I have a maxwell 1200F 2.7V ultracapacitor which I charge using a 16V supply and a voltage divider in parallel with the capacitor. It's a really slow charge.
I also have a 4V supply which has higher current ability. I want to charge the cap directly from it.
I was wondering if it is safe to charge it up directly with that supply. It will be under constant supervision and I will disconnect it as soon as it hits 2.5V. Is this safe?
If you think it can be done, can you post some references and sources for further reading? I don't want a supercap blowing up in my face...
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\$\begingroup\$ That should work as far as the capacitor is concerned, but the PSU might not like it. It all depends on that; it might draw too much current from it. \$\endgroup\$– JodesCommented Apr 13, 2015 at 11:13
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\$\begingroup\$ What about a power supply at 24V 6A. The capacitor has maximum rated voltage only of 2.7V \$\endgroup\$– AvZCommented Apr 13, 2015 at 11:14
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\$\begingroup\$ If you're measuring the voltage at the capacitor, and cut it off when the voltage reaches 2.7v or less, then it will not be damaged. However, the PSU will see it as a short (unless you have a resistor as described in my answer) and it may burn out, or cut out, etc etc) \$\endgroup\$– JodesCommented Apr 13, 2015 at 11:21
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\$\begingroup\$ The only further reading you need to do is find out if your PSU can basically be short circuited. Either it was designed to be able to do it, or it wasn't. If it wasn't, then it's unsafe to do so. Find the PSU's datasheet, or contact the manufacturer. Otherwise, use a resistor as per my answer. It's not to do with what the supercap can handle. As long as the voltage across the capacitor doesn't exceed its rating, it will be fine. As you said, you'd be monitoring the voltage, so that side is fine, unless you exceed the capacitor's current rating - but that won't happen with yours. \$\endgroup\$– JodesCommented Apr 13, 2015 at 17:32
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\$\begingroup\$ @Jodes I read somewhere that applying higher voltage across the cap can cause the electrolyte to boil and go boom. \$\endgroup\$– AvZCommented Apr 13, 2015 at 18:55
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1 Answer
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That should work as far as the capacitor is concerned, but the PSU might not like it. It all depends on that; it might draw too much current from it. But what you could do is buy a resistor to put in series. You can use the following two formulas to work out the resistance and power rating they'd need to be:
P = I V
R = V / I
So if you want to use the 4v PSU and it has a 2A rating, the resistor you should use is:
R = 4v / 2A = 2 ohms (or higher)
And the resistor's power rating should be: P = 4v x 2A = 8W (or higher).
The 4v PSU sounds like the best choice, because less power would be wasted in the resistor while charging the capacitor. Also, since it has a high current rating, it should be faster than the other PSUs. [However, as the capacitor charges more fully, it will slow down charge rate to about a third. (4-2.5)/4 = 37% speed.]
If you had a 150v PSU, you would waste about 98% of power but it would maintain 99% of charging speed.
To help you decide on a PSU: for the fastest charging speed, you want a PSU with the highest current rating. (So a 24v, 2A PSU would charge at the same speed as a 4v, 2A PSU). But the closer the PSU's voltage is to 2.7v, the less power you will waste on the resistor. So if you used a 24v, 2A PSU and a 4v, 2A PSU with resistors as calculated above, they would charge it at the same rate. But the 24v PSU would consume a lot more power doing so, wasting it through the resistor.
simulate this circuit – Schematic created using CircuitLab
You could use the following circuit to stop at a given voltage:
That should charge the capacitor up to about 2.3v. (A bit of safety margin is a good thing)
