How to calculate the value of a smoothing capacitor

Tony Stewart · Dec 1, 2015

The Electrician said:
How does a current*current*resistance = voltage?

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

Voltz said:
Assuming 10% Ripple is acceptable does anyone know the formula for the value needed for a capacitor for smoothing. And does it make a difference if the capacitance is greater than this, and if so what difference, I've tried searching but have got 2 or 3 different formulas, need help - thanks in advance

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%

Does this answer your question?

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.

MrAl · Dec 2, 2015

Tony Stewart said:
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%

Does this answer your question?

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.

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.

Hubert · Dec 2, 2015

Tony Stewart said:
Fairchild uses the similar but different approach that I use computing duty cycle of charge time. The duty cycles of half and full wave are computed for the fill cycle instead of half cycle, which looks wrong, because 30% for half wave and 15% pulsed twice is about the same time total ON time per sine cycle, hmmm....

They use an efficiency estimate instead of ESR and use RMS inputs instead of peak.

Comments?

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.

Ratchit · Dec 2, 2015

Voltz said:
Assuming 10% Ripple is acceptable does anyone know the formula for the value needed for a capacitor for smoothing. And does it make a difference if the capacitance is greater than this, and if so what difference, I've tried searching but have got 2 or 3 different formulas, need help - thanks in advance

You might want to peruse this link **broken link removed**

Ratch

MrAl · Dec 3, 2015

Hubert said:
Am I right seeing a big mistake in that formula?
View attachment 95751
From my point of view this results in
View attachment 95752

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

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.

Hubert · Dec 3, 2015

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.

MrAl · Dec 3, 2015

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.

Colin · Dec 3, 2015

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

Tony Stewart · Dec 3, 2015

MrAl said:
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.

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..

MrAl · Dec 4, 2015

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

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.

MrAl · Dec 4, 2015

Tony Stewart said:
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..

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.

Tony Stewart · Dec 4, 2015

Ratchit said:
First of all, are you computing transients. I am not. Am I supposed to do so?

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

Tony Stewart · Dec 4, 2015

MrAl said:
Hi Tony,

I guess you were using your own worked out formula, which is good for comparison.

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

crutschow · Jul 28, 2016

audioguru said:
Capacitor values are uf, not mF.

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

How to calculate the value of a smoothing capacitor

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