Verifying my capacitor calculation

Question

I just received in the mail a 0-60 V power supply that has the capability of outputting 24 A. I needed a high-voltage, high-current power supply to help design and test high-powered LED arrays for grow lights.

The issue that I'm having is that the power coming out of the PSU is not the cleanest. I'm getting a ripple of around 350 mV peak-to-peak. So I'm thinking of adding in a capacitor to help smooth out the voltage.

I do realize there is a bit more than just adding in a capacitor. However, just to stay on topic I only want to talk about the capacitor specifications.

I used the formula C = I/(2·f) · y

Which broken down is: C = Expected Amperage / (2 × the input voltage frequency) × (the desired ripple voltage)

Here is the actual calculation: C = 24 A/120 hz x 0.1 V

so C = 0.021F or 21000 μF

That would mean I would need a 60+ V, 21000 μF capacitor. Please correct me if I'm wrong, but that would be a dangerous capacitor that could cause harm. Due to safety concerns, I would greatly appreciate if someone could please verify if my calculations are correct.

Answer 1

The formula you have used tells you what the peak current into the capacitor will be for a given ripple voltage and frequency. It doesn't mean that there will be any filtering effect (although it's likely there will be some effect). However, it will not guarantee that there will be a significant reduction in the ripple voltage if that's what you are looking for.

Verifying my capacitor calculation

The basic formula for a capacitor is this: \$I = C\cdot\frac{dv}{dt}\$ and it can be re-arranged:

• Assume a triangle wave voltage ripple of \$V_{P-P}\$

• The voltage will change in half a cycle of the triangle wave

• This is equivalent to \$\frac{1}{2\cdot f}\$ where \$f\$ is the ripple frequency

• This means that \$\frac{dv}{dt}\$ becomes \$V_{P-P}\cdot 2\cdot f\$

• Hence, \$C = \frac{I}{V_{P-P}\cdot 2\cdot f}\$

So, for a given ripple voltage across the capacitor, you can calculate the peak current into the capacitor. The mistake you are making is assuming that there will be any significant attenuation. I mean, if the supply can supply 24 amps, then a capacitor of 21,000 μF will draw that current when there is a ripple voltage of 350 mVp-p.

What might be required is a two component filter; a series resistor followed by the capacitor. However, when you do the maths, to get maybe a 10:1 reduction in ripple voltage without using a resistor that's the size of a water-cooled house-brick, the capacitance will need to be over 1 farad.

This is all very theoretical and, you might argue that your power supply may have some output impedance but, it's not going to be much. After all, if it has an output impedance of 0.1 ohm, at 24 amps, it will dissipate a power of 57.6 watts so, it seems unlikely that this will be the case.

I think your best bet is forget about solving the ripple problem and live with the problem by designing your LED circuits not to be bothered by it.