Prove that the gain is independent of bias current and voltages.

CS stage with diode connected load.
For CS Stage with diode, if the variation of ($\mu = g\,m_b/g\,m$) with the output voltage is neglected then prove that the gain is independent of bias current and voltage.

-In General, $A_v=-g\,m_1\,R_D \hspace{3cm}$ -(1)

-To calculate equivalent Resistance of $m_2$

For SSM,
$(gm_2+gmb_2 ) V_x+\frac{V_x}{r_0}=1x$

$\therefore \frac{V_x}{I_x}=\frac{1}{gm_2+gmb_2+r_0^{-1}} \approx \frac{1}{gm_2+gmb_2}$

$A_v=-gm_1R_D=-gm_1(\frac{1}{gm_2+gmb_2})$

$\therefore A_v=-\frac{gm_1}{gm_2}(\frac{1}{1+\mu}) \hspace{2cm} ....\eta=\frac{gmb_2}{gm_2} $

but, $gm=\sqrt{2\mu_n\,C_{ox} (\frac{W}{L}) I_D}$

$ \therefore A_v=-\frac{\sqrt{2\mu_n\,C_{ox} (\frac{W}{L})_1 I_D}}{\sqrt{2\mu_n\,C_{ox} (\frac{W}{L})_2 I_D}}*\frac{1}{1+\eta} $

$\hspace{2cm}=-\sqrt{\frac{(\frac{W}{L})_1}{(\frac{W}{L})_2}}*\frac{1}{1+\eta} $

Neglecting $\eta$,
then $A_v=-\sqrt{\frac{(\frac{W}{L})_1}{(\frac{W}{L})_2}} \hspace{2cm}$

• Gain is independent of bias current and voltage.