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Chapter 6

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.

6.16

Step-by-Step

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.