Find out the cut-off frequency directly from the transient response of buck converter

Question

In Chapter 4 of the book Designing Control Loops for Linear and Switching Power Supplies A Tutorial Guide, the author designed a buck converter using a PID controller. The cut-off frequency of the buck converter is 10 kHz, and the phase margin is 80°. The transient response of the buck converter is as follows:

Then the author said:

If you look carefully, the oscillations do not correspond to a 10 kHz signal (our crossover point) but to the 1.2 kHz LC network resonant frequency.

Because the system is a second-order system, I calculated the resonant frequency is 1.2 kHz using the equation:

$$f_0=\frac{1}{t_d \sqrt{1-ξ^2}}$$

What puzzles me is that from what I understand, the author means that if you design a good system, the resonant frequency of the transient frequency should be the cut-off frequency. Could anyone explain why?

Answer 1

In this example, I wanted to show that if you blindly compensate a voltage-mode buck converter using a PID by placing two complex zeroes for compensating the complex double poles of the \$LC\$ network then you end up with an oscillating response as you show.

What dictates the small-signal response of a switching converter operated in closed loop is its output impedance. For an open-loop voltage-mode-controlled buck converter, this output impedance \$Z_{out,OL}\$ is resistive, then inductive, resonates and then capacitive to eventually land on the capacitor ESR. Once you close the loop, the stepped output current - which is seen as a perturbation - is rejected by the sensitivity function \$S=\frac{1}{1+T(s)}\$ in which \$T(s)\$ is the compensated loop gain. The closed-loop output impedance \$Z_{out,CL}\$ becomes \$Z_{out,CL}=\frac{Z_{out,OL}}{1+T(s)}\$. This is what the below graph shows:

However, because of the flaw in the compensation scheme, you can see the original \$LC\$ filter peaking is not damped at all and shows up despite the loop closure. This is because there is not enough loop gain at \$f_0\$ and any excitation in the output reveals the oscillatory response you've shown. This is not a loop instability - the Bode plot shows adequate margins - but the system does not have enough gain at the resonant frequency and cannot fight the perturbation: no gain, no feedback!

The proper compensation strategy for the buck operated in voltage mode is to select a crossover frequency that is 3-5 times higher than \$f_0\$, making sure the system will have enough gain to fight the \$LC\$ oscillations. When this is the case, the closed-loop output impedance no longer shows peaking as shown below:

The violet curve shows the open-loop output impedance while the red curve illustrates a closed-loop output impedance with a very low crossover, well below \$f_0\$: the response of this converter will obviously ring considering the untouched peaking at \$f_0\$. Now, push crossover to 4 kHz and the black curve shows how the new closed-loop output impedance looks like without peaking anymore, bringing a good transient response this time.