How do I calculate ripple current for a capacitor?

Other Hardware General Electronics
1

I need to pick out some capacitors for one of my circuits, and one parameter I don't understand well is the ripple current.

I've been told ripple current can be a concern with DC-DC converters, but I've so far been unable to find any tutorial or video that explains exactly how I determine what my ripple current and ESR ratings should be.

In this video for example, the guy tests a voltage regulator at around 5m30s:

And he says there's an output ripple of 30mV peak to peak.

Now, if I have that information, and I know my regulator is putting out, say 3A, what do I do with that information to calculate the ripple current?

Or is the ripple current unrelated to how much current the regulator is putting out? Actually now that I think about it, that kinda makes sense.

So with this cap:

Would I go:

I = E / R

.030V (Ripple) * .0.35 ohms (ESR) = 0.86A

Seems like a sane result, so I think I'm on the right track?

2

Ripple current is just the rms current through the capacitor. Its usually important to
limit this to prevent overheating due to ohmic or other current-dependent losses.

So you have to calculate the rms current for the circuit in question. For a DC-DC
converter or other switching circuit the current is approximately a rectangular wave,
so the calculation is in two parts, for each part of the wave - average the squares of
currents over time, then take the overall square root. Typically in a switch mode
circuit the capacitor is charging up at the difference between input current and load
current for one part of the cycle, then discharging to provide the load current for
the other part of the cycle.

3

Because the capacitor is supplying the output current for at least half of the cycle, you may approximate the ripple current as the same as the output current but given that it is not constant, you should at least double the estimate.

Now you can combine that with your ESR to get the power - P = I2R to get the power that will be dissipated. Offhand, I would imagine you want to have that much less than a watt and all the more so for a small capacitor.

4

Paul:

What you're saying doesn't make sense.

I went to Digikey and searched for surface mount capacitors, and the highest ripple current rating for any 100uF cap, like you'd find on the output of any standard 5A buck regulator used in remote controlled vehicle applications, is only 1A:
http://www.digikey.com/short/78f7m5

Because the capacitor is supplying the output current for at least half of the cycle, you may approximate the ripple current as the same as the output current but given that it is not constant, you should at least double the estimate.

If you're saying that I should approximate the ripple current as the output current of he regulator, er, make that double, how could anyone build a 5A switching regulator with a single surface mount cap on the output, if said cap would need to be rated for 10A of ripple?

Now you can combine that with your ESR to get the power - P = I2R to get the power that will be dissipated.

Why bother calculating this? Capacitors have no wattage or power dissipation ratings.

Offhand, I would imagine you want to have that much less than a watt and all the more so for a small capacitor.

I prefer not to guess in my designs. Guessing means I ship a hundred boards and then six months later capacitors start exploding. And I could not even begin to guess what constitutes a "small" capacitor versus a "large" one. I don't even know if you're referring to physical size or capacity.

5

scswift:
What you're saying doesn't make sense.

Sadly, sometimes doesn't. :smiley:

scswift:
I went to Digikey and searched for surface mount capacitors, and the highest ripple current rating for any 100uF cap, like you'd find on the output of any standard 5A buck regulator used in remote controlled vehicle applications, is only 1A:

I won't dispute that.

scswift:
If you're saying that I should approximate the ripple current as the output current of he regulator, er, make that double, how could anyone build a 5A switching regulator with a single surface mount cap on the output, if said cap would need to be rated for 10A of ripple?

Actually, I had boost or half-wave transformer devices in mind. For a buck regulator the ripple is less as the inductor current varies between certain proportions of the load current so the ripple would be somewhat less than the load current and the safety factor wodul propbably then be to assume the ripple is the same as the load current.

scswift:

Now you can combine that with your ESR to get the power - P = I2R to get the power that will be dissipated.
Why bother calculating this? Capacitors have no wattage or power dissipation ratings.

OK, then it doesn't matter how hot they get when you make that calculation. Fair enough.

scswift:
I prefer not to guess in my designs. Guessing means I ship a hundred boards and then six months later capacitors start exploding.

You are probably more fussy than the manufacturers of those standard 5A buck regulator used in remote controlled vehicle applications.

scswift:
And I could not even begin to guess what constitutes a "small" capacitor versus a "large" one. I don't even know if you're referring to physical size or capacity.

Perhaps a large one is one in a can and a small one is SMD. Anyway, if they do not have wattage or power dissipation ratings and do not heat up, it doesn't matter.

6

http://www.eetimes.com/document.asp?doc_id=1273335

Now you can combine that with your ESR to get the power - P = I2R to get the power that will be dissipated.

Why bother calculating this? Capacitors have no wattage or power dissipation ratings.

You know this doesn't mean that they have infinite power ratings, just that the sellers don't make any effort to make this information available.

Page 3, top left:

CURRENT LIMITATION (HIGH FREQUENCY)
At frequencies in the 10 kHz to several hundred kilohertz range, the power dissipation becomes the limiting factor. The
following formula gives the maximum permissible ripple current for a sinusoidal wave form:
(9)
Pmax is the maximum power dissipation the capacitor can tolerate. The ESR value in the formula is the maximum ESR
of the capacitor at the required frequency. This can be determined by measuring capacitors and determining a
maximum value by using the mean value and adding 3 or more standard deviations. Some manufacturers specify the
maximum impedance at 100 kHz or 1 MHz. Either value may be used in ripple current calculations.

Power dissipation limits calculated for the most popular surface mount types of solid tantalum capacitors are:

Molded Case Chip (293D):
CASE SIZE
A
B
C
D
E
PMAX Watts
0.075
0.085
0.110
0.150
0.165

There are different kinds of capacitors that tolerate higher ripple currents, or high pulse, or high dVdT, etc. Not all caps are either ceramic or electrolytic.

7

This app note discusses published capacitor ratings c/w calculations to help determine what's best suited for a given application.

8

OK, then it doesn't matter how hot they get when you make that calculation. Fair enough.

I'm not sure if you're being facetious or not, but I understand that it matters how hot the capacitor gets. The thing is, knowing the power that needs to be dissipated is useless without also knowing how much heat the capacitor can dissipate. You said I want the power dissipated to be much less than a watt, but that's awfulyl vague. I would imagine a giant through hole capacitor and a tiny SMD capacitor would have very different abilities to dissipate heat. At least if it were a small ceramic cap I might be able to say, "Well, this same size resistor is rated for 1/8W, so maybe the capacitor is as well." but that would just be a guess.

You are probably more fussy than the manufacturers of those standard 5A buck regulator used in remote controlled vehicle applications.

Well I'm sure that's true since I get most of these regulators from China and they're like $2 each, less than what I'd pay for the regulator chip itself.

Still, there are a lot of big names in the hobby industry and I don't expect that their regulators tend to explode after an hour of use. Which I would expect to happen if I used a 2A rated cap on a 5A output - if in fact what you said about all the current being output having to pass through the cap is true.

For a buck regulator the ripple is less as the inductor current varies between certain proportions of the load current so the ripple would be somewhat less than the load current and the safety factor wodul propbably then be to assume the ripple is the same as the load current.

So now we're back where we started. It seems you're saying it IS the ripple I need to worry about.

So to restate my original question: How do I calculate the ripple current through the cap?

Now you can combine that with your ESR to get the power - P = I2R to get the power that will be dissipated.

See, right there. If R is the ESR, and I is the ripple current, Ohm's law still holds, right? So if I don't know the current through the cap, but I DO know the ripple voltage, then I should be able to calculate the ripple current with the ripple voltage and the ESR, right?

9

Now you can combine that with your ESR to get the power - P = I2R to get the power that will be dissipated.

Why bother calculating this? Capacitors have no wattage or power dissipation ratings.

You know this doesn't mean that they have infinite power ratings, just that the sellers don't make any effort to make this information available.

Yes, I am aware of that. What I'm saying is if there's no rating, I don't know how much the cap can dissipate, so why bother calculating how much power it will need to dissipate?

It's all well and good to say that it should be much less than a watt and use that as a general guideline, but who came up with that guideline, and how do I know it's correct?

Also, let's say I have a tiny ceramic cap and a smd electrolytic next to eachother. Given that tiny ceramic resistors can only hand around 1/8 of a watt, "less than a watt" might blow that capacitor up. Yet nowhere in that equation is there any factor regarding the capacitor's capacity, so am I to assume any 0.1uF cap on the regulator's output has to be rated to take a full 5A like the electrolytic? That doesn't make sense.

10

polymorph:
There are different kinds of capacitors that tolerate higher ripple currents, or high pulse, or high dVdT, etc. Not all caps are either ceramic or electrolytic.

I can't figure out how to copy and paste or quote that table you listed there, but I just wanted to point out that those dissipation ratings aren't "much less than a watt". Those are more like "much less than a tenth of a watt."

This is why I'm looking for exact figures and equations. If I just went by the "much less than a watt" assumption there's a good chance I'd exceed the ratings of all those caps by a great deal.

11

I have one other question. If I were to put two 1A rated caps on the output of a regulator, would that allow me to handle 2A, or would both the capacitors end up having to dissipating the same amount of current as before?

If so, how does that work? The information I've been presented with so far would seem to indicate it wouldn't. But I feel it must, because a larger cap can handle more current because it has a larger area to dissipate the heat.

12

scswift:
I have one other question. If I were to put two 1A rated caps on the output of a regulator, would that allow me to handle 2A, or would both the capacitors end up having to dissipating the same amount of current as before?

You would have two capacitors in parallel sharing the current according to their respective capacitances and ESR. Presuming both were from the same batch their complex impedances should be close enough for them to share the current equally so each would see half the ripple current and their corresponding power dissipation would be one quarter that of a single capacitor in the same situation.

Of course I was being facetious. You said what I was saying didn't make sense. You do not have to believe me, so I was waiting for a few other people to support me in applying simple (or maybe not so obvious) electrical theory - and locate the references which clearly had to exist. My personal guess on the power ratings (about a tenth of a watt) appears to have been spot on in the event.

13

scswift:
Also, let's say I have a tiny ceramic cap and a smd electrolytic next to each other. Given that tiny ceramic resistors can only hand around 1/8 of a watt, "less than a watt" might blow that capacitor up. Yet nowhere in that equation is there any factor regarding the capacitor's capacity, so am I to assume any 0.1uF cap on the regulator's output has to be rated to take a full 5A like the electrolytic? That doesn't make sense.

That is because the large electrolytic smooths the voltage and so the ripple voltage ( hence current ) that the ceramic is subjected to is a lot smaller.

This picture shows what happens to a stepping motor driver when there is insufficient smoothing from the main capacitor.

Stepping driver.jpg
Stepping driver.jpg764×1000 210 KB

14

Grumpy Mike has illustrated why several kinds of capacitors are used to smooth the output of a switch mode regulator.

We are trying to help you, but you need to read the links we've given you.

15

That is because the large electrolytic smooths the voltage and so the ripple voltage ( hence current ) that the ceramic is subjected to is a lot smaller.

That's what I suspected from the beginning, but Paul contradicted the notion of the ripple current being related to the ripple voltage when he said:

you may approximate the ripple current as the same as the output current

Also, I still don't have an equation I can use to calculate the ripple current from the ripple voltage.

16

polymorph:
Grumpy Mike has illustrated why several kinds of capacitors are used to smooth the output of a switch mode regulator.

We are trying to help you, but you need to read the links we've given you.

I read those papers. I found the first one on my own before I even posted this thread. I read a whole bunch of others as well. I didn't see anything which answers my question. That first paper just seems to cover power dissipation.

What I'm trying to figure out is when I should choose caps with low ESR and a specific ripple current rating. At this point I still don't have a clue how much current is going to flow through my capacitor if I have 50mV of ripple on a 5V 5A supply. All I do know is that if I have a big cap and a little cap the big one will take most of the ripple. Which makes sense, but I still don't know what that big cap's ripple current rating needs to be.

17

Take this link you sent me for example:

It explains how to calculate the ripple current on a buck converter using it's duty cycle. But that's useless to me. I don't need to know how to calculate that. I'm not designing a buck regulator. I'm sticking bulk capacitors around a board. Some are at my power input which is supplied by a buck regulator, but I have no idea what its duty cycle is it's in a wall adapter. I've also got caps by a DC motor driver. Am I supposed to use it's PWM frequency in place of the regulator's duty cycle in that equation? I don't know. I have an LED driver too. I guess it also has a PWM frequency which may or may not be what I'm supposed to plug into that equation.

What I was hoping to do was just stick a scope on my board, see that I have X microvolts of ripple in my supply, calculate what ripple current rating a 100uF cap would need if placed in that location, and maybe the ESR too if that was related and important, and then I could breathe easy knowing my cap wouldn't explode.

I'm just so confused right now. I feel I'm being fed conflicting information, and on top of that I'm being told how to calculate power dissipation when I never even asked how to calculate that and none of the caps on Digikey even have power dissipation ratings. What some of the more expensive ones do have are current and ESR ratings. Though the vast majority don't even have those.

18

Grumpy_Mike:
That is because the large electrolytic smooths the voltage and so the ripple voltage ( hence current ) that the ceramic is subjected to is a lot smaller.

Well here's another thing I don't understand. How does the electricity decide to flow into the larger cap first, to reduce the ripple the smaller one sees?

I would assume both caps would see the same dip in voltage, and both would react simultaneously to correct for it.

In other words, let's say you had a power rail that was oscillating between 4.95V and 5.05V. If you connect the small cap, it tries to stabilize the voltage, but the current being drawn is so high that it's voltage is just swinging wildly back and forth. Now you connect the large cap, and with it's larger capacity it can supply the necessary current, so the voltage doesn't drop as much...

And ah, I think I get what's going on now. The small cap stops swinging so wildly back and forth because the large cap is preventing the voltage swing from happening in the first place.

This also makes it easy to understand what happens when you have two caps that are identical. The capacitances add up the capacitors are capable of supplying more current, and thus more able to prevent the voltage from dipping, and so each dips half as far each cycle.

This of course assumes the cycles or spikes in voltage are short enough that the capacitors don't drain completely.

If they did... I suppose I'd calculate how much current the capacitors are supplying based on their Farads? Or... would they never drain completely because even if the voltage dips to 4.95V and stays there, one side of the capacitor is at 4.95V and the other side is still at 0V, so technically it's oscillating with only a fraction of it's full capacity?

Now I've gone and confused myself again.

19

scswift:
It explains how to calculate the ripple current on a buck converter using it's duty cycle. But that's useless to me. I don't need to know how to calculate that. I'm not designing a buck regulator.

Ah, well that may be the problem, as in your first post here you said

scswift:
I've been told ripple current can be a concern with DC-DC converters, but I've so far been unable to find any tutorial or video that explains exactly how I determine what my ripple current and ESR ratings should be.

Which quite reasonably implied that you were in fact, designing a DC-DC converter of some sort, itself.

scswift:
I'm sticking bulk capacitors around a board. Some are at my power input which is supplied by a buck regulator
...
What I was hoping to do was just stick a scope on my board, see that I have X microvolts of ripple in my supply, calculate what ripple current rating a 100uF cap would need if placed in that location, and maybe the ESR too if that was related and important, and then I could breathe easy knowing my cap wouldn't explode.

Quite a different question entirely. Clearly your external regulator module already has an output reservoir capacitor which is (hopefully) rated to absorb the ripple current. Whatever further capacitors you provide on your board will merely be supplementing that capacitor and in fact, in general will not be needing to be that large as their purpose is not to suppress ripple on the power supply, but to absorb transients (and those relatively infrequent, compared to the supply ripple) produced by your board.

scswift:
Or is the ripple current unrelated to how much current the regulator is putting out? Actually now that I think about it, that kinda makes sense.

And you indeed had an inkling that it was not so much of a problem, The problem was that you had not properly explained what it was you had in mind.

scswift:
I'm just so confused right now. I feel I'm being fed conflicting information, and on top of that I'm being told how to calculate power dissipation when I never even asked how to calculate that and none of the caps on Digikey even have power dissipation ratings. What some of the more expensive ones do have are current and ESR ratings. Though the vast majority don't even have those.

Then you do have the power ratings, because by the formula I previously quoted, Power = I2·R. You were given the current, and you were given the resistance, so you were given the power. Why do you suppose they publish ESR anyway? Well, actually, it is a performance figure as it limits how effectively they regulate, but why does the "maximum" ripple current matter? Only because it generates heat in the capacitor. A fuse works because the current generates heat; too much current and it melts. Too much heat will destroy the capacitor.

Now as to ripple voltage, this is less relevant because the relationship between voltage and ripple current is much more difficult to calculate - we are talking about a capacitor, not a resistor; the impedance is not resistive and the ESR is only a (generally small) part of it. This actually works to your advantage; what you calculate if you divide the ripple voltage by the ESR is actually always greater than the actual ripple current, but you will find it almost impossible to figure out how much greater.

As to your last question, it is quite irrelevant how much charge or voltage the capacitors contain at any instant - the current flow relates only to the change in voltage.
Charge = Capacitance x Voltage = Current x Time
Q=C·V=I·t

20

Why do you suppose they publish ESR anyway?

I don't know. I barely have a grasp on what ESR is in the first place, despite reading up on it. I had no idea if it was related to the ripple current rating in any way or not.

Then you do have the power ratings, because by the formula I previously quoted, Power = I2·R. You were given the current, and you were given the resistance, so you were given the power. Why do you suppose they publish ESR anyway?

Great. But what use is that to me? Why would I want to calculate the power the capacitor can dissipate?

Yes, I know; heat will destroy a capacitor. But if I have the capacitor's ripple current rating, and I can calculate the ripple current, I can tell if my capacitor is going to blow up based on that alone.

So why perform an extra step of converting those to power?

And if I don't have the ripple current, how would I even calculate the power the cap is going to have to dissipate anyway?

Now as to ripple voltage, this is less relevant because the relationship between voltage and ripple current is much more difficult to calculate - we are talking about a capacitor, not a resistor; the impedance is not resistive and the ESR is only a (generally small) part of it. This actually works to your advantage; what you calculate if you divide the ripple voltage by the ESR is actually always greater than the actual ripple current, but you will find it almost impossible to figure out how much greater.

Okay. This is useful information.

As to your last question, it is quite irrelevant how much charge or voltage the capacitors contain at any instant - the current flow relates only to the change in voltage.
Charge = Capacitance x Voltage = Current x Time
Q=C·V=I·t

I kinda guessed that. But the equation is useful. Which means I'll surely forget it by tomorrow.

I'm still lost as far as how one should go about calculating ripple current. As you correctly surmised I assumed the ripple current and voltage were related according to Ohm's law.

So what do I do here? Do people just do like you said and divide the ripple voltage by the ESR and say "Well, I know it's not gonna blow up..." and leave it at that?

Or am I supposed to do some calculation with the frequency of the ripple and the impedance, and derate the result?