电子电路分析与设计--模拟电子技术(答案)第11章

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1、thMicroelectronics:CircuitAnalysisandDesign,4editionChapter11ByD.A.NeamenProblemSolutions______________________________________________________________________________________Chapter1111.1(a)CMRR=∞⇒υ=Aυ=(250)(1.5sinωt)(mV)dBoddυ=0.375sinωt(V)o4250(b)CMRR=80dB⇒CMRR=10=⇒A=0.

2、025dBcmAcmυ=()250(0.0015sinωt)()+0.025(3sinωt)oυ=0.45sinωt(V)o250(c)CMRR=50dB⇒CMRR=316.2=⇒A=0.791dBcmAcmυ=()250(0.0015sinωt)()+0.791(3sinωt)oυ=2.75sinωt(V)o______________________________________________________________________________________11.2(a)(i)υ=−gRυ=−()()(150.7+0.

3、1sinωt)o1m1υ=−3.5−0.5sinωt(V)o1(ii)υ=−gRυ=−()()(150.7−0.1sinωt)o2m2υ=−3.5+0.5sinωt(V)o2(iii)υ−υ=−1.0sinωt(V)o1o2(b)Δ()υ−υ=(0.7+0.1sinωt)−(0.7−0.1sinωt)=0.2sinωt12−0.5(c)(i)A==−2.5d10.2+0.5(ii)A==+2.5d20.2−1(iii)A==−5d30.2____________________________________________________

4、__________________________________11.3(a)Neglectdcbasecurrents−0.7−(−3)I=I+I=0.2mA=⇒R=11.5kΩEC1C2ERE3−1.2υ=1.2V,⇒R==18kΩo1C0.1(c)ForV=0⇒V=0.7VCBCE3=I()18+0.7+2I()11.5−3CCSoI=0.1293mACυ=υ=3−(0.1293)(18)=0.673Vo1cm+υ=−2.3Vcm−So−2.3≤υ≤0.673Vcm_________________________________

5、_____________________________________________________thMicroelectronics:CircuitAnalysisandDesign,4editionChapter11ByD.A.NeamenProblemSolutions______________________________________________________________________________________11.4a.1020.7−()II=⇒=1.01mA118.5I1.011II==⇒≅1.

6、01mACC222211++ββ()1+()100101()⎛⎞100⎛⎞1.01II=⇒⎜⎟⎜⎟≅0.50mACC44⎝⎠101⎝⎠2VVCE22=−−−⇒=()00.7(5)CE4.3VVVCE44=−⎡⎤⎣⎦50().52()()−−⇒=0.7CE4.7Vb.ForVV=⇒=2.5V−0.72.51.8V+=CE44C51.8−II=⇒=1.6mACC442⎛⎞11+β⎛⎞01IICC24+=⇒⎜⎟()22⎜⎟()()1.6IC2=3.23mA⎝⎠β⎝⎠100II≈=3.23mA12C1020.7−()RR=⇒=2.66kΩ113.2

7、3______________________________________________________________________________________11.5a.Neglectingbasecurrents300.7−II==400Aμ⇒=R⇒=R73.25kΩ13110.4V=⇒=10VV9.3VCE11C159.3−RR=⇒=Ω28.5kCC0.2b.()1000.026()r==13kΩπ0.250rQ03()==Ω125k0.4WehaveβR()10028.5()CAA==⇒=62dd22()rRπ++B(

8、)1310⎧⎫⎪⎪βRC⎪⎪1A=−⎨⎬cmrRπ+B⎪⎪21r0()+β1+⎪⎪rR+⎩⎭πB⎧⎫()10028.5()⎪⎪⎪⎪1=−⎨⎬⇒A=−0.113cm1310+⎪⎪2