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Old 10-19-2017, 07:08 PM   #52
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Default Re: How to Recondition (Reform) Electrolytic Capacitors and Why

Originally Posted by acercanto View Post
Is 250mA too much, even for such a big cap as this?
Yes, probably.

Trying to guess how much current is "too much" or "too little" can prove a little hard. This is one reason why I suggested in my first post to use a series resistor of 1 kOhms or greater - it simplifies the process. And after all, that was the recommendation in some Panasonic datasheets too.

With a 1-kOhm resistor, the initial current will be limited as follows:
Supply Voltage --------- Current Limit
10 V ------------------- 10 mA
20 V ------------------- 20 mA
30 V ------------------- 30 mA
100 V ------------------ 100 mA

However, this initial current above will decrease as the capacitor chargers up and the voltage across its terminal increases. So in the first case with 10V supply voltage: once the capacitor charges up to 5V, the 1-kOhm will limit the current to 5 mA, and once the capacitor charges up to 9V, the 1-kOhm resistor will limit the current down to 1 mA. However, if the capacitor is leaky for some reason and it want to draw 2 mA of leakage current, then the voltage across its terminals will never be able to go past 8V with that 10V supply and 1-kOhm resistor.

That said, most capacitor datasheets give a worst-case leakage current formula for each capacitor series. It is usually given as:
I = 0.01 * C * V
I = 0.03 * C * V
where I is the capacitor's internal leakage current in uA, C is the capacitance of the capacitor in uF, and V is the voltage across the capacitor.

I didn't want to include that formula in my first post, as I wanted to make the cap reforming process as simple as possible for people. However, I do recommend to run through this calculation, as it will give a worst-case scenario of what to expect for the leakage current from any given capacitor.

In your case with the 19V, 18000 uF capacitors, the maximum leakage current for each capacitor should be:
I = 0.03 * 18000 * 19 = 10260 uA = 10.26 mA

Thus, if any of your reformed capacitors draw more than about 11 mA after fully charging them to 19V, they may be too out-of-spec to use. As for this:
Originally Posted by acercanto View Post
Is that first one dead, or can I revive it?
See how much current it draws after charging up. If it's 10x more than the leakage current above, I definitely wouldn't use it.

Originally Posted by acercanto View Post
Without any special equipment, is there any way to test the ESR of them?
You can do a "spark" test. Charge the cap to at least 5V and short its leads on a metal surface. Do beware that for a large 18000 uF cap like that, you *will* get a lot of sparks and possibly pit whatever metal you touch. So keep away from flammable surfaces and substances. And probably a good idea to wear safety glasses/goggles too, just in case some hot piece of metal try to go anywhere.

Last edited by momaka; 10-19-2017 at 07:13 PM..
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