# LC filter out of the loop if crossover frequency is below resonant frequency?

Referring to slides 78-84 of http://powersimtof.com/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202009.pdf, the RLC network rings alone and the oscillations are not due to the loop.

It is my understanding that Bode plots are always valid for stability assessments for minimum-phase systems and say it all about stability. The compensated buck in question doesn't contain RHP poles, RHP zeroes or time delays and thus is a minimum-phase system. So anything that has to do with stability should be seen in the Bode plot in this case.

From the magnitude and phase it looks like the open-loop gain never really approaches the -1+0jw point (if I visually map them into a Nyquist plot), so the system should be very stable. I understand the gain at the resonant frequency drops well below 1 and the system cannot fight the fast disturbances. It is mentioned the dc feedback is there but the ac feedback is gone. But the RLC is still physically in the loop, so why am I not seeing this instability somewhere in the Bode plot? Is the Bode plot not valid here and if not, why not if all the conditions for a minimum-phase system are fulfilled? Does a Bode plot not say everything about stability for minimum-phase systems?

I am glad you picked up this slide because it can generate confusion with a perfectly-stable converter. Actually, the step response of a switching converter operated in open- or closed-loop, depends on its output impedance $$\Z_{out}\$$. We assume here small-signal excitation and a linearized system of course. Let's have a look.

For the following illustrations, I have used a buck operated in voltage-mode control available from my free ready-made templates simulating on SIMPLIS and its demo version Elements. Below is the typical control-to-output transfer function of a buck converter: You see a resonance located at 576 Hz and it makes sense considering a 100-µH inductance with a 680-µF capacitor. The peaking depends on the various losses in the circuit (load current, switching losses, diode dynamic resistance and so on). Now, if I plot the open-loop output impedance, this is what I have: Nothing mysterious, this is the output impedance of the $$\LC\$$ filter: the inductor ohmic loss $$\r_L\$$ dominates at dc, then the inductance shows up until resonance occurs. After the peak, the output capacitance decreases the impedance which flattens out to $$\r_C\$$ at high frequencies. If you step this open-loop converter with a current source, you obtain a damped oscillatory response centered around the 5-V dc output: If we now close the loop, we can show that perturbations such as the input voltage or the output current are rejected by the loop gain $$\T_{OL}(s)\$$. Below is an excerpt from APEC 2012 seminar: You can see how the open-loop output impedance is affected by the sensitivity function which changes its value according to the open-loop gain $$\T\$$. So if there is a large gain, then the output impedance is small and the drop on the output remains acceptable. On the contrary, if the loop gain is very small, then there is no action on the output impedance. And this is what happens when you shape the loop gain according to the compensation strategy (the poles and zeroes you place in the compensator): you force the gain to crossover at a certain frequency $$\f_c\$$. Before this value, there is gain and the output impedance is under control by the loop but beyond $$\f_c\$$, alas, there is no gain anymore and the converter runs open-loop in ac.

Remember, "no gain - no feedback" so if the loop is open in ac, you have the response of the $$\LC\$$ filter I gave in the beginning and there is nothing the converter can do. The output is stable, it regulates the dc voltage but it can't fight the ringing which is not a loop issue per se.

If I now close the loop with a 5-kHz crossover frequency and compare the closed-loop output impedance with the previous open-loop plot, this is what I see: The impedance is almost flat and if I excite this converter with a load step, the ringing has disappeared: In 2019, I presented another APEC seminar in which I actually built a converter purposely wrongly compensated by an integrator, as in the document you referred to. As you can see in the measured Bode plot, the loop is perfectly stable but, as in the above example, the stepped response rings because of the absence of gain at resonance: The loop is closed in dc but open in ac, hence the poor response. And it has nothing to do with loop stability ^_^ Hope this answer helps you understand this phenomenon.

• Thank you. So stability is responsiveness to perturbations. The loop gain ToL(s) in sensitivity function 1/(1+Tol(s)) then determines how well different perturbations are suppressed. One perturbation type is an output current step, which is tranferred into Vout by Zoutol(s)/(1+T(s)). As the inductor and capacitor are physically present in open loop, they would be seen in the Bode plot but only as a peak well below 0dB and beyond fcrossover. So at fres Zoutol cannot be decreased by 1/(1+T(s)) as |T(s)| <1 and the disturbance is not surpressed. Is that correct?
– Hyp
Aug 18 at 11:09
• And another type of disturbance is a step of Vref, which is suppressed by Tol(s)/(1+Tol(s)). And Vref as loop input seems to be used to derive the quality factor for 2nd-order systems based on phase margin. The quality factor then determines the time response to a Vref step. Other disturbances such as a Vin step would affect Vout depending on where and how they are introduced into the loop and how Tol(s) affects them.
– Hyp
Aug 18 at 11:29
• @Hyp, a step on $V_{ref}$ is unlikely to happen as, per definition, it is a stable dc source, silent in ac. So yes, a switching regulator will respond to changes in $V_{in}$ and $I_{out}$ unless you purposely sweep the reference voltage and would like the output to follow. Keep in mind that my example is just there to highlight the necessity to crossover after $f_0$ in this voltage-mode buck converter and not before as incorrectly suggested in Fig. 16 of this doc. If my answer pleases you, thank you to acknowledge it. Aug 18 at 11:54
• I understand Vref in normally stable but I interpreted from here, sl.34 that the Q-PM analogy ends up having T/(1+T), which multiplies Vref. Unlike Vin, which is multiplied by Asc/(1+T) acc. to this, p.1. So I assumed the analogy uses Vref, not Vin as input signal. Is this correct?
– Hyp
Aug 18 at 12:53
• Correct, most of the stability studies in the literature assume $V_{ref}$ as the stimulus input which is not the case for our switching regulators where it is 0 V in ac. This is what I explained in my 2012 seminar where I showed that the output impedance truly mattered in the end. I too was confused when I started looking at it many years ago. Aug 18 at 13:47