 BR k= Ω

Also ( )( )

( )100 4.3

12.4 101CQ

E

IR

= =+

which yields 4.23 ER k= Ω

TYU6.16 (a)

dc analysis: For 2 1 ,EQI mA=

( )2100 1 0.990 101CQI mA⎛ ⎞= =⎜ ⎟⎝ ⎠

21

1 0.0099 1 101

EQEQ

II mA

β= = =

+

and 1

10.0099 0.000098

1 101EQ

BQ

II mA

β= = =

+

and ( )( )1 100 0.000098 0.0098 CQI mA= =

and ( )( )1 1 0.000098 10B BQ BV I R= − = −

or 1 0.00098 0BV = − ≅

so 1 0.7 EV V= −

and 2 1.4 EV V= −

Now 1 1 2 0.0098 0.990 1 CQ CQI I I mA= + = + ≅

then ( )( )5 1 4 1 OV V= − =

and ( )( )

2

1

1 1.4 2.4

1 0.7 1.7 CEQ

CEQ

V V

V V

= − − =

= − − =

(b) Small-signal transistor parameters:

( )( )

( )( )

11

11

22

22

1 2

100 0.026265

0.0098

0.0098 0.377 0.026100 0.026

2.63 0.990

0.990 38.1 /0.026

T

CQ

CQm

T

T

CQ

CQm

T

o o

Vr kI

Ig mA

V

Vr kI

Ig mA V

Vr r

π

π

β

β

= = = Ω

= = =

= = = Ω

= = =

= = ∞

(c) Small-signal voltage gain: From Figure 4.73 (b)

( )1 1 2 2

1 2

o m m C

s

V g V g V RV V V

π π

π π

= − += +

and

12 1 1 2 1 2

1 1

1ππ π π π π

π π

β⎛ ⎞ ⎛ ⎞+= + ⋅ = ⋅⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

mVV g V r V rr r

Then

1 1 2 2 11

1π π π

π

β⎡ ⎤⎛ ⎞+= − + ⋅ ⋅⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

o m m CV g V g r V Rr

Also

( )

1 2 11

21

1

1

1 1

π π ππ

ππ

π

β

β

⎛ ⎞+= + ⋅⎜ ⎟⎝ ⎠

⎡ ⎤⎛ ⎞= + +⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

sV V r Vr

rVr

or

( )1

2

1

1 1π

π

π

β=

⎛ ⎞+ + ⎜ ⎟

⎝ ⎠

sVVrr

Now

( )

( )

21 2

1

2

1

1

1 1

π

π

π

π

β

β

⎡ ⎤⎛ ⎞+ + ⋅⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦= = −⎛ ⎞

+ + ⎜ ⎟⎝ ⎠

m m Co

vs

rg g RrVA

V rr

so

( ) ( ) ( )

( )

2.630.377 38.1 101 4265

2.631 101265

⎡ ⎤⎛ ⎞+ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦= −⎛ ⎞+ ⎜ ⎟⎝ ⎠

vA

or 77.0= −vA

(d) ( ) ( )( )1 21 265 101 2.63iR r rπ πβ= + + = +

or 531 iR k= Ω

TYU6.17 (a) dc analysis: For 1 2 3 100 R R R k+ + = Ω

11 2 3

12 0.12 100

CCVI mAR R R

= = =+ +

Now ( )( )1 2

1 1 1

0.5 0.5 0.25 0.25 4 4.25

E CQ E

C E CEQ

V I R VV V V V

= = == + = + =

and

2 1 2 4.25 4 8.25 C C CEQV V V V= + = + =

So 2 12 8.25 7.5 0.5

CC CC

CQ

V VR k

I− −= = = Ω

Also ( )1 1 0.25 0.7 0.95 B E BEV V V on V= + = + =

And

( )3 31

1 2 3

0.95 12100B CC

R RV V

R R R⎛ ⎞

= ⋅ ⇒ =⎜ ⎟+ +⎝ ⎠

which yields 3 7.92 R k= Ω

We have ( )2 1 4.25 0.7 4.95 B C BEV V V on V= + = + =

and

2 32

1 2 3B CC

R RV V

R R R⎛ ⎞+

= ⋅⎜ ⎟+ +⎝ ⎠

or

( )2 7.924.95 12

100R +⎛ ⎞= ⎜ ⎟

⎝ ⎠

which yields 2 33.3 R k= Ω

Then 1 100 33.3 7.92 58.8 R k= − − = Ω

(b) Small-signal transistor parameters:

( )( )1 2

100 0.0260.5

T

CQ

Vr rIπ πβ

= = =

or 1 2 5.2 r r kπ π= = Ω

and

1 20.5

0.026CQ

m mT

Ig g

V= = =

or 1 2 19.23 mA/V= =m mg g

and 1 2o or r= = ∞

(c) Small-signal voltage gain: We have

( )2 2o m C LV g V R Rπ= − Also

22 2 1 1

2m m

Vg V g V

rπ

π ππ

+ =

or 2

2 1 1 1 and 1m s

rV g V V Vπ

π π πβ⎛ ⎞

= =⎜ ⎟+⎝ ⎠

We find

( )

( )

22 1

2

1

1

ov m C L m

s

m C L

V rA g R R g

V

g R R

π

β

ββ

⎛ ⎞= = − ⎜ ⎟+⎝ ⎠

⎛ ⎞= − ⎜ ⎟+⎝ ⎠

We obtain

( )( ) 10019.23 7.5 2101vA ⎛ ⎞= − ⎜ ⎟⎝ ⎠

or 30.1= −vA

TYU6.18 (a) dc analysis: with vs = 0

( )

1 2

2

1 2

125 30 24.2

30 12125 30

TH

TH CC

R R R k

RV VR R

= = = Ω

⎛ ⎞ ⎛ ⎞= ⋅ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

or 2.32 THV V=

Now ( ) ( )1TH BQ TH BE BQ EV I R V on I Rβ= + + +

or

( )( )2.32 0.7 0.0250

24.2 81 0.5BQI mA−= =+

and ( )( )80 0.025 2.00 CQI mA= =

Also

( ) ( )

1

8112 2 2 0.580

ββ

⎡ ⎤⎛ ⎞+= − +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

⎡ ⎤⎛ ⎞= − + ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

CEQ CC CQ C EV V I R R

or 6.99 CEQV V=

Power dissipated in RC: ( ) ( )22 2.0 2 8.0 RC CQ CP I R mW= = =

Power dissipated in RL: 0 0LQ RLI P= ⇒ =

Power dissipated in transistor:

( )( ) ( )( )0.025 0.7 2.0 6.99 14.0 Q BQ BEQ CQ CEQP I V I V

mW= += + =

(b) With ( )18cossv t mVω=

( )( )80 0.0261.04

2.0T

CQ

Vr kIπβ= = = Ω

We can write

( ) cosce C L Pv R R V trπ

β ω=

Power dissipated in RL:

( ) ( )

( )( )

2 2

2

3

1 12

1 1 80 2 2 0.0182 1.042 10

ceRL C L P

L L

v rmsp R R V

R R rπ

β⎡ ⎤= = ⋅ ⋅ ⎢ ⎥

⎣ ⎦

⎡ ⎤= ⋅ ⋅ ⎢ ⎥× ⎣ ⎦

or 0.479 RLp mW= Power dissipated in RC: Since 2 ,C LR R k= = Ω we have 8.0 0.479 8.48 RCp mW= + = Power dissipated in transistor: From the text, we have

( )

( )( ) ( )

2 2

223 3 3

3

2

80 0.0182 10 6.99 2 10 21.04 10 2

π

β

−

⎛ ⎞ ⎛ ⎞≅ − ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= × − × ×⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠

PQ CQ CEQ C L

Vp I V R Rr

or 13.0 Qp mW=

TYU6.19 (a) dc analysis:

( )

1 2

2

1 2

53.8 10 8.43

10 553.8 10

TH

TH CC

R R R k

RV VR R

= = = Ω

⎛ ⎞ ⎛ ⎞= ⋅ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

or 0.7837 THV V=

Now 0.7837 0.7 0.00993

8.43BQI mA−= =

and ( )( )100 0.00993 0.993 CQI mA= =

We have CEQ CC CQ CV V I R= −

or ( )2.5 5 0.993 CR= −

which yields 2.52 CR k= Ω

(b) Power dissipated in RC:

( ) ( )22 0.993 2.52RC CQ CP I R= = or

2.48 RCP mW= Power dissipated in transistor:

( )( )0.993 2.5Q CQ CEQP I V≅ = or

2.48 QP mW= (c) ac analysis: Maximum ac collector current:

( ) ( )0.993 cosci t mAω= average ac power dissipated in RC:

( ) ( ) ( )2 21 10.993 0.993 2.522 2RC Cp R= =

or 1.24RCp mW=

Now 1.24Fraction 0.25

2.48 2.48= = =

+ +RC

RC Q

pP P