## Chapter 6

## Q. 6.6

A student considers only C_{G D} in Fig. 6.13(a) so as to obtain a one-pole response, reasons that the voltage gain drops by 3 dB (by a factor of = \sqrt{2}) at the pole frequency, and concludes that a better approximation of the Miller effect should multiply C_{G D} by 1 + g_m R_D \sqrt{2}. Explain the flaw in this reasoning.

## Step-by-Step

## Verified Solution

Setting C_{GS} and C_{DB} to zero, we obtain

\frac{V_{out}}{V_{in}}(s) =\frac{(C_{G D}s − g_m)R_D}{\frac{s}{ω_0}+ 1} (6.31)

where ω_0 = R_S(1+ g_m R_D)C_{G D} + R_DC_{G D}. We note that C_{G D} is multiplied by 1+ g_m R_D in this exact analysis. So where is the flaw in the student’s argument? It is true that the voltage gain in Fig. 6.13(a) falls by \sqrt{2} at ω_0, but this gain would be from V_{in} to V_{out} and not the gain seen by C_{G D}. The reader can readily express the transfer function from node X to V_{out} as

\frac{V_{out}}{V_x}(s) =\frac{(C_{G D}s − g_m)R_D}{R_DC_{G D} + 1} (6.32)

observing that this gain begins to roll off at a higher frequency, namely, at 1/(R_DC_{G D}). Thus, the multiplication of C_{G D} by 1 + g_m R_D is still justified.