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There is an example in Razavi's textbook (Design of Analog CMOS Integrated Circuits 2nd edition) demonstrating the Miller effect/approximation, shown below.

miller effect

As you can see the example claims that the Miller output capacitance is given by \$C=\frac{C_F}{A+1}\$. However, I think that the correct value for this capacitor should be \$C=C_F (1+A^{-1})\$, since the Miller approximation describes the output impedance as \$\frac{Z_0}{1-A^{-1}}\$, where \$Z_0\$ is the impedance before the transformation.

This isn't listed in the most current errata that I'm aware of. I'm also aware that part of the point of this particular example is to show that the Miller approximation can introduce errors like extra poles/fewer zeros, but either way I would just like to confirm if the capacitor value here is an error or not. (Contrary to the figure above, the rest of the example actually uses the value \$C=\frac{C_F}{1+A^{-1}}\$, which still seems incorrect.)


Update: Prof. Razavi has confirmed to me via email that these are typos. As of this update the errata haven't yet been updated.

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2 Answers 2

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You're correct, it's a typo on his book. Which version do you have? he didn't correct it in the 2016 one?

We can prove it easily by writing a KCL equation on his simplified schematic (I used Vin instead of "X"):

$$ \frac{V_{out}-V_{in}}{Z_{C_f}} = I_{out} $$

If we let \$V_{in}=-\frac{V_{out}}{A}\$ and \$Z_{C_f}=\frac{1}{sC_f}\$ and do some arithmetic manipulations, we arrive at:

$$ \frac{V_{out}}{I_{out}}=\frac{1}{sC_f(1+A^{-1})} $$

In short, \$C_f\$ is scaled by a factor of \$\frac{A+1}{A}\$, or, in other words, its value remains almost the same when seen from the output at low frequencies and assuming \$A\$ is large.

EDIT:

2 Sanity checks.

  1. With LTSpice

I setup the following schematic using an ideal gain of 100 and a coupling capacitor of 100pF.

schematic

These are the results I got:

AC results

The one with "V(out_1)/V(in1)" is the approximation I used with the formula I derived above.

I realized I used positive feedback in the examples above, but that's not important as the aim was to just only estimate the poles, not designing a proper amplifier.

  1. Razavi himself writes this formula in one his video lectures (the second he does is already included in the link)
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  • \$\begingroup\$ Thanks for helping to confirm - and yeah, the inside cover of this book lists the edition as 2017. It would be great to know if there are any more up-to-date errata for the book from the publisher or somewhere else, but I haven't been able to find any. Unfortunately the errata I shared is undated. \$\endgroup\$
    – Halleff
    Apr 28, 2023 at 21:52
  • \$\begingroup\$ @knzy I heard Razavi is quite fond of receiving corrections to his book when it comes to typos as the one you found. Maybe you can try his UCLA e-mail? \$\endgroup\$
    – Designalog
    Apr 28, 2023 at 22:04
  • \$\begingroup\$ Thanks for the suggestion and additional details, I'll try emailing him about this, perhaps once some of the discussion in this post dies down. On that subject, quick clarification about your update - in the schematic, are the values of C3 and C5 \$C_C/(A+1)\$? Maybe this is related to your note about using positive FB, but these should be \$C_C(A+1)\$, no? \$\endgroup\$
    – Halleff
    Apr 29, 2023 at 20:27
  • \$\begingroup\$ @knzy you're right, see my edit. Now the one with correct formula fits pretty well the true response until the zero happens. \$\endgroup\$
    – Designalog
    Apr 30, 2023 at 8:10
  • \$\begingroup\$ Sounds good, and the graph with all three transfer functions is easier to compare as well. Although in retrospect, I think the transfer function \$V_O/V_X\$ (which is what's being depicted in the plots) for the Miller approximation case shouldn't depend on the value of the input Miller capacitance anyway since once \$C_F\$ is split the input and output are uncoupled. \$\endgroup\$
    – Halleff
    Apr 30, 2023 at 18:19
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The circuit of Example 6.4 can be transformed into the Miller equivalent circuit with component values that give identical transfer functions and capacitances seen from input and output for both circuits, original and Miller-equivalent.

The input Miller impedance is placed at the amplifier input and calculated as usual, with the feedback impedance including the amplifier's output resistance: \$Z_{fb}=1/(jωC_f) + R_{out}\$, \$Z_{M}=Z_{fb}/(1+A)\$. The output Miller impedance \$Z_{Mout}\$ is a capacitor and a resistor connected in series. \$Z_{Mout}\$ is connected to the node Vout, where the output resistor Rout and the feedback capacitor Cf are connected. \$Z_{Mout}\$ replaces the capacitor \$C_{F}/(A+1)\$ shown in Figures 6.9(a) and (c). The calculation of an output Miller impedance value is complicated because of non-zero output resistance. The idea is to calculate the output voltage \$V_{out}\$ in Figure 6.9(a) and use the equality of currents through \$R_{out}\$ and the output Miller impedance \$Z_{Mout}\$. The calculation gives \$Z_{Mout}=1/(jωC_f(1+A^{-1}))-R_{out}/(1+A)\$. Notice the negative resistance of the output Miller impedance. You can see the Miller equivalent circuit in the LTspice schematics below.

The AC analysis runs produce identical graphs of transfer function for both circuits, original and Miller-equivalent.

Figure

Summing up, the transfer function of the Miller equivalent circuit and capacitances seen from input and output exactly replicate those of the original circuit with feedback. The input Miller impedance is \$Z_{M}=Z_{fb}/(1+A)\$ where \$Z_{fb}=1/(jωC_f) + R_{out}\$; the output Miller impedance is \$Z_{Mout}=1/(jωC_f·(1+A^{-1})) - R_{out}/(1+A)\$.

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  • \$\begingroup\$ I never understood why Miller's approximation (as said by Razavi) was called a theorem. If anyone wants to use the theorem strictly, then you'd also need to compute the gain as a function of frequency as well. And this so-called "theorem" removes the zero out of the transfer. Hence my resistance to call it "theorem". \$\endgroup\$
    – Designalog
    May 1, 2023 at 8:18
  • \$\begingroup\$ A purely resistive component \$R_{out}\$ of the feedback impedance \$Z = 1/(jωC)+R_{out}\$ gives you the coveted frequency-dependent contribution into the gain. Agreed, it is overkill to pile up complex impedances for presumably intuitive circuit analysis, but there comes SPICE to the rescue. \$\endgroup\$
    – V.V.T
    May 1, 2023 at 20:43
  • \$\begingroup\$ As for the "theorem", you may call it any way, but a good sign is that it makes you notice how seemingly insignificant circuit transformation can distort the pole/zero arrangement, and this realization introduces you to a pole splitting technique. The good textbook might have a typo, but even its typos urge astute students to study in depth. \$\endgroup\$
    – V.V.T
    May 1, 2023 at 20:57

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