Ib=10-0.7/100k=0.093mA
Ic=beta*Ib=50*Ib=4.65mA
Vce=10-Ic*Rc=10-4.65mA*1K=5.35V
Vth=VCC*(R1/R1+R2)=10*[10/(10+10)]=5V
Rth=100K//100K=50K
IB=(5-0.7)/50K
IC=βIB=[4.3/50K]*50=4.3mA
VCE=VCC-ICRC=10-4.3=5.7V
修改 1F 算法
Ib=(10-0.7)/100k - 0.7/100K = 8.6/100 mA
Ic=beta*Ib=50*Ib= 50 * (8.6/100) = 4.3 mA
Vce=10-Ic*Rc=10-4.3mA * 1K = 5.7 V
20.如【圖 9】所示之電路,VCC=10 V,R1=R2=100 kΩ,RC=..-阿摩線上測驗