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电子线路 非线性部分 第五版 谢嘉奎 课后答案[1-5章].pdf

发布：2019-05-09约5.15万字共27页下载文档
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1-2 PCM 1-3 P = 1000 W C 40 70 PD PC = 40 P = P / = 2500 W P = P P =1500 W C1 D1 o C C1 D1 o = 70 P = P / =1428.57 W P = P P = 428.57 W C2 D2 o C C2 D2 o PD (PD1 PD2) = 1071.43 W PC (PC1 PC2) = 1071.43 W 1-6 3DD325 1-2-1 a VCC = 5 V PL PD C 1 RL = 10 Q 2 R = 5 I 1 I == III 3 R = 5 Q L BQ ccmm CQCQCQ L 1 4 R = 5 QQ L (1) R = 10 (( V = V I R ) Q L CE CC C L VCEQ1 = 2.6V IICCCQQQ111 === 220220220mAmAmA IBQ1 = Ibm = 2.4mA V = V V = (2.6 0.2) V = 2.4 cm CEQ1 CE(sat) V Icm = I CQ1 = 220 mA 1 P V I 264 mW PD = VCC ICQ1 = L cm cm 2 1.1 W = P / P = 24 C L D (2) R = 5 V = V I R L CE CC C L IBQ (1) IBQ2 = 2.4mA Q2 VCEQ2 = 3.8V ICQ2 = 260mA
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