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第 3 章 多级放大电路 PowerPoint PPT Presentation


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第 3 章 多级放大电路. 3.1 多级放大电路的耦合方式 3.2 多级放大电路的动态分析 3.3 直接耦合放大电路. 输入. 推动级. 输出级. 输入级. 第二级. 3 .1 多级放大电路的耦合方式. 输出. 耦合方式:信号源与放大电路之间、两级放大电路之间、放大器与负载之间的连接方式。. 常用耦合方式:直接耦合、阻容耦合、变压器耦合、光电耦合。. 静态:保证各级有合适的 Q 点. 对耦合电路的要求. 动态 : 传送信号. + V CC. R B. R B. R C. R 1. C 2. C 3. C 1. C 4. T 3.

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第 3 章 多级放大电路

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3

3

3.1

3.2

3.3


3

3.1

Q

:


3

+VCC

RB

RB

RC

R1

C2

C3

C1

C4

T3

T1

T2

RS

R2

RE1

ui

RL

RE2

RE2

uo

us

CE

3.1.1


3

A

f


3 1 2

3.1.2


3


3

3.1.3


3

Q

3.1.4


3

:

3.2


3

+VCC

RB

R1

RC

C1

T2

C3

C2

T1

+

+

RS

R2

RE1

ui

uo

+

RE2

us

CE

RL

-

-

-

3.2


3

+VCC

RB

R1

RC

C1

T2

C3

C2

T1

+

+

RS

R2

RE1

ui

uo

+

RE2

us

CE

RL

-

-

-

1

----


3

+VCC

+VCC

RB

R1

RC

ICQ2

ICQ1

T2

IBQ1

+

+

IBQ2

T1

UCEQ2

UCEQ1

-

-

R2

RE1

RE2

1

----


3

+VCC

RB

R1

RC

C1

T2

C3

C2

T1

+

+

RS

R2

RE1

ui

uo

+

RE2

us

CE

RL

-

-

-

2


3

+

+

RS

+

+

RC

RB

R1

R2

RE2

-

-

RL

-

-

2

1


3

+

+

RS

+

+

RC

RB

R1

R2

RE2

-

-

RL

-

-

2

1


3

+

+

RS

+

+

RC

RB

R1

R2

RE2

-

-

RL

-

-

2

2


3

+VCC

RC2

R2

R1

(+24V)

C2

82k

1M

10k

C1

C3

T2

T1

RL

RS

10k

20k

R3

RE2

RE1

CE

8k

27k

43k

NEXT


3

+VCC

RC2

R2

R1

(+24V)

C2

82k

1M

10k

C1

C3

T2

T1

RL

RS

10k

20k

R3

RE2

RE1

CE

8k

27k

43k

:

RS

R1

RC2

R2

R3

RL

RE1


3

RS

R1

RC2

R2

R3

RL

RE1

(1) Ri=

: RL1= Ri2 = R2 // R3//rbe2rbe2= 1.7k

Ri=1000//(2.9+511.7) 82k

(2) Ro= Ro2 = RC2= 10k


3

RS

R1

RC2

R2

R3

RL

RE1

(3) :


3

+VCC

R1

RC1

RC2

R2

T1

T2

uo

ui

RE2

3.3

3.3.1

1 :Q

RE2:Q


3

+VCC

R1

RC1

RC2

R2

T1

T2

uo

ui

RE2

3.3

3.3.1

1 :Q

NPNPNP:Q


3

+VCC

R1

RC1

RC2

R2

T1

T2

uo

ui

uo

RE2

t

0

2 :

ui uo


3

(1)

Q

----

(2)

Q

----


3 3 2

+

V

CC

R

R

c1

c2

+

-

u

o

R

R

b1

b2

+

+

T

T

u

u

1

2

i1

i2

-

-

R

e

-

V

EE

3.3.2

1:

T1T2

1=2

UBE1=UBE2

rbe1= rbe2

Rc1=Rc2

Rb1=Rb2

VCC=VEE

IB1IB2 VEE


3

+

V

CC

R

R

c1

c2

+

-

u

o

R

R

b1

b2

+

+

T

T

u

u

1

2

i1

i2

-

-

R

e

-

V

EE

2.

(1)

ui1 = ui2

= uic

uC1 = uC2

uo=0

= uoc

<1)


3

+

V

CC

R

R

c1

c2

+

-

u

o

R

R

b1

b2

+

+

T

T

u

u

1

2

i1

i2

-

-

R

e

-

V

EE

2.

(2)

ui1 = ui2

uid = ui1 ui2

uC1 = -uC2

uo= (uC1uC1)(uC2 +uC2 )

=2 uC1

(>1)

= uod


3

2.

ui1 ui2

(3)

:ui1 = 10 mV, ui2 = 6 mV

:ui1 = 8 mV + 2 mV

ui2 = 8 mV 2 mV


3

+

V

CC

R

R

c1

c2

+

-

u

o

R

R

b1

b2

+

+

T

T

u

u

1

2

i1

i2

-

-

R

e

-

V

EE

(4)

ui1= ui2= 0

uO= UCQ1 - UCQ2= 0

T ICQUCQ

UCQ1 = UCQ2

uO=(UCQ1 + UCQ1)(UCQ2 + UCQ2 )=0


3

RE

+

V

CC

R

R

c1

c2

+

-

u

o

R

R

b2

b1

+

+

T

T

u

u

1

2

i1

i2

-

-

R

e

-

V

EE

Q

(4)

UE

IE1,

2Re

ui1= ui2= 0

UE

IC

IRe= 2IC

T

IC

IB

UBE

RE


3

(KCMR)

CMRR Common Mode Rejection Ratio

KCMR =

KCMR(dB) =

()

:Ad=-200

Ac=0.1

KCMR=20 lg (-200)/0.1 =66 dB

KCMR


3

+

V

CC

R

R

c1

c2

+

-

u

o

R

R

b1

b2

+

+

+

+

u

u

T

T

u

u

o2

o1

1

2

i1

i2

-

-

-

-

R

e

-

V

EE

3.


3

+

V

CC

IC

IC

R

R

c1

c2

u

+

-

o

R

R

RL

b1

b2

+

+

IB

IB

T

T

u

u

1

2

i1

i2

-

-

IRe

R

e

-

V

EE

1


3

RL

T1

T2

+

V

CC

R

R

c1

c2

u

+

-

o

R

R

RL

b1

b2

+

+

+

+

+

u

T

T

u

1

2

i1

i2

-

-

-

-

R

e

-

u

i2

-

V

EE

+

-

a.

(UCQ2)(UCQ1)


3

+

V

CC

R

R

c1

c2

u

+

-

o

ib1

ib2

RL

R

R

ic2

ic1

b1

b2

+

+

+

+

+

u

T

T

u

1

2

i1

i2

-

-

-

-

R

e

-

iRE

u

i2

-

V

+

EE

-

a.

ib1 = - ib2

Re

ic1= - ic2

iRE= ie1+ie2= 0

ib1, ic1

ui1

ui2

uRE= 0

RE

ib2, ic2


3

+

+

u

od

2

-

u

od

+

u

od

2

-

-

i

R

b1

b

+

+

R

r

b

u

i

L

be

R

i1

b1

c

2

-

u

id

-

R

L

r

u

R

be

i2

2

b

c

i

-

+

b2

i

R

b2

b


3

i

R

b

b

+

+

R

r

b

u

i

L

be

R

i1

b

c

2

-

u

id

-

R

i

L

r

b

u

R

be

i2

2

b

c

i

-

+

b

i

R

b

b

R

R

i

o

+

+

u

od

2

-

u

od

+

u

od

2

-

-


3

+

V

CC

R

R

c1

c2

u

+

-

o

R

R

RL

b1

b2

+

+

+

u

T

T

u

1

2

i1

i2

-

-

-

R

e

-

V

EE

=

u

0

o

u

u

c1

c2

C

C

1

2

u

oc

u

u

+

-

i1

i2

b.

KCMR =

uc1uc2


3

2

T2


3

Re


3

3


3

4

T1T1T2


3


3

KCMR =


3

+VCC

RC

RC

uo

RB

RB

T2

T1

ui1

ui2

RE

VEE

4.

1


3

+VCC

ic2

ic1

RC

RC

uo

RB

RB

T2

T1

ib2

ib1

ui1

ui2

R

R

E

IC3

R1

T3

R3

R2

-VEE

2

ReT3Re


3

ic2

ic1

+VCC

RC

RC

uo

iC

RB

RB

T2

IB3

T1

Q

IC3

ib2

ib1

ui1

ui2

R

R

uCE

E

UCE3

IB3

IC3

R1

UCE3

T3

IE3

I2

R3

R2

-VEE

T3 :

T3

rce3 1M

UB3UE3 IE3 IC3


3

______

______ ____________


3

3


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