A zener diode exhibits a constant voltage of 5.6 volts for currents greater than five times the knee current. Izk is specified to be 1 ma. the zener is to be used in the design of a shunt regulator fed from a 15 volts supply . the load current varies over range of 0 ma to 15mA . find a suitabe value for resistor R. what is the meaximum power dissipation of the zener diode.

Given Information:  

Voltage across zener =  Vz = 5.6 V

Knee current = Izk = 1mA

Supply voltage = V = 15 V

Load current = IL = 15mA

Required Information:  

Resistor = R = ?

Power dissipation of zener = P = ?

R = 470 Ω

P = 0.112 Watts

Explanation:

R = supply voltage - zener voltage/Total current

Where Total current = zener current + Load current

zener current = 5*knee current = 5*1 = 5 mA

Total current = 5 mA + 15 mA = 20 mA

R = (15 - 5.6)/0.02

R = 470 Ω

There maximum power will be dissipated in zener when load current is zero so all the current will flow through the zener

Iz = (15 - 5.6)/470

Iz = 0.02 A

P = IzVz

P = 0.02*5.6

P = 0.112 Watts

Therefore, the maximum power dissipation of zener is 0.112 Watts.

answer:

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