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I recently came across a rather nice straight-to-the-point application note by RickTek (AN028) on how to calculate approximately the compensation network of a PCM buck regulator.

Link : https://www.richtek.com/~/media/AN%20PDF/AN028_EN.pdf

However I think I need some explanations. The application note gives an example of design calculations.

At step 4 (page 9), the following equation is given for finding stability margin at crossover frequency \$f_c = 34kHz\$ : $$\Phi_{M} = \Phi_{fc} + 180 - 90 + atan(f_c/f_z)\cdot 180/\pi - atan(f_c/f_p) \cdot 180/\pi $$

At step 5, gain is calculated at crossover frequency : $$ G_A = -G_{fc} - 20\cdot log(V_{ref}/V_{out}) + 20\cdot log(ceil(f_c/f_p))-20log(ceil(f_z/f_c)) $$

My questions :

  • What are \$\Phi_{fc}\$ and \$G_{fc}\$ expressions? How are they calculated in order to obtain the given values for \$\Phi_M\$ and \$G_A\$? The application note doesn't mention where these expressions come from and how they are computed it seems.
  • From where those two equations come from? It's obviously taken from some sort of transfer function, I'm just not sure which?

It would be logical those equations would be derived from the given open loop transfer function \$G_d(s)\$ of a PCM buck converter at \$s = j\omega = j2\pi f_c \$ (i.e. at crossover frequency), but I'm not sure how to pass from this expression to the equations mentioned above.

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  • \$\begingroup\$ This question relies on the link staying active so that people can refer to the text to understand how your formulas are derived. If that link fails (as they can quite often), your question also fails so, maybe you can paste the relevant sections into your question or improve your question in some way so that it doesn't rely on the link. \$\endgroup\$
    – Andy aka
    Commented Apr 28 at 19:59
  • \$\begingroup\$ I agree with you. I will enhance the question once I understand better. My current confusion somewhat makes it harder to understand what's revelant and what isn't. I took care to at least mention the application note number, which is probably more time-proof than a link in the meantime. \$\endgroup\$
    – Yannick
    Commented Apr 28 at 20:04
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    \$\begingroup\$ I recommend you take a look at my APEC 2021 seminar, you may find your answers in it. Your first equation simply determines how far you are from 180° or 360°. You start from the phase of the power stage at crossover, then add the phase of the inverting integrator, then the phase brought by the zero and pole of the compensator. This is what I show on slide 18. For the second expression, it seems to be the needed mid-band gain to crossover at 34 kHz. \$\endgroup\$
    – Verbal Kint
    Commented Apr 28 at 20:35

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What are \$\Phi_{fc}\$ and \$G_{fc}\$ expressions? How are they calculated ...

Using Equation 1 \$G_d(s)\$ with \$s=j\omega\$

\$\Phi_{fc}\$ is the \$\angle G_d(j\omega)\$ when \$\left|G_d(j\omega)\right|=0dB\$

\$G_{fc}\$ is the \$\left|G_d(j\omega)\right|\$when \$\angle G_d(j\omega)=-180^\circ \$

From where [do] those two equations come from?

Then the uncompensated phase margin is
\$\Phi_{Mu} = \Phi_{fc}+180^\circ\$

The gain in (Eq 12), \$G_A\$ is the gain at the phase crossover, so looks more like a gain margin. So the uncompensated gain (margin) in dB is
\$G_{Au}dB = -G_{fc}dB\$

Then the compensated phase (margin) is: $$ \Phi_{M} = \Phi_{Mu} - 90 + \text{atan}(f_c/f_z) 180/\pi - \text{atan}(f_c/f_p) 180/\pi $$

and the compensated gain (margin) is: $$ G_A = -G_{Au} - 20\text{log}(V_{ref}/V_{out}) + 20\text{log}(ceil(f_c/f_p))-20\text{log}(ceil(f_z/f_c)) $$

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    \$\begingroup\$ I will accept the answer, but the method simply doesn't work or is not clear enough to apply it properly. Gives capacitor values in the nF range which is much too high and limits bandwidth and thus transient response. Capacitors in the pF range work better. Besides I never got able to get the desired crossover frequency starting when having Rcomp=inf and Ccomp=0. Tested in LTSPICE with LT8646s. What you said makes sense though in the context. An application node AN149 made by LT gives much better insights without risking giving equations that prove useless. \$\endgroup\$
    – Yannick
    Commented May 15 at 1:26

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