CMOS模拟集成电路习题答案-Allen版(181)

CMOS模拟集成电路习题答案-Allen版(181)

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时间:2019-05-27

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1、ndCMOSAnalogCircuitDesign(2Ed.)HomeworkSolutions:9/20/20021Chapter1HomeworkSolutions1.1-1UsingEq.(1)ofSec1.1,givethebase-10valueforthe5-bitbinarynumber11010(b4b3b2b1b0ordering).FromEq.(1)ofSec1.1wehaveNb-1+b-2+b-3+...+b-N=b-iN-12N-22N-3202∑N-i2i=1-1-2-3

2、-4-5110101´2+1´2+0´2+1´2+0´2=++++248163216+8+0+2+02613===3232161.1-2ProcessthesinusoidinFig.P1.2throughananalogsampleandhold.Thesamplepointsaregivenateachintegervalueoft/T.15141312111098Amplitude765432101234567891011Sampletimes__tTFigureP1.1-21.1-3Digit

3、izethesinusoidgiveninFig.P1.2accordingtoEq.(1)inSec.1.1usingafour-bitdigitizer.ndCMOSAnalogCircuitDesign(2Ed.)HomeworkSolutions:9/20/20022111115111014110113121100111010109100010008Amplitude70110650101400113200100010101234567891011Sampletimes__tTFigureP1

4、.1-3Thefigureillustratesthedigitizedresult.Atseveralplacesinthewaveform,thedigitizedvaluemustresolveasampledvaluethatliesequallybetweentwodigitalvalues.Theresultingdigitizedvaluecouldbeeitherofthetwovaluesasillustratedinthelistbelow.SampleTime4-bitOutpu

5、t01000111002111031111or11104110151010601107001180010or0001900101001011110001.1-4Usethenodalequationmethodtofindvout/vinofFig.P1.4.ndCMOSAnalogCircuitDesign(2Ed.)HomeworkSolutions:9/20/20023ABR1R2vinR3v1gmv1R4voutFigureP1.1-4NodeA:0=G1(v1-vin)+G3(v1)+G2(

6、v1-vout)v1(G1+G2+G3)-G2(vout)=G1(vin)NodeB:0=G2(vout-v1)+gm1(v1)+G4(vout)v1(gm1-G2)+vout(G2+G4)=0G1+G2+G3G1vingm1-G20vout=G1+G2+G3-G2gm1-G2G2+G4vGout1(G2-gm1)=vGin1G2+G1G4+G2G4+G3G2+G3G4+G2gm11.1-5Usethemeshequationmethodtofindvout/vinofFig.

7、P1.4.R1R2viniaR3v1gmv1R4vouticibFigureP1.1-50=-vin+R1(ia+ib+ic)+R3(ia)ndCMOSAnalogCircuitDesign(2Ed.)HomeworkSolutions:9/20/200240=-vin+R1(ia+ib+ic)+R2(ib+ic)+voutvoutic=R4ib=gmv1=gmiaR3vout0=-vi+Rin+R1a+gmiaR3+R3ia4voutvout0=-vi+Rg+vin+R1

8、a+gmiaR3+R2miaR3+Rout44R1vin=ia(R1+R3+gmR1R2)+voutR4R1+R2+R4vin=ia(R1+gmR1R3+gmR2R3)+voutR4R1+R3+gmR1R3vinR1+gmR1R3+gmR2R3vinvout=R1+R3+gmR1R3R1/R4R1+gmR1R3+gmR2R3(R1+R2+R4)/R4vinR3R4(1-gmR2)vout=22(R1+R3+gm

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