Ch15p

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  • 1. Chapter 15Exercise SolutionsEX15.1For the circuit shown in Figure 15.71 f 3dB =2 RCor11RC === 3.979 106 2 f3dB 2 ( 40 103 )For R = 75 KThenC = 5.311011 = 53.1 pFWe have C3 = 1.414C = 75.1 pFC4 = 0.707C = 37.5 pFEX15.2 1 fC =CReqor 11C= =f c Req (105 )( 20 106 )C = 0.5 pFEX15.3 C130Low-frequency gain: T = == 6 C2 5fC C2 (100 10 )( 5 10 ) 3 12 f 3dB = = f 3dB = 6.63 kHz 2 CF 2 (12 1012 )EX15.4 1 f0 = 2 3RC 11RC = = = 6.13 1062 f 0 3 2 (15 10 ) 33Let C = 0.001 F = 1 nFThen R = 6.13 k so R2 = 8R = 49 kEX15.511 f0 = C =2 RC2 f 0 R 1C= C 0.02 F2 ( 800 ) (104 )R2 = 2 R1 = 2 (10 ) R2 = 20 kEX15.6

2. R1 VTH = VH R1 + R2 R1 2= (12) R1 + 20 2 ( R1 + 20 ) = 12 R140 = 10 R1 R1 = 4 kEX15.7 R1 VTH VTL = (VH VL ) R1 + R2 R1 0.10 = (10 [ 10]) R1 + R2 R2 20R1+== 200 2 = 199 R1 0.10R1 R2 VS = VREF R1 + R2 R 1 VREF = 1 + 1 VS = 1 + (1) VREF = 1.005 V R2 199 VH VBE ( on ) VI=R + 0.110 0.7 0.7R + 0.1 == 43 k 0.2R = 42.9 kEX15.8At t = 0 , let v0 = 5 so v X = 2.5. For t > 0 t v X = 10 + ( 2.5 10 ) exp rX When v X = 5.0, output switches t 5.0 = 10 12.5 exp 1 rX t 10 5 5.0exp 1 = = rX 12.5 12.5 t 12.5 12.5 exp + 1 = t1 = rX ln t1 = rX ( 0.916 ) rX 5.0 5.0 During the next part of the cycle t v X = 5 + ( 5 [ 5]) exp rX When v X = 2.5, output switches 3. t 2.5 = 5 + 10 exp 2 rX t 5 2.5 2.5exp 2 == rX 10 10 t 10 10 exp + 2 = t2 = rX ln t2 = rX (1.39 )rX 2.5 2.5 1Period = t1 + t2 = T = ( 0.916 ) + (1.39 ) rX = 2.31rX Frequency = 2.31rXrX = ( 50 103 )( 0.01 10 6 ) = 5 10 4 s f = 866 Hzt1 ( 0.916 )Duty cycle = 100% = 100% Duty cycle = 39.7% t1 + t2( 0.916 ) + (1.39 )EX15.9a.rX = RX C X R1 10 vY = v0 = (12 ) = 1.2 V R1 + R2 10 + 90 R1= = 0.10 R1 + R2 0.7 1 + V /VP 1 + 12 T = rX ln = rX ln 1 1 (0.10) T = 50 106 = rX ln [1.18] = (0.162) rX50 106RX = RX = 3.09 k (0.1 106 )(0.162)b.Recovery time t v X = VP + (1.2 VP ) exp rX When v X = V , t = t2 4. t 0.7 = 12 + ( 1.2 12 ) exp 2 rX t 12 0.7exp 2 == 0.856 rX 13.2 1 t2 = rX ln = ( 0.155 ) rX 0.856 rX = ( 3.09 103 )( 0.1 106 ) = 3.09 104 t2 = 48.0 sEX15.10T = 1.1 RCT = 75 106Let C = 10 nFThen75 106R== 6.82 K(1.1) (10 109 )EX15.11 11 f == f = 802 Hz0.693 ( RA + 2 RB ) C ( 0.693) 20 + 2 ( 80 ) 103 ( 0.01 106 ) R + RB 20 + 80Duty cycle = A 100% = 100% Duty cycle = 55.6%RA + 2 RB 20 + 2 ( 80 )EX15.121 VP2a.P= 2 RLVP = 2 RL P = 2 ( 8 )(1) VP = 4 VVP 4 IP = = I P = 0.5 ARL 8b.VCE = 12 4 = 8 V I C 0.5 ASo P = I C VCE = ( 0.5 )( 8 ) P = 4 WEX15.13VP = 2 RL PL = 2 ( 8 )(10 ) = 12.65 V V PS = VS P RL PS R2 (10 ) ( 8 )VS == VP12.65VS = 19.9 VEX15.14 dV0 dV dVLine regulation = += 0 Z+ dVdVZ dVNow 5. dV0 10 = 1 + = 2dVZ 10 dVZ rZ 10 + == = 0.00227dV rZ + R1 10 + 4400So Line regulation = ( 2 )( 0.00227 ) = 0.004540.454%EX15.15V1 V0 V11 1 V = V1 + = 010 10 10 10 10 2 V VV1 = 0 V0 = 2V1 V1 = 0 10 102V0 V1 V0 V0 A0 L (VZ V1 ) ++ =0 10RL R0V0 V0 V0 A0 LVZ V1 A0 LV1++ = 10 RL R0 R0 10R0 V0 A V= 0L 02(10) 2 R0V0 V 1000 ( 6.3) V0 (1000 ) V0 + I0 + 0 =10 0.50.5 20 2 ( 0.5 )V0 [0.10 + 2.0 0.05 + 1000] + I 0 = 12, 600V0 (1002.05) + I 0 = 12, 600For I 0 = 1 mA V0 = 12.5732For I 0 = 100 mA V0 = 12.4744V0 ( NL ) V0 ( FL )Load reg = 100% V0 ( NL ) 12.5732 12.4744= 100%12.5732Load reg = 0.786%EX15.16a. 6. VZ 3VBE ( on )IC 3 =R1 + R2 + R3 5.6 3 ( 0.6 ) 3.8IC 3 = = I C 3 = 0.482 mA 3.9 + 3.4 + 0.576 7.88I I C 4 R4 = VT ln C 3 IC 4 0.482 I C 4 (0.1) = (0.026) ln IC 4 By trial and error I C 4 = 0.213 mAVB 7 = 2(0.6) + (0.482)(3.9) VB 7 = 3.08 Vb. R13 V0 = VB 8 = VB 7 R13 + R12 2.23 (5) = 3.08 2.23 + R12 ( 2.23)( 5 ) = ( 3.08 )( 2.23) + ( 3.08) R1211.15 = 6.868 = 3.08R12 R12 = 1.39 kTYU15.11 f 3dB =2 RC11RC === 1.59 105 2 f3dB 2 (104 )Let C = 0.01 F R = 1.59 kThenC1 = 0.03546 FC2 = 0.01392 FC3 = 0.002024 F11T = =6 6 f 20 1+ 1+ f3d 10 T = 0.124 or T = 18.1 dBTYU15.2 11 f 3dB = RC = 2 RC2 f3dB1RC == 3.18 106 2 ( 50 103 )Let C = 0.001 F = 1 nF R = 3.18 kThen 7. R1 = 2.94 kR2 = 3.44 kR3 = 1.22 kR4 = 8.31 k1T = 0.01 = 8 f 1 + 3 dB f 82 f 1 41 + 3 dB = = 10 f 0.01 2 f3 dB f 3dB 10 f = f 15.8 kHz f 10TYU15.3 11-pole T = 3.87 dB 2 12 1+ 10 12-pole T = 4.88 dB 4 12 1+ 10 13-pole T = 6.0 dB 6 12 1+ 10 14-pole T = 7.24 dB 8 12 1+ 10 TYU15.4 1Req =fC C1 1or f C C === 2 107 Req 5 106If C = 10 pF fC = 20 kHzTYU15.5 11 f0 == f 0 65 kHz2 6 RC 2 6 (10 )(100 1012 )4R2 = 29 R = 29 (104 ) R2 = 290 kTYU15.6 8. 1 1 f0 == f 0 = 7.12 MHz CC (109 ) 22 L 1 2 2 (10 ) 6 9 C1 + C2 2 10 C2 = gm RC1 C2 11gm = = g m = 0.25 mA / V C1 R 4 103We have k W g m = 2 (VGS VTh ) 2 L k 20 A / V 2 , VGS VTh 1 V W 0.25 103So = = 12.5 L ( 20 106 ) (1)and a value of W / L = 12.5 is certainly reasonable.TYU15.7R VTH = 1 VL R2 R R0.10 = 1 ( 10 ) 1 = 0.010 R2 R2Let R1 = 0.10 k then R2 = 10 kTYU15.8a. R2 10 VS = VREF = ( 2) R1 + R2 1 + 10 VS = 1.82 V R1 1 VTH = VS + VH = 1.82 + (10 ) R1 + R2 1 + 10 VTH = 2.73 V R1 1 VTL =VS + VL = 1.82 + ( 10 ) R1 + R2 1 + 10 VTL = 0.91 Vb. 9. TYU15.9 R VS = 1 + 1 VREF R2 R R VTH = VS 1 VL and VTL = VS 1 VH R2 R2 R Hysteresis Width = VTH VTL = 1 (VH VL ) R2 R R 2.5 = 1 ( 5 [ 5]) = 10 1 R2 R2 RSo 1 = 0.25R2Then R VS = 1 = 1 + 1 VREF = (1 + 0.25)VREF VREF = 0.8 V R2 ThenVTH = 1 ( 0.25 )( 5 ) VTH = 0.25 VVTL = 1 ( 0.25 )( 5 ) VTL = 2.25 VTYU15.10 R1 10 1vX = v0 = v0 = v0 R1 + R2 10 + 20 3 10t = 0, vX = 3 10 t v X = 10 + 10 exp 3 rX 10Output switches when v X = 3 10. 10 t1 = 10 13.33 exp 3 rX t 10 3.33 6.67exp 1 = = rX 13.33 13.33 t 13.33exp + 1 = 2 rX 6.67t1 = rX ln (2) = (0.693)rXT = 2(0.693)rX 1f =2(0.693)rXrX = RX C X = (104 )( 0.110 6 ) = 110 3 f = 722 Hz Duty cycle = 50%TYU15.11 R1 20 = == 0.333 R1 + R2 20 + 40rX = RX C X = (104 )( 0.01 106 ) = 1 104 0.7 1 + V / VP 1+ 8 = (1 10 ) ln T = 48.9 s4T = rX ln 1 1 0.333 Recovery time 11. R1 20 vY = v0 = (8) = 2.667 V R1 + R2 20 + 40 t 0.7 = 8 + ( 2.667 8 ) exp 2 rX t 8 0.7exp 2 = = 0.6844 rX 10.66 1 t2 = rX ln t2 = 37.9 s 0.685 TYU15.121 f = ( 0.693)( RA + RB ) C 1RA + RB = ( 0.693) fCLet C = 0.01 F,f = 1kHz1RA + RB == 1.443 105 ( 0.693) (103 )( 0.01106 ) RA + RBDuty cycle = 55 = 100% RA + 2 RB55 =(1.443 10 ) (100 )5(1.443 10 ) + R5BRB= (1.443 10 ) (100 55) R5= 118 k so RA = 26.2 k B55TYU15.13 v01 R2 30 a. = 1 + = 1 + = 2.5 vI R1 20 v02R50 = 4 == 2.5 vI R3 201 VL2 1 [12 (12)]2(b)P= = = 240 mW2 RL 2 1.2OrP = 0.24 W12c.= V pi = 4.8 V2.5