Chapter 8 Exercise Solutions EX8.1

a. 30 V, 30 , 10CC CE C C C CEV V I R I V= = − =

1Maximum power at 152

10 10 215 3

2So 15 30 22.5 Maximum Power 10 W3

CE CC

CCE

L L

V V

IV

R R

= =

= = =

= − ⇒ = Ω ⇒ =

b.

( )( )

,max15 V, 2 A15

0 15 2 7.5 Maximum Power 1 7.5 7.5 W

CC C

CE C L

L L

V IV I R

R R

= == −

= − ⇒ = Ω ⇒ = =

EX8.2 (a) ( )( )8 2.4

19.2 CT PT

θ°

Δ = =Δ =

(b) 853.7

23.0 W

T PTP

P

θ

θ

Δ =Δ= =

=

EX8.3

( )( )Power 1 12 12 wattsD DSi v= ⋅ = = c.

( ) ( )sink amb snk amb

sink sink25 12 4 73 CT T PT T

θ −= + ⋅= + ⇒ = °

b. ( )( )

case sink case snk

case case73 12 1 85 CT T PT T

θ −= + ⋅= + ⇒ = °

a. ( )( )

dev case dev case

dev dev85 12 3 121 CT T PT T

θ −= + ⋅= + ⇒ = °

EX8.4

( )( )( )

,max ambdev case

D,rated

,max amb,max

dev case case snk snk amb

,max

case amb ,max case snk snk amb

case

200 25 3.5 C/W50

200 25 29.2 W3.5 0.5 2

25 29.2 0.5 2 98 C

J

JD

D

D

T TP

T TP

P

T T PT

θ

θ θ θ

θ θ

−

− − −

− −

− −= = = °

−=

+ +−= ⇒ =

+ += + += + + ⇒ = °

EX8.5

a. 10 4 60 mA0.1DQ DQI I−= ⇒ =

b.

( )( ) ( )9 60 0.050 2.7 V min 4 2.7 1.3 V10ds DSv v⎛ ⎞= − = − ⇒ = − =⎜ ⎟⎝ ⎠

So maximum swing is determined by drain-to-source voltage. ( )2 2.5 5.0 VPPV = × =

c. ( )

( )( )

22 2.51 1 31.25 mW2 2 0.1

10 60 600 mW

31.25 5.2%600

PL L

L

S DD DQ

L

S

VP PR

P V I

PP

η η

= ⋅ = ⋅ ⇒ =

= ⋅ = =

= = ⇒ =

EX8.6 Computer Analysis EX8.7 No Exercise Problem EX8.8 a.

( )

0

0 0

2

2

1

BBI GSn

GSnI

Dn n GSn TN

DnGSn TN

n

GSn GSn Dn

o Dn o

Vv v v

dvdvdv dv

i K v V

iv VK

dv dv didv di dv

= + −

= +

= −

= +

= ⋅

1 1 1So 2

GSn

Dn n Dn

dvdi K i

= ⋅ ⋅

At 0 0, 0.050 ADnv i= =

So 1 1 1 520.2 0.050

GSn

Dn

dvdi

= ⋅ ⋅ =

Dn L Dpi i i= +

For a small change in ( )0 L Dn Dpv i i i→ Δ = Δ − −Δ

So 12Dn Li iΔ = Δ

or 0 0

1 1 1 1 1 0.0252 2 2 20

Dn L

L

di didv dv R

= ⋅ = ⋅ = ⋅ =

Then ( )( )0

5 0.025 0.125GSndvdv

= =

Then 0

1 0.125 1.125Idvdv

= + = and 0 1 0.8891.125v v

I

dvA Adv

= = ⇒ =

b. 0For 5 V, 0.25 A , and 0L Dn Dpv i i i= = = =

( )( )

0 0

0 0

0

0

0

1 1 1 1 1 1 2.242 20.2 0.251 0.0520

2.24 0.05 0.112

1 0.112 1.112

1 0.8991.112

GSn GSn Dn

Dn

GSn

Dn n Dn

Dn L

GSn

I

v vI

dv dv didv di dv

dvdi K idi didv dv

dvdvdvdv

dvA Adv

= ⋅

= ⋅ ⋅ = ⋅ ⋅ =

= = =

= =

= + =

= = ⇒ =

EX8.9 a.

( ) ( ) ( )

( ) ( )( )( )

( )( )

22

2 2

1 and 10 8 800 1.5 k

18 22.5 mA10 8

100 0.0260.116 kΩ

22.50.116 101 0.8 80.9 kΩ

b E E L

i TH b

CCQ

L

b

R r R R a RR R R

VIa R

r

R

π

π

β ′ ′= + + = = = Ω= Ω =

= = =

= =

= + =

( )( ) ( ) ( )( )

2

1 2 1

80.91.5 80.9 80.9 1.5 1.5 80.9 1.53 kΩ

80.9

1

THTH TH TH

TH

TH CC TH CC

RR R R

R

RV V R VR R R

= = ⇒ − = ⇒ =+

⎛ ⎞= = ⋅ ⋅⎜ ⎟+⎝ ⎠

( )( ) ( )( )

( ) ( )( )

11

2

2

2 2

22.5 0.225 mA1000.7 1 1.53 18 0.225 1.53 0.7 26.4 kΩ

26.4 1.5326.426.4 1.53 1.53 26.4 1.62 kΩ

QBQ

THBQ

TH

II

VI RR R

RR

R R

β= = =

−= ⇒ = + ⇒ =

=+− = ⇒ =

b. ( )( )( )( )

( )( )

( )( )

0

0

0.9 0.9 18 16.2 V0.9 0.9 22.5 20.25 mA

16.2 1.62 V1010 20.25 203 mA

1 1.62 0.203 0.164 W2

E CC

E CQ

EP

E P

L L

v Vi I

vv Va

i ai I

P P

= = == = =

= = ⇒ =

= = ⇒ =

= ⇒ =

EX8.10 a.

( )3

1 213

exp ln

5 100.026 ln 0.6225 V 0.62252 10

CBEC SQ BE T

T SQ

BE D D

IVI I V VV I

V V V−

−

⎛ ⎞⎛ ⎞= ⇒ = ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞×= = ⇒ = =⎜ ⎟×⎝ ⎠

Bias

13

Bias

0.6225exp0.026

0.62255 10 exp0.026

12.5 mA

D SDI I I

I

−

⎛ ⎞= = ⎜ ⎟⎝ ⎠

⎛ ⎞= × ⎜ ⎟⎝ ⎠

=

b. 022 V, 26.7 mA

0.075LV i= = =

1st approximation:

( )

( )

3

13

3

13

13

26.7 mA, 0.444 mA

26.7 100.026 ln 0.66612 10

12.5 0.444 12.056 mA

12.056 100.026 ln 0.62165 10

2 1.243 V2 0.5769

0.57692 10 exp 0.866 mA0.026

Cn Bn

BE

D

D

D

EB D BE

P

i i

V

I

V

VV V V

ic

−

−

−

−

−

≅ =

⎛ ⎞×= =⎜ ⎟×⎝ ⎠= − =

⎛ ⎞×= =⎜ ⎟×⎝ ⎠== − =

⎛ ⎞= × =⎜ ⎟⎝ ⎠

2nd approximation:

( )

26.7 0.866 27.6 mA

60 27.6 27.1 mA61

0.452 mA12.5 0.452 12.05 mA

En L CP En

Cn Cn

Bn

D D

i i i i

i i

iI I

= + = + ≅ =

⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

== − ⇒ =

( )

( )

3

13

3

13

13

27.1 100.026 ln 0.6664 V2 10

12.05 100.026 ln 0.6215 V5 10

2 1.243 V1.243 0.6664 0.5766 V

0.57662 10 exp 0.856 mA0.026

BEn BEn

D

DD

EB EBp

CP CP

V V

V

VV V

i i

−

−

−

−

−

⎛ ⎞×= ⇒ =⎜ ⎟×⎝ ⎠⎛ ⎞×= =⎜ ⎟×⎝ ⎠

== − ⇒ =

⎛ ⎞= × ⇒ =⎜ ⎟⎝ ⎠

c.

( )

( )

0

3

13

3

13

1010 V, 133 mA0.075

133 mA 131 mA

2.18 mA 12.5 2.18 10.3 mA

10.3 100.026 ln 0.61755 10

2 1.235 V

131 100.026 ln 0.7074 V2 10

1.235 0.7074

L

En L CN

Bn D D

D

DD

BEn BEn

EBp

V i

i i i

i I I

V

V

V V

V

−

−

−

−

= = =

≅ = ⇒ =

= ⇒ = − ⇒ =

⎛ ⎞×= =⎜ ⎟×⎝ ⎠=

⎛ ⎞×= ⇒ =⎜ ⎟×⎝ ⎠= −

13

0.5276 V

0.52762 10 exp 0.130 mA0.026

EBp

CP CP

V

i i−

⇒ =

⎛ ⎞= × ⇒ =⎜ ⎟⎝ ⎠

EX8.11 No Exercise Problem EX8.12 a.

0 3

1 11

0 , 0.7 V12 0.7 11.3 45.2 mA

0.25

I B

R R

v v v

I IR

= = =−= = ⇒ =

1 3

31 1 3 1

1 1 1

1 1 2

11 2 1 2

If transistors are matched, then

1

1 11 11 41

45.2 44.1 mA1.024

44.1 1.08 mA1 41

E E

ER E B E

R E E

E E E

EB B B B

i iii i i i

i i i

i i i

ii i i i

β

β

β

=

= + = ++

⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠

= ⇒ = =

= = = ⇒ = =+

b.

0

0 0

3 3 3

3 1

1 1 1

4

For 5 V 5 V5 0.625 A8

0.6250.625 A, 15.2 mA41

12 5.75.7 V 25.2 mA0.25

1025.2 15.2 10.0 mA 0.244 mA41

5 0.7 4.3 V

I

E B B

B R

E E B

B

v v

i i

i i i

v i

i i i

v

= ⇒ =

= ⇒ =

≅ = ⇒ =

−= ⇒ = =

= − ⇒ = ⇒ = =

= − =

( )2 2

2

2 1

4.3 1265.2 mA

0.2565.2 1.59 mA41

1.59 0.244 1.35 mA

R E

B

I B B I

I i

i

i i i i

− −= = ≅

= =

= − = − ⇒ =

c. 0 625 4631.35I I

I

iA Ai

= = ⇒ =

From Equation (8.54) ( ) ( )( )

( )1 41 250

6412 2 8I

L

RA

Rβ+

= = =

TYU8.1

For ( ) ( )D240, max 1.2 A I max20DS DV I= = = =

For ( )0 max 24 VD DSI V= ⇒ =

( )

( ) ( ) ( )( )

Maximum power whenmax

12 V and2

max0.6 A max 12 0.6 7.2 Watts

2

DSDS

DD D

VV

II P

= =

= = ⇒ = =

TYU8.2

Maximum power at center of load line

( )( )max max0.05 10 0.5 WP P= ⇒ = TYU8.3

a. ( )( )7.5 7.556.3 mW

Q CEQ CQ

Q

P V IP

= ⋅ ==

b. ( )

( )( )

22 6.51 1 21.1 mW2 2 1

15 7.5 113 mW

21.1 18.7%113

56.3 21.1 35.2 mW

PL L

L

S S

L

S

Q

VP PR

P P

PP

P

η η

= ⋅ = ⋅ ⇒ =

= ⇒ =

= = ⇒ =

= − =

TYU8.4

a. ( )( )21 202 2 8 25 20 V 25 V

2 0.8P

L P L L P CC CCL

VP V R P V V VR

= ⋅ ⇒ = = ⇒ = ⇒ = ⇒ =

b. 20 2.5 A8

PP P

L

VI IR

= = ⇒ =

c.

( )( )( )

( )( )

2

2

4

25 20 2019.9 12.5 7.4 W

8 4 8

CC P PQ

L L

Q Q

V V VPR R

P P

π

π

= −

= − = − ⇒ =

d. 20 62.8%4 4 25

P

CC

VV

π πη η= = ⋅ ⇒ =

TYU8.5

a. ( )( )

22 41 80 mW2 2 0.1

PL L

L

VP PR

= ⋅ = ⇒ =

b. 4 40 mA0.1

PP P

L

VI IR

= = ⇒ =

c.

( )( )( )

( )( )

2

2

4

5 4 463.7 40 23.7 mW

0.1 4 0.1

CC P PQ

L L

Q Q

V V VPR R

P P

π

π

= −

= − = − ⇒ =

d. 4 62.8%4 4 5

P

CC

VV

π πη η= = ⋅ ⇒ =

TYU8.6 a.

( )

1 2

2

1 2 1

21 12 8 mA2 1.5

1

8 0.107 mA75 1

CC CCCQ

L L

TH

TH CC TH CC

CQ TH BEBQ

TH E

V VIR R

R R R

RV V R VR R R

I V VIR Rβ β

⎛ ⎞≅ ⋅ = = =⎜ ⎟

⎝ ⎠=

⎛ ⎞= = ⋅ ⋅⎜ ⎟+⎝ ⎠

−= = = =+ +

( ) ( )( )( )( )

( )

( ) ( )( )

1

11

22 2

2

Let 1 76 0.1 7.6 k1 7.6 12 0.7

0.1077.6 7.6

1 91.2 2.33 39.1 k

39.1 7.6 39.1 7.6 7.6 39.1 9.43 k39.1

TH ER R

R

RR

R R RR

β= + = = Ω

⋅ −=

+

⋅ = ⇒ = Ω

= ⇒ − = ⇒ = Ω+

b.

( ) ( )( ) ( )

( )( )

2 21 10.9 0.9 8 1.5 38.9 mW2 2

12 8 96 mW

96 38.9 57.1 mW

38.9 40.5%96

L CQ L L

S CC CQ

Q S L Q

L

S

P I R P

P V I

P P P P

PP

η η

= ⋅ = ⇒ =⎡ ⎤⎣ ⎦

= = =

= − = − ⇒ =

= = ⇒ =

TYU8.7

( )( ) ( )

3 4 5

3 4 5 5

3 4 4 5 4 4

3 3 4 3 3 5 4 3 3

11 1

E E C C

E C B

E B B

E B B B

I I I II I II I I

I I I I

ββ β β

β β β β β β

= + += + += + + += + + + +

If 4β and 5β are large, then 3 4 5 3E BI Iβ β β≅ So that composite current gain is 3 4 5β β β β≅