电子线路 非线性部分 第五版 谢嘉奎 课后答案

电子线路 非线性部分 第五版 谢嘉奎 课后答案

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

电子线路 非线性部分 第五版 谢嘉奎 课后答案

Vcm = VCEQ3 vCEmin = 1.2 VIcm = iCmax CQ3 = 240 mA

1VcmIcm2

(4) RL = 5 PL

PL

1-7

= 0I CEO = 0

PLmax(a):PLmax(b):PLmax(c)144mWPD = VCC ICQ3 = 2.3 W= PL/ PD = 6.26Icm = I CQ3 = 460 mAVcm = VCCVCEQ3 =2.25 V 517.5mWPD = VCC ICQ3 = 2.3 W= PL/ PD = 22.51VcmIcm2VCC

VCE(sat)

(1)

Icm = I CQVcm = VCEQ Icm

VCEQ Vcm = VCC IcmRC CCI

2IcmRL= VCC 2Vcm

Vcm

PLmax(a)121VCC31VCC3RL

PD1VCC 18RLVCCICQVCCIcm

PLmax(a)

C21VCC 3RL2vCE = VCC iQ(R)RLCDRLCCQ 1RC2VCEQ = VCC ICQRC AB1VCC31VCC3 RL

(2)

Vcm = VCEQ = VCC/2IcmICQPD1 6CF VCC 2RL

1VcmIcm221VCC

8RLQ PLmax(b)PDVCCICQ2VCC 2RL