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## Q. 6.7

The circuit shown in Fig. 6.16(a) is a special case where $R_D → ∞$. Calculate the transfer function (with λ = 0) and explain why the Miller effect vanishes as $C_{DB}$ (or the load capacitance) increases.

## Verified Solution

Using (6.30) and letting $R_D$ approach infinity, we have

$\frac{V_{out}}{V_{in}}(s) = \frac{C_{G D}s − g_m}{R_Sξ ^s2 + [g_m R_SC_{G D} + (C_{G D} + C_{DB})]s}$                                                      (6.36)

= $\frac{C_{G D}s − g_m}{s[R_S(C_{GS}C_{G D} + C_{GS}C_{DB} + C_{G D}C_{DB})s + (g_m R_S + 1)C_{G D} + C_{DB}]}$

As expected, the circuit exhibits two poles—one at the origin because the dc gain is infinity [Fig. 6.16(b)]. The magnitude of the other pole is given by

$ω_{p2} ≈\frac{ (1 + g_m R_S)C_{G D} + C_{DB}}{R_S(C_{G D}C_{GS} + C_{GS}C_{DB} + C_{G D}C_{DB})}$                                  (6.37)

For a large $C_{DB}$ or load capacitance, this expression reduces to

$ω_{p2} ≈ \frac{1}{R_S(C_{GS} + C_{G D})}$                                              (6.38)

indicating that $C_{G D}$ experiences no Miller multiplication. This can be explained by noting that, for a large $C_{DB}$, the voltage gain from node X to the output begins to drop even at low frequencies. As a result, for frequencies close to $[R_S(C_{GS} + C_{G D})]^{−1}$, the effective gain is quite small and $C_{G D}(1 − A_v) ≈ C_{G D}$. Such a case is an example where the application of the Miller effect using low-frequency gain does not provide a reasonable estimate.