# How to calculate the value of a smoothing capacitor

Discussion in 'General Electronics Chat' started by Voltz, Apr 20, 2010.

1. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

Yes those higher current thermistors are sometimes used now as input current limiters.

Well, the K you are talking about is different than the K i was talking about:
K1=2*f*R*C
K2=w*R*C

My K (K2 above) is based on a measurement after C has been chosen.
Your K (K1 above) does seem to be related to the 1/percentdroop as you mentioned, which is interesting. The two K's are of course related.
Your ESR calculation from post 34 though brings up a small question...
You stated:

but did you really mean:

by any chance?

We could investigate either of these further too if you like to find out why the percent ripple comes into play and also why it gets squared, unless you already know why it gets squared.

2. ### Tony StewartWell-Known MemberMost Helpful Member

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Thanks for the correction that should be Watts of power loss. I must remind myself to triple check

Ripple^2...
If we look at the efficiency and power ratios then V^2/ESR becomes important in addition to droop with ESR<=RLoad*PercentRipple^2 ok

Keep in mind although peak current is inverse to droop , similar to duty cycle and diodes must be sized properly with caps for power loss. Very high current rated Bridge will also produce higher Cap ripple current....

I am not sure of the industry standard for "low" and ultra low but the figure of merit would be the series R C product. which increases with √V rating and affected by cap design

I suggest for series RC in Alum. electrolytic

gen. purpose. = msec range
ultralow ESR = μs range
ceramic RF = 1 ns ~1 us time constant.

Cap losses exceed diode sized at 800mV@1A for 1A avg load. resulted in 1.68 drop in bridge and overall droop to 93.86V

Here I added a series C to show ripple and avg DC separately.

Scope traces in order left to right are

In Vac , Load V , Load W, Cap W, Diode W

It seems I chose a rather good general purpose cap. with a ESR*C =T series time constant of 33us. Quality has improved over the decades by orders of magnitude.
I forgot, my experience tells me, if a cap gives rated ESR, it is already a low ESR type, otherwise they are rated in ripple amps for caps.
e.g. Polyester caps for motor start do not have ESR ratings but have ripple current specs and are generally cheaper.
e.g. bridge caps rated for ripple current at twice line frequency, normally do not have ESR guaranteed or listed

Going from load = 100 ohm ; Vmin ~94 , Vavg=96V
to 10 ohm , Vmin ~69, Vavg= 85V

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3. ### MrAlWell-Known MemberMost Helpful Member

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Hello again,

Well the part i was concentrating on for the moment was the:

for the sake of the ripple, not for the efficiency. In other words we rearrange this into:

so we can estimate the effect the ESR has on the ripple percent by taking the square root sqrt(ESR/RLoad) and that helps choose based on ripple, at least that is what it looks like.
But now the ratio ESR/RLoad makes sense because we want a measure that compares ESR to the load, but where does the squared percent ripple come from. I guess this comes to:

So this allows us to compare the (1 ohm ESR power)/(1 ohm Load power) ratio to the (ESR)/(RLoad) ratio, as an estimate, or just the ESR power to the Load power as unity for the least case, also as an estimate.

It is starting to look like a partial differential equation would describe this nicely, except it is difficult to find the analytical form for the full wave plus series R plus cap ESR. The cap ESR brings in immense complexity to the output equation and that muddies up the whole solution, unless there is a simpler solution of course. The solution i found takes up a half page filled with all the variables and their powers and sines and cosines and exponentials. It looks worse than a bowl of alphabet soup

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5. ### Tony StewartWell-Known MemberMost Helpful Member

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Nonlinear analysis always looks like jumble but essential for a good simulation.

For me the reduction to linear ratios is more important to find, for design rules, such as load regulation is the ratio of ESR/Rload so 1% max is a good rule of thumb for dynamic loads and 10% for lossy static loads at low line frequency maybe reasonable or desirable to reduce ratings/cost on cap ESR and Diode Imax. Otherwise the caps appear as shorts for energy storage at low f or noise suppression at high frequency.

Thus when moving to a preferred SMPS design with PFC, at > 1k line f, the cap is downsized by same ratio and is used for both at optimum choices.

The analysis started in this thread was for the ideal case where caps are not treated as shorts and ESR was negligible at low current. Still use RC/2f as the figure of merit , FOM, or K for droop. But other factors may be compromised such as efficiency and peak/avg ratio which are reliability stress factors.

My fave quote in my career in design and test Engineer was always, trust but verify by testing.

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6. ### The ElectricianActive Member

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Can you redo this with the cap ripple current calculated and shown?

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7. ### Tony StewartWell-Known MemberMost Helpful Member

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I hope the Max, Min values are readable for each trace
the 100m was intended as ESR for Cap but could also be external.
The diode was chosen as 800mV @ 1A and reads 893mV at 49A so is probably rated much higher ( no part numbers are used in Falstad sim. but these are like Schottky but Vth=0 square law.)
.. to make diode model accurate I should have added ESR for diode ~ 1/ W max !! which could be 100m in practice for 10W diode and thus reduce Cap Imax

You can edit the values here.
select part then right click. Add trace delete trace, add part, wire etc. Javascript will use lots of CPU power as calculations are usually >8 digit resolution

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8. ### The ElectricianActive Member

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The circuit I asked about is the one with the 100 ohm load, not the 10 ohm load. Also, can Falstad calculate the RMS value of the capacitor current?

9. ### Tony StewartWell-Known MemberMost Helpful Member

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No to RMS but yes to peak or instaneous VAR or AMPs and shape of triangle can be converted to RMS x Duty cycle,D by convention.

Trace for 100 mOhm shows 13.2Wpk = > Ipk = √1.32 A to rms = 2Ipk√(D/3)

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10. ### The ElectricianActive Member

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OK. I now see that the Cap W in your post #122 images are peak power, not average power. It would improve clarity of your results if peak powers were labeled as such.

I don't think this is a good figure of merit; the smaller, the better, right? Consider two capacitors each having 100 milliohm ESR, one of 1000 uF and one of 2000 uF. The 1000 uF cap will have the smaller, and therefore "better" figure figure of merit, if that figure of merit is R C; I don't think that's what is wanted, is it?

It seems to me that a good figure of merit would be ESR/ωC, or just the plain old dissipation factor for the cap, DF.

Also, a decent aluminum electrolytic, even a "not low ESR" one, will have an ESR much lower than ωC at grid frequencies. The ripple will be mostly determined by 1/(ωC) (more precisely, the impedance Z = ESR + 1/(jωC)) with the ESR a minor part of Z.

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11. ### Tony StewartWell-Known MemberMost Helpful Member

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I agree but Dissipation factors are not always provided.

Sometimes just ripple current , or ESR or both.

For a given Working Voltage, Mfg and model, I see that ESR*C is constant over a decade range in C values.

Then comparing different models or Mfg, once you have RC for one value of C, you can expect about the same of other values in same V range and type.

Thus you need to compare RC at the same value for a) energy storage value or b) Holdup time

So should ten C/10's perform better than one C1 of same family . Maybe.

- although ESR should be same for both, if in same family , but if different types, the distributed smaller body parts may have lower ESL and higher SRF or another model of C/10 may be much lower ESR*C product so ten C/10's might perform better.

I find the same applies to batteries and CCA or charge rate * Ah capacity for figures of merit.

For these e-caps ESR*C is constant for a given V which translates to a constant dissipation factor.

12. ### The ElectricianActive Member

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One could use the R C product as a stand-in for dissipation factor if one wished.

The P-P ripple voltage and ripple current isn't much affected by the ESR.

I simulated your circuit with 100 ohm load using 1N5406 rectifiers and got these results:

With ESR = 0, P-P ripple is 4.279 volts, RMS ripple current is 3.046 amps, peak capacitor current is 15.37 amps.

With ESR = .1 ohms, P-P ripple is 4.178 volts, RMS ripple current is 2.683 amps, peak capacitor current is 10.942 amps.

If I increase the capacitor to 4000 uF, I get these results:

With ESR = 0, P-P ripple is 2.214 volts, RMS ripple current is 3.457 amps, peak capacitor current is 18.642 amps.

With ESR = .05 ohms, P-P ripple is 2.166 volts, RMS ripple current is 2.952 amps, peak capacitor current is 13.658 amps.

P-P ripple voltage is hardly affected by the ESR of the capacitor for ordinary aluminum electrolytics. It's the capacitance that mainly determines the P-P ripple voltage.

13. ### Tony StewartWell-Known MemberMost Helpful Member

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But .1 ohms * 10.942A = 1.09V is 26% of 4.178V

Again 0.05 ohms * 13.658 amps =0.68V which is 32% of 2.166V

32% may not be much at 100 Ohms @ 100V from ESR (assuming I calculated right)
but the percentage rises significantly at high currents.

14. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

I found something similar to that which i thought i mentioned but let me repeat a little differently.

When we have a certain ESR, it's ok for certain lower load currents, but if we try to keep that same ESR for higher load currents, for 10 times the load current the ripple voltage will go up by about 100 percent once the capacitor value is adjusted for the new current (cap value10 times as before, and we will always increase the cap value or we'll see a huge ripple increase).
We expect to have to raise the capacitor value by 10 times, but not changing the ESR means the ripple voltage goes up just because of the incorrect ESR. The corrected ESR would bring us back down to the same ripple we had before.
So the ESR does affect the ripple voltage, but of course it is only a problem if it is not the right ESR for the load current. This is also part of why i suggested using 10 of the same capacitors for 10 times the current rather than trying to use one capacitor with the same ESR. Not only do we get 1/10 the ESR (if wired correctly of course) but we also get 10 times the ripple current rating (if wired very carefully).

This is all assuming that we have some standard we must follow, such as no more than 10 percent ripple for any power supply. Obviously if we dont have that kind of standard and more ripple is acceptable, then we are playing a different ball game altogether where we dont have to worry about the ESR as much.

One other little thing i thought i should mention is that as the power supply current rating goes up, at some point an input inductor is used to smooth out the current to the rectifiers and caps or they might see a lot of repetitive high peak currents, as well as create a pretty nasty power line power factor issue. This would also help to reduce ripple.

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15. ### The ElectricianActive Member

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What I'm going to say is only approximate because the waveshapes involved are not sinusoids, but this should do for a start.

The ripple voltage is due to the voltage drop caused by the ripple current flowing through the impedance of the capacitor. The impedance of the capacitor has two components--the reactance of the capacitor at 100 Hz, and the resistance of the ESR.

For the circuit under discussion here, the reactance of the 2000 uF cap at 100 Hz is .7958 ohms, and the ESR is .1 ohm. They are in series, so the ripple voltage is the sum of the voltage drops across the reactance of the cap (.7958 ohms) and the ESR (.1 ohms). Since the reactance is 8 times as large as the ESR one might think that the voltage across the reactance would be 8 times larger than the voltage across the ESR, and this would be true if the voltages added arithmetically. But they don't add arithmetically, they add vectorially (if the ripple current were a perfect sinusoid, and approximately so in the real case).

The voltage drop across the perfect 2000 uF cap in series with the .1 ohm ESR would only increase about .8% because those voltages don't add arithmetically; they add vectorially: Vtotal = sqrt(Vcap^2 + Vesr^2).

The reactance of the 2000 uF cap at 100 Hz is .7958 ohms, with ESR=0 (this is also the impedance of the cap with ESR = 0). If the ESR is .1 ohms, then the impedance of the cap is .802 ohms, less than a 1% increase due to the ESR of .1 ohm, not a 12.5% increase.

The ripple current isn't a perfect sinusoid, so the increase in ripple voltage is more than .8%. To find out what it really is requires solving the non-linear rectifier circuit, which is what a simulation does. The simulation shows that the voltage across the ESR is only about 4% of the total P-P ripple voltage.

Furthermore, you are making a false assumption when you say "32% may not be much at 100 Ohms @ 100V from ESR (assuming I calculated right) but the percentage rises significantly at high currents."

This would be true if the voltage across the reactance of the capacitor remained the same as at lower currents. But the ripple current flows through the capacitor's reactance in series with the ESR. The voltage across the reactance goes up in the same proportion as the voltage across the ESR. Therefore, the percentage of the ripple voltage due to the ESR remains the same at all currents.

Also, you'll notice that the simulation with the 2000 uF cap showed that with ESR=0, the P-P ripple voltage was 4.279 volts, and with ESR=.1 ohm, the P-P ripple voltage was 4.178 volts.

The ripple voltage decreased with the addition of .1 ohms of ESR, because the increased impedance of the capacitor due to the non-zero ESR reduced the ripple current enough to more than compensate for the extra ripple voltage drop across the ESR.

So, all other things being equal, increased ESR doesn't necessarily increase ripple voltage, and when it does, it doesn't increase it much.

16. ### MrAlWell-Known MemberMost Helpful Member

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Hello again,

Electrician:
I think we are talking about a situation that is different than what you are talking about. This is pretty clear because the results i see either way i do it are very different from what you seem to be suggesting. What i am saying and i think Tony is saying is that the ESR matters at some point, whatever that point it. The problem then is to find out what point that is not whether or not an incremental change in ESR has an effect or not. A small incremental change in ESR wont matter, it's the combination of increased load current and previous value of ESR that shows the problem, or looking at it another way, with a larger current an incremental change in ESR will matter quite a bit, if it is larger enough, and even 0.15 makes quite a difference.

17. ### The ElectricianActive Member

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The ESR has two effects that have been discussed in this thread. There are heating effects from ripple current flowing through the ESR, and there is an increase in ripple voltage due to increase in ESR. I'm not addressing the heating effect, only the ripple voltage.

When you say "...ESR matters...", what do you mean? Matters in what way? Matters in its effect on ripple voltage, or something else?

If you mean with respect to ripple voltage, sure, ESR matters, in that the voltage drop across the ESR increases when the ripple current increases, but so does the voltage drop across the capacitor's reactance. The percentage of the ripple voltage attributable to the ESR with respect to the ripple voltage attributable to the cap reactance does not change with current level, does it?

The voltage drop across the ESR rises only in proportion to the current through it, not in proportion to the square of the current through it.

Tony said "...the percentage rises significantly at high currents". Analysis and simulation doesn't bear this out.

18. ### MrAlWell-Known MemberMost Helpful Member

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Hello again,

Well see for my experiments the capacitor reactance does in fact change. The 1/jwC changes, but at first not the ESR. The 1/jwC goes down by 10 times because the capacitor value goes up by 10 times. In your experiment it does not change, but in mine it does. This makes a difference in the ripple voltage even thought we bring C up by 10 times, so there is no other explanation for the increase in ripple now...it must be the ESR.

Let me restate the rationale for this.

First, we know from wRC=K that to keep ripple the same with a decrease in R we must increase C by the same factor, so R_new=R_old/10 means we must make C_new=C_old*10 when we make ILoad_new=ILoad_old*10. If we have any hope of keeping the ripple the same we must at least do this.
The problem is, that doesnt work, at least not exactly. It works to a large extent, but not as much as we really want to see happen. When we do this we still see an increase in ripple voltage by 2 times, so instead of 10 percent ripple we now see 20 percent ripple, and that is because the ESR is now too high for the intended application specifications. Once we lower ESR to the right value however, then we see nearly the same ripple voltage as we did before. So we attribute the extra ripple to the cap ESR, and fixing that fixes the problem.
Can this be achieved by increasing the cap value even more instead? Probably, but then we'll see more dissipation also which we probably dont want (subject to application).

Here are some numbers:

100vac, 0.1 ohm series resistance, ESR=0, 13 ohms load, ripple percent=2.49

So to get back to around 2.5 percent ripple voltage we had to lower ESR by 10 times even though we increased the cap by 10 times.

BTW this is at 60Hz but i could switch to 50Hz if that's more desirable.

19. ### The ElectricianActive Member

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We should all be using the same circuit, same frequency, and calculate the percent ripple on the same reference (peak DC voltage or average DC voltage).

Tony was using loads of 100 ohms and 10 ohms; why are you using 13 ohms?

Please post the circuit you're using and list all the relevant details; tell what you change from experiment to experiment.

20. ### Tony StewartWell-Known MemberMost Helpful Member

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As tanδ = ESR/Xc is the normal data-sheet parameter given for computing losses at 2x line rate , 120 Hz or 100Hz , I decided to compare as many product lines as I could find in Alum. elect category and found some unusual common trends.

1. The best case tan δ seems to be near 50~100 V ratings , in the 2nd example 0.12 or 12% @120Hz.
These values rise with both increasing V ratings and progressive higher at lower Working Voltage ratings.

2. In the 1st example, the footnote indicates that tan δ increases 2% per nF above 1nF.

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21. ### The ElectricianActive Member

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Conrad Hoffman pointed this out some years ago. This is a fact worth noting. The best DF is typically to be found in the 50 to 100 volt range. He especially has noted that very low voltage electrolytics, such as 6.3 or 10 volts seem to be more failure prone.