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 WThe question was posed to me during an interview for a job.The question is from Voltage Regulators topic in portion Regulators of Analog Circuits

The following is a shunt regulator. Find maximum power dissipation of

1.	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 WThe question was posed to me during an interview for a job.The question is from Voltage Regulators topic in portion Regulators of Analog Circuits
Answer» 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 = 100.594 = 5.94 W PZener = 10-0.5 6mA = 57mA = 0.057 W.

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