A common-source amplifier with a drain resistance, RD = 4.7 kΩ is powered using a 10 V power supply. Assuming that the transconductance, gm, is 520 μA/V, the voltage gain of the amplifier is closest to:

The correct answer is Option 2: – 2.44

Step-by-Step Explanation:

1. Understand the given circuit parameters:

For a basic common-source (CS) amplifier without a source resistance (or with the source bypassed to ground), the key parameters provided are:

Drain resistance, $R_D=4.7\text{ kΩ}=4700\text{ Ω}$

Transconductance, $g_m=520\text{ μA/V}=520\times {10}^{\mathrm{–6}}\text{ A/V}$

2. Formula for the Voltage Gain of a Common-Source Amplifier:

The small-signal voltage gain ($A_v$) of an unbypassed/ideal common-source MOSFET amplifier (assuming infinite channel length modulation resistance $r_o$) is given by:

$A_v=–g_m·R_D$

The negative sign indicates a 180° phase shift between the input signal at the gate and the output signal at the drain.

3. Calculate the Voltage Gain:

Substitute the given numerical values into the gain formula:

$A_v=–(520\times {10}^{\mathrm{–6}}\text{ A/V})\times \left(4700\text{ Ω}\right)$

$A_v=–2.444$

Rounding to two decimal places, the voltage gain of the amplifier is – 2.44.