What is the function of this capacitor ?

[huu_tri0101]

What does it do ? Can I change it with a higher(uF and Voltage) capacitor ?
Thank for reading.

[M0BSW]

I know you can put  in higher voltage ones , don't think you want to mess around with the  uf,  I been told that you can increase the working volt a bit, I had a monitor that had blown caps I left the capacitance the same , but changed the 16v to 25v. And it works just fine, sorry I can't be of more help, learning myself.

[Excavatoree]

It's always safest to stay with the same capacitance value.  However, if you know the circuit, and know the consequences of changing, you may be able to use something else.  In this case, it appears the capacitors are on the output of the power supply.  These could be a bit larger, but if you don't have the experience to know how much larger is OK, you shouldn't change.   Someone will be along to reply that knows more than I do, I'm sure.

As for voltage rating, I always understood that you could go to a higher voltage rating without consequence - eventually the capacitor would be too large to fit.  However, I'll defer to those who know more than I do.

[mariush]

You can increase the voltage rating of the capacitors, the minimum rating just has to be more than what the power supply outputs. The voltage the capacitors will ever see is the 18-20v - that's the maximum voltage the power supply will output, it's an 18v 5A power supply.  But it's always safer to use capacitors with at least about 20-30% margin...

They're using 50v rated capacitors because being of larger size, they can handle more current. Jamicon otherwise is just average, they're not special at any characteristic.

Otherwise, increasing the capacitance is unlikely to make much of a difference. Increasing it too much will make the power supply react slower to changing the output voltage, if you turn the knob to lower the voltage, it will take a bit more time to actually see that new voltage on the output (as the capacitors will have to discharge). The less current you pull from the power supply, the more time it's going to take. 

I'm not sure about the following (I didn't study such power supplies a lot) but the capacitors might also be required there to be a sort of protection for the moment you actually turn on the power supply.

At that point, the pass transistors and the opamps that control the voltage and current may not be yet fully set up so for a few microseconds/ms the power supply may output the full 18v (or some value depending if relays are used to connect several transformer secondaries) and the capacitors act like a short and basically store that energy not letting it appear on the output.

[IanM]

Just my $0.02: Replace them with the same capacitance value of 220uf and any voltage rating of >=50.

[qno]

Why do you want to replace them?

[Flávio V]

Probably those jamicons are a capacitive filter to reduce the ripple voltage....i might be wrong but i think i am right...the one i had for free had 4 electros and 1 ceramic in the same place as those so i am almost sure that they are for that...

[alm]

These caps are probably directly connected to the output. You can set the current limit to 0.01A, but if you short the outputs, the caps will dump all their charge almost instantly. Especially if you get nice low ESR caps. This could easily be tens of amps until the cap is empty. Increasing the capacitance is not a good idea, if the supply needs that much capacitance to get ripple down to an acceptable level, it's poorly designed. Good power supplies usually have only a small amount of capacitance on the output.