I'm confused about the following discussion of the Miller effect/approximation from Razavi (image below). I don't understand the premise of the question. If one reasons that the gain seen by the capacitor at the pole would have dropped by \$3 dB\$ at the pole frequency (even if as the example points out, it doesn't), wouldn't one then decide that a better approximation would be to multiply \$C_{GD}\$ by \$1+g_m R_D \frac{1}{\sqrt2}\$ to reflect the reduced gain?

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  • \$\begingroup\$ If one reasons that the gain seen by the capacitor at the pole What are you trying to say here? I think you're mixing things up. The Miller effect makes /$C_{GD}/$ appear as if it is a much larger capacitor. That shifts the frequency of the pole down. The -3dB point is the frequency point in the transfer function that corresponds with that pole's frequency. Realize that at DC (0 Hz) nothing changes. You mention a factor \$\frac{1}{\sqrt2}\$ which corresponds to -3 dB but since the gain at DC does not change, you must not include \$\frac{1}{\sqrt2}\$... \$\endgroup\$ – Bimpelrekkie Feb 11 at 19:42
  • \$\begingroup\$ The factor \$\frac{1}{\sqrt2}\$ or -3 dB comes from \$\frac{1}{\omega s}\$. Compare it to the transfer function of a first order RC lowpass filter, there we also do not add a factor \$\frac{1}{\sqrt2}\$ even though the attenuation at the cutoff frequency is -3 dB (a factor \$\frac{1}{\sqrt2}\$) \$\endgroup\$ – Bimpelrekkie Feb 11 at 19:42
  • \$\begingroup\$ @Bimpelrekkie I agree the DC gain is unaffected. By "gain seen by cap at the pole", I mean that there are two "gains" in this question: the gain from \$V_{in}\$ to \$V_{out}\$ and the gain from node \$X\$ to \$V_{out}\$. The second one (call it \$A_c\$) is what I'm calling the gain seen by the cap and that's the one that actually should get used in the miller approximation. The example points out that the miller approximation uses the DC gain instead of the gain evaluated at frequency of interest. \$\endgroup\$ – knzy Feb 11 at 19:54
  • \$\begingroup\$ My confusion is why - knowing that \$A_c\$ actually decreases as frequency increases - one might try to adjust for the reduced gain by increasing it instead of decreasing it. The example starts out with a train of thought that's incorrect for one reason (pointed out at the end), but my point is that I don't understand a secondary, supposedly correct part of that train of thought. \$\endgroup\$ – knzy Feb 11 at 19:55
  • \$\begingroup\$ OK, \$A_c\$ is factor by which \$C_{GD}\$ is made larger and \$A_c\$ is simply the gain at DC. The pole seen in the transfer function of \$X\$ to \$V_{out}\$ also shifts down in frequency with a factor \$A_c\$. knowing that Ac actually decreases as frequency increases Why does your \$A_c\$ decrease over frequency? My \$A_c\$ is simply the gain (from \$X\$ to \$V_{out}\$) at DC, \$A_c\$ (= \$gm * R_D\$) is constant over frequency. The transfer function of \$X\$ to \$V_{out}\$ does change over frequency. \$\endgroup\$ – Bimpelrekkie Feb 11 at 20:20

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