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### Transcript of Ch15p

Chapter 15 Exercise Solutions EX15.1 For the circuit shown in Figure 15.7

31

2dBfRCπ

=

or

( )6

33

1 1 3.979 102 2 40 10dB

RCfπ π

−= = = ××

For 75 KR = Then

115.31 10 53.1 pFC −= × =

3

4

We have 1.414C 75.1 pF0.707C 37.5 pF

CC

= == =

EX15.2

1C

eq

fCR

=

or

( )( )5 6

1 110 20 10

0.5 pFc eq

Cf R

C

= =×

=

EX15.3

Low-frequency gain: 1

2

30 65

CTC

= − = − = −

( )( )( )

3 122

3 312

100 10 5 106.63 kHz

2 2 12 10C

dB dBF

f Cf fCπ π

× ×= = ⇒ =

×

EX15.4

0

63

0

12 3

1 1 6.13 102 3 2 (15 10 ) 3

fRC

RCf

π

π π−

=

= = = ××

Let 0.001 μF 1 nFC = =

Then 6.13 kR = Ω so 2 8 49 kR R= = Ω EX15.5

( )( )( )

00

4

2 1 2

1 12 2

1 0.02 F2 800 10

2 2 10 20 k

f CRC f R

C C

R R R

π π

μπ

= ⇒ =

= ⇒ ≅

= = ⇒ = Ω

EX15.6

( )

1

1 2

1

1

1 1

1 1

2 (12)20

2 20 1240 10 4 k

TH HRV V

R R

RR

R RR R

⎛ ⎞= ⎜ ⎟+⎝ ⎠⎛ ⎞

= ⎜ ⎟+⎝ ⎠+ =

= ⇒ = Ω

EX15.7

( )

[ ]( )

( )

( )

1

1 2

1

1 2

2 2

1 1

2

1 2

1

2

0.10 10 10

201 200 1990.10

11 1 1 1.005 V199

on0.1

10 0.7 0.70.1 43 k0.2

42.9 k

TH TL H L

S REF

REF S REF

H BE

RV V V VR R

RR R

R RR R

RV VR R

RV V VR

V V VI

R

R

R

γ

⎛ ⎞− = −⎜ ⎟+⎝ ⎠

⎛ ⎞= − −⎜ ⎟+⎝ ⎠

+ = = ⇒ =

⎛ ⎞= ⎜ ⎟+⎝ ⎠

⎛ ⎞ ⎛ ⎞= + = + ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

− −=

+− −+ = = Ω

= Ω

EX15.8 At 0 ,t −= let 0 5v = − so 2.5.Xv = − For 0t >

( )10 2.5 10 exp XX

tvr

⎛ ⎞−= + − − ⎜ ⎟⎝ ⎠

When 5.0,Xv = output switches

( )

1

1

11 1

5.0 10 12.5 exp

10 5 5.0exp 12.5 12.5

12.5 12.5exp ln 0.9165.0 5.0

X

X

X XX

tr

tr

t t r t rr

⎛ ⎞= − −⎜ ⎟

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

⎝ ⎠⎝ ⎠

During the next part of the cycle

[ ]( )5 5 5 exp XX

tvr

⎛ ⎞= − + − − −⎜ ⎟

⎝ ⎠

When 2.5,Xv = − output switches

( )

( ) ( )

( )( )

2

2

22 2

1 2

3 6 4

2.5 5 10 exp

5 2.5 2.5exp 10 10

10 10exp ln 1.392.5 2.5

1Period 0.916 1.39 2.31 Frequency2.31

50 10 0.01 10 5 10 s 866 Hz

Du

X

X

X XX

X XX

X

tr

tr

t t r t rr

t t T r rr

r f− −

⎛ ⎞− = − + −⎜ ⎟

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

⎝ ⎠⎝ ⎠

= + = = + = ⇒ =⎡ ⎤⎣ ⎦

= × × = × ⇒ =

( )( ) ( )

1

1 2

0.916ty cycle 100% 100% Duty cycle 39.7%

0.916 1.39t

t t= × = × ⇒ =

+ +

EX15.9 a.

( )

[ ]

10

1 2

1

1 2

6

6

6

10 12 1.2 V10 90

0.10

0.711 / 12 ln ln 1 1 (0.10)

50 10 ln 1.18 (0.162)50 10 3.09 k

(0.1 10 )(0.162)

X X X

Y

PX X

X X

X X

r R C

Rv vR RR

R R

V VT r r

T r r

R R

γ

β

β

=

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

= =+

⎡ ⎤++ ⎢ ⎥⎡ ⎤= = ⎢ ⎥⎢ ⎥− −⎣ ⎦ ⎢ ⎥

⎢ ⎥⎣ ⎦= × = =

×= ⇒ = Ω×

b. Recovery time

( 1.2 ) exp X P PX

tv V Vr

⎛ ⎞= + − − −⎜ ⎟

⎝ ⎠

When ,Xv Vγ= 2t t=

( )

( )

( )( )

2

2

2

3 6 42

0.7 12 1.2 12 exp

12 0.7exp 0.85613.2

1 ln 0.1550.856

3.09 10 0.1 10 3.09 10 48.0 s

X

X

X X

X

tr

tr

t r r

r t μ− −

⎛ ⎞= + − − −⎜ ⎟

⎝ ⎠⎛ ⎞ −− = =⎜ ⎟⎝ ⎠

⎛ ⎞= =⎜ ⎟⎝ ⎠

= × × = × ⇒ =

EX15.10

6

1.1 RC75 10

TT −

== ×

Let 10 nFC = Then

( )( )6

9

75 10 6.82 K1.1 10 10

R−

×= =×

EX15.11

( ) ( ) ( ) ( )

( )

3 6

1 1 802 Hz0.693 2 0.693 20 2 80 10 0.01 10

20 80Duty cycle 100% 100% Duty cycle 55.6%2 20 2 80

A B

A B

A B

f fR R C

R RR R

−= = ⇒ =

+ + × × ×⎡ ⎤⎣ ⎦+ += × = × ⇒ =

+ +

EX15.12

a.

( )( )

212

2 2 8 1 4 V

4 0.5 A8

P

L

P L P

PP P

L

VPR

V R P V

VI IR

= ⋅

= = ⇒ =

= = ⇒ =

b. 12 4 8 V0.5 A

CE

C

VI

= − =≈

So ( )( )0.5 8 4 WC CEP I V P= ⋅ = ⇒ = EX15.13

( )( )

( ) ( )2

2 2 8 10 12.65 V

10 812.65

19.9 V

P L L

PS S

L

SS

P

S

V R P

VP VR

P RV

VV

πππ

= = =

⎛ ⎞= ⎜ ⎟

⎝ ⎠

= =

=

EX15.14

Line 0 0regulation Z

Z

dV dV dVdVdV dV+ += = ⋅

Now

0

1

101 210

10 0.0022710 4400

Z

Z Z

Z

dVdV

dV rr RdV +

⎛ ⎞= + =⎜ ⎟⎝ ⎠

⎛ ⎞= = =⎜ ⎟+ +⎝ ⎠

So Line ( )( )regulation 2 0.00227 0.00454= = 0.454% EX15.15

( )

( ) ( )( )

0 1 011

0 01 0 1 1

0 0 10 1 0

0

0 0 0 0 0 11

0 0 0

0 0 0

0

00 0 00

0

1 110 10 10 10 10

2 210 10 2

010

10 10

2(10) 21000 6.3 1000

10 0.5 0.5 20 2 0.5[0.10 2.0 0.05

L Z

L

L Z L

L

L

V V VV V

V VV V V V

V A V VV V VR R

V V V A V A VVR R R R

V A VR

VV V VI

V

− ⎛ ⎞= ⇒ + =⎜ ⎟⎝ ⎠

⎛ ⎞ = ⇒ = ⇒ =⎜ ⎟⎝ ⎠

− −−+ + =

+ + − = −

= −

+ + − = −

+ − + 0

0 0

1000] 12,600(1002.05) 12,600

IV I

+ =+ =

For 0 01 mA 12.5732I V= ⇒ = For 0 0100 mA 12.4744I V= ⇒ =

( ) ( )( )

0 0

0

NL12.5732 12.4744 100%

V VV

−= ×

−= ×

=

EX15.16 a.

( )

( )

31 2 3

3 3

34 4

4

44

3 on

5.6 3 0.6 3.8 0.482 mA3.9 3.4 0.576 7.88

ln

0.482(0.1) (0.026) ln

Z BEC

C C

CC T

C

CC

V VI

R R R

I I

II R VI

II

−=

+ +−

= = ⇒ =+ +

⎛ ⎞= ⎜ ⎟

⎝ ⎠⎛ ⎞

= ⎜ ⎟⎝ ⎠

By trial and error 4

7 7

0.213 mA

2(0.6) (0.482)(3.9) 3.08 VC

B B

I

V V

=

= + ⇒ =

b.

( )( ) ( )( ) ( )

130 8 7

13 12

12

12

12 12

2.23 (5) 3.082.232.23 5 3.08 2.23 3.08

11.15 6.868 3.08 1.39 k

B BR V V V

R R

RR

R R

⎛ ⎞= =⎜ ⎟+⎝ ⎠

⎛ ⎞=⎜ ⎟+⎝ ⎠

= += = ⇒ = Ω

TYU15.1

( )

3

54

3

12

1 1 1.59 102 2 10

dB

dB

fRC

RCf

π

π π−

=

= = = ×

Let 0.01 F 1.59 kC Rμ= ⇒ = Ω Then

1

2

3

0.03546 F0.01392 F0.002024 F

CCC

μμμ

===

6 6

3

1 1

2011 10

0.124or 18.1 dBd

Tf

f

T T

= =⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

= = −

TYU15.2

( )

33

63

1 12 2

1 3.18 102 50 10

dBdB

f RCRC f

RC

π π

π−

= ⇒ =

= = ××

Let 0.001 F 1 nF 3.18 kC Rμ= = ⇒ = Ω Then

1

2

3

4

2.94 k3.44 k1.22 k8.31 k

RRRR

= Ω= Ω= Ω= Ω

83

8 243

23 3

10.01

1

11 100.01

10 15.8 kHz10

dB

dB

dB dB

Tf

f

ff

f ff f

f

= =⎛ ⎞+ ⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞+ = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞≅ ⇒ = ⇒ ≅⎜ ⎟

⎝ ⎠

TYU15.3

2

4

6

8

11-pole 3.87 dB12110

12-pole 4.88 dB12110

13-pole 6.0 dB12110

14-pole 7.24 dB12110

T

T

T

T

= ⇒ −⎛ ⎞+ ⎜ ⎟⎝ ⎠

= ⇒ −⎛ ⎞+ ⎜ ⎟⎝ ⎠

= ⇒ −⎛ ⎞+ ⎜ ⎟⎝ ⎠

= ⇒ −⎛ ⎞+ ⎜ ⎟⎝ ⎠

TYU15.4

1eq

C

Rf C

=

or 76

1 1 2 105 10C

eq

f CR

−= = = ××

If 10 pF 20 kHzCC f= ⇒ = TYU15.5

( )( )( )

0 04 12

42 2

1 1 65 kHz2 6 2 6 10 100 10

29 29 10 290 k

f fRC

R R R

π π −= = ⇒ ≅

×

= = ⇒ = Ω

TYU15.6

0 09 2

61 29

1 2

2

1

23

1

1 1 7.12 MHz(10 )2 (10 )22 10

1 1 0.25 mA / V4 10

m

m m

f fC CL

C C

C g RC

Cg gC R

ππ−

−−

= = ⇒ =⎛ ⎞ ⎡ ⎤

⋅⋅⎜ ⎟ ⎢ ⎥×+ ⎣ ⎦⎝ ⎠

=

= ⋅ = ⇒ =×

We have

( )

( )( )

2

3

6

22

20 A / V , 1 V0.25 10So 12.520 10 1

m GS Th

GS Th

k Wg V VL

k V VWL

μ−

′⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

′ ≅ − ≅×= =

×

and a value of / 12.5W L = is certainly reasonable. TYU15.7

( )

1

2

1 1

2 2

0.10 10 0.010

TH LRV VR

R RR R

⎛ ⎞= −⎜ ⎟

⎝ ⎠⎛ ⎞

= − − ⇒ =⎜ ⎟⎝ ⎠

Let 1 0.10 kR = Ω then 2 10 kR = Ω TYU15.8 a.

( )

( )

( )

2

1 2

1

1 2

1

1 2

10 21 10

1.82 V

11.82 101 10

2.73 V

11.82 101 10

0.91 V

S REF

S

TH S H

TH

TL S L

TL

RV VR R

V

RV V VR R

V

RV V VR R

V

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠=

⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠=

⎛ ⎞ ⎛ ⎞= + = + −⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠=

b.

TYU15.9

1

2

1 1

2 2

1

and

S REF

TH S L TL S H

RV VR

R RV V V V V VR R

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Hysteresis 1

2

Width ( )TH TL H LRV V V VR

⎛ ⎞= − = −⎜ ⎟

⎝ ⎠

[ ]( )1 1

2 2

2.5 5 5 10R RR R

⎛ ⎞ ⎛ ⎞= − − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

So 1

2

0.25RR

=

Then

1

2

1 1 (1 0.25) 0.8 VS REF REF REFRV V V VR

⎛ ⎞= − = + = + ⇒ = −⎜ ⎟

⎝ ⎠

Then ( )( )( )( )

1 0.25 5 0.25 V

1 0.25 5 2.25 VTH TH

TL TL

V V

V V

= − − − ⇒ =

= − − ⇒ = −

TYU15.10

10 0 0

1 2

10 110 20 3

100,3

1010 10 exp 3

X

X

XX

Rv v v vR R

t v

tvr

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

= = −

⎛ ⎞⎛ ⎞= + − − −⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

Output switches when 103Xv =

( )( )

1

1

1

1

4 6 3

10 10 13.33 exp 3

10 3.33 6.67exp 13.33 13.33

13.33exp 26.67

ln (2) (0.693)2(0.693)

12(0.693)

10 0.1 10 1 10 722 Hz Duty cycle 50%

X

X

X

X X

X

X

X X X

tr

tr

tr

t r rT r

fr

r R C f− −

⎛ ⎞= − −⎜ ⎟

⎝ ⎠⎛ ⎞ −− = =⎜ ⎟⎝ ⎠⎛ ⎞

+ = ≅⎜ ⎟⎝ ⎠

= ==

=

= = × = × ⇒ = ⇒ =

TYU15.11

( )( )

( )

1

1 2

4 6 4

4

20 0.33320 40

10 0.01 10 1 10

0.711 / 8 ln 1 10 ln 48.9 s1 1 0.333

X X X

PX

RR R

r R C

V VT r Tγ

β

μβ

− −

⎛ ⎞= = =⎜ ⎟+ +⎝ ⎠

= = × = ×

⎡ ⎤++ ⎢ ⎥⎛ ⎞= = × ⇒ =⎢ ⎥⎜ ⎟− −⎝ ⎠ ⎢ ⎥

⎢ ⎥⎣ ⎦

Recovery time

( )

10

1 2

2

2

2 2

20 (8) 2.667 V20 40

0.7 8 2.667 8 exp

8 0.7exp 0.684410.66

1 ln 37.9 s0.685

Y

X

X

X

Rv vR R

tr

tr

t r t μ

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠⎛ ⎞

= + − − −⎜ ⎟⎝ ⎠

⎛ ⎞ −− = =⎜ ⎟⎝ ⎠

⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

TYU15.12

( )( )

( )

10.693

10.693

Let 0.01 F, 1kHz

A B

A B

fR R C

R RfC

C fμ

=+

+ =

= =

( )( )( )

( )( )( )( )( )

53 6

5

5

5

1 1.443 100.693 10 0.01 10

Duty cycle 55 100%2

1.443 10 10055

1.443 10

1.443 10 100 55118 k so 26.2 k

55

A B

A B

A B

B

B B A

R R

R RR R

R

R R R

−+ = = ×

×

+= = ×+

×=

× +

× −= ⇒ = Ω = Ω

TYU15.13

a. 01 2

1

02 4

3

301 1 2.520

50 2.520

I

I

v Rv Rv Rv R

⎛ ⎞ ⎛ ⎞= + = + =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

= − = − = −

(b) 2 21 1 [12 ( 12)] 240

2 2 1.2L

L

VP mWR

− −= ⋅ = ⋅ =

Or 0.24 P W=

c. 12 4.8 V2.5 piV= =