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

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?

• Shouldn't the control loop cut-off frequency be significantly lower than the LC resonant frequency in order to prevent wild and excessive output voltage perturbations? Oct 15, 2022 at 11:45
• Is that with the open loop, or closed loop? Because only for the former it is correct to say what is quoted from the book. What does the book say (maybe further on)? Oct 15, 2022 at 12:05
• @verbalkint - this one might be for you. Oct 15, 2022 at 13:20
• This is a voltage-mode control? Oct 15, 2022 at 14:11
• If you have ringing at the LC frequency, you don't have enough open loop gain at that frequency. @TimWilliams I'm guessing you're right, that this is voltage-mode control. Oct 15, 2022 at 17:26

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.