Really lost understanding the Buck Converter steady state analysis - Page 1
Forum section: electronics / beginners

author: LoveLaika
author_rank: Frequent Contributor
timestamp: May 27, 2022, 03:29:53 pm
content: I'm self-studying power electronics using Fundamentals of Power Electronics. I'm really confused about some topics involving analysis of the buck converter, so I wanted to ask for help here. It's not the most ideal, but here's somereference slidesto help explain my point and to follow along. Sorry, this may seem like a stream-of-consciousness, but it's kind of hard to explain as a simple question.In chapter 2 (slide 39), we examine the buck converter. Using small-ripple approximation (SRA), we can get a graph of the inductor voltage at steady state in terms of the input/output DC voltage (slide 47, though I think there's a contradiction here explained later). Now, following the equation of the inductor, this voltage waveform shows that there's an AC-component of the inductor current centered around the DC value. This AC part gets passed to the capacitor. Ideally, at steady state, the capacitor has no current flowing in or out of it. However, because of the inductor current's AC component, the capacitor current gets that AC component as shown in Slide 74 (though the average value of the cap current is 0 over 1 period). This changing cap current leads to a small AC-voltage at the capacitor's voltage, and thanks to that, we can calculate the estimated voltage ripple given the estimated current ripple (Slide 77). However, since there's a voltage ripple, assuming that the components are well selected, all of the AC-voltage-ripple should go from the cap to the inductor, and that would cause further change in the inductor current ripple.From all this, my question is how does it all reconcile together? I apologize for the long and rather long-winded explanation, but it feels that with how one affects the other (Inductor AC current->Capacitor AC Current->Capacitor AC Voltage->Inductor AC Voltage->Inductor AC Current), it feels like a constant looping cycle, especially when you need to take it into consideration when selecting components as shown in Slide 48. This is causing me to be a bit confused about some of the problems. Also, speaking of which, I think I might be misunderstading something. From Slide 47, you see the inductor's voltage waveform, but at steady state, the inductor DC voltage should be 0, like how it shows in Slide 78 (though there is an AC-component to the inductor voltage caused by the capacitor, it is centered around the DC inductor voltage of 0, and it averages out to 0 over one period T). How do all of these things reconcile with each other if they seem to contradict each other?

author: magic
author_rank: Super Contributor
timestamp: May 27, 2022, 03:46:40 pm
content: What you wrote isn't wrong.Inductor voltage is the difference between its input side (which switches between Vin and ground) and its output side (which is Vout±ripple). In sensible designs the AC component is approximately Vin peak-peak in amplitude and Vout ripple doesn't affect it much, being much smaller than Vin. Hence it is often neglected in such calculations. Similarly, it is neglected that due to the circular feedback which you found the waveforms aren't actually perfectly linear squares and triangles.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: May 27, 2022, 05:10:01 pm
content: Thanks for replying. If I may, I would like to follow up on your answer and clarify my understanding. I'm assuming (following the book), that the AC-ripple of anything is small enough to neglect away. To put it another way....assuming steady-state, we assume that the inductor voltage can be written in terms of input and output voltage (at least the DC-component). However, following the inductor equation, integrating the inductor's voltage shows that there is an AC component to the inductor current (though, it averages out to 0 over 1 period). That AC component flows through the capacitor, which then in turn creates an output voltage ripple. However, following your comments, are we assuming that this output ripple (caused by the inductor ripple) is so small that we can just ignore it?The way I'm seeing it, Inductor current (as an example) is made up of three parts:Inductor DC componentInductor AC component, from integrating Inductor VoltageAnother AC component, caused by capacitor voltage ripple creating a current rippleIt's that third term that seems to cause this 'circular feedback'. However, assuming steady-state, we just ignore it because it is so small? If that's the case, then the only AC ripples would come from only one component: the inductor (for the ripple current), and the capacitor (for the ripple voltage). Hence, you can just simply use the single integration of the one component to solve for the appropriate ripple (as demonstrated in Slide 48, using only the inductor voltage integration to solve for the ripple current). Whereas that set of slides (from Slides 74-78) show how the SRA assumption isn't ideal and showcases what needs to be accounted for? When would you take this into consideration? (As an aside, that circuit shown at Slide 78 represents two LC-filters, which are 2nd order by themselves, so why would the additional filter cause SRA to not be useful? Would it be easier analyze it with an ideal switch as shown in the attached picture?)With that being said, what is the graph of the inductor voltage at slide 78 trying to represent? I can see it's the inductor's ripple voltage centered around the DC component of 0 (because at steady state, inductor acts as a short, so no DC voltage), but how does that contrast with what we saw back at slide 47 where we saw a steady-state voltage across the inductor (yes, sure, different circuits, sure, but wouldn't principal still hold)? When switching, the inductor current cannot change instantaneously, but the inductor voltage can change all it wants.

author: magic
author_rank: Super Contributor
timestamp: May 27, 2022, 09:32:48 pm
content: I see that page 43 of your PDF explicitly states that they assume "small ripple" and ignore its contribution.Quote from: LoveLaika on May 27, 2022, 05:10:01 pmI suppose we could say so. Of course the exact thing that matters is the "differential" voltage across the inductor's ends, but the differential voltage is the difference of voltages at individual ends and its AC variation is the sum/difference of individual AC waveforms.If output ripple is at least 10x smaller than Vin, ignoring it results in <10% error in inductor differential voltage and peak-to-peak ripple current.Quote from: LoveLaika on May 27, 2022, 05:10:01 pmThat's how they calculate it.Pages 74-77 are simply calculation of output ripple voltage in ordinary buck converter. That could be useful just to know how much capacitance is needed to achieve acceptable ripple level. Not sure what they mean on page 78 because labels on the plot don't correspond exactly with anything on the schematic and the text is rather laconic.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 28, 2022, 01:39:34 am
content: Right, it's an effective approximation because:1. We don't want loud ass power supplies, we have enough filtering (C) to get the ripple down;2. The filter cutoff frequency is a small fraction of f_SW.So during a cycle, very little happens at the filter output (>30dB attenuation, say), and if we like, we can approximate that waveform as a parabolic wave (think: two parabolic sections glued together, so as to approximate a sine wave), and do a small correction to inductor current as a result, and this is fine.The output will be strongly phase shifted (180°) to the switch node, so we don't even have to worry much about superposition, it's just... expected square voltage making triangle current ripple, plus a small cosine (270°) correction to bring in the output ripple.And yeah, technically there are corrections to the corrections. It's not corrections all the way down; it's... well, you can write it that way, in fact, but easier to solve the simultaneous equations and be done with it -- if you must.This is a greatly welcome alternative to calculating the whole thing, which involves a piecewise source (square wave) driving a 2nd order (or worse) differential system; we can do this in the frequency domain fine as well, but neither is going to be particularly simple, and we don't care a damn about the exact waveforms' value over time (i.e. what exact functions and parameters are in v(t), i(t)), as long as it's doing the basics -- so, the approximation also captures our intent, in a sense....To say nothing of cap ESR, of course; but that adds some triangle voltage to the output, so, parabolic current to the inductor. This is a higher order effect (1st not 2nd) so could be worth including, but again, the same premise applies: as long as the current is accurate within, like, even 10% is precise for this -- even just, more that the slope is within say 2:1 of what it should be, and is smooth so peak-current control for example is applicable -- and again, as long as output ripple is usefully low, we don't care, it's fine, all those correction terms (or, for exact solution: the simultaneous equations solving for the corrections of corrections) are down in the noise, we don't care.Aside: the idea of adding correction factors, and corrections on corrections, is a similar process as solving for the value of a geometric series. If we takeS = 1 + x + x^2 + ...then we haveS - 1 = x + x^2 + x^3 + ...S - 1 = x(1 + x + x^2 + ...)S - 1 = xSSolving,S = 1 / (1 - x)(which is true for -1 < x < 1.)It's like saying, if you know one fraction, say 1/10 = 0.1, and want to know one-off from that say 1/11, and want to avoid doing a whole-ass division, then, you can get closer by subtracting the ratio of the divisor (i.e., 11 is about 1/10th larger than 10, so subtract 1/(10*10), etc.). But that underestimates the result (1/(0.09) ~ 11.1...), so add back in the same but even littler, but then...... and so on. Which, for a simple decimal expansion like this (one off from a nice round power of 10), it's pretty apparent why the decimal expansions of 1/9 and 1/11 repeat so simply (0.111..., 0.0909...).With respect to differential equations, it might be that we can solve for them easily (in this case, we can -- easy being a relative term!), or it might be that we can't feasibly solve the simultaneous equations at all, and our only tool is an infinite series, an approximation, or a perturbation method. The functions being operated upon, are obviously much more complex, but the basic mechanics are still common to them.Tim

author: LoveLaika
author_rank: Frequent Contributor
timestamp: May 31, 2022, 03:14:13 pm
content: Thanks again for replying. Going over the notes and waveforms again, I see now that I think my problem is that I assumed SRA initially, but then I don't assume it later on (at least when I make assumptions about problems).Starting with the Buck Converter (and later on, when deriving the Cuk), the slides start off with figuring out the instantaneous inductor voltage and the capacitor current at steady state. SRA assumes that the ripple for those equations is much, much smaller than the DC component. Therefore, at steady state, you can neglect it for the DC terms. However, that only applies for inductor voltage/capacitor current terms. When you integrate the appropriate equations, you find that inductor CURRENT and capacitor VOLTAGE has a ripple. (I'm not sure why we don't assume zero ripple here; maybe it's due to the fact that inductor current/cap voltage cannot change instantaneously?) Realistically, these ripples will affect the other parts (forming that circular feedback loop mentioned before); however, SRA assumes their contributions to the changes be very small, so if there is ripple, it would only be caused by the corresponding component (for a lack of a better term).Take the boost converter analysis. Applying SRA, there's no inductor voltage ripple or capacitor current ripple, but we see that there's capacitor voltage ripple (and therefore, output voltage ripple). Whereas with the buck and cuk converter, there is no output voltage ripple. From this, it seems fine to assume SRA, but realistically, there's always going to be ripple.

author: LoveLaika
author_rank: Frequent Contributor
timestamp: May 31, 2022, 03:20:05 pm
content: Thanks for the reply. I've been going over my notes and such, and I think what I was confused about was maybe the meaning of SRA. SRA assumes that the ripple is negligible compared to the DC component. However, that doesn't mean that ripple doesn't exist. As I wrote out in my previous reply, we assume inductor voltage/cap current ripple is 0, but that doesn't mean that inductor current/cap voltage ripple is 0. With the Buck and Cuk, applying SRA shows that there is the aforementioned ripple when deriving the steady state graphs. It's just that the effects of the calculated ripple are small. Otherwise, the inductor current ripple would pass through the capacitor, causing changes in the cap voltage, which get passed to the inductor voltage, and so forth (a circular feedback cycle). And solving it would be very painful.