电子术 第3章放大电路基础 例332(1)求差模输入电压v、共模输入电压n x(2)若A=-50、Ame= n1式叫9v求输出电压,及K0s R R C 101Ⅴ OO 1。M=x(,如[解](1)可将任意输入信号分解为 R 共模信号差模差模信号 i1=1.01=.00+0.01(v) EE =0.991.00-0.01(V id=ln-u2=101-0.99=0.02(V) C d i=(u1+u12)/2 L (2) Mod=Aue 50×0.02=-1() loc= A.u=-005×1=-0.05(V) Auduid+ Aulic=-1.05(V) 50 CM(dB)=2010=20g 60(dB) 0.05uod = Auduid (2) 若 Aud = – 50、 Auc = – 0.05 求输出电压uo，及 KCMR 1.01 V 0.99 V [解] 可将任意输入信号分解为 共模信号和差模信号之和 (1) ui1 = 1.01 = 1.00 + 0.01 (V) ui2 = 0.99 = 1.00 – 0.01 (V) 例 3.3.2 (1)求差模输入电压 uid 、共模输入电压 uic ui1 V1 +VCC V2 VEE RC RC REE uo ui2 uC1 uC2 uid = u i1 – u i2 uic = (ui1+ ui2 ) / 2 = 1.01 – 0.99 = 0.02 (V) = 1 (V) 2 i d 1 ui1 = uic + u 2 id 1 ui2 = uic − u (2) = – 50  0.02 = – 1 (V) uoc = Aucuic = – 0.05  1 = – 0.05 (V) uo = Auduid + Aucuic= –1.05 (V) c d CMR (dB) 20lg u u A A K = 0.05 50 = 20lg = 60 (dB) 共模信号 差模信号 第 3 章 放大电路基础