Power supply ripple calculation

Thread Starter

ischonfeld

Joined Jun 22, 2019
63
I'm trying to calculate a reasonable value for a filter capacitor on a full wave rectifier setup. I'm expecting a load of about 4 amps. I know it's backwards, but I started with a 4700uf capacitor just to see what the ripple voltage would be.

My application can tolerate a decent amount of ripple - the actual output of this circuit will be fed into a Recom DC-DC switching converter that can take a pretty wide range of DC input voltage and produce 12V (essentially a more efficient version of an L7812 regulator).

I ran an LTspice simulation (only the second time I've tried LTspice):

Ripple Sim.jpg

The simulation shows a ripple voltage variation of about 2V.

I also calculated the ripple using the following formula:

dV = (I * dt)/C (4 * .0083)/ .0047 = 7.1V

I'm certainly missing something here, but I can't see what. I know that according to the simulation, the average current is closer to 3.4 A in this setup, but that doesn't account for the big difference between the sim and the calculation.

What aren't I seeing?

Thank you.
 

AnalogKid

Joined Aug 1, 2013
12,043
There is an approximation equation that can be used without calculus. If you assume a constant current load, then the standard integral equation for a capacitor reduces to

EC = it

E - voltage change across the filter cap - the peak-to-peak ripple voltage
C - filter capacitor value on farads
i - discharge current.
t - ripple period - 1/120 with a bridge circuit in the US

With a 4700 uF cap and a 5 A load in the US:

E - (i x t) / C = (5 / 120) / .0047 = 8.9 V p-p

With a 3.4 A average load:

E - (i x t) / C = (3.25 / 120) / .0047 = 5.76 V p-p

With a resistive load and not-very-deep ripple, this can get you to within 10% of the theoretical value. But with a downstream voltage regulator and a constant current load on it, the accuracy is much better.

ak
 
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