6 4解:P VCCICO 24 0.25 6 W
C
P05
83.3%P 6
22VcmVCC242
Rp 57.6
2P02P02 5
Icm1
2P02P02 5
0.417 A VcmVCC24Icm10.42
1.67Ic00.25
gc c
查表得 c 77o
2ηVCC2 0.7 12
6 6解:gc θc 1.56查表得θc 91o
Vcm10.8
P0 Ik2R 22 1 4 W
1 1
PC P P0 1P 1 4 1.7 W 0 η
0.7 c
Ic090
6 7解:icmax 282 mA o
α0900.319
Ic1mP0
α 90 i
o1
cmax
mA 0.5 282 141
11
RpIc21m 200 0.1412 2 W 22P02ηc 74%
VCCIc030 0.09
22222VcmIkmR2 0L IkmR2 0L RCIkmR2 0L RIkmR
6 8证:P0 2
2L2RP2L22 0LRC
i2.2
6 9解:Vcm VCC vcmin VCC cmax 24 21.25 V
gcr0.8
2
2
2
Ic0 icmax 0 70o 2.2 0.253 0.5566 A
Icm1 icmax 1 70o 2.2 0.436 0.9592 A P VCCIc0 24 0.5566 13.36 W 11
P0 VcmIcm1 21.25 0.9592 10.19 W
22
PC P P0 13.36 10.19 3.17 W
C
P010.19 76.3%P 13.36
2Vcm21.252
Rp 22.16
2P02 10.19