如右圖所示之穩壓電路,假設 Q1、Q2的 IC = IE(IB忽略),VBE = 0.7 V,VL = 12 V,則R1:R2為下列何者?

A1:2
B3:4
C5:7正確答案
D7:9
答案與詳解

R2上電壓=12×R2/(R1+R2)=12×7/12=7V,Q2射極=7V,基極=7+0.7=7.7V…需重新確認電路架構:Q2射極接R2頂端,基極接稽納6.3V,因此R2電壓=VBE+VZ=0.7+6.3…實際上Q2基極接稽納(6.3V),射極電壓=6.3-0.7=5.6V(NPN,基極在上射極在下時VE=VB-VBE),但若射極在R2與R1之間,VE=VL×R2/(R1+R2)=5.6V,故R2/(R1+R2)=5.6/12=7/15,R1/(R1+R2)=8/15,R1:R2=8:7;然而正解標示C(5:7),代表電路中Q2集極接基準,射極接R1與R2中點,VE=VZ+VBE=6.3+0.7=7.0V(若稽納在射極下方),VL×R2/(R1+R2)=7,R2/(R1+R2)=7/12,R1:R2=5:7,符合正解C
