A student designs a zener shunt regulator to produce a regulated output of 10 V

Question

A student designs a zener shunt regulator to produce a regulated output of 10 V using a zener diode specified to have a 10V drop at a test current of 25mA. Incremental resistance of the zener is 10 ohm and minimum diode current = 2mA. Given that supply voltage is 20V plusminus 25%, and load current can vary between 0mA and 20mA, find Vzo for the zener diode Required series resistance R for zener diode Line regulation Change in Vout corresponding to plusminus 25% change in unregulated supply Load rgulation Change in Vout corresponding to 0 - 20mA change in load current Zener diode power rating to handle the worst case power dissipation

Explanation / Answer

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 voltageacross series resistr R is (20-10) = 10 ..... also current flowing thru zener is also equal to currrent flowingthru R... R=voltage acrooss R/ current thru R=10/25*10-3=.4k c)applying voltage divider rule, Vo=Vi*Rz/R+Rz=±1*10/400+10=±24.4 mV so line regulation=Vo/Vi=24.4 mV/V d)25% of supply voltage is 5V means change is ±5V so applying voltage divider rule, Vo=Vi*Rz/R+Rz=±5*10/400+10 =±121.9 mV e)now change in o/p voltage is given by, Vo=Rz*Iz=10*(±1*10-3)=±10mV load regulation=Vo/Iz=±10 mV/mA f)load current is increased by 20 mA zener current decreases bysamee Iz=-20 mA so Vo=Rz*Iz=10*(-20*10-3)=-200 mV g)maximum power disssipation takes place when current thru zenerdiode 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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