Chapter 11 Exercise Solutions EX11.1

( )

( )( )1 2

1 2 1 2

on 0.7 V

0.5 mA10 0.5 10 5 V

E BE E

C C

C C C C

v V v

I Iv v v v

= − ⇒ = −

= == = − ⇒ = =

EX11.2

2 1 0.991 exp

11 exp0.99

1exp 10.99

1ln 1 119.5 mV0.99

C

Q d

T

d

T

d

T

d T d

iI v

V

vV

vV

v V v

= =⎛ ⎞

+ ⎜ ⎟⎝ ⎠

⎛ ⎞+ =⎜ ⎟

⎝ ⎠⎛ ⎞

= −⎜ ⎟⎝ ⎠

⎡ ⎤= − ⇒ = −⎢ ⎥⎣ ⎦

EX11.3 a.

( )( )1 2

1 2 1

0 0.7 V

0.25 8 2 V 3 V 3.7 VE

RC C C EC

v v v

V v v v

= = ⇒ =

Δ = = ⇒ = = − ⇒ =

b. 1 2 12.5 V 3.2 V 6.2 VE ECv v v v= = ⇒ = ⇒ =

c. 1 2 12.5 V 1.8 V 1.2 VE ECv v v v= = − ⇒ = − ⇒ = EX11.4 Let 1 ,QI mA= then 1 2 0.5 CQ CQI I mA= =

1 20.5 19.23 /

0.026m mg g mA V= = =

At 22 2

1, 2

cC d m C

d

vv A g Rv

= =

So, ( ) 2 21150 19.23 15.6 2 C CR R k= ⇒ = Ω

At 11 1

1, 2

cC d m C

d

vv A g Rv

= = −

So, ( ) 1 11100 19.23 10.4 2 C CR R k− = − ⇒ = Ω

If 10 and 10 ,V V V V+ −= + = − dc biasing is OK. EX11.5

( )

( )( )( )

( )( )( )( )( )

( )( )( ) ( )

0.8 122 0.0262

0.05941 101 0.8 100

110.026 100

0.0594 1sin 59.4sin

Q C

Tcm

Q O

T

o cm cm

I RV

AI R

Vv A v t mV t V

ββ

ω ω μ

⎡ ⎤⎛ ⎞ −− ⎢ ⎥⎜ ⎟⎝ ⎠ ⎣ ⎦= = = −+

++

= = − = −

EX11.6

(a) ( )( )40 , 050 40 2.0

d cm

o

v V vv mV

μ= == ⇒

(b) 40 200 d cmv V v Vμ μ= = Assuming 0cmA >

( )( ) ( )( )

105060 20log

0.0550 40 0.05 200

2.010

cm

cm

o

o

AAvv mV

=

== +=

EX11.7

a. Diff. Gain 4Q C

dT

I RA

V=

For 1 2 5 Vv v= = ⇒Minimum collector voltage

2 5 V 15 5 10 V2Q

C C

Iv R= ⇒ ⋅ = − = or 20 VQ CI R = for max. dA

Then ( ) ( )20 max 192

2 0.026d dA A= ⇒ =

b. If 0.5 mA, 40 kΩQ CI R= =

( )

( )( )( )( )

( )( )

0

10

21

1

Then

202 0.026

0.199 201 0.5 100

10.026 200

192and 20log 59.7 dB0.199

Q C

Tcm

Q

T

cm cm

dB dB

I RV

AI R

V

A A

C M RR C M RR

ββ

⎛ ⎞−⎜ ⎟⎝ ⎠=+⎡ ⎤

+⎢ ⎥⎣ ⎦

⎛ ⎞−⎜ ⎟⎜ ⎟⎝ ⎠= ⇒ = −

⎡ ⎤+⎢ ⎥

⎣ ⎦⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

EX11.8

( )( )( )

( )( )

2 1

100 0.02610.4 K

0.252 10.4 101 0.5 122 K

id E

T

CQ

id

R r R

VrI

R

π

π

β

β

= + +⎡ ⎤⎣ ⎦

= = =

= + =⎡ ⎤⎣ ⎦

EX11.9

( )( )( )

( )

2 1

9.62 1010

2 1 9.62

m Cd

m E

E

g RAg R

R

=+

=+⎡ ⎤⎣ ⎦

( )1 9.62 4.81 0.396 KE ER R+ = ⇒ =

( ) ( )( )2 1 2 10.4 101 0.396

100.8 Kid E

id

R r R

Rπ β= + + = +⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

=

EX11.10

( )

( )( )( )( )

( ) ( )( )( )

241 3 4

12

4 4

24 4 4

24 4

2

4

10

10 0.1 80 0.8

10 8 1.6 0.64

8 11.8 4.88 0

11.8 11.8 4 8 4.881.81 V

2 8

GSn GS TN

GS GS

GS GS GS

GS GS

GS

VI K V VR

V V

V V V

V V

V

−= = −

− = −

− = − +

− − =

± += =

( )( )

( )( )

1

1 2

21 1

21 1

01 02

10 1.81 0.102 mA80

0.102 0.0512 mA2

0.0512 0.050 0.8 1.81 V5 0.0512 40 2.95 V

Q

D D

n GS TN

GS GS

I I

I I

K V V

V Vv v

−= = =

= = =

= −

= − ⇒ == = − =

( )

( ) ( )

( )( )

( ) ( )

( )

1 1

01 1 1

4 4

1 4

:1.81 0.8 1.01 V

max sat2.95 1.01 1.81

max 3.75 V :

1.81 0.8 1.01 Vmin sat 5

1.81 1.01 5min 2.18 V

2.18 3.75 V

cm DS GS TN

cm DS GS

cm

cm DS GS TN

cm GS DS

cm

cm

Max v V sat V V

v v V V

vMin v V sat V V

v V V

vv

= −= − =

= − += − +

== −= − == + −= + −= −

− ≤ ≤

EX11.11

( ) ( )( )( ) ( )

( )( )

1 2max max 1 /

2 2

1 5 5

n Qf f

d f D d

K Ig g mA V

A g R A

= = ⇒ =

= = ⇒ =

EX11.12

21 1 12 2 2

n nDd d

Q Q Q

K Ki v vI I I

⎛ ⎞= + ⋅ − ⎜ ⎟⎜ ⎟

⎝ ⎠

Using the parameters in Example 11.11, 20.5 / , 1 ,n QK mA V I mA= = then

( ) ( )21 1 0.5 0.50.90 1

2 2 1 2 1D

d dQ

i v vI

⎛ ⎞= = + ⋅ − ⎜ ⎟⎜ ⎟

⎝ ⎠

By trial and error, 0.894 dv V= EX11.13 a.

( )

( ) ( )( )

202

2

404

4

02 04

0.5 9.615 mA/V2 2 0.026

125 500 kΩ0.2585 340 kΩ

0.259.615 500 340 1946

Qm

T

A

C

A

C

d m d

Ig

VVrIVrI

A g r r A

= = =

= = =

= = =

= = ⇒ =

b. ( )( )

02 04

9.615 500 340 100 644

d m L

d d

A g r r R

A A

=

⎡ ⎤= ⇒ =⎣ ⎦

c. ( )( )150 0.026

15.6 kΩ0.25

2 31.2 kΩ

T

CQ

id id

VrI

R r R

π

π

β= = =

= ⇒ =

d. 0 02 04 0500 340 202 kΩR r r R= = ⇒ =

EX11.14

( )( ) ( )( )

( )

( ) ( )

( )( )

2 24 5

4 5

5 4

4 4

4 4

2

80 8010 0.5 0.33 0.52 2

2 61 62

10 10.5 6 0.50.33 2

6.0 5.252 0.8754 V80 10 0.8754 0.5 53.37 A2

REF GS GS

GS GS

GS GS

GS GS

GS GS

REF

I V V

V V

V V

V V

V V

I μ

= − = −

+ =

= −

− = − −

= ⇒ =

= − =

Also 53.37 AQ REFI I μ= =

( )( )

( )

( ) ( )( )

1 8

1 8

802 2 10 53.37 0.2066 mA/V2

1 1 1873.7 K0.053370.02

20.2066 1873.7 1873.7

194

m n Q

o oD

d m o o

d

g k I

r rI

A g r rA

λ

⎛ ⎞= = ⇒⎜ ⎟⎝ ⎠

= = = =⎛ ⎞⎜ ⎟⎝ ⎠

= ==

EX11.15

( )( )

( )

2

1 1 2

400 500 101000 0.8 /

2

0.080.8 2 0.1 402

d m o O

m m

m n DQ

A g r Rg g mA V

g K I

W W WL L L

== ⇒ =

=

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= ⇒ = =⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠

EX11.16

( )( )

( ) ( )( )

6 6 7

7 7 6

47

6

11

1 100 101 1.01 10

e b b

c b b

c

b

I I II I III

ββ β β

β β

= + == = +

= + = = ×

EX11.17

( )

( )( )

( )( )

4

3001

1

1 1 9.803 A1 1100 0.026

265.2 K0.009803

265.2 300 5.596 K101

101

100 0.0262.626 K

0.992.626 5.596 10

10181.4 10 80.7

AOA

CA EA

A

OA

B OAO

B

O

O

rR

I I

r

R

r RR

r

R

R

π

π

π

π

ββ

β

β μβ β

β

+=

+

⎛ ⎞= ⎜ ⎟+⎝ ⎠⎛ ⎞⎛ ⎞

= ⇒⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠

= =

+= =

+=

+

= =

+=

= = Ω

EX11.18

( )

( ) ( )

1 11

12

2 2

10 0.7 100.6 32.2 K

ln

0.60.2 0.026 ln 143 0.2

Q TQ

I RR

II R VI

R R

− − −= = ⇒ =

⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠⎛ ⎞= ⇒ = Ω⎜ ⎟⎝ ⎠

.

( )

6 1 3

1 1

2

4

0

10 0.710.7 0.1 4 67 K

0.7 4 3.3 V3.3 1.4 1.9 V

R

C C CE

C C

o

E

I I R

I R VR R

vv

= ⇒ =

= + −= + ⇒ =

= − + == − =

44 4 4

4

3 2 4

1.9 3.17 K0.6

1.43.3 1.4 3 4.9 V

ER

C O CE

vI R RR

v v v

= ⇒ = ⇒ =

= − += − + =

( )

5 4 55

5 6

7 7

10 4.90.6 8.5 K

4.2 0.74.9 0.7 4.2 V 5.83 K0.6

0 102 K

5

R R

E

I I RR

v R

R R

−= = = ⇒ =

−= − = ⇒ = =

− −= ⇒ =

EX11.19 ( )

( )( )

( ) ( )

( )( )( )

( )( )

( )( )

2 3 4

4

22

34

2 2

3 5 6 6 7

5

6

1100 0.026

4.333 K0.6

100 0.026433.3 K

0.6433.3 101 4.333 871 K

(1 1

100 0.0264.333 K

0.6100 0.026

0.52 K5

i

T

R

i i

i

R r r

r

VrI

R R

R r R r R

r

r

π π

π

π

π π

π

π

β

β

β β

= + +

= =

≈ = =

= + ⇒ =

= + + + + +⎡ ⎤⎣ ⎦

= =

= =

( ) ( )( )

( )

( )

( ) ( ) ( )( )( )

3

3

1 2

1

42 5 3

1 2

4.333 101 5.83 0.52 101 2

21.0 MΩ

0.1 3.846 mA/V2 0.026

3.846 67 871 119.642

0.6 8.5 21000 98.0372 2 0.026

119.64 98.037 11,729

i

i

md C i m

d

Ri

T

d

R

R

gA R R g

A

IA R RV

A A A

= + + +⎡ ⎤⎣ ⎦=

= = =

= =

⎛ ⎞= = =⎜ ⎟⎝ ⎠

= ⋅ = =

EX11.20

( )( )

( )( )

6 12

12

1 12 2 10 10 0.2 10

79.6 kHz

12

51.98 from Example 11.201

2 51.98 0.2 10

15.3 GHz

zO O

z

peq O

eq

p

p

fR C

f

fR C

R

f

f

π π

π

π

−

−

= =× ×

=

=

= Ω

=×

=

TYU11.1

( ) ( )1 2

1 2

2 0.005sin 0.5 0.005sin 1.5 0.010sin V

22 0.005sin 0.5 0.005sin 1.25 V

2

d

d

cm

cm

V V Vt t V t

V VV

t t V

ω ω ω

ω ω

= −= + − − ⇒ = +

+=

+ + −= ⇒ =

TYU11.2

1 2 1 2

1 2

For 4 V Minimum 4 V

1 mA2

10 4 6 kΩ1

C C

QC C

C C

v v v vI

I I

R R

= = + ⇒ = =

= = =

−= ⇒ =

TYU11.3 From Equation (11.41)

m O

C

C

g RCMRRR

R

=⎛ ⎞Δ⎜ ⎟⎝ ⎠

For | 75 5623.4dBCMRR dB CMRR= ⇒ =

Then ( )( ) ( )3.86 100

5623.4 10CR

= ⋅Δ

Or 0.686 KCRΔ = TYU11.4 From Equation (11.49)

1 2

2

O m

m

m

R gCMRRg

g

+=

⎛ ⎞Δ⎜ ⎟⎝ ⎠

For | 90 dB 31622.8dBCMRR CMRR= ⇒ =

Then ( )( ) ( )1 2 100 3.86

31622.8 3.862 mg

+⎛ ⎞= ⎜ ⎟Δ⎝ ⎠

Or 0.0472 mA/VmgΔ = or 0.0472 0.0122 1.22%3.86

m

m

gg

Δ= = ⇒

TYU11.5 For 1 2 1 25 V min 5 VC Cv v v v= = ⇒ = = So 1 10 5 0.25 20 kΩC C C CI R R R= − = ⇒ =

( )( )( )

4 34

0.5 2096.2 Let 0.5

4 4 0.02696.295 db 5.62 10 1.71 10

5.62 10

Q Cd d Q

T

dB cm cm

I RA A I mA

V

C M RR C M RR A A −

= = ⇒ = =

= ⇒ = × ⇒ = ⇒ = ××

( )

( )( )( )

( )( )( )( )

3

0

3

0

21.71 10

11

0.5 202 0.026

1.71 10201 0.5

10.026 200

Q C

Tcm

Q

T

I RV

AI R

V

R

ββ

−

−

⎛ ⎞⎜ ⎟⎝ ⎠= = ×

+⎡ ⎤+⎢ ⎥

⎣ ⎦⎡ ⎤⎢ ⎥⎣ ⎦ = ×

⎡ ⎤+⎢ ⎥

⎣ ⎦

5 30 01 19.33 1.125 10 5.82 10 kΩ

5.82 MΩR R+ = × ⇒ = ×

=

( )0 04 4 2 4We have 1 mR r g R rπ⎡ ⎤= +⎣ ⎦

( ) ( )

04

4

2

125 250 kΩ0.5

0.5 19.23 mA/V0.026

200 0.02610.4 kΩ

0.5

A

Q

Qm

T

T

Q

VrI

Ig

V

VrIπ

β

= = =

= = =

= = =

( )( )

( )( )

4 2 4

2 2

2 2

2

2

5.820 250 1

19.23 22.28

1.159 kΩ

10.41.159

10.4

mg R r

R r

R r

RR

π

π

π

⎡ ⎤= +⎣ ⎦=

=

=+

( ) ( )( )

( )( ) ( ) ( )

2 2

1

12 1 3

3 3

10.4 1.16 1.16 10.4 1.30 kΩ

Let 1

ln

10.5 1.304 1 0.026 ln 0.634 kΩ0.5

Q TQ

R R

I mA

II R I R VI

R R

− = ⇒ =

=

⎛ ⎞− = ⎜ ⎟⎜ ⎟

⎝ ⎠⎛ ⎞− = ⇒ =⎜ ⎟⎝ ⎠

If ( )3 0.7 VBEV Q ≅

( )1 3 1

10 0.7 1019.3 18.7 kΩ

1R R R

− − −+ = = ⇒ ≅

TYU11.6 a.

( )( ) ( )( ) ( )

0

1 2

1 2

0 0

0.505sin 0.495sin0.01sin

0.505sin 0.495sin2 2

0.50sin60 0.01sin 0.5 0.5sin 0.85sin V

d d cm cm

d

cm

v A v A vv v v t t

tv v t tv

tv t t v t

ω ωω

ω ω

ωω ω ω

= += − = −=

+ += =

== + ⇒ =

b.

( )

( )( ) ( )( ) ( )

1 2

1 2

0 0

0.5 0.005sin 0.5 0.005sin0.01sin

20.5 0.005sin 0.5 0.005sin

20.560 0.01sin 0.5 0.5 0.25 0.6sin V

d

cm

v v vt t

tv vv

t t

v t v t

ω ωω

ω ω

ω ω

= −= + − −=

+=

+ + −=

== + ⇒ = +

TYU11.7

a. ( )1 2 1 2

/ 2 1 6.62 A1 151

QB B B B

II I I I μ

β= = = ⇒ = =

+

b. ( )( )

( )( ) ( )

150 0.0263.9 kΩ

1

2 2 3.9 7.8 kΩ10sin mV

1.28sin A7.8 kΩ

T

CQ

id

db b

id

VrI

R rtVI I t

R

π

π

β

ωω μ

= = =

= = =

= = ⇒ =

c. ( ) ( )( )

( )02 1 2 151 50 15.1 MΩ

3sin 0.199sin A15.1 MΩ

icm

cmb b

icm

R RV tI I tR

βω ω μ

≅ + = ⇒

= = ⇒ =

TYU11.8

( ) ( )

( )

( )2 2

8 4 max 2 /

max2

42 2 /2

d f D

f f

n Qf

n n

A g Rg g mA V

K Ig

K K mA V

== ⇒ =

=

⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

TYU11.9 From Example 11-10, 0.587 mAQI =

( )( ) ( )

( )( )( ) ( )( )

2

4 0 04

2

0.1 0.58716 2.74

2 21 1For , 85.2 kΩ

0.02 0.5872 2 0.1 2.71 1

0.342 mA/V

n Qd D d

Q

m n GS TN

K IA R A

M R RI

g K V Vλ

= ⋅ = ⋅ ⇒ =

= = ⇒ =

= − = −=

( )( )( )( )0

10

0.342 160.0923

1 2 1 2 0.342 85.2

2.7420log 29.4 dB0.0923

m Dcm cm

m

dB dB

g RA Ag R

C M RR C M RR

−−= = ⇒ = −+ +

⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

TYU11.10

( )( ) 0

0 0

1 1 2 22

60 dB 100011000 1 2 2 0.1 0.22

2000 1 0.4 5 MΩ

n Q o

dB

CMRR K I R

CMRR C M RR

R

R R

⎡ ⎤= + ⋅⎣ ⎦

= ⇒ =

⎡ ⎤= + ⋅⎣ ⎦= + ⇒ ≅

TYU11.11

( )4 2 4 41o o o m oR r r g r= + +

Assume 100 REF OI I Aμ= = and 10.01 Vλ −=

( ) ( )2 41 1 1

0.01 0.1o oD

r r MIλ

= = = ⇒ Ω

Let ( ) 2all devices 0.1 /nK mA V=

Then ( )( )4 2 2 0.1 0.1 0.2 /m n Dg K I mA V= = =

( )( )( )1000 1000 1 0.2 1000 202 oR M= + + ⇒ Ω

Now 1 20.05 1 1.707 0.1

DGS GS TN

n

IV V V VK

= = + = + =

( )1 1 1.707 1 0.707 DS GS TNV sat V V V= − = − =

( ) ( )1 1 1So min 4 4 1.707 0.707o GS DSv V V sat= + − + = − +

( ) ( )1 min 3 10 10 0.05 140 Ωo D D D Dv V I R R R k= = − = − ⇒ = For a one-sided output, the differential gain is:

( )( )

( )( )

1 11 where 222 0.1 0.05 0.1414 /

1 0.1414 140 9.902

d m D m n D

d d

A g R g K I

mA V

A A

= =

= =

= ⇒ =

The common-mode gain is: ( )( ) ( )( )( ) ( )

2 2 0.1 0.1 1400.0003465

1 2 2 1 2 2 0.1 0.1 202000n Q D

cm cmn Q o

K I RA A

K I R

⋅ ⋅= = ⇒ =

+ ⋅ + ⋅

Then 1020log 89.1 ddB dB

cm

ACMRR CMRR dB

A= ⇒ =

TYU11.12

a. ( ) ( )( )5

0.5 15.3 nA1 180 181

QB

II

β β= = ⇒

+

So 0 15.3 nAI = b. For a balanced condition

( ) ( )4 3 3 5 4

2 2 2 2

1.4 V

10 1.4 0.7 9.3 EC EC EB EB EC

CE C E CE

V V V V V

V V V V V

= = + ⇒ =

= − = − − − ⇒ =

TYU11.13

( )

( )

( )( )

02 04

202

2

404

4

2

0.2 1.923 mA/V4 4 0.026

120 1200 kΩ0.180 800 kΩ0.1

2 1.923 1200 800 1846

d f

Qf

T

A

C

A

C

d d

A g r r

Ig

VVrIVrI

A A

=

= = =

= = =

= = =

= ⇒ =

TYU11.14

( ) ( )( )( )( )

( )1 1

5 5

10 0.1 10 0.9 5 0.7 5 9.3 10.3

0.9

Q REF

REF REF

REF

P I I

I I mA

R R kI

= + − −

= + ⇒ =− − −

= = ⇒ = Ω

ln

0.026 0.9ln 0.571 Ω0.1 0.1

REFQ E T

Q

E E

II R VI

R R k

⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

( ) ( )( )

22

2

44

4

2 4

120 2.4 0.0580 1.6

0.050.05 1.923 /

0.0261.923 2400 1600 90 158

Ao

C

Ao

C

m

d m o o L d

Vr MIVr MI

g mA V

A g r r R A

= = ⇒ Ω

= = ⇒ Ω

= =

= = ⇒ =

TYU11.15 a.

0 02 04

02

04

0 0

120 1.2 MΩ0.180 0.8 MΩ0.11.2 0.8 0.48 MΩ

R r r

r

r

R R

=

= =

= =

= ⇒ =

b. ( ) ( )( ) ( )

( ) ( ) ( )

02 04

02 04

02 04

open circuit 2

with load 2

1For with load open circuit 0.48 MΩ2

d f

d f L

d d L L

A g r r

A g r r R

A A R r r R

=

=

= ⇒ = ⇒ =

TYU11.16

( )( )

( )

2 4

2 12

2 0.1 12 1130.1 0.01 0.015

nd

Q

d

KAI

A

λ λ= ⋅

+

= ⋅ ⇒ =+

TYU11.17

( )( )1

75For the MOSFET, 25 24.26 A101

2 2 20 24.26 44.05 A/V

For the Bipolar, 100 25 75 A

D

m n D

E

I

g K I

I

μ

μμ

= − =

= = =

= − =

( )

( )( )

( ) ( ) ( )( )( )( )

2

2

1 2

1

00 75 74.26 A101100 0.026

35.0 K0.07426

0.07426 2.856 mA/V0.026

44.05 1 2.856 3511 1 0.04405 35

1.75 mA/V

C

m

m mCm

mCm

I

r

g

g g rg

g rg

π

π

π

μ1⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

= =

+⎡ ⎤+ ⎣ ⎦= =+ +

=

TYU11.18 From Figure 11.43

( )( )04

0 04 0

80 160 k0.5

150 160 k 24 M

r

R r Rβ

= = Ω

≅ = Ω ⇒ = Ω

From Figure 11.44

( )( )( )( ) ( )( )

( )( )( ) ( )( )( )( )

06 06

2

6

04

0 6 06 04 0

1 1 160 k0.0125 0.5

0.5 0.5 1 2 V2 2 0.5 2 1 1 /160 k

1 160 150 160 3.840 M

D

GS GS

m n GS TN

m

r rI

V Vg K V V mA VrR g r r R

λ

β

= = ⇒ = Ω

= − ⇒ == − = − == Ω= = ⇒ = Ω

TYU11.19 From Equation (11.126)

( ) ( )( )( )

( )( )11

11 3

11

011

2 1 2 121 120 0.0261.51 M

0.5

120 0.0266.24 k

0.5

6.24 0.1 0.0984 k

0.5 19.23 mA/V0.026120 240 k0.5

Ti i

Q

T

Q

E

Qm

T

A

Q

VR R

I

VrI

R r RI

gVVrI

π

π

β β

β

+= = ⇒ = Ω

= = = Ω

′ = = = Ω

= = =

= = = Ω

( )( )( )

11 011 11Then 1240 1 19.23 0.0984

694 k

C m ER r g R′= += +⎡ ⎤⎣ ⎦= Ω

( )( )

( ) ( )( )8

8

8 8 4

7 11 8

120 0.0261.56 k

21 1.56 121 5

607 kThen 694 607 324 k

T

C

b

L C b

VrI

R r R

R R R

π

π

β

β

= = = Ω

= + + = += Ω

= = = Ω

( ) ( )7

80 4

11 7

7

0 0

0.5Then 324 31152 2 0.026

||1

120 240 k0.5

694 240 178 k1.56 1785 || 5 1.48 1.14 k

121

Qv L v

T

C C

AC

Q

IA R A

V

r ZR R

Z R RVRI

Z

R R

π

β

⎡ ⎤⎛ ⎞= = ⇒ =⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠ ⎣ ⎦⎛ ⎞+

= ⎜ ⎟+⎝ ⎠=

= = = Ω

= = Ω+⎛ ⎞= = ⇒ = Ω⎜ ⎟

⎝ ⎠

TYU11.20

( )

7

37 7

2

0.510 104 k2 0.026

Qv L

T

L L

IA R

V

R R

⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞

= ⇒ = Ω⎜ ⎟⎜ ⎟⎝ ⎠