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GATE 2016 EE – Question 22

Analog and Digital Electronics · Amplifiers: biasing, equivalent circuit and frequency response · 1 mark · Numerical answer

A transistor circuit is given below. The Zener diode breakdown voltage is 5.3 V as shown. Take base to emitter voltage drop to be 0.6 V. The value of the current gain $\beta$ is ________.

A 10 V supply feeds a 4.7 kΩ resistor, which connects to the transistor base and to a 5.3 V Zener diode to ground, carrying 0.5 mA. The collector connects to the 10 V supply through a 220 Ω resistor, and the emitter connects to ground through a 470 Ω resistor.

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Correct answer: 18 to 20

Explanation

The base is held at the Zener voltage 5.3 V, so the emitter is at $5.3 - 0.6 = 4.7$ V and the emitter current is $\frac{4.7}{470} = 10$ mA. The 4.7 kΩ resistor carries $\frac{10 - 5.3}{4.7\text{k}} = 1$ mA, of which 0.5 mA goes through the Zener, so the base current is $1 - 0.5 = 0.5$ mA. The collector current is $10 - 0.5 = 9.5$ mA, so $\beta = \frac{9.5}{0.5} = 19$.