In the circuit below, assume that the long channel NMOS transistor is biased in saturation. The small signal transconductance of the transistor is \(g_m\). Neglect body effect, channel length modulation, and intrinsic device capacitances. The small signal input impedance \(Z_{in}(j\omega)\) is:
Show Hint
When analyzing circuits with reactive components and active devices, combine the reactive impedances with the contributions of transconductance. Use the \(j\omega\)-domain for accurate calculations of frequency-dependent impedances.
Step 1: Understand the circuit components.
The input impedance \(Z_{in}(j\omega)\) is determined by the combined effects of the capacitors \(C_1\) and \(C_L\), as well as the NMOS transistor's small signal behavior, which includes a dependent current source characterized by its transconductance \(g_m\).
Step 2: Calculate the impedance contributions.
- The capacitors contribute reactive impedances:
\[
Z_{C_1} = \frac{1}{j\omega C_1}, \quad Z_{C_L} = \frac{1}{j\omega C_L}.
\]
- The NMOS transistor adds a term proportional to \(-g_m\), influenced by the interaction of the two capacitors. The total input impedance is the combination of these contributions:
\[
Z_{in}(j\omega) = \frac{-g_m}{C_L C_1 \omega^2} + \frac{1}{j\omega C_1} + \frac{1}{j\omega C_L}.
\]
Step 3: Confirm the expression.
The derived expression represents the total impedance of the circuit, including the reactive components of the capacitors and the transconductance contribution. This matches the given option:
\[
\boxed{\frac{-g_m}{C_L C_1 \omega^2} + \frac{1}{j\omega C_1} + \frac{1}{j\omega C_L}}.
\]
Final Answer:
\[
\boxed{{(1) } \frac{-g_m}{C_L C_1 \omega^2} + \frac{1}{j\omega C_1} + \frac{1}{j\omega C_L}}
\]
