Subject: CMOS VLSI Design

Topic: Single Stage Amplifier

Difficulty: Medium

For MOSFET in saturation,
$\hspace{2cm}I_D=\frac{1}{2}\mu_n C_{OX}\frac{W}{L}(V_{GS}-V_{th})^2$

where, $(V_{GS}-V_{th})$ -> $I_D$ dependant on square of overdrive voltage.

• In some applications, square law dependence of $I_D$ on overdrive voltage introduces excessive non-linearity.

• Solution: using diode connected load.

• Another approach: Using source degeneration resistor.

• As $V_{in}$ increases, $I_D$ increases, voltage drop across $R_S$ increases, i.e a fraction of $V_{in}$ appears across $R_S$ rather than as a gate source overdrive, Therefore leading to a smoother variation of $I_D$.

• Voltage gain: $A_v$
  $\hspace{3cm} $A_v$ =\frac{\partial V_{out}}{\partial V_{in}}=-\Big( \frac{\partial I_{D}}{\partial V_{in}} \Big)R_D \hspace{2cm}$..........(1)

let, $\hspace{1cm}G_m=-\Big( \frac{\partial I_{D}}{\partial V_{in}} \Big)\hspace{2cm}$ => equivalent transconductance of circuit.

$\therefore\,\,\, A_v=-G_m\,R_D\hspace{6cm}$ ...........(2)

$G_m=\frac{\partial I_{D}}{\partial V_{GS}}\frac{\partial V_{GS}}{\partial V_{in}}$

but, $V_{GS}=V_{in}-I_DR_S$

$\therefore \,\,\, \frac{\partial V_{GS}}{\partial V_{in}}=1-R_S\Big( \frac{\partial I_{D}}{\partial V_{in}} \Big) $

$\therefore \,\,\,G_m= \frac{\partial I_{D}}{\partial V_{GS}}\Big( 1-R_S\frac{\partial I_{D}}{\partial V_{in}} \Big) $

$ \,\,\,G_m=g_m(1-R_S*G_m)$

Since, $\frac{\partial I_{D}}{\partial V_{GS}}$ is transconductance of $M_1$.

$\therefore \,\,\,G_m=\frac{g_m}{1+g_mR_S}$

from (2),
$A_v=\frac{-g_mR_D}{1+g_mR_S} \hspace{3cm}$

.......i.e $A_v$ decreases with $R_S$.

$A_v=\frac{-R_D}{\frac{1}{g_m}+R_S}$