The figure belows shows a circuit which uses zener diode to
Solution
a) For V1=6 V
i) R2=1k ohm assume that zener diode is not there and find V2=V1*1/(1+10)=0.5455 V it is less than Vz i.e 4.7 V
Zener is not in break down region so V2=0.5455 V
ii)
R2=10k ohm assume that zener diode is not there and find V2=V1*10/(10+10)=3 V it is less than Vz i.e 4.7 V
Zener is not in break down region so V2=3 V
iii)
R2=100k ohm assume that zener diode is not there and find V2=V1*100/(100+10)=5.45 V it is greater than Vz i.e 4.7 V
Now Zener is in break down region so V2=Vz=4.7 V
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b) For V1=-6 V
i) R2=1k ohm assume that zener diode is not there and find V2=V1*1/(1+10)=-0.5455 V it is less than Vf(forward bias voltage) i.e -0.7 V
so V2=-0.5455 V
ii)
R2=10k ohm assume that zener diode is not there and find V2=V1*10/(10+10)=-3 V it is greater than Vf(forward bias voltage) i.e -0.7 V so V2=-0.7 V forward Conduction mode
iii)
R2=100k ohm assume that zener diode is not there and find V2=V1*100/(100+10)=-5.45 V it is greater than Vf(forward bias voltage) i.e -0.7 V so
V2=-0.7 forward Conduction mode
