GATE 2017 EC – Question 51

Analog Circuits · BJT and MOSFET Amplifiers · 2 marks · Numerical answer

For the DC analysis of the Common-Emitter amplifier shown, neglect the base current and assume that the emitter and collector currents are equal. Given that $V_T = 25$ mV, $V_{BE} = 0.7$ V, and the BJT output resistance $r_o$ is practically infinite. Under these conditions, the midband voltage gain magnitude, $A_v = |v_o/v_i|$ V/V, is ________.

A common-emitter amplifier with $V_{CC} = 12$ V, a base divider of $R_1 = 73\ \text{k}\Omega$ and $R_2 = 47\ \text{k}\Omega$, collector resistor $R_C = 2\ \text{k}\Omega$, emitter resistor $R_E = 2\ \text{k}\Omega$ bypassed by $C_E = 100\ \mu$F, and a load $R_L = 8\ \text{k}\Omega$ coupled through $C_2$.

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Correct answer: 128

Explanation

The base voltage is $12 \times \frac{47}{73 + 47} = 4.7$ V, so the emitter is at $4.7 - 0.7 = 4.0$ V and $I_E \approx I_C = \frac{4.0}{2\ \text{k}\Omega} = 2$ mA. The transconductance is $g_m = \frac{2\ \text{mA}}{25\ \text{mV}} = 80$ mS. The emitter is bypassed, so the midband gain is $g_m(R_C \parallel R_L) = 0.08 \times 1.6\ \text{k}\Omega = 128$.