# Why do we use closed-loop feedback in buck converters?

My power electronics professor spent most of his life in the time domain and in application-specific theory (SMPS-only vocabulary).

For me, a buck converter is basically a square wave (generated by a switch) passed through an LC filter.

The square wave's fundamental frequency and its harmonics are filtered out, and all that's left is the DC component. I understand that closed-loop feedback is necessary if the Vin changes - because you'll need to increase or decrease your duty cycle since your square wave's amplitude (ergo DC component) has changed. Is that it?

I also understand that the load impedance might affect the corner frequency of the filter. Is that why we need closed loop feedback? How is this issue mitigated?

Which other components of the (supposedly) DC signal does the control loop eliminate? Ripple? (Isn't this just a matter of filter quality?)

• For guaranteed CCM, you can sometimes get away with straight Vin to duty-cycle control. Problem is that you seldom can guarantee this and closed loop comes with almost no cost compared to the other buck components. – winny Oct 17 '17 at 11:08

The main thing you are missing is that what is put into the LC filter is not necessarily always a square wave. It is when the buck converter is in continuous mode, but unless you know that to always be the case, you can't assume the square wave input to the filter as you are.

In continuous mode, the output voltage is ideally the input voltage times the duty cycle. However, it's not that simple in the real world. Even if the input voltage stays constant, there is the DC resistance of the inductor to consider, the voltage across the switch, and the voltage across the diode from ground during the pulse low time.

The latter can be mitigated by synchronous rectification, but that isn't perfect either. At the least, there is the voltage drop across whatever is being used as the synchronous rectifier switch. Synchronous rectification timing is also usually made conservative, meaning it errs on the side of staying on a little too short rather than too long. The cost of turning off a little early is more voltage drop at the end of the flyback part of the pulse. However, the cost of turning on too late is shoot thru, which rapidly decreases efficiency, and risks damaging parts.

I have seen pre-regulation power supplies which were fixed duty cycle buck switchers. In one case, it was used to drop a 48 V distribution voltage down to a rough 12 V, which was distributed locally and dropped to the final regulated voltages by other power supplies. It didn't matter if the 12 V varied a bit.

A general purpose power supply has to be designed to handle low load too. Below some load for any switching frequency, a buck switcher can't maintain continuous mode. Some OEM supplies simply state a minimum load is required.

More general purpose supplies fall back to discontinuous mode. In that case your fixed square wave assumption fails. Now there are really 3 parts to the cycle. At the start, the input to the LC filter is actively driven high. When that stops, the flyback part begins, which drives the input actively low. Then there is the third phase in discontinuous mode where you consider the input effectively high impedance. The function of duty cycle to output voltage is not longer linear.

• I did not consider DCM at all. – Andrew Pikul Oct 17 '17 at 12:19
• " Below some load for any switching frequency, a buck switcher can't maintain continuous mode. Some OEM supplies simply state a minimum load is required. " This is only true of asynchrnous buck converters. A synchrous converter can maintain CCM down to zero load but doing so comes at the cost of higher losses. – Peter Green Oct 17 '17 at 17:39
• @Peter: In that case current actually flows backwards thru the inductor, something that is usually carefully avoided. At that point you don't really have a buck converter anymore, but a "DC transformer". – Olin Lathrop Oct 17 '17 at 17:51

A buck converter can indeed be visualized as a low-impedance square-wave generator feeding a low-pass filter combining an inductor $L$ and a capacitor $C$. However, as you can imagine, when the power switch closes, $V_{in}$ is not the value applied to the left-side inductor terminal. The input source undergoes a voltage drop inherent to the power switch $r_{DS(on)}$ and the inductor ohmic loss $r_L$. As a result, the on-state inductor voltage is not $V_{in}-V_{out}$ but less than that as shown in the left-side picture:

During the off-time, in continuous conduction mode or CCM, the left terminal of the inductor does not drop to 0 V but to the diode forward drop which forces the node to swing below ground. Therefore, when you apply the volt-second balance law for the inductor, you realize that the full output voltage formula including these losses differs from the simple one in CCM, $V_{out}=DV_{in}$. You could further complicate the expression by including the diode recovery time and the switch turn- on and off losses.

Practically speaking, as you said, a CCM-operated buck converter with 0 parasitics and operated at a constant input voltage would not need a loop to maintain its output operating point. However, as you can see, several parasitic elements affect the dc transfer function and a control loop needs to correct the control voltage forcing the output voltage to meet the target. The load resistance will affect the corner frequency but very marginally actually, involving $r_L$ and $r_C$. The loop is there to truly make the regulator (the setpoint is fixed) immune to external perturbation like the input voltage and the output current. See the below picture:

You see the effect of the loop on several parameters:

• the output voltage: obviously you want a precisely-regulated $V_{out}$ so you need gain in your loop (no gain, no control system) to a) reduce as much as possible the static error b) ensure a fast-reacting system to a sudden power demand c) make the system robust to external perturbations.
• the output impedance: as you can see, the output impedance is hampered by all the parasitics like the $r_{DS(on)}$, the ohmic losses etc. The small-signal response to a step is dictated by the output impedance. You thus want this impedance to be of sufficiently-low value to make sure the output drop when the load current changes remains reasonable. The gain of the loop will work to reduce the output impedance by the sensitivity function $S=\frac{1}{1+T(s)}$ in which $T$ is the loop gain.
• same for the other perturbation, $V_{in}$. When you have $V_{out}=DV_{in}$ you can see that if you differentiate $V_{out}(V_{in})$ with respect to $V_{in}$ you obtain $D$. That means that any static change in the input voltage will be propagated to the output by $D$. Not very good. Again, the addition of the loop will improve this input voltage rejection or audiosusceptibility by the sensitivity function.
• I was unaware of a sensitivity function. – Andrew Pikul Oct 17 '17 at 12:20
• You can have a look at a seminar I gave in 2012 at an APEC conference: cbasso.pagesperso-orange.fr/Downloads/PPTs/… The sensitivity function is analyzed at the end with the concept of modulus or magnitude margin. – Verbal Kint Oct 17 '17 at 12:43

You are assuming Switch Mode Power Supply (SMPS) uses Pulse Width Modulation (PWM) to pass an average voltage level and the LC filter removes the switching part to leave that average voltage. However that is not how they work.

An SMPS uses PWM to pass energy from a source to store it onto a capacitor so that the voltage level on that capacitor is as defined by the feedback circuitry.

As the load changes, and requires more or less energy, the SMPS changes how fast that energy is transferred to keep that capacitor at the target voltage. If the load goes away completely the PWM can actually stop.

If your load is fixed and your input supply is also fixed, then some steady state PWM operation will happen, but that is actually quite rare. If you attempt it without feedback, ANY difference in load or source will cause the output voltage to drift one way or the other over time since the energy transfer will be either too high or too low.

• Actually, in continuous mode, the output does roughly equal the input times the duty cycle. – Olin Lathrop Oct 17 '17 at 11:28
• @OlinLathrop yes true, roughly, but roughly means without feedback you can never guarantee the level you need. – Trevor_G Oct 17 '17 at 11:32
• I thinks that's the OP's question though. He sees that ideally output voltage is input voltage times duty cycle, and wonders why feedback is needed. The two main answers are 1: To adjust for the real world non-idealities, and 2: Because it doesn't work at all in discontinuous mode. – Olin Lathrop Oct 17 '17 at 11:37
• @OlinLathrop yes, good point. – Trevor_G Oct 17 '17 at 11:46

The above answer is really great.Thanks.

Among the non-ideality effect from switch and diodes and DCM, I think one of the reason is the transient response. To have fast transient response, you must have a high cross over frequency which makes response fast. But the LC filter actually cut the 0dB in several kHz. Usually you want your cross over frequency to be as high as possible but cant over half of the switching frequency as the matter of Nyquist rate. So you need the feedback the give you some gain so that you can make cross over frequency to be about hundred something kHz.

If one designed a buck-mode switcher with synchronous switches rather than diodes, and if the switches could convey current in both directions, then the supply would convey power from the supply cap to the load cap when its output voltage is less than half the input voltage, and from the load cap to the supply cap when it's greater, thus achieving somewhat sloppy (but perhaps useful) regulation. If nothing draws current from the load cap, then a system driven with 50% duty would stabilize toward a mode which would:

1. Feed current from the load cap to the supply cap for the first quarter of each cycle, using stored energy in the inductor to drive the current against the potential difference.

2. Feed current from the supply cap to the load cap during the next quarter, while charging the inductor with energy from the potential difference.

3. Continue sourcing current into the load cap (with the supply disconnected) during the next quarter, using stored up energy in the inductor.

4. Draw current from the load cap (again with the supply disconnected) during the last quarter, storing that energy in the inductor.

If the switches can all operate in both directions, the system would be be stable in this pattern. If, however, one or both switches can only operate in one direction, any energy transferred into the inductor from the source would have to either get transferred to the load cap or be dissipated as heat somewhere. The amount of energy the inductor receives from the source in an "on" cycle will depend upon how much current was flowing through it initially, but if initial current can't be negative the energy received in an on cycle will have a non-trivial minimum. If there is nowhere for that energy to go, the "on" time must be cut back.