Subject: CMOS VLSI Design

Topic: Single Stage Amplifier

Difficulty: Difficult

Using Miller's theoram,

$C_{GS}$ can be shifted to input and output side.

Modified circuit using Miller's theoram,

Input => Total capacitance seen from (X) to ground,
$\hspace{1cm}=C_{GS}+(1-A_v)C_{GD}\hspace{3cm}$.........$A_v=-g_mR_D$
$\therefore\,\,$ Input cutoff frequency,
$\hspace{1cm}\omega_{in}=\frac{1}{R_{in}C_{in}}=\frac{1}{R_S[C_{GS}+(1+g_mR_D)C_{GD}]}$

Output => Total capacitance w.r.t ground,
$\hspace{1cm}=C_{DB}+(1-A_v^{-1})C_{GD}$
$\hspace{1cm}\approx C_{DB}+C_{GD}$
$\hspace{1cm} \therefore\,\omega_{out}=\frac{1}{R_D(C_{DB}+C_{GD})}$

$\therefore\,\,\,T.F=\frac{V_{out}}{V_{in}}(s)=\frac{-g_mR_D}{\Big( 1+\frac{S}{\omega_{in}} \Big)\Big( 1+\frac{S}{\omega_{out}} \Big)}$

Calculations of input impedance:

• In high speed applications, input impedance of signal stage amplifier is important.
• From below figure,

$Z_i=\frac{1}{\Big[ C_{GS}+(1+g_mR_D)C_{GD} \Big]s}$

• But at high frequency, effect of output node must be taken into account, ignoring $C_{GS}$ for the moment and using following circuit,

$(I_x-g_mV_x)\frac{R_D}{1+R_DC_{DB}S}+\frac{I_x}{C_{GD}S}=V_x$

$\therefore\,\,\,\, \frac{V_x}{I_x}=\frac{1+R_D(C_{DB}+C_{GD})S}{C_{GD}S(1+g_mR_D+R_DC_{DB}S)}$