电子线路 非线性部分 第五版 谢嘉奎 课后答案
1-2 PCM
1-3 PD
= 70= 40 P= 1000 WPC40 70PD1 = Po/
PD2 = Po/
PD
PC
VCC = 5 V
1PC1 = PD1 Po=1500 W PC2 = PD2 Po = 428.57 W =1428.57 W(PD1 PD2) = 1071.43 W (PC1 PC2) = 1071.43 W = 2500 W1-6 1-2-1aQ3DD3252RL
= 5
4RL = 5 IBQ
QPLPD1
I
cm I1RL = 103RL = 5Q
(1) RL = 10 (VCE = VCC ICRL)
VCEQ1 = 2.6VIIBQ1 = Ibm
= 2.4mA
Vcm = VCEQ1VCE(sat) = (2.6 0.2) V = 2.4
VIcm = I CQ1 = 220 mA
1VcmIcm264mWPD = VCC ICQ1 = 2
1.1 W= PL/ PD = 24(2) RL = 5 VCE = VCC ICRL
IBQ(1)IBQ2 = 2.4mAQ2
VCEQ2 = 3.8VICQ2 = 260mA
Vcm = VCCVCEQ2 =1.2 VIcm = I CQ2 = 260 mA PL
PL
(3)1VcmIcm156mWPD = VCC ICQ2 = 1.3 W= PL/ PD = 122RL = 5 Q(1)
Q3VCEQ3 = 2.75VICQ3= 460mAIBQ3 = 4.6mA Ibm = 2.4mA
vCEmin= 1.55ViCmax= 700mA Q