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 option is – 2.44.

To find the voltage gain of a common-source amplifier, we can analyze its small-signal equivalent circuit. For a basic common-source amplifier, assuming the transistor's output resistance ($r_o$) is much larger than the drain resistance ($R_D$), the voltage gain ($A_v$) is given by the formula:

$A_v=-g_m{R}_D$

where:
• $g_m$ is the transconductance of the transistor.
• $R_D$ is the drain resistance.
• The negative sign indicates a 180-degree phase shift (signal inversion) between the input and the output.

From the given data, we have:
• Transconductance, $g_m=520\text{μA/V}=520\times {10}^{-6}\text{A/V}$
• Drain resistance, $R_D=4.7\text{kΩ}=4700\text{Ω}$

Substituting these values into the voltage gain equation:

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

Calculating the value:

$A_v=-2.444$

Therefore, the voltage gain of the amplifier is closest to – 2.44.