模拟电子技术基础习题解答第3章

第3章

多级放大电路第一部分：自测题四、解：（1）T3管的集电极电流C3

E3

Z

BEQ3I

＝

I

＝

（U

－U

）/

Re3/2=0.15mA＝0.3mAIE1＝IE2＝

IC3（2）静态时：UO

=

ICQ

4

RC

4

+(-6)

>

0ICQ

4IRC

2RC

20ICQ4＝VEE

/

Rc4＝0.6mAC

2

e

4+UEB

4UR

=

URC

2IR

=

IC

2

-

IB

4b=

ICQ

4

Re4

+

0.7=

0.6

·0.5

+

0.7=1VI=

IC

2

-

C

4

200=

0.15

-

0.6

=

0.147mAc2RC2IRU

R=

C20.14=

1

»

6.8kWEQ4bb'be4EQ2bb'be2IIr

=

r

+

(1+

b)

26mV

»

9kWr

=

r

+

(1+

b)

26mV

»

35kWbe22

ru1=

bRc2

∥

Ri

2Ae4be4»

-18.18=

-u

2r

+

(1+

b)Rb

Rc4A2

rbe2=

b{Rc2

∥[rbe4

+

(1+

b

)Re4

]}»18.95Au

2

»－345Au

=

Au1第二部分：习题3.1共射-共基共射-共射共射-共射共集-共基共源-共集共基-共集3.2

(a)Ri

2be11u1R

+

r=

-

b1R2

∥

Ri

2Abe2

2

32

3u

2r

+

(1+

b

)R

(1+

b

)R

=ARi

2

=

rbe2

+

(1+

b2

)R3Ri

=

R1

+

rbe1be2be11ur

+

(1+

b2

)R3(1+

b2

)R3R

+

r=

-

b1{R2

∥[rbe2

+

(1+

b2

)R3

]}A21+

brbe2

+

R2Ro

=

R3

∥3.2

(b)1

2

i

2rbe1

+

(1+

b1

)(R2

∥

Ri

2

)

(1+

b

)(R

∥

R

)

Au1

=Ri

2

=

R3

//

rbe

2be2ru

2=

-

b2

R4Abe1

be2u(-

b2

R4

)=r

+

(1+

b1

)(R2

∥

R3

∥

rbe2

)

r(1+

b1

)(R2

∥

R3

∥

rbe2

)ARi

=

R1

∥[rbe1

+

(1+

b1

)(R2

∥

Ri

2

)]Ri

=

R1

∥[rbe1

+

(1+

b1

)(R2

∥

R3

∥

rbe2

)]Ro

=

R4u1R1

+

rbe1=

-

b1

(R2

∥

Ri

2

)ARo

=

R3Ri

2

=

rbe2

+

(1+

b2

)rdu

2b2

R3rbe2

+

(1+

b2

)rd=

-Au

u1

u

2A

A=

ARi

=

R1

+

rbe1Ro

=

R8be2ru

2=

-

b2

R8AAu1

=

-gm

(R4

∥

Ri

2

)Ri

2

=

R6

//

R7

//

rbe

2Au

2Au

=

Au1Ri

=

R3

+

R1

∥

R23.4rbe1b

RAu1

=

-12R

RR3T1T2Ri

=

R1

∥

R2

∥

rbe1

»

0.93kW21»

-11

=i

2rbe1rbe21+

b-

bbe2=125ru

2=

b2

R3AAu

2

»

-125Au

=

Au1Ro

=

R3

=

3kWR

2R3R

4R1T2T1R5be1ru1(R1∥

Ri2

)=

-

b1Arbe1=

-

b1

(R1∥

rbe2

)

»

-50»

-422

4rbe2b

R

Au

2

=

-Au

2

»

2100Au

=

Au1»1.2kWRi

=

(R5

+

R2

∥

R3

)

∥

rbe1Ro

=

R4

=1kW3.5be1ru1(R3

∥

Ri2

)=

-

b1A»

-107ru

2=

-

b2

R4Arbe1=

-

b1

(R3

∥

rbe2

)

»

-62be2Au

=

Au1

Au

2

»

6634Ri

=

R1

∥

rbe1

»1.5kWRo

=

R4

=

2kWAu

=

Au1

Au

2

»

-6Au1

=

-gm

(R2

∥

Ri

2

)

»

-6Au1

=

-gm

(R2

∥

Ri

2

)Ri

2

=

rbe

+

(1+

b)R4»1=u

2rbe

+

(1+

b)R4(1+

b)R4A»

43Wrbe

+

R21+

bRo

=

R4

∥Ri

=

R1

=10MW解：（1）RW的滑动端在中点时Ad的表达式为WcI

bed)2rRDuOA

=

=

-Dub

(R

+（2）RW的滑动端在最右端时2c1uIu左u

=

AberA=

-

b

(Rc

+

Rw

)u左2c

2u

=

A

(-

uI

)u右berA=

-

bRcu右uo

=

uc1

-

uc

222u左

u右uuuo

=

A

I

-

A

（-

I

）2uIu右u左=

(

A

+

A

）rbe

2=

-b

(2Rc

+

Rw

)

uIRwrbe=

-

2

uIb

(Rc

+

)beI

O

dW

)

2

rRuu=

-A

=b

(Rc

+3.6解：RW滑动端在中点时T1管和T2管的发射极静态电流：EQBEQ2RW+

2IEQ

Re

=

VEEU

+

IEQ»

0.517mARWI

=

VEE

-U

BEQ＋2R2

eEQbe

bb'Ir

=

r

+

(1+

b)

26mV

»

5.18kW2bed»

-97RWA

=

-r

+

(1+

b)b

RcRi

=

2rbe

+

(1+

b)RW

»

20.5kW3.82IC=15mV=

uI1

+

uI2解：uuId

=

uI1

-uI2

=10mVbecd»

-672rA

=

-b

R3.9DuO

=

AduId

»

-0.67V由于电路的共模放大倍数为零，故△uO仅由差模输入电压和差模放大倍数决定。解：eEQCQCQ2CQ12RV

-UI

=

I

=

I

»

I

»

EE BEQ

=

0.265mACRLCQRVCC

-UC1Q

UC1Q=

I

+VCQCCRLC1QI

»3.23VRC

+RL-

RC

RLRC

+RLU

=UCQ2

=

VCC

=15V△uO＝uO－UCQ1≈－1.23VEQbe

bb'Ir

=

r

+

(1+

b

)

26mA

»

5.1kW2(Rb

+

rbe

)Ad

=

-

L

»

-32.7b

R'dOI»

37.6mVADuu

=若uI=10mv，则DuO

=

AduI

»

-0.327VuO

=

UCQ1

+

DuO

»

2.9V3.10be1

1

be2Ld)2Rr

+

(1+

b

)rA

»

-b1b2

(Rc

∥Ri

=

2[rbe1

+

(1+

b1

)rbe2

]3.11解：差模放大倍数和输入电阻分别为Ad＝－gmRD＝－40Ri＝∞3.13Au1DuI

2rbe1=

DuO1

=

b1R2

∥

Ri

2+

(1+

b4

)R5Ri

2

=

rbe44

5be4I2Au

2r

+

(1+

b

)R=

-Du=

DuO2b4

R6

∥

Ri3Ri3

=

rbe5

+

(1+

b5

)R7be5I3Au

3(1+

b5

)R7==

DuO3IOuu1

Au

2

Au

3=

ADuDu

r

+

(1+

b5

)R7DuA

=Ri

=

rbe1

+

rbe256b