电子线路 非线性部分 第五版 谢嘉奎 课后答案

Icm = ICQ1=0.3AVcm = VCC =15 V

PLmax1VcmIcm22.25W PD

VCCICQ5082.5 4.5W PLmax

PD2.254.550%nRL

RL

(2)RL(RLVCCIbmRLRLQ2ICQEF ICQQPLmax) 2.25 W RL(IbmIcmVcm PD = VCC ICQ= 9 W PLmax/ PD = 25% RLRL< 50

IcmRL

RL> 50 Q2VCCIbm

PL Q2

3

3DD303

VCE(sat) = 0 ICEO = 0REIbmVcmIcm (3)VCC =30 VEFPLPL= 2.25 W WPD = VCC ICQ= 9 W (4)Ibm= 6 mAQ3 1-9 VCC

1RLVCC PCmaxRLiCmax = 2 ARL = 8 RLn2PCM = 30 W

Pomax ICM = 3 AV (BR)CEO= 60 V

(1)

Pomax

PCmax

1VcmIcm211VCCiCmax2278.5% 30W ) 1VcmIcm211VCCiCmax222Pomax = 30 W 50% 15W 0.2Pomax = 6 W(

2Vcm/2PomaxRL 2VCC/2Pomax30 15