Determine Capacitor Value for Smoothing DC Ripple

Thread Starter


Joined Feb 18, 2023
Hi Ladies / Gents - I am trying to understand this concept a little better.

C = I / (2 x f x Vpp)
"Vpp = the minimum ripple (the peak to peak voltage after smoothing) that may be allowable or OK for the user, because practically it's never feasible to make this zero, as that would demand an unworkable, non-viable monstrous capacitor value, probably not feasible for anybody to implement.",feasible%20for%20anybody%20to%20implement.

Reason being - I need to smooth the output (post rectifier) on a 300w / 48v Permanent Magnet Alternator.
It is being fed directly into a step down buck converter which will provide 12v for an inverter.
I want the system designed to deliver 250w of the "300w" capacity and for the capacitors to be sized appropriately.

The mathematical equations required for precise calculation are confusing to the uninitiated - I am hoping someone can simplify it in practical terms..

What would be a good way to understand this at a basic level?
The idea is to calculate the capacity to optimal value to reduce ESR but also mitigate DC ripple from the rectifier.

Is this necessary or will the capacitors on the buck converter "do the job" in smoothing the output?

PMA goes up to about 60V at full rpm so planning to use a 150v / 150uf electrolytic (if this makes sense to do.)
The buck converter is a 75v to 2.5v adjustable type (600w)

Many thanks in advance for any help!


Joined Jan 27, 2019
Have you measured the actual output waveform of the buck converter?

Since the value of the capacitor is proportional to \( \mathsf{V_{PP}} \)why not do the filtering on the output of the buck converter to reduce the required cap size?


Joined Aug 7, 2020
Is it a three-phase alternator?
The equation, usually written Vpp=It/C, is an approximation for single-phase supplies. t should really be the time when no diode of the rectifier is conducting. i.e. just less than 10ms for a 50Hz supply.
The peak-to-peak ripple of a single phase supply is equal to V peak.
For a 3-phase supply it is equal to 0.134Vpeak, so that already reduces the capacitance requirement by a factor of 7.
A buck regulator might even be able to cope with that amount of ripple with no extra smoothing.

Don't forget that adding a lot of capacitance increases the peak currents, and the power dissipated as copper losses in the windings is then proportional to the peak current, so less may be better than more, if you want to get the full power output without it overheating.


Joined Nov 6, 2012
Going through 2 series connected Buck-Converters is going to waste roughly ~40%
of the total available Power from the Alternator as HEAT-Dissipation.

300-Watts X 60% = ~180-Watts of available Power.

300-Watts / 2.5-Volts = 120-Amps,
are You trying to build a Welder ?

Thread Starter


Joined Feb 18, 2023
@ Ya’akov
Only just received it - Will be testing it out soon.
The load (inverter) will have its' own smoothing caps onboard (most likely)
As BobTPH put it - Buck converter will likely need a stable DC input.
All I want to know is which capacitor value would be best for smoothing DC ripple from current generated within the windings of a 48v / 300w / 3 phase PMA (the rest of the system is planned out)
Using 1N5408 x6 as 3p Rectifier (unless someone can suggest something more appropriate)

@ BobTPH
Not sure what you mean by "both"
Do you mean both input and output of the buck converter?
As stated the inverter board will likely handle this part of it without modification.

@ Ian0
It's a 3 phase, yeah. Thanks for the elaboration.
Only aiming for 80% as full power as is not needed. (200-250w / 300 is OK)
Thanks for the note re. converter may be able to manage the ripple by itself.
I will have to put it through its' paces practically and find out - I suppose it can't damage anything.

@ LowQCab
Not sure what you mean?
Never, would I commit such a travesty - lol

Am only planning to use one adjustable converter (provides a range of voltages)
(No current limit pot. on this but it's not needed)

Feeding in 40-60V from PMA / 3P rectifier and want 12v @ about 20A out the other end,
as efficiently as possible.

Included in the circuit will be 100v / 10A V/A LCD Shunt, and a 3 way switch (for dump load)
There will be minimal heat loss in this system.


Joined Aug 7, 2020
There are circuits that can cope with a lot of ripple on the input - think PFC front end!
It would have an additional advantage in that it keeps the current sinusoidal thus reducing the power loss in the windings.
Unfortunately, the PFC front end circuit is a boost converter, so the voltage you get out of it has to be more than the peak voltage.
Any buck regulator worth its salt should cope with 14% ripple.