In the nMOS feedback amplifier (capacitors act as shorts at the signal frequency): \(V_{DD}=5\,\text{V}\), \(R_D=1~\text{k}\Omega\), \(R_S=1~\text{k}\Omega\), \(R_1=200~\text{k}\Omega\), \(R_2=300~\text{k}\Omega\). Given \(\mu_n C_{ox}=1~\text{mA/V}^2\), \(W/L=2\), \(V_T=1~\text{V}\), and the DC drain voltage is \(4~\text{V}\). Find \(V_{\mathrm{out}}/V_{\mathrm{in}}\) (magnitude).
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With unbypassed \(R_S\): \(A_v\simeq-\dfrac{g_mR_D}{1+g_mR_S}\). Use \(g_m=2I_D/V_{ov}\) when a square-law MOS model applies.
Step 1: Small-signal parameters. Bias gives \(I_D=(5-4)/R_D=1~\text{mA}\).
Overdrive from square law: \(I_D=\frac{k_n}{2}V_{ov}^2\) with \(k_n=\mu_n C_{ox}(W/L)=2~\text{mA/V}^2\Rightarrow V_{ov}=1~\text{V}\).
Hence \(g_m=2I_D/V_{ov}=2~\text{mS}\). Step 2: Gain with source degeneration (no \(r_o\)).
Gate is driven by \(V_{in}\) (coupling cap short; supplies are AC ground), so
\[
v_{gs}=\frac{v_{in}}{1+g_mR_S},\qquad v_o=-g_mR_D\,v_{gs}.
\]
Thus
\[
\frac{V_o}{V_{in}}=-\frac{g_mR_D}{1+g_mR_S}
=-\frac{(2~\text{mS})(1~\text{k}\Omega)}{1+(2~\text{mS})(1~\text{k}\Omega)}
=-\frac{2}{3}=-0.6667.
\]
Magnitude \(\approx 0.67\). Final Answer: \(|V_{\mathrm{out}}/V_{\mathrm{in}}| \approx 0.67\)
