Using the parameters of Table 3.1-2, we find that the saturation voltage of M2 is



$$

V_{d s 1}(\mathrm{sat})=\sqrt{\frac{2 I_1}{K_N^{\prime}\left(W_1 / L_1\right)}}=\sqrt{\frac{200}{110 \cdot 10}}=0.4264 \mathrm{~V}

$$





Now, using the active equation of M2, we set $I_2=100 \mu \mathrm{~A}$ and solve for $W_2 / L_2$.



$$

\begin{aligned}

100 \mu \mathrm{~A} & =K_N^{\prime}\left(W_2 / L_2\right)\left[V_{d s 1}(\mathrm{sat}) \cdot V_{d s 2}-0.5 V_{d s 2}^2\right] \\

& =110 \mu \mathrm{~A} / \mathrm{V}^2\left(W_2 / L_2\right)\left[0.426 \cdot 0.0426-0.5 \cdot 0.0426^2\right] \mathrm{V}^2=1.883 \times 10^{-6}\left(W_2 / L_2\right)

\end{aligned}

$$





Thus,



$$

100=1.883\left(W_2 / L_2\right) \quad \rightarrow \quad \frac{W_2}{L_2}=53.12

$$





Now if M2 should become saturated, the value of the output current of the mirror with 100 $\mu \mathrm{A}$ input would be $531 \mu \mathrm{~A}$ or a boosting of 5.31 times $I_1$.