ch11_solutions

ch11_solutions - 11-342 CHAPTER 11

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Unformatted text preview: 11-342 CHAPTER 11 11.1vO=vS1MΩ1MΩ+5kΩ1000(291kΩ1kΩ+0.5Ω| Av=vOvS=990 or 59.9 dBiS=vS1MΩ+5kΩand iO=990vS1kΩ| Ai=iOiS=9901000106=9.9x105or 120 dBAP=AvAi=990 9.9x105(29=9.8x108or 89.9 dB | vS=vOAV=5V990=5.05 mV11.2vO=vS5kΩ5kΩ+5kΩ31.6(291kΩ1kΩ+1kΩand Av=vOvS=7.91 or 18.0 dBvs=vOAv=10V7.91=1.27 VSince Routhas the same value as RL, the powerdissipated in Routis also 0.5W. The power dissipated in Ridwill bePI=Vid22Rid=VS222Rid=VS28Ridwhere VS=VO7.91and VO=2 0.5W(291000Ω(29=31.6VPI=428 5000(29=0.4mW. The total power dissipated in the amplifier isP = 500mW +0.4mW = 500 mW.11.30.99mV≥1mVRidRid+50kΩ⇒Rid≥4.95 MΩ11.4Io=2 100W(2950Ω=2Aand Io2Rout2≤5W or Rout≤2.5Ω11.5vid=vOA=10V105=0.1 mV| 10VA≤10-6Vrequires A≥107or 140 dB11.6vid=vOA=15V106=15 mV| 15VA≤10-6Vrequires A≥15x106| i+=15mV1MΩ=15 pA11.711-343 a( 29Av= -R2R1= -220kΩ4.7kΩ= -46.8 | 20log 46.8(29=33.4 dB| Rin=R1=4.7 kΩ| Rout=0 Ωb( 29Av= -R2R1= -2.2MΩ47kΩ= -46.8 | 20log 46.8(29=33.4 dB| Rin=R1=47 kΩ| Rout=0 Ω11.8a( 29Av= -R2R1= -120kΩ12kΩ= -10.0 | 20log 10.0(29=20.0 dB| Rin=R1=12 kΩ| Rout=0 Ωb( 29Av= -R2R1= -300kΩ150kΩ= -2.00 | 20log 2.00(29=6.02 dB| Rin=R1=150 kΩ| Rout=0 Ωc( 29Av= -R2R1= -220kΩ4.3kΩ= -51.2 | 20log 51.2(29=34.2 dB| Rin=R1=4.3 kΩ| Rout=0 Ω11.9Av= -R2R1= -8200Ω910Ω= -9.01 | VO= -9.01 0.05V(29= -0.451V| vot( 29= -0.451sin 4638t(29VIs=VsR1=0.05V910Ω= 54.9mA | ist( 29=54.9sin 4638t(29mA11.10Av=1+R2R1=1+680kΩ8.2kΩ=83.9 | 20log 83.9(29=38.5 dB| Rin= ∞| Rout=0 Ω11.11First printing should refer to Fig. 11.49. a( 29Av=1+R2R1=1+120kΩ24kΩ=6.00 | 20log 6.00(29=15.6 dB| Rin= ∞| Rout=0 Ωb( 29Av=1+R2R1=1+300kΩ15kΩ=21.0 | 20log 21.0(29=26.4 dB| Rin= ∞| Rout=0 Ωc( 29Av=1+R2R1=1+360kΩ4.3kΩ=84.7 | 20log 84.7(29=38.6 dB| Rin= ∞| Rout=0 Ω11.12Av=1+R2R1=1+8200Ω910Ω=10.0 | VO=10.0 0.05V(29=0.500V| vot( 29=0.500sin 9125t(29V11.13vO= -R3R1v1-R3R2v2= -51kΩ1kΩv1-51kΩ2kΩv2= -51v1-25.5v2vOt( 29= -51 0.01sin 3770t(29-25.5 0.04sin10000t(29vOt( 29=0.510sin 3770t-1.02sin10000t(29V and v-t( 29≡0.11.1411-344 a( 29Avnom=1+R2R1=1+47kΩ0.18kΩ=262 | 20log 262(29= 48.4 dBRin=10kΩ+ ∞= ∞| Rout=0 Ωb( 29Avmax=1+47kΩ1.1(290.18kΩ0.9(29=320 | Avmin=1+47kΩ0.9(290.18kΩ1.1(29=215Avmax-AvnomAvnom=320-262262=0.22 | Avmin-AvnomAvnom=215-262262= -0.18c( 29Tolerances: +22%, -18% d( 29320215=1.49 :1(e) function count=c; c=0; for i=1:500, r1=180*(1+0.2*(rand-0.5)); r2=47000*(1+0.2*(rand-0.5)); a=1+r2/r1; anom=1+47000/180; if (a>=0.95*anom & a<=1.05*anom), c=c+1; end; end c Executing this function twenty times yields 44% . 11.15(a) Using Eq. 11.36:Av= -R2R1= -10RR= -10b( 29R2=v2i2v1==11R=110 kΩ| R1=v1i1v2==R=10 kΩ11.16V1=3.2 V| V2=3.1 V| V+=3.1V10RR+10R=2.82 V| V-=V+=2.82 VVO= -R2R1V1-V2(29= -10RR3.2-3.1(29= -1.00 V11.17Av= -R4R31+R2R1= -10kΩ10kΩ1+100kΩ2kΩ = -51vO=Avv1-v2(29= -51 2...
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This note was uploaded on 01/17/2011 for the course ECE 3040 taught by Professor Hamblen during the Fall '07 term at Georgia Tech.

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ch11_solutions - 11-342 CHAPTER 11

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