课件与实验-非线电

课程安排第一章：1.1第二章：University

of

Science

and

Technology

of

China1.2(2)1.3(2)1.42.12.22.32.42.52.62.72.102.122.142.18i

5uu2

0.5u2(mA)UQ

3.5V,uUQ

0.8cost(V)解:I

5U

U2

0.5U3

8.3mAQQU

3.5VG

IQ

8.3mA

2.375mSg

i

5

2u1.5u2

6.375msu

u

UQ

u

UQ将u

UQ

0.8

cost(V)代入i5uu20.5u3(mA)并计算得:I

4

1.6U

0.192

1.2U2

5.292mA1

I1m110.8

5.292

6.615msGU1.2

（2）University

of

Science

and

Technology

of

ChinaUniversity

of

Science

and

Technology

of

Chinau

0.5

cos

21.5105t

2

cos

2

2

105t(V)t

21.5105t,t

2

2105t1

2u

0.5

cos1t

2

cos2t2i

5u

u解:设计算得直流分量:2.125mA1

:

2.5mA;2

:

10mA;21

:

0.125mA;22

:

2mA;1

2

:1mA;1.3

（2）21

21

2

5(0.5

cost

2

cost)

(0.5

cos

t

2

cost)1

2

:

1mACj

20(1

0.25u)0.5

(pF);u(1,3)C

C

j

20

20

1

(1

0.25u)

0.5L

10H2

10

106

20

1(1

0.25u)0.51012u

1,

f

8.17

106

Hz,u

3,

f

8.49106

Hz.1

12

LCf

HZ1.4

（2）University

of

Science

and

Technology

of

China解得:（2）静态时:

0.98

4910.02

IC2

IB2,

IE2

(1

)IB22(IC2

2IB2

)

UBE

0.5IE2

6IK

IC2

2.05mAUBE

UCEOUrUniversity

of

Science

and

Technology

of

ChinaKES

ESUBE

UCEOI

IeUr

I

eCEOrESIU

U

ln

IK

778.85mV解得:（1）恒流源:2.1x

Ui

20UrUBE0IE0

IESKeUr

I

(x)

I0I

(20)

0.08978

e20BE

r

0University

of

Science

and

Technology

of

ChinaUCE

UBE

322mvESIU

U

[ln

IK

ln

I

(20)]

321.52mv解得:查表得:2.1（3）由Urx

Ui

10m1r

02I1

xU

xI

x

IE0G所以:查表得:2.2（1）由1LC

107网络调谐于基波频率QT

RC

500University

of

Science

and

Technology

of

China2I1

xxI

xx10

0.18972Gm1

7.3mst

10

4.74

cos10

7to

m1

i

Lu

t

10

G

UR

cosTHD

Dx/

QT

1.27%Vm30r

0i2I3

xU

xI

xU I

x

IE0

2I3

x

IE0G所以:2.2（2）由1LC

3107网络调谐于三次谐波频率QT

RC

507o

m3

i

Lu

t

10

G UR

cos

t

10

3.12

cos

310

t

3231TTHD

Qn33n

4.2%n1

I

x n

9

In

xVn

3nUniversity

of

Science

and

Technology

of

Chinan2n231

n

n

9BBCCB1

B2RB2V

3VU

R

RRB

RB1

||

RB2

33.33KBBBEQU

U1.07mAU

2.3vUEQI

RER

B1IEQgmQ

U

40.3mSrUx

U

88.502I1(x)x15

0.12881xI

(x)r0Ux

Ui

15查表得:I

(15)

0.1039e15m1mQx

xI

(x)0G

(x)

g[1

ln

I0(x)]

2I1(x)

5.94msu0(t)

Gm1

(x)Rui

2.32

cos

w0t(v)UUniversity

of

Science

and

Technology

of

China( 0

)2P

2

2.69mW

RL电阻分压式偏置电路:2.32.4

E0I

IES0exp()IxEBUurIE0

IK5k

UEm1mQ0

r0xI

(x)

u

xI

(x)G

(x)

g

2I1

(x)

IE0

2I1

(x)

13.487msu0(t)

9v

Gm1

(x)

RL

ui

cos0t

9

5.26

cos0tV静态时，x

=10对9V电压源进行等效IK

10V/

5k

2mAR

5kUEB

UE

570.2mVIE0

1.886mA解得又所以University

of

Science

and

Technology

of

China01LC

108

rad/

sQT

0RC10rU

1001

101u'

(t)

VI

a(2)R

cos108t

10

4.02

cos108t(v)oCCK

1

L(4.02

cos108t)

2.68cos108t(v)6060

30ou

(t)

THD

1

D(x)

0.25%QTC1C2C1

C2University

of

Science

and

Technology

of

China

20PFCKxUi

2

I

2mA2.5恒流源偏置差分放大电路:输出端RLC谐振回路:2.7当Ui

0.5V时，VD1Uo

0VD2截止当2V

Ui

0.5V时，VD1导通VD2截止3o

iU

2U

1

2

cost

1

3

3当

Ui

2V时，VD1

VD2导通2

2o

iU

1U

3

cost-1.5-1-0.500.51 1

522

533.54-10-0.510.51.5232.5Uo

/VUi

/VoU

/V-

1

-

0

.

5t/0

0

.

5

1-

3University

of

Science

and

Technology

of

China-

2-

101232.10（1）由题，自生负偏压场效应管放大电路：u

U

U

cos107

tGS

1

1DPD

u

GS0I

iDSS

I

6mA

00PI

6mAI

1.8mA

0.3

由再次查表并内插计算：

112.71

0.4705Uo

1IPRL

11.32V0

BWQTQT

20.01University

of

Science

and

Technology

of

China

1.49%QTTHD

D2.10（2）由导通角：1Ucos

UP

UQ由I1

IP1

2.823mA1U

2.886VuDS

uo

15

11.32

costiD

I0

I1

costTi0D

DSTP

1u

dt平均功率0

1

22)d1.8

2.823cos

(15

11.32

cosPU1University

of

Science

and

Technology

of

ChinaP

11.05mWA

Uou

3.922.12uGS

2

2

cos1tGSuGS22cos1t1g(t)

iD|

2.5

2.5

costu1University

of

Science

and

Technology

of

Chinapup

uQ又g

5ms,

arccosu1

180o

所以g

2.5ms回路调谐于频率f0

f1

f2,u0

20

0.625cos

2

465

10

t3（1）因为是自生负偏压，所以1

22

2IF

f1

f2I

I

1

gU

1

2.5

0.2

0.25mAV2.12（2）选频网络的带宽

BW

f0

/

QT

500KHz

2

f2输出电压u0包含f1,f1

f2,f1

f2三个频率分量3

3GS1U1University

of

Science

and

Technology

of

China导通角

arccosUp

UQ

120o对于f1频率分量I

I

(120o)

4.801mA

平方律D1p

1

pDSSI

I

10mA自生负偏压

u

8

8

cost|4uGSU1U1

cos1t

5(1

uGS)g

iDGSug

g

(120o)

2.6815ms

折线律1

p

12IF

1

2I

1

gU

0.40225mA0u

2012[10.11847cos(2250310

t

)]cos

2710

t

V4f1

f2,

f1

f2落在通带边缘，4

42其对应幅度为

1

对应的相位差偏移和－

2.12对于f1

f2频率分量采用时变参量分析1University

of

Science

and

Technology

of

China20

D1

L

2故：u

20[I R

IIFRL

cos(2ft

]cos

2)ftV4pg

5msU

r

UCO（1）周期信号由电容耦合到非线性电阻未加输入电压时，IK给C充电io

IK

IS

exp2.14C

iUU

U

IK

IS

exp

r

加输入电压后：io稳态时，io

0

9

exp

260mV

0rUC

1

exp

0Uroi

IK

IS

expUC

580mV故：uo

ui

UCr

U

10

U

10-580-8403ou

/

mV30

t/

s（2）dUiUniversity

of

Science

and

Technology

of

China

dUC

12.18（1）VD为指数律二极管

7ii

t

4

cos10

tR

2k,L

0.5H,C

0.02F

107

rad/

s1LC0Q

CR

400假设Q

足够高T流过VD直流分量uo

t

Uo1

costiD

IK

2mA基波线性负载方程：Uo1

0.3268Vo1

iRI

I

Uo1IK

I0xIo1

2I1xUo126mVx由指数律：GNS

Io1

/Uo1

11.74ms

16.34University

of

Science

and

Technology

of

ChinaTNSGQ

Q0

G

G2.18（2）

7ii

t

4

cos10

t假设Q

足够高T非线 件导通角o1

arccosUUIo1

Ii

o1RGNR

Io1

/Uo1