Electronic – Bode plot of poles that are close to each other

Calculated values for the poles and zeros look about right, except for the missing integrator pole at the origin. Maybe it would be best to break the loop up into pieces. Write the transfer function of the error amp, from \$V_{\text{out}}\$ to error amp output to start. You should get something like:

\$\frac{\text{EA}_{\text{out}}}{V_{\text{out}}}\$ = \$\frac{\text{R2 } \text{EA}_{\text{gm}}}{\text{R1}+\text{R2}}\$ \$\frac{\text{Cs } \text{Rs } s+1}{s (\text{Cp}+\text{Cs}) \left(\frac{\text{Cp } \text{Cs } \text{Rs } s}{\text{Cp}+\text{Cs}}+1\right)}\$

Which for the values shown in the schematic (except, using R1=10kOhms and R2=1kOhm, since values for those were not shown) would look like:

Then write the output filter response. It would be like this:

\$\frac{V_{\text{out}}}{V_{\text{sw}}}\$ = \$\frac{\text{Co } \text{esr } \text{Ro } s+\text{Ro}}{s^2 (\text{Co } \text{esr } \text{Lo }+\text{Co } \text{Lo } \text{Ro })+s (\text{Co } \text{esr } \text{Ro }+\text{Lo})+\text{Ro}}\$

Which would look like this:

Of course the total loop would also have the modulator gain which would be ~ \$V_{\text{in}}\$ / \$V_{\text{ramp}}\$. Also left out are the Nyquist poles.

You can see that the loop put together will be unstable unless another zero is added to the error amp response. It might also be necessary to damp the LC filter since with such a small amount of esr the Q is quite high. Under damped filters are a common problem with this type of regulator.

About Infinite Gain Margin, and what it means when Phase doesn't cross -180.

I don't think infinite gain margin is meaningful or realizable. A situation like that shows a limitation or inaccuracy of the model: Poles or loop delays that haven't been included, like Nyquist poles and higher frequency poles that exist in the error amp (both of which have been left out here). Or, it might show an error of interpretation, that the plot is not of Phase, but Phase margin, which wouldn't cross -180 but instead 0 degrees.

First check the transfer function for poles and zeros. In this case the numerator is a constant, so there are no zeros. The poles are the zeros of the denominator. By inspection we see that they are at s=0, s=-0.1, s=-100. For each pole we get additional -20dB/dec and additional -90 degrees phase shift.

The gain starts to change at the corresponding positive value of the pole. E.g. for s=-100 we will see a change at w=100. The phase changes from about one decade before this frequency to about one decade above this frequency by -90 degrees. Exactly at the corresponding frequency the change because of this pole is -45 degrees.

Before we can draw the diagram we need to find a starting point. This is a little bit tricky because of the 1/s term.

We can rewrite the equation and deal with the parts separately $$ G(s) = \dfrac{48000}{s(s+0.1)(s+100)} = \frac 1s \cdot \dfrac{48000}{(s+0.1)(s+100)} $$ Now we have a 1/s-term that is infinity at zero and 1 at w=1. It will decrease by 20dB per decade. Now we look at the other part $$ \dfrac{48000}{(s+0.1)(s+100)} $$ For s=0 the gain is 48000/(0.1*100) = 4800. At w=0.1 it will start to decrease by 20dB/dec, at w=100 it will start to decrease by 40dB/dec.

Now we know everything about the components and can construct the Bode plot.

The result looks as shown below. The 1/s term is red. The other term is blue and the sum of these two is yellow.

Starting at w=1, we have 0dB for the 1/s term.

The other term has 20*log10(4800) ~ 73dB left to the corner frequency and at w=1 it is 20dB below that value, so we have 53dB.

This is our first point! At this point we have a slope of -40dB/dec. This slope will extend one decade to the left (so we have 93dB there) and then continue with -20dB/dec. To the right it will extend up to w=100 and then continue with -60dB/decade. At w=100 the magnitude is 53dB - 2decades*40dB/dec = -27dB.