如圖所示之電路，其中電晶體之參數為 $\beta =100$，$V_A=\infty$，$V_T=26\mathrm{mV}$ 且 $V_{BE(on)}=0.7\mathrm{V}$，求其小信號電壓增益值為何？

正確答案。由圖可見電路為共射極組態，集極接3.6kΩ，輸出端透過耦合電容接5kΩ負載，射極接1kΩ。直流分析：VB = 10×(2.2k)/(10k+2.2k) ≈ 1.803V，VE = VB - 0.7 ≈ 1.103V，IE ≈ IC = 1.103/1k ≈ 1.103mA。gm = IC/VT = 1.103m/26m ≈ 42.4mA/V，rπ = β/gm = 100/42.4m ≈ 2.36kΩ。交流RC_eff = 3.6k//5k = (3.6×5)/(3.6+5) = 18/8.6 ≈ 2.09kΩ。有射極退化增益：Av = -gm·RC_eff/(1 + gm·RE) ≈ -42.4m×2.09k/(1+42.4m×1k) = -88.6/43.4 ≈ -2.04 ≈ -2.02