Chapter 5 Exercise Problems EX5.1

1αβ

α=

−

For 0.9800.980, 491 0.980

α β= = =−

For 0.9950.995, 1991 0.995

α β= = =−

So 49 199β≤ ≤ EX5.2

3

200120

CBOCEO n

BVBV

β= = or 40.5 CEOBV V=

EX5.3

( )

( )( )( )( )

( ) ( )( ) ( )( )

2 0.7 6.5 200

120 6.5 0.78 5 0.78 4 1.88

0.0065 0.7 0.78 1.88 1.47

BB BEB

B

C C

CE CC C C CE

B BE C CE

V V onI A

RI I A I mA

V V I R or V VP I V on I V mW

μ

β μ

− −= = ⇒

= = ⇒ == − = − == + = + ≅

EX5.4

( )

( )( )( )

5 0.7 2.8 or 4.62 325

80 4.615 0.369 2 5 0.369 which yields 8.13

EB BBB B

B

C B C

EC C C

V V on VI I A

RI I I mA

V R R k

μ

β

+ − − − −= = =

= = ⇒ == = − = Ω

EX5.5

(a) ( )

( )( )( )

( )( )

2 0.7 5.91 220

100 5.91 0.591 1 0.597

10 0.591 4 or 7.64

BB BEB

B

C B

E B

CE CC C C CE

V V onI A

RI I A mAI I mA

V V I R V V

μ

β μβ

− −= = ⇒

= = ⇒= + == − = − =

(b) 6.5 0.7 26.4 220BI Aμ−= ⇒

( )( )

Transistor is biased in saturation mode, so 0.2

10 0.2 or 2.45 4

2.45 0.0264 2.48

CE CE

CC CEC C

C

E C B

V V sat VV V sat

I I mAR

I I I mA

= =− −= = =

= + = + ≅

EX5.6 For 0 0.7 ,IV V≤ < Qn is cutoff, 9 OV V=

When Qn is biased in saturation, we have ( )( )( )100 0.7 4

0.2 9 5.1 200I

I

VV V

−= − ⇒ =

So, for 5.1 ,IV V≥ VO = 0.2 V EX5.7

( )( ) ( )( )

8 0.71 30 76 1.2

BB BEB

B E

V V onI

R Rβ− −= =

+ + +

or 60.2 BI Aμ=

( )( )( )

( )( ) ( )( )

75 60.23 4.52

1 4.58

12 4.517 0.4 4.577 1.2

C B

E B

CE CC C C E E

I I A mA

I I mAV V I R I R

β μβ

= = ⇒

= + == − −= − −

or 4.70 CEV V= EX5.8 For VC = 4 V and ICQ = 1.5 mA,

( )

10 4 4 1.5

1 101 1.5 1.515 100

CC C

CQ

E C

V VR R kI

I I mAββ

+ − −= = ⇒ = Ω

⎛ ⎞+ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

( )

( )

also

0.7 10Then 6.14

1.515

BEE

E

E E

V on VI

R

R R k

−− −=

− − −= ⇒ = Ω

EX5.9

( )

3 0.7 0.25

9.2 k75 0.25 0.2467 mA76

0.7 3 03 0.7 2 6.89 k

0.2467

EQE

E

CQ

CEQ CQ C

C C

IR

R

I

V I R

R R

−= =

= Ω

⎛ ⎞= =⎜ ⎟⎝ ⎠

− + + − =+ −= ⇒ = Ω

EX5.10

( )5 on 2E E EB B BI R V I R= + + −

(a) 1805 2 0.7 2 0.9859 mA41

0.962 mA

E E

C

I I

I

⎛ ⎞+ − = + =⎜ ⎟⎝ ⎠

=

(b) 1806.3 2 1.2725 mA61

1.25 mA

E E

C

I I

I

⎛ ⎞= + =⎜ ⎟⎝ ⎠

=

(c) 1806.3 2 1.6657 mA101

1.64 mA

E E

C

I I

I

⎛ ⎞= + =⎜ ⎟⎝ ⎠

=

(d) 1806.3 2 1.97365 mA151

1.94 mA

E E

C

I I

I

⎛ ⎞= + =⎜ ⎟⎝ ⎠

=

EX5.11

( ) 4 0.71.0

BB EBE E

E

V V onI R

R− −= ⇒ =

or 3.3 ER k= Ω

( )( )

( )( )

0.992 1 0.992 1.0 0.992 or 8

0.992 1 5

C E

B E C B

CB C C CC

I I mAI I I I A

V I R V

αμ

= = == − = − == − = −

or 4.01 CBV V= − EX5.12

( )( ) ( )5 1.5 0.215

CE

C

V V V satR

Iγ

+ − + − += =

or 220 R = Ω

( )

15 1 15

5 0.7 4.3 1

B

I BEB

B

I mA

v V onR k

I

= =

− −= = = Ω

( )( ) ( )( )( )

1 0.7 15 0.2 3.7 B BE C CEP I V on I V

mW= += + =

EX5.13 (a) For V1 = V2 = 0, All currents are zero and VO = 5 V. (b) For V1 = 5 V, V2 = 0; IB2 = IC2 = 0,

1

1 1

5 0.7 4.53 0.95

5 0.2 8 0.6

0.2

B

C C R

O

I mA

I I I mA

V V

−= =

−= ⇒ = =

=

(c) For V1 = V2 = 5 V, IB1 = IB2 = 4.53 mA; IR = 8 mA, IC1 = IC2 = IR/2 = 4 mA, VO = 0.2 V EX5.14

5 5O C C B C

O B C

v i R i Rv i R

ββ

= − = −Δ = − Δ

( )BB I BEB

B

IB

B

V v V oni

Rvi

R

+ Δ −=

ΔΔ =

Then O C

I B

v Rv R

βΔ −=

Δ

Let 100, 5 , 100 C BR k R kβ = = Ω = Ω

So ( )( )100 5

5100

o

I

vv

−Δ= = −

Δ

Want Q-point to be ( ) ( ) ( )2.5 5 100 5o BQv Q pt I− = = −

Then 0.70.005 , 0.005100

BBBQ BQ

VI mA I −= = =

so 1.2 BBV V=

( )( )Also, 100 0.005 0.5 CQ BQI I mAβ= = = EX5.15

2.5 5CEQ CQ CV I R= = −

5 2.5or 10 0.25CR k−= = Ω

0.25 0.002083 120

CQBQ

II mA

β= = =

5 0.7Then 2.06 0.002083B BR R M−= ⇒ = Ω

EX5.16 (a)

( )

1 2

2

1 2

9 2.25 1.8

2.25 59 2.25

TH

TH CC

R R R k

RV VR R

= = = Ω

⎛ ⎞ ⎛ ⎞= ⋅ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

or 1.0 THV V=

(b) ( )

( ) ( )( )1 0.7

1 1.8 151 0.2TH BE

BQTH E

V V onI

R Rβ− −= =

+ + +

( )( )or 9.375

150 9.375 BQ

CQ BQ

I AI I A

μβ μ

== =

( )

( )( ) ( )( )

or 1.41 1 1.42

55 1.41 1 1.42 0.2

or 3.31

CQ

EQ BQ EQ

CEQ CQ C EQ E

CEQ

I mAI I I mA

V I R I R

V V

β== + ⇒ == − −= − −=

(c) For 75β =

( )( )( )( )

1 0.7 17.6 1.8 76 0.2

75 17.6

BQ

CQ BQ

I A

I I A

μ

β μ

−= =+

= =

( ) ( )( )or 1.32

1 76 17.6 CQ

EQ BQ

I mAI I Aβ μ

== + =

( ) ( ) ( )( ) or 1.34

5 1.32 1 1.34 0.2EQ

CEQ

I mAV

== − −

or 3.41 CEQV V= EX5.17

( )( )or 2.5 5 1 0.2

which yields

CEQ CC CQ C E

CQ

V V I R R

I

≅ − +

≅ − +

( )( ) ( )( )( )

2.08 ,2.08 0.0139 150

0.1 1 0.1 151 0.2

CQ

CQBQ

TH E

I mAI

I mA

R Rβ

β

=

= = =

= + =

( ) ( )

2

1 2 1

1

or 3.02

1Now

1so 3.02 5

TH

TH CC TH CC

TH

R k

RV V R VR R R

VR

= Ω

⎛ ⎞= ⋅ = ⋅ ⋅⎜ ⎟+⎝ ⎠

=

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

or ( )( ) ( )( ) ( )( )( )1

1 3.02 5 0.0139 3.02 0.7 151 0.0139 0.2R

= + +

We obtain 1 13 R k= Ω and then 2 3.93 R k= Ω EX5.18

( ) ( )( )

1 2150, ||5 0 5

15 0.0333

1505

1

TH

CQ

CQBQ

TH BEBQ

TH E

R R R

I mA

II mA

V V onI

R R

β

β

β

= =−= =

= = =

− − −=

+ +

( )( )Set 0.1 1TH ER Rβ= +

( )

( )

( )( )( )

2

1 2

2

1 2

2

1 2

We have 10 5

10 0.7Then 0.0333

1.1 151 0.2

which yields 0.1806

TH

BQ

RVR R

RR R

I

RR R

⎛ ⎞= −⎜ ⎟+⎝ ⎠

⎛ ⎞−⎜ ⎟+⎝ ⎠= =

⎛ ⎞=⎜ ⎟+⎝ ⎠

( )( )( )

( )

1 2

1 2

1

Now 0.1 151 0.2 3.02

so 0.1806 3.02

THR RR k

R RR k

⎛ ⎞= = = Ω⎜ ⎟+⎝ ⎠

= Ω

We obtain 1 16.7 R k= Ω and 2 3.69 R k= Ω EX5.19

( )5 5 54.5 0.5

ECQCQ

C E

V V VI

R R

+ −− − − − −≅ =

+ +

so 1 , andCQI mA=

1 0.00833 120

CQBQ

II mA

β= = =

( )( ) ( )( )( )0.1 1 0.1 121 0.5TH ER Rβ= + = or 6.05 THR k= Ω We can write ( )onEQ E EB BQ TH THV I R V I R V+ = + + +

We have ( )1

1 10 5TH THV RR

= ⋅ − and if we let 1 ,EQ CQI I mA≅ =

then we have ( )( ) ( )( ) ( )( )1

15 1 0.5 0.7 0.00833 6.05 6.05 10 5R

= + + + −

which yields 1 6.91 R k= Ω and 2 48.6 R k= Ω EX5.20

( )

( ) ( )

1

11

1

2 21 0.25 1 0.2625 40

0 0 0.7 50.2625

or 16.38

Q

BE

I I mA

V on VR

IR k

β−

⎛ ⎞ ⎛ ⎞= + = + =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

− − − − −= =

= Ω

For 3 , then 2.3 CEO COV V V V= =

( )40 0.25 0.2439 1 41

5 2.3 11.07 0.2439

CO Q

COC

CO

I I mA

V VR kI

ββ

+

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

EX5.21

50 ||100 33.3 THR k= Ω

( )

( )( )( )1

1 1

50 10 5 1.67 50 100

1.67 0.7 511.2

33.3 101 21.12 , 1.13

TH TH

B

C E

V V V

I A

I mA I mA

μ

⎛ ⎞= = − = −⎜ ⎟+⎝ ⎠− − − −

= ⇒+

= =

( )( )1 1 1

1 1

5 1.13 2 5 2.74 3.25 0.51

E E E

CE C

V I R VV V V V

= − = − = −= ⇒ =

2Now 0.51 0.7 1.21 EV V= + =

( )

2 2

2

1 1 2

1

2 2 2 2

2

5 1.21 1.90 18.8 2

1.88 1.12 0.0188 1.10

5 0.51 4.08 1.10

2.51.21 2.5 1.29

1.29 51.97

1.88

E B

C

R C B

C

EC C E EC

C

I mA I A

I mAI I I mA

R k

V V V VV

R k

μ−= = ⇒ =

== − = − =

−= = Ω

= ⇒ = −= − = −

− − −= = Ω

EX5.22

1 2 312We find 240

0.05k R R R= Ω = + +

( )( )1

3

Then 0.5 2 0.7 1.7 1.7 34 0.05

BV V

R k

= + =

= = Ω

( )( )2

2

2

Also 0.5 2 4 0.7 5.7 5.7 1.7 4

4so 80 0.05

B

R

V VV V

R k

= + + =Δ = − =

= = Ω

1

2

and 240 80 34 126 1 4 4 9 C

R kV V

= − − = Ω= + + =

2 12 9Then 6 0.5

CC

CQ

V VR kI

+ − −= = = Ω

Test Your Understanding Exercises TYU5.1

1βα

β=

+

For 7575, 0.986876

β α= = =

For 125125, 0.9921126

β α= = =

TYU5.2

( )1E BI Iβ= +

so 0.7801 81.250.00960

E

B

II

β+ = = = then 80.3β =

Now 80.25 0.9877

1 81.25βα

β= = =

+

( )( )0.9877 0.78 0.770 C EI I mAα= = = TYU5.3

0.990 991 1 0.990

αβα

= = =− −

Now 2.15 21.5 1 100

EB

II Aμβ

= = ⇒+

and ( )( )0.990 2.15 2.13 C EI I mAα= = =

TYU5.4

150Ao

C C

Vr

I I= =

For 0.1 1.5 C oI mA r M= ⇒ = Ω For 1.0 150 C oI mA r k= ⇒ = Ω For 10 15 C oI mA r k= ⇒ = Ω TYU5.5

1 CEC O

A

VI I

V⎛ ⎞

= +⎜ ⎟⎝ ⎠

At 1 , 1 CE CV V I mA= =

(a) For 75 ,AV V= 11 1 0.9868 75C O OI I I mA⎛ ⎞= = + ⇒ =⎜ ⎟

⎝ ⎠

Then, at 10 CEV V=

( ) 100.9868 1 1.12 75CI mA⎛ ⎞= + =⎜ ⎟

⎝ ⎠

(b) For 150 ,AV V= 11 1 0.9934 150C O OI I I mA⎛ ⎞= = + ⇒ =⎜ ⎟

⎝ ⎠

At 10 ,CEV V= ( ) 100.9934 1 1.06 150CI mA⎛ ⎞= + =⎜ ⎟

⎝ ⎠

TYU5.6

CBOCEO

BVBV

n β= so ( )3 100 30 139 CBOBV V= =

TYU5.7 (a) For ( )0.2 0,I BE B CV V V on I I= < ⇒ = = 5 and 0OV V P= = (b) For 3.6 ,IV V= transistor is driven into saturation, so

( ) 3.6 0.7 4.53 0.64

I BEB

B

V V onI mA

R− −= = = and

( ) 5 0.2 10.9 0.440

CEC

C

V V satI mA

R

+ − −= = =

Note that 10.9 2.414.53

C

B

II

β= = < which shows that the transistor is indeed driven into saturation. Now,

( ) ( )( )( ) ( )( )4.53 0.7 10.9 0.2 5.35

B BE C CEP I V on I V satmW

= += + =

TYU5.8 For 0 0.7 BC OV V V= ⇒ =

Then 5 0.7 9.77 0.44CI mA−= = and 9.77 0.195

50C

BII mAβ

= = =

Now ( ) ( )( )0.195 0.64 0.7I B B BEV I R V on= + = + or 0.825 IV V=

( )( )( ) ( )( )

Also 0.195 0.7 9.77 0.7 6.98 B BE C CEP I V on I V

mW= += + =

TYU5.9

10 6.34 0.915 4

CC

C

V VI mAR

+ − −= = =

And ( ) ( )0.7 10

10BE

EE

V on VI

R

−− − − − −= = or IE = 0.930

Now 0.915 0.98390.930

C

E

II

α = = = and 0.9839 611 1 0.9839

αβα

= = =− −

Also 0.930 0.915 15 B E CI I I Aμ= − = − ⇒ and ( )6.34 0.7 7.04 CE C EV V V V= − = − − = TYU5.10

( )( )( )( )

( )( )( )

10 0.7 1.16 mA8

1.1625 22.8 A51

50 22.8 A 1.14 mA10 1.1625 8 0.7 V

10 1.1397 4 10 5.440.7 5.446.14 V

E

B

C

E E E

C C C

EC

EC

I

I

IV V I RV I R

VV

μ

μ+

−= =

= =

= == − = − == − = − = −= − −=

TYU5.11

( )BB B B BE E EV I R V on I R= + +

or ( ) ( )1BB B B BE B EV I R V on I Rβ= + + +

Then ( )

( ) ( )( )2 0.7

1 10 76 1BB BE

BB E

V V onI

R Rβ− −= =

+ + +

or 15.1 BI Aμ= Also ( )( )75 15.1 1.13 CI A mAμ= = and ( )( )76 15.1 1.15 EI A mAμ= =

( )( ) ( )( )Now

8 2 1.13 2.5 1.15 1 6.03 CE CC BB C C E EV V V I R I R

V= + − −= + − − =

TYU5.12

2 5 2 5 CE E E EV . V V . V I R= ⇒ = = We have ( )BB B B BE EV I R V on V= + +

so ( ) 5 0 7 2 5

10BB BE E

BB

V V on V . .IR

− − − −= =

or 0 18 BI . mA= Then ( )( )101 0 18 18 2 EI . . mA= =

And 2 5 0 138 13818 18E

.R . k

.= = Ω ⇒ Ω

TYU5.13

( )BB E E EB B BV I R V on I R= + +

2.22.2 0.0431 51E BI mA I mA= ⇒ = =

and ( )50 2.2 2.16 1 51C EI I mAβ

β⎛ ⎞ ⎛ ⎞= ⋅ = =⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠

Then ( )( ) ( ) ( )2.2 1 0.7 0.0431 50BBV = + + or VBB = 5.06 V Now ( )( )5 5 2.2 1 2.8 EC E EV I R V= − = − = TYU5.14 (a) For 0,Iv = 0, 12 , 0B C Oi i v V P= = = =

(b) For 12 ,Iv V= ( ) 12 0.7 47.1

0.24I BE

BB

v V oni mA

R− −= = =

( ) 12 0.1 2.38 5

CC CEC

C

V V sati A

R− −= = =

0.1 Ov V= and

( ) ( )( )( ) ( )( )0.0471 0.7 2.38 0.1 0.271 B BE C CEP i V on i V sat

W= += + =

TYU5.15

(a) For 2.5 ,CEQV V= 5 2.5 1.25 2CQI mA−= =

1.25 12.5 100

CQBQ BQ

II I Aμ

β= = ⇒ =

Then 5 0.7 344 0.0125BR k−= = Ω

(b) IBQ is independent of .β

For 5 11 , 2 2CEQ CV V I mA−= = =

2 1600.0125

C

B

II

β β= = ⇒ =

For 5 44 , 0.5 2CEQ CV V I mA−= = =

0.5 400.0125

C

B

II

β β= = ⇒ =

So 40 160β≤ ≤ TYU5.16

5 0.7 0.005375 800BQI mA−= =

For ( )( )75, 75 0.005375CQ BQI Iβ β= = = Or 0.403 CQI mA=

For ( ) ( )150, 150 0.005375CQIβ = = Or 0.806 CQI mA= Largest Smallest CQ CEQI V⇒

For 5 1150, 4.96 0.806CR kβ −= = = Ω

For 5 475, 2.48 0.403CR kβ −= = = Ω

For a nominal 0.604 CQI mA= and 5 2.52.5 , 4.14 0.604CEQ CV V R k−= = = Ω

Now for 0.403 ,CQI mA= ( )( )5 0.403 4.14 3.33 CEQV V= − =

For 0.806 ,CQI mA= ( )( )5 0.806 4.14 1.66 CEQV V= − = So, for 4.14 ,CR k= Ω 1.66 3.33 CEQV V≤ ≤ TYU5.17 (a)

( )100 1 0.99 1 101

1 9.90 1 101

CQ EQ

EQBQ

I I mA

II A

ββ

μβ

⎛ ⎞ ⎛ ⎞= ⋅ = =⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠

= = ⇒+

( )( )0.0099 50

or 0.495 B BQ B

B

V I R

V V

= − = −

= −

( )3

14

0.99 10ln 0.026 ln3 10

or 0.630

CQBE T

S

BE

IV V

IV V

−

−

⎛ ⎞ ⎛ ⎞×= =⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠=

( )( )( )

Then 0.495 0.630 1.13 10 0.99 5 5.05

Then 5.05 1.13 6.18

E B BE

C

CEQ C E

V V V VV VV V V V

= − = − − = −= − =

= − = − − =

(b)

( )( )

( )

( )

( )( )( )

3

14

11 , 0.0196 1 51

0.0196 50 0.98

50 1 0.98 1 51

0.98 100.026 ln 0.629 3 10

0.98 0.629 1.61 10 0.98 5 5.1

5.1 1.61 6.71

EQEQ BQ

B

CQ EQ

BE

E

C

CEQ C E

II mA I mA

V V

I I mA

V V

V VV V

V V V V

β

ββ

−

−

= = = =+

= − = −

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

= − − = −= − == − = − − =

TYU5.18

1 121Q Q

B

I II

β= =

+

and

( ) ( )

( )

( )

( ) ( )

( ) ( )

20 0.165121

0.165 0.7

120 0.9921 121

5 0.992 4 53.97 5

0.165 0.7 3.97 5

3.805 5.7

QB Q

E Q

CQ EQ Q Q

C CQ C Q

Q

ECQ E C

Q Q

Q

IV I

V I

I I I I

V I R II

V V V

I I

I

ββ

⎛ ⎞= =⎜ ⎟⎝ ⎠

= +

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

⎡ ⎤ ⎡ ⎤= + − −⎣ ⎦ ⎣ ⎦= − +

Then 3 5.7 3.805 QI= − which yields 0.710 QI mA=