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GATE 2021 EE – Question 33

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

In the BJT circuit shown, beta of the PNP transistor is 100. Assume $V_{BE}=-0.7$ V. The voltage across $R_C$ will be 5 V when $R_2$ is ________ k$\Omega$. (Round off to 2 decimal places.)

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Correct answer: 16.70 to 17.70

Explanation

$I_C=\frac{5\text{ V}}{3.3\text{ k}\Omega}=1.515$ mA and $I_E=\frac{101}{100}I_C=1.530$ mA. So $V_E=12-1.2\text{ k}\times1.530\text{ mA}=10.164$ V and $V_B=V_E-0.7=9.464$ V. The current in $R_1$ is $\frac{12-9.464}{4.7\text{ k}}=0.540$ mA and the base current is $\frac{I_C}{100}=15.2\ \mu$A, so the current in $R_2$ is $0.555$ mA. Then $R_2=\frac{9.464}{0.555\text{ mA}}=17.06\ \text{k}\Omega$.