电子线路 非线性部分 第五版 谢嘉奎 课后答案
2-2 (1)
(2)
2-4
VCC
C Po2VCE(sat)VCC2(VCC2RL2VCE(sat))2VCC = 18 VVCE(sat) = 0.5 VRL = 50 PDPo
(1) v
A
IcmVcmRL2(VCCLVcm2(VCC2VCE(sat))2VCE(sat))
IC0
Po
PD1VcmIcm2VCCIC01Icm 2VCE(sat))2 2VCE(sat)) 2(VCC2RL2VCC(VCC2RL
VCC
CPo/PD2VCE(sat)VCC
(2) PD22VCC(VCCRL2VCE(sat))1.24W