# Why does output voltage vary with input voltage at steady state in a switching converter?

All switching converters have feedback which varies the duty cycle so that the output voltage is maintained at a specific voltage. However, the feedback is not perfect and there is still some variance in the output voltage.

I am particularly curious about the line regulation - the ability for the converter to maintain the output voltage with a varying input voltage. Albeit, it is normally very small, what actually causes the line regulation error? Shouldn't the feedback be able to perfectly compensate for the change the input voltage by adjusting the duty cycle?

I can understand why the absolute desired voltage might be off from the actual voltage because of inaccuracies such as the reference voltage used for the error amplifier not being perfect. But I cannot understand why a change from 3V input gives a slightly different output voltage than a 6V input, when all the PWM controller has to do is modify its duty cycle in a continuous analog fashion, without any discrete steps.

I attached, as an example, a line regulation plot from a TI TPS62840 buck converter module. Any insight would be greatly appreciated.

• First thing I'd like to make sure of. Are you asking about why there is a $\pm 400\:\text{ppm}$ variation on the output over various output loads when the input varies $\pm 2\times 10^6\:\text{ppm}$? Maybe I'm just reading the chart wrong.
– jonk
Nov 20, 2022 at 20:53
• More or less, yes. I know it sounds silly given that the output voltage change is so small, but I want to understand why there is any change at all if the feedback is capable of adjusting itself in a continuous analog fashion Nov 20, 2022 at 21:09
• It's just that no matter how much gain you create, it's always finite! There will always be some kind of small variation like this. It's just not possible to create ideal circuits. One could consider integrating the error and pushing. But this would greatly retard the response and make it sluggish to the point of uselessness. Reality just isn't perfect. Engineering is about making the right trade-offs.
– jonk
Nov 20, 2022 at 21:11

Any insight would be greatly appreciated.

Any feedback system that performs "regulation" compares its output with a known fixed value and, attempts to minimize the errors. Thus the output becomes the same (or a fixed ratio) to the known fixed value (aka the demand) thus: -

Output equals demand

That's the simple theory but, to drive the output towards the demand requires an error and, what you are trying to do ideally is minimize the error. Do you see the problem?

You cannot have infinite gain and so, you accept that have sufficient gain to create a steady-state error in the output that is acceptable i.e. it doesn't precisely cause output to equal demand but, we "live with" the error.

If you look at your figure 4 (in the question) you could work how much the error amplification is. For an input range of 2.5 to 6.5 volts (4 volts delta), the output shifts from 1.7994 volts to 1.8006 volts (a delta of 1.2 mV). Thus, you can say that the error amplifier has a net gain of 4 volts ÷ 1.2 mV i.e. 3333.33.

There are more complexities to this but that's a broad estimation of the DC error gain.

Then there are dynamics to consider; if the input voltage has a lot of ripple voltage on it, the error amplifier gain is inevitably smaller (think of the open loop gain of an op-amp and how it starts falling above a few hertz). So when you say this....

Shouldn't the feedback be able to perfectly compensate for the change the input voltage by adjusting the duty cycle?

....There are things to consider.

• The first is that any input change won't drive the error amplifier to produce a steady-state error immediately because of the limited dynamics of the error amplifier.
• The second thing to consider is that a switching regulator is fundamentally delayed by up to one switching cycle. In other words it can't begin compensation immediately. This of course can cause a significant disturbance in the output if the error amplifier is not "well-tempered" and the input disturbance is profound.
• The third thing to consider in a switching regulator is that there is an enormous phase shift caused by the LC circuits (used to transfer energy). The phase shift is 180° and occurs in a very small part of the spectrum. This plays havoc with error amplifiers and, considerable efforts have to be made to "compensate" the error amplifier to prevent a nice voltage regulator turning into an oscillating chip fryer. This phase change could, turn negative feedback into positive feedback.

So, in the real world, the error amplifier becomes severely hampered. To help it we use input and output capacitors to stop things moving too fast and, sometimes we use feed-forward. Feed-forward of the input supply voltage can result in a direct change of the duty cycle in some regulators so that the "error control system" is given a head-start and doesn't have to work so hard.

It's a tricky old business given also that regulators have a very wide load current that throws a big spanner in the works regarding the error amplifier.

• Andy, thanks a lot for your insightful answer and quick response! I guess one thing that still confuses me is how you are computing the error amplifier gain. The error amplifier is not seeing the input voltage of the converter as its input, it is seeing the output of the converter as its input. And its output is some voltage that drives the PWM signal on or off. I guess what you are computing is the system gain. I need to review some control theory and why going to infinite gain results in zero steady state error, but I think you have put me on the right track. Nov 20, 2022 at 21:45
• @JohnAllen Think of what happens when you have no error control; if the input changes, the output changes proportionately (a not very respectable regulator of course). To be more accurate in my calculation I should have used the output error change when no error reduction feedback is present. So, mid-scale for 4.5 volts in you'd get 1.8 volts out and, with 2.5 volts in you'd get 1 volt out and, with 6 volts in you'd get 2.4 volts out. In other words, my calculation should strictly have been a delta on the output of 1.4 volts with no error feedback reducing to 1.2 mV with error feedback...... Nov 20, 2022 at 21:53
• So, I over-egged it a little. Nov 20, 2022 at 21:54

Shouldn't the feedback be able to perfectly compensate for the change the input voltage by adjusting the duty cycle?

You cannot eliminate the error entirely, but you can make a regulator which reduces the error to an arbitrarily small value. Precision isn't free though: you'll pay with input/load range (an arbitrarily precise regulator will only be stable with an essentially constant input and load) and speed (the time it takes for the output voltage to reach the target value will be arbitrarily long).

In practice an error of 0.03% is already negligible for many applications.