\runParam{I_D = [30, 20, 10, 40] #\paramUnits{\mu A} }
\runParam{V_omin = [0.5, 0.4, 0.3, 0.6] #\paramLatex{V_{o,min}}}
\runParam{R_out = [60, 72, 80, 50]#\paramUnits{M \Omega}}
\runParam{V_tn = [0.3, 0.25, 0.35, 0.4] }
\runParam{unCox = [160, 250, 300, 200]# \paramUnits{\mu A/V^2} \paramLatex{\mu_n C_{ox}}}
\runParam{lambn = [0.04, 0.025, 0.05, 0.03]#\paramUnits{\mu m/V} \paramLatex{\lambda_n'}}

\question[6] Consider the wide-swing current mirror shown below where the desired output current is  \valU{I_D}.  Given that $M_1$ and $M_2$ are identical in size and the minimum output voltage is \valU{V_omin}, find the length of the transistors such that the current mirror output resistance is \valU{R_out}.

NMOS: $\val={V_tn}; \val={unCox}; \val={lambn}$

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 \hbox{ \hspace{-100pt}  \incProbLTspice[1.0\textwidth]{currMirror01a}}
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\end{flushleft}

\begin{blankSpace}
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\begin{solutions}
\textbf{Solution}

\val={V_omin} $= 2V_{ov}$; \run={V_ov = V_omin/2} for both transistors\\
\run()={g_m = 2*I_D/V_ov} for each transistor\\
$R_{out}\approx g_mr_o^2$; \run()={r_o = sqrt(R_out/g_m)}\\
$r_o = L/(|\lambda_n'|I_D)$\\
\run()={L = r_o*abs(lambn)*I_D#\paramUnits{m}}\\
\hlite{\val={L}}

\end{solutions}

\begin{answers}
\textbf{Answer}

L = \valU{L}
\end{answers}