# Capacitor sizing for power supply

Discussion in 'General Electronics Chat' started by Mad Professor, Nov 4, 2009.

Apr 15, 2009
133
1
Good Day All.

Can someone please explain to me how I go about working out the right size capacitor(s) to use in a simple DC Power Supply.

The power supply is 34volts DC and can supply upto 10amps.

I am going to be driving 3 stepper motors pulling around 8 to 9 amps total.

I want to pick the best sized capacitor(s) to try and take the as much load off the transformer.

Is it best to have one big capacitor or a bank of smaller capacitors?

Best Regards.

2. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
50/60Hz?
fullwave/halfwave rectifier?
Rectifier configuration, i.e. two/four rectifiers?
transformer secondary voltage (AC)?
How much ripple can you tolerate? (Vp-p)?
Are you using a regulator or not?

3. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283

Apr 15, 2009
133
1
MikeML: Input voltage is 240vac 50Hz, Output voltage is 27vac, and I am using a fullwave bridge rectifier.

beenthere: I will have a look at the link.

5. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
You only answered half my questions.

27Vac means that the capacitor will charge to a peak value of 1.414*27 - 2*Vf (forward drop of the rectifiers). Vf is about 1V for most large silicon rectifiers. So unloaded voltage will be ~ 38 - 2 = 36V.

Under load, the capacitor will charge to a value less than the 36V, mostly due to the impedance of the transformer (caused by winding resistance and non-ideal coupling). Hard to know what that drop will be because you gave no transformers specs, but I'm guessing about a 3V drop if the transformer is not overloaded. So assume 33V.

During a half cycle of the 50Hz AC waveform, the capacitor will be discharging due to the 10A load current. The time until the next peak which recharges the capacitor will be slightly less than 10msec, so the voltage sag can be computed thus:

C*ΔV = q = I*t, where ΔV is the voltage sag (ripple), C is the capacitance in Farads, q is the charge lost in Coulombs, I is Amps, t is time in seconds.

So, if you will tolerate a ripple of 5V, C = I*t/ΔV = 10 * 10e-3 / 5 = 2e-3 F or 20,000 uF. 2V ripple would require 50,000 uF.

So the regulation is not very good, starting out at 36V no-load, and ending up at a minimum voltage at the lowest part of the ripple waveform of 33-5 = 28V in one case and 31V in the other.

Sep 30, 2009
4,092
1,673