何丰主编低频电子电路习题参考答案

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1、第一章：半导体基础元件与非线性电路1-1解：根据题意得：v⎛⎞vTiI=−⎜⎟e1s⎜⎟⎝⎠⎧2.16−=11.16(0.02V)vi⎛⎞⎪vT=−⎜⎟eV1=⎨5.84()0.05所以，I⎜⎟s⎝⎠⎪⎩45.81()0.1V1-2解：根据题意得：TIm/A2DT1TT〉21VV/1-4解：根据题意得：二极管的模型如下：IIDVV=0.7VDo()nV⑴、电路图如下所示：R=3kΩ1R=2kΩ2V=5v1VV=5vD2VV+55+12假设，VD断开，这时电流I==3=2mARR12++()32×10VIAB=−R11+V=()−23×+5=−

2、1V故VD断开，流过其电流为零。即ID=0IIDS=−⑵、电路图如下所示：Rk=2ΩRk=3Ω12I1I3I210V5V10+5I==33mA()23+×10所以二极管两端的电压VIAB=−R11+V=(−23)×+10=4V故二极管导通⎧VV=−IRD111所以可以得到⎨⎩VI=−R−VD222VV−10−0.7Im==1D=4.65A1R21VV+0.7+5Im=−D2=−=−1.9A2R22∴I31=+II2=4.65−1.9=2.75mA1-6解：根据题意得：++V+-VDvRvi0−−假设二极管V断开DVV=+VDi当V=−2V时，

3、VwD=−21+.0sint<0v=0二极管截止，ov0t当V=0V时，VD=sinwt>0时，vv0=iVD=sinwt<0时，v0=0v0t当V=2V时，VwD=+2sint>0vw0=2s+intv02t1-8:解：根据题意得：题图如下：VDC6V25Ω+5.1kΩvi−先做直流分析：VDIDQ6V5.1kΩVVDD−Do()n60−.7Im===1.04ADQR5.12Vr==T25ΩjIDQ再做交流分析+25Ω+viv0R2=Ω5.1krj=25Ω−−1由于Rr2//j=Ω25vvom==im2.5mV21-12解：根据题意得：IV

4、VV=0.6on⎧⎪VvDi1=−−VDo()n⎧VvD1=−i⎨⎨Vv=−V⎩Vv=⎪⎩Di2Do()nD2ivVi<−Do()n=−0.6VVD1导通vVi>Do()n=0.6VVD2导通⑴、vVi<−0.6时，VD1导通，VD2截止，情况如下所示：I−1kΩVD()on+++vvi0−−1kΩvV+=2IiDo()nvV=−+I=1()v−VoiDo()n2D()on⑵、vVi>Do()n=0.6VVD2导通VD1截止，情况如下：I+1kΩVD()on++−vvi0−−1kΩ1vv0=−(ivD()on)+vD()on2vvi−Do()

5、n=2I1=+vv()iDo()n2vv=+IoDo()n⑶、当−≤vvDo()ni≤vD(on)时，即−0.6Vv≤≤i0.6V时，VD1、VD2截止∴vvoi=所以，输出电压和输入电压的关系图如下：v01/20.6−0.610.6vi−0.6⑵、解：根据题意得+10V10kΩDD1B3CA+D2VvIED410kΩ−10kΩ−10V解：根据题意得：假设DDDD断开1234可得vvD1=10−IVVD2=10VD3=10VvvDI4=10+当vI>0时，D4优先导通可得vvEI=−0.7此时电路图如下所示+10V10kΩDD1B3CA+vI

6、+VED4710kΩ0.−−10kΩ−10VvvD1=10−IvVD2=10vvD3=−0E=0.7−vID2先导通故又可得到电路图如下+10V10kΩ+V7D1B0.−D3CA+VvI+VED3.710kΩ0−−10kΩ−10V10−0.7vvD1=10−IvVC=×10=4.6520vvD3=−=CEv5.35−vIvVA=4.65+=0.75.35vvD1=−AIv=5.35−vI可得vVD1>0.7时，04<4.65V，此时，D1D3截止。D2D4导通vV=4.65o

7、v<0的情况同理可以推测I最终得到的图形如下：v04.65−4.654.65vI−4.651.13解：根据题意得：R++RLvv0i−−1.2VVI−22−=VI1.2VI−10max−52=0RRR0.8VVI−22V0.8V−10−=II−55=minRRR得出：VI=18.75VRL=0.5kΩR+++VV2=10RLvvi−0−−第二章：半导体受控器件基础2-6解：根据题意得：⑴、VVBE==0.7VCE11.7V处于放大区域（放大状态）硅⑵、VVBE=−0.2VVCE=−0.2处于饱和区锗锗硅NPNPNP-10−5.7+4+4.2e

8、+9c−5bcbeccbbNPNPNPee2-7解：根据题意得：VV=−3.2+2.5=−0.7BE放大VV=−72+.5=−4.5CEVV=−V=−3.2+2.5