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Miller Compensation


\[ \begin{gather} \frac{V_o}{I_i} = \frac{(sC_f - g_m)R_1 R_2}{1+sX(s) + s^{2}Y(s)} \\\\ X(s) = R_1C_1 + R_2C_2 + R_1C_f(1+g_mR_2)+R_2C_f \\\\ Y(s) = (C_1C_2 + C_1C_f + C_2C_f)\ R_1R_2 = \binom{\{C_1, C_2, C_f\}}{2}\cdot R_1R_2 \end{gather} \]
  • DC gain
\[ \begin{gather} A_0 = A(0) = -g_mR_1R_2 \end{gather} \]
  • zero \(z = g_m/C_f\)
    by definition, when
\[ \begin{gather} A(s)\big|_{s=z} = A(z) = 0 \end{gather} \]

implies there is no (AC)current go through \(R_D\), thus we can easily have

\[ \begin{gather} (V_i - V_o)sC_f= V_i\ g_m \\\\ \implies sC_f=g_m \end{gather} \]
  • poles
    consider case without \(C_f\)
\[ \begin{gather} 1+sX_o(s)+s^{2}Y_o(s) = (1+sR_1C_1)(1+sR_2C_s) \\\\ \implies \left\{ \begin{aligned} X_o(s) &=R_1C_1 + R_2C_2 \\\\ Y_o(s) &= R_1R_2C_1C_2 \end{aligned} \right. \end{gather} \]

now consider \(C_f\)

\[ \implies \left\{ \begin{aligned} X(s) &= R_1C_1' + R_2C_2' = R_1(C_1 + C_{f-miller})+ R_2(C_2 +C_f) \\\\ Y(s) &= (C_1C_2 + C_1C_f + C_2C_f)\ R_1R_2 \end{aligned}\right. \]

Pole Splitting

\[ \begin{gather} D(s) = \left(1+\frac{s}{\omega_{p1}'}\right)\left(1+\frac{s}{\omega_{p2}'}\right)= 1+X(s)s+Y(s)s^{2} \end{gather} \]

assume \(\omega_{p1}'\) dominates (\(\omega_{p1} \ll \omega_{p2}\)), then

\[ \begin{gather} D(s) \approx 1 + \frac{s}{\omega_{p1}} + \frac{s^{2}}{\omega_{p1}\omega_{p2}} \\\\ \implies \left\{ \begin{aligned} \omega_{p1} &= \frac{1}{R_1C_1 + R_2C_2 + R_1C_f(1+g_mR_2)+R_2C_f} \approx \frac{1}{g_mR_1R_2C_f} \\\\ \omega_{p2} &= \frac{g_mC_f}{C_1C_2+C_1C_f+C_2C_f} \end{aligned}\right. \end{gather} \]