Question

The following is a shunt regulator. Find maximum power dissipation of the Zener diode and transistor, given that the source voltage varies from 20-40V, the Zener breakdown voltage is 5V, the output voltage is 10V, the resistance R1=50Ω, R2=20kΩ, β=99, VBE=0.5V.

(a) PTransistor = 5.12 W, PZener = 0.051 W

(b) PTransistor = 0.41 W, PZener = 0.57 W

(c) PTransistor = 5.94 W, PZener = 0.057 W

(d) PTransistor = 6.22 W, PZener = 5.66 W

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The question is from Voltage Regulators topic in portion Regulators of Analog Circuits

Answer 1

Correct answer is (c) PTransistor = 5.94 W, PZener = 0.057 W

Explanation: Output voltage = 10V = VCE

Current across R1 = VS-VO/R1

For maximum power dissipation, supply is maximum

I = 40-10/50 = 0.6A

Current I = IZ + IC = IB + βIB = 100IB

IB = 6mA

IC = 0.594 A

PTransistor = 10*0.594 = 5.94 W

PZener = 10-0.5 * 6mA = 57mA = 0.057 W.