高频电子线路课后习题答案(第五版)张肃文(9)

4 18证明：1 Ib1 yieVbe1 yreVce1 1

Ic1 yfeVbe1 yoeVce1 2 Ib2 yieVbe2 yreVce2Ic2 yfeVbe2 yoeVce2

yieVce1 yre Vcb2 Vce1 yreVcb2 yie yre Vce1 3 yfeVce1 yoe Vcb2 Vce1 yoeVcb2 yfe yoe Vce1 4

2 3 得Ic2 yfeVbe1 yie yre yoe Vce1 yreVcb2

Vce1

Ic2 yreVcb2 yfeVbe1

yie yre yoe

5

Ic2 yreVcb2 yfeVbe1

yie yre yoe

5 代入 4 Ic2 yoeVcb2 yfe yoe

Ic2

2 yfe yfe yoe yieyoe yreyfe yoe

Vbe1 Vcb2 6 yie yre yfe 2yoeyie yre yfe 2yoe

yf

yfe yfe yoe yfeyie yre yfe 2yoe

由 1 乘 yfe yoe 与 4 乘yre后相加得

Ib1 yfe yoe Ic2yre yie yfe yoe Vbe1 yreyoeVcb2由 6 代入消去Ic2得Ib1

2 yie yieyre yieyfe 2yieyoe yreyfe yre yoe yre Vbe1 Vcb2

yie yre yfe 2yoeyie yre yfe 2yoe

2

yieyoe yreyfe yoe

yo yre

yie yre yfe 2yoe

2yie yieyre yieyfe 2yieyoe yreyfe

yi yie

yie yre yfe 2yoe

yr

yre yoe yre y y yre reoeyie yre yfe 2yoeyfe

略

同理可证2

2

4 22解：vbn 4kTrb fn 4 1.38 10 23 273 19 70 200 103 0.226 10 12 V2

2

ien 2qIE fn 2 1.6 10 19 10 3 200 103 0.64 10 16 A2

0

f

1 f

2

0.95 10 10 1 500 106

6

2

0.95

2

icn 2qIC 1 0 fn 2 1.6 10 19 10 3 1 0.95 200 103 0.32 10 17 A2