Question

Calculate the circuit output resistance in the common drain amplifier with \[ r_d=100~k\Omega,\qquad g_m=990~\mu S \] and \[ R_L=10~k\Omega. \]

• A. \(322~\Omega\)
• B. \(332~\Omega\)
• C. \(990~\Omega\)
• D. \(10~k\Omega\)

Hint:

For a source follower, output resistance is very low and is approximately equal to \[ \frac{1}{g_m} \] when other resistances are much larger.

Answer 1

The Correct Option is B

Concept: A common drain amplifier is also called a source follower. Its output resistance is approximately \[ R_o=\left(\frac{1}{g_m}\right)\parallel r_d \parallel R_L \] where \[ g_m=\text{transconductance}. \]

Step 1: Calculate \(1/g_m\).
\[ g_m=990\times10^{-6}S \] \[ \frac{1}{g_m} = \frac{1}{990\times10^{-6}} \] \[ = 1010\Omega \]

Step 2: Calculate parallel combination.
Since \[ r_d=100k\Omega \] and \[ R_L=10k\Omega \] are much larger than \(1010\Omega\), \[ R_o = 1010\parallel100000\parallel10000 \] \[ R_o\approx332\Omega \] \[ \boxed{R_o=332\Omega} \]