The input bias current in a current mirror changes systematic gain error mainly through changes to the Drain Source voltages.
For the MOS current mirror the output voltage is

$$
\begin{aligned}
& V_{o v}=V_{\mathscr{W}}-V_t \\
& V_{o v}=0.7-0.6 \\
& V_{o v}=0.1 \mathrm{~V}
\end{aligned}
$$


We know that $I_o=\frac{k^{\prime}}{2} \frac{W}{L_{\text {eff }}}\left(V_{G S}-V_t\right)^2$
Where

$$
\begin{aligned}
& k^{\prime}=\mu_n C_{o x} \\
& k^{\prime}=\mu_n \times \frac{\varepsilon_{o x}}{t_{o x}}
\end{aligned}
$$


We know that

$$
\begin{aligned}
& \mu_n=450 \mathrm{~cm}^2 / \mathrm{V}-\mathrm{s} \\
& \varepsilon_{o x}=3.9 \times 8.86 \times 10^{-14} \mathrm{~F} / \mathrm{cm}^2 \\
& t_{o x}=80 \times 10^{-8} \mathrm{~cm}
\end{aligned}
$$


Then

$$
\begin{aligned}
& k^{\prime}=450 \times \frac{3.9 \times 8.86 \times 10^{-14}}{80 \times 10^{-8}} \\
& k^{\prime}=194.366 \times 10^{-6}
\end{aligned}
$$

And

$$
\begin{aligned}
& V_{o v}=V_{\circledast}-V_t \\
& V_{o v}=0.1 \quad(\text { from }(1))
\end{aligned}
$$


Then

$$
I_o=\frac{k^{\prime}}{2} \frac{W}{L_{\text {eff }}} V_{o v}^2
$$


Given output current is $I_o=50 \mu \mathrm{~A}$
Now,

$$
\begin{aligned}
& 50 \times 10^{-6}=\frac{194.366 \times 10^{-6}}{2} \times \frac{W}{L_{\text {eff }}} \times(0.1)^2 \\
& \frac{W}{L_{\text {eff }}}=\frac{50 \times 10^{-6} \times 2}{194.366 \times 10^{-6} \times(0.1)^2} \\
& \frac{W}{L_{\text {eff }}}=51.449 \ldots \ldots \text { (3) }
\end{aligned}
$$


We know that resistance $r_0$ is

$$
r_o=\frac{L_{\mathrm{eff}}}{I_D}\left(\frac{d X_d}{d V_{D S}}\right)^{-1}
$$

$$
r_0=\frac{L_{\text {eff }}}{I_D\left(\frac{d X_d}{d V_{D S}}\right)}
$$


We know that $\frac{d X_d}{d V_{D S}}=0.02 \mu \mathrm{~m} / \mathrm{V}$

$$
\begin{aligned}
& r_0=\frac{L_{\text {eff }}}{50 \times 10^{-6} \times 0.02 \times 10^{-6}} \\
& r_0=\frac{L_{\text {eff }}}{1 \times 10^{-12}} \quad \ldots \ldots \text { (4) }
\end{aligned}
$$


The small signal output resistance $R_o$ is

$$
\begin{aligned}
R_o & =\frac{\Delta V_{\text {out }}}{\Delta I_{\text {out }}} \\
R_o & =\frac{1 \mathrm{~V}}{(0.0002)(50 \mu)} \\
R_o & =100 \mathrm{M} \Omega
\end{aligned}
$$


The cascode current mirror does not suffer from formula $\beta_{\xi}$ effects.
The small signal output resistance is

$$
\begin{aligned}
& R_o=r_{o 2}\left[1+\left(g_{m 2}+g_{m b 2}\right) r_{o 1}\right]+r_{o 1} \\
& R_o \approx r_{o 2} g_{m 2} r_{o 1}
\end{aligned}
$$

$$
\begin{aligned}
& R_o \approx r_{o 2} r_{o 1} \sqrt{2 I_D \mu_n C_{o x} \frac{W}{L_{\text {eff }}}} \\
& R_o \approx r_{o 2} r_{o 1} \sqrt{2 I_D k^{\prime} \frac{W}{L_{\text {eff }}}} \\
& R_o=\left(\frac{L_{\text {eff }}}{10^{-12}}\right)^2 \sqrt{2 \times 50 \times 10^{-6} \times 194.366 \times 10^{-6} \times 51.449} \text { (from (3),(4)) } \\
& R_o=\left(\frac{L_{\text {eff }}}{10^{-12}}\right)^2 \times 1 \times 10^{-3} \\
& \left(L_{\text {eff }}\right)^2=\frac{R_o \times 10^{-24}}{1 \times 10^{-3}} \\
& \left(L_{\text {eff }}\right)^2=\frac{100 \times 10^6 \times 10^{-24}}{10^{-3}} \text { (from (5)) } \\
& L_{\text {eff }}=0.316 \times 10^{-6} \mathrm{~m} \\
& L_{\text {eff }} \approx 0.32 \times 10^{-6} \mathrm{~m}
\end{aligned}
$$


The channel length is given by

$$
\begin{aligned}
& L_{\text {eff }}=L_{\text {dran }}-2 L_d-X_d \\
& L_{\text {eff }}=L_{\text {dran }}-2 L_d \quad\left(\because X_d=0\right) \\
& L_{\text {drann }}=L_{\text {eff }}+2 L_d \\
& L_{\text {dirm }}=0.32 \times 10^{-6}+2 \times 0.09 \times 10^{-6} \\
& L_{\text {dran }}=0.5 \times 10^{-6} \mathrm{~m} \ldots \ldots(7)
\end{aligned}
$$


From (3) we have,

$$
\begin{aligned}
& \frac{W}{L_{\text {eff }}}=51.449 \\
& W=51.449 \times L_{\text {eff }} \\
& W=51.449 \times 0.32 \times 10^{-6} \quad(\text { from }(6)) \\
& W=1.646 \times 10^{-5} \\
& \therefore W=16.46 \mu \mathrm{~m}
\end{aligned}
$$