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Filename: Microelectronics 4th Neaman chpt2 (1).pdf

### Transcript of Microelectronics 4th Neaman Chpt2 (1)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 22.1 1000 (a) For I > 0.6 V, O = ( I 0.6 ) 1020 For I < 0.6 V, O = 0

1000 (b) (ii) O = 0 = [10 sin ( t )1 0.6] 1020 0. 6 Then sin ( t )1 = = 0.06 ( t )1 = 3.44 0.01911 rad 10 Also ( t )2 = 180 3.44 = 176.56 0.9809 rad Now

O (avg ) ==

1 1 O (t )dt = T 0 2

T

2

[10 sin x 0.6]dx0

1 2

0.9809 0.6 x 10 cos x 0.01911 0.01911 0.9809

1 [( 10 )( 0.9982 0.9982 ) 0.6(0.9809 0.01911 )] 2 O (avg ) = 2.89 V

=

1000 (iii) O ( peak ) = 10 sin 0.6 = 9.2157 V; i d (max ) = 9.2157 mA 1020 2 (iv) PIV = 10 V ______________________________________________________________________________________

2.2

v0 = vI vD i v vD = VT ln D and iD = 0 IS R v v0 = vI VT ln 0 IS R

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.3 1 (a) S = 120 2 = 16.97 V (peak) 10 O ( peak ) = 16.27 V

(b) i D ( peak ) =

16.27 = 8.14 mA 2 (c) O = 16.97 sin t 0.7sin ( t )1 = 0. 7 = 0.04125 ( t )1 = 2.364 16.97 ( t )2 = 180 2.364 = 177.64

177.64 2.364 %= 100% = 48.7% 360

(d)

O (avg ) =

1 2

0.9869

0.01313

[16.97 sin x 0.7]dx

0.9869 0.9869 1 = 0. 7 x ( 16.97 ) cos x 2 0.01313 0.01313 1 [( 16.97 )( 0.99915 0.99915) 0.7(0.9738 )] = 2 O (avg ) = 5.06 V

5.06 = 2.53 mA 2 2 ______________________________________________________________________________________

(e) i D (avg ) =

O (avg )

=

2.4 (a) R (t ) = 15 sin t 0.7 9 = 15 sin t 9.7

( t )1 = sin 1 9.7 = 40.29 0.2238

( t )2

R (avg ) ==

1 2

0.7762

0.2238

[15 sin x 9.7]dx

0.7762 0.7762 1 1 9.7 x ( 15) cos x = 2 [( 15)( 0.7628 0.7628 ) 9.7(0.5523 )] 2 0.2238 0.2238 R (avg ) = 0.9628 V

i D (avg ) = 0.8 =

0.9628 R = 1.20 R

(b) 139.71 40.29 %= 100% = 27.6% 360 ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________2.5

R= R (b) R (t ) = 15 sin t 9.7

(a) i ( peak ) =

R ( peak )

15 9.7 = 4.417 1. 2

( t )1 = 0.2238 ; ( t )2 = 0.776210.7762

R (avg ) =

Or from Problem 2.4, R (avg ) = 2(0.9628) = 1.9256 V (avg ) 1.9256 = = 0.436 A i D (avg ) = R R 4.417 (c) 139.71 40.29 %= 100% = 27.6% 360 ______________________________________________________________________________________

0.2238

[15 sin x 9.7]dx

2.6 (a) S ( peak ) = 12 + 0.7 = 12.7 V

N 1 120 2 = = 13.4 N2 12.7 (b) R =12 = 60 0 .2 VM 12 = 6667 F C= 2 fRV r 2(60 )(60 )(0.25)

(c) PIV = 2 S (max ) V = 2(12.7 ) 0.7 = 24.7 V ______________________________________________________________________________________2.7

v0 = vS 2V vS ( max ) = v0 ( max ) + 2Va.

v ( max ) = 25 V vS ( max ) = 25 + 2 ( 0.7 ) = 26.4 V For 0 N1 160 N = 1 = 6.06 N 2 26.4 N2 v ( max ) = 100 V vS ( max ) = 101.4 V For 0 N1 N 160 = 1 = 1.58 N 2 101.4 N2

b.

PIV = 2vS ( max ) V = 2 ( 26.4 ) 0.7 From part (a) PIV = 2 (101.4 ) 0.7 or PIV = 52.1 V or, from part (b) or PIV = 202.1 V ______________________________________________________________________________________2.8

(a)

vs (max) = 12 + 2(0.7) = 13.4 V 13.4 vs ( rms ) = vs (rms) = 9.48 V 2

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b)Vr = C= (c) 2VM 1 + Vr 2 (12 ) 12 1 + = 150 0.3 id , peak = 2.33 A id , peak = VM R

VM VM C = 2 f RC 2 f Vr R 12 C = 2222 F 2 ( 60 )( 0.3)(150 )

______________________________________________________________________________________2.9

(a)

vS ( max ) = 12 + 0.7 = 12.7 V vS ( rms ) =Vr =

vS ( max ) 2

vS ( rms ) = 8.98 V

(b)

VM V 12 C = M = fRC fRVr ( 60 )(150 )( 0.3)

or

C = 4444 F

VM 12 12 1 + 4 1 + 4 = 2Vr 150 2 ( 0.3) or iD , max = 4.58 A (c) For the half-wave rectifier ______________________________________________________________________________________ iD , max = VM R2.10 (a) O ( peak ) = 10 0.7 = 9.3 VVM 9.3 = 620 F fRV r (60 )(500 )(0.5) (c) PIV = 10 + 9.3 = 19.3 V ______________________________________________________________________________________

(b) C =

2.11 (a) 10.3 O 12.3 V

(b) Vr =Vr =

VM 12.3 = = 0.586 V fRC (60 )(1000 ) 350 10 6

(

)

10.3 = 0.490 V (60)(1000) 350 10 6 So 0.490 V r 0.586 V

(

)

VM 12.3 = 513 F fRV r (60)(1000 )(0.4) ______________________________________________________________________________________

(c) C =

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________2.12

(a) S ( peak ) = 8.5 2 = 12.02 VVOmax

( )

= 12.02 0.7 = 11.32 V

VM 11.32 = = 0.03773 F 2 f RV r 2(60 )(10 )(0.25) (c) PIV = 2 S ( peak ) V = 2(12.02 ) 0.7 = 23.34 V

(b) C =

______________________________________________________________________________________2.13 (a)

vs ( peak ) = 15 + 2 ( 0.7 ) = 16.4 V vs ( rms ) =C=

16.4 2

= 11.6 V

(b) ______________________________________________________________________________________2.14

VM 15 = = 2857 F 2 f RVr 2 ( 60 )(125 )( 0.35 )

______________________________________________________________________________________2.15 (a) S = 12.8 V

N 1 120 2 = = 13.3 N2 12.8 (b) R =12 = 24 0 .5 V r = 3% V r = (0.03)(12 ) = 0.36 V

C=

VM 12 = = 0.0116 F 2 fRV r 2(60 )(24 )(0.36 )VM R 12 = 1 + 2(12 ) 24 0.36

(c) i D ( peak ) =

1 + 2V M Vr i D ( peak ) = 13.3 A

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (d) i D (avg ) =1 2V r V M VM R 1 + 2 2V M Vr 1 = 2(0.36 ) 12 1 + 12 24 2 2(12 ) 0.36

i D (avg ) = 0.539 A (e) PIV = 12.8 + 12 = 24.8 V ______________________________________________________________________________________

2.16 (a) S = 9 + 2(0.8) = 10.6 V

N 1 120 2 = = 16 N2 10.6 (b) R =9 = 90 0. 1 VM 9 = 4167 F C= 2(60)(90)(0.2) 2 fRV r VM R 1 1 + 2V M Vr 2V r V M VM R 9 = 1 + 2(9 ) = 3.08 A 90 0. 2 2V M Vr 1 = 2(0.2 ) 9 1 + 9 90 2 2(9 ) 0. 2

(c) i D ( peak ) = (d) i D (avg ) =

i D (avg ) = 0.1067 A (e) PIV = S (max ) V = 10.6 0.8 = 9.8 V

1 + 2

______________________________________________________________________________________2.17

For vi > 0V = 0

Voltage across RL + R1 = vi

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ RL 1 v0 = vi = vi RL + R1 2 Voltage Divider

______________________________________________________________________________________2.18vi > 0, (V = 0 )

For

a. R2 || RL v0 = vi R2 || RL + R1 R2 || RL = 2.2 || 6.8 = 1.66 k 1.66 v0 = vi = 0.43 vi 1.66 + 2.2

v0 ( rms ) = 3.04 V 2 b. ______________________________________________________________________________________

v0 ( rms ) =

v0 ( max )

2.19

(a) I L =

3 .9 = 0.975 mA 4 20 3.9 II = = 1.342 mA 12 I Z = I I I L = 1.342 0.975 = 0.367 mA PZ = I Z V Z = (0.367 )(3.9) = 1.43 mW3.9 = 0.39 mA 10 I Z = 1.342 0.39 = 0.952 mA

(b) I L =

PZ = (0.952 )(3.9 ) = 3.71 mW ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________2.20 (a)IZ = 40 12 = 0.233 A 120 P = ( 0.233)(12 ) = 2.8 W

(b) So

IR = 0.233 A, IL = (0.9)(0.233) = 0.21 A 12 0.21 = RL = 57.1 RL

P = ( 0.1)( 0.233)(12 ) P = 0.28 W (c) ______________________________________________________________________________________2.21 (a) PZ = I Z V Z 4 = I z (15.4 ) I Z (max ) = 259.74 mA So 15 I z 259.74 mA

(b) I I =

60 15.4 = 297.33 mA 0.15 So I L (max ) = 297.33 15 = 282.33 mA

I L (min ) = 297.33 259.74 = 37.59 mA

15.4 = 54.55 0.28233 15.4 R L (max ) = = 410 0.03759 So 54.55 R L 410 ______________________________________________________________________________________

Then R L (min ) =

2.22 a.

20 10 I I = 45.0 mA 222 10 IL = I L = 26.3 mA 380 I Z = I I I L I Z = 18.7 mA II =

b.PZ ( max ) = 400 mW I Z ( max ) = I L ( min ) = I I I Z ( max ) = 45 40 I L ( min ) = 5 mA = RL = 2 k (c)I Z ( max ) = 40 mA I L ( min ) = 57.1 40 = 17.1 mA RL = 10 RL = 585