# How to calculate the value of a smoothing capacitor

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

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

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Thankyou for pointing out my contradiction. I said Voltage when I meant Power,

We know when RC=1 in a linear filter the exponential step response is ~60% of full scale (1-1/e)*100% for a square wave RC=T=1/2f of rectified f for load =R and shunt cap C,
This is assuming= Dissipation Factor = in Cap and Series R in source and diodes =0

so we can easily compute or use my results for RC vs % Ripple
RC= 1T , 60% pp ripple
RC=2T, 38%
RC=3T, 28%
RC=4T, 23%
RC=5T, 17%
RC=6T, 14%
RC=7T, 12%
RC=8T, 11%
RC=9T, 10%
RC=10T, 9%
RC=30T, 3%
RC=50T, 1.8%
RC=100T, 0.9%

The next question is how does ESR of Cap and series diodes and secondary coil affect %ripple?

The Cap size is now easy to calculate but the dissipation factor at 100Hz ( or 120Hz) must be designed by component selection with the rated RMS ripple current and extra series resistance to reduce ripple.
Occasionally diode turn off results in voltage noise which can be suppressed with small RF caps across each diode.

You can compute step pulse voltage ratio from impedance ratio of series to shunt ESR ratio.

2. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

Most of us assumed you meant power not voltage on the right side.

But are you looking for a simplified answer or you want the full blown *exact* analysis?
Of course this means you have to specify what you are allowing to be assumed, such as zero line impedance to keep things at least a little simple. The normal assumptions are zero line inductance, zero line resistance, and diodes that conduct with no voltage drop, and no secondary effects like radiation, skin effect. Variations on that include some line resistance and some diode drop, but the program i created and posted earlier uses a spice diode model which is probably an overkill. The spice diode model does make a difference, but i still think it is an overkill.

3. ### HubertNew Member

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Am I right seeing a big mistake in that formula?

From my point of view this results in

edit:
No there is no mistake - it was my fault to mix with dimensions of Coulomb and Farad.

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5. ### RatchitWell-Known Member

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You might want to peruse this link http://powerelectronics.com/power-management/analyzing-full-wave-rectifiers-capacitor-filters

Ratch

6. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

How did you arrive at that conclusion: V=sqrt(V^2-V) ? Maybe you can show some work.

If we take the limit of the right side as V goes toward infinity we do get V=V, so that may suffice as an approximation, although i wont say for sure until i see some of your work.

7. ### HubertNew Member

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Hi Al.

It was MY big fault. But the mix started allready in der Fairchild formula, they used V for voltage. Usually in Germany we use U as formula symbol for voltage and later I was looking C as Coulomb what leads me to A•s unstead of A•s/V. Shame on me. That's all.

8. ### MrAlWell-Known MemberMost Helpful Member

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Hello again,

Ok, no problem

I almost forgot that i had created a program version that includes calculations with the capacitor ESR also.
Because of the increased complexity this takes longer to calculate, but it includes the entered cap ESR as well as calculates the percent ripple and the diode highest voltage.
This should probably be thought of as a "Beta" version, as it can be optimized and improved much more.

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9. ### ColinMember

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You are missing out on one important fact.
At 24v AC you have an enormous voltage "up your sleeve" to reduce the ripple electronically.

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

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I assume this is for a full bridge and R1 means Rin and I suspect you check the Rin/ESRcap ratio here.
A with target Vdc min or avg. and ripple as input rather than output might be useful.

I did a quick test for 50Hz to compare with my table for T= 10ms, Rin=0.1 ESR=0.1
I arbitrarily chose 76Vac R=100 gave ~100Vdc avg so I chose RC=9T= 90ms to get 10% ripple and your calc. obtained.~9%.

Mine forces you to know the load impedance, and choose the capacitance based on ripple, with no care about actual voltage or actual current or power used. ( assumes wise choices)

Yours does a excellent job at showing the rise in ripple with ESR.

Filter , TS% , Mr Al%
RC=7T, 12% 11%
RC=8T, 11% 10%
RC=9T, 10% 9%

Generally ESR * C is a constant for a family and size, so a range of ESR*C families exist from 100us for low ESR electrolytic to 1000 us for std .ESR and many others above and blow this range..

11. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

Not sure what you mean there, but the program does not to ESP, it just does ESR
In other words, it does not do any circuit under the sun, just a full wave bridge rectifier with filter cap and various other standard elements. It can be made to do more, but the rectifier was the main goal at the time.
If you have a circuit suggestion anyway though im sure we'd all be happy to hear it.

12. ### MrAlWell-Known MemberMost Helpful Member

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Hi Tony,

The program basically does the raw network calculation but then uses the results to calculate the ripple based on the highest and lowest wave points and that is also used to calculate the percent ripple. So it's almost like a simulation more than some worked out short cut calculation.
It can however be made faster by using more analytical calculations but i just didnt get to do that yet.
I guess you were using your own worked out formula, which is good for comparison.

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

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Yes you should be considering transients for power dissipation (Ipk/Iavg )²* = Ppk/Pload and we know Ipk is inversely related to % ripple on voltage.

ESR must be tightly controlled and yet there is a tradeoff between load regulation, %ripple and MTBF in caps at line frequencies.

This is one reason why electrolytics heat up (and dry up over time) at high ripple currents w.r.t rated level and why series R and L are added to improve MTBF by reducing Pk/avg ripple current ratios in Caps. It is also a big advantage to increase switching frequency and duty cycle at full power near 50% so charge and discharge currents are equal, unlike at line rates where this is not the case.

For example this 100Vp sine input 1K load = 0.94A avg DC load (90.57 to 98.69mA)=~ 10 Watts has a peak supply of 112W (affects VAR rating on transformer)
and has

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

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Note that I used Falstad to derive the results and also follows the simple 1st order LPF response ( 1-1/e^k) for k=RC/T and RC=8T or 9T for 10% ripple and T is 10ms for FW 50Hz and 16.7ms for FW rectified 60Hz

Last edited: Dec 4, 2015
15. ### crutschowWell-Known MemberMost Helpful Member

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μF is certainly common but it's easier to write 10mF instead of 10,000μF