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