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Response Details:
for a zener diode,
Vz=Vzo+IzRz
a)10=Vzo+25*10
so we gret,
Vzo=9.75 V
b)now voltage drop across zener diode is 10 V, so the voltage across series resistr R is (20-10) = 10 .....
also current flowing thru zener is also equal to currrent flowing thru R...
R=voltage acrooss R/ current thru R=10/25*10-3=.4 kΩ
c)applying voltage divider rule,
ΔVo=ΔVi*Rz/R+Rz=±1*10/400+10=±24.4 mV
so line regulation=ΔVo/ΔVi=24.4 mV/V
d)25% of supply voltage is 5V means change is ±5V
so applying voltage divider rule,
ΔVo=ΔVi*Rz/R+Rz=±5*10/400+10 =±121.9 mV
e)now change in o/p voltage is given by,
ΔVo=Rz*ΔIz=10*(±1*10-3)=±10 mV
load regulation=ΔVo/ΔIz=±10 mV/mA
f)load current is increased by 20 mA zener current decreases by samee
ΔIz=-20 mA
so ΔVo=Rz*ΔIz=10*(-20*10-3)=-200 mV
g)maximum power disssipation takes place when current thru zener diode is maximum
here max current is 25 mA
so power dissipation is,
P=I2*Rz=25*10-3*25*10-3*10=6.25 mW
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