Archive for September 14th, 2009

Maxwell 10 F Ultracapacitor: First Charge

Maxwell PC10 Ultracapacitors

Maxwell PC10 Ultracapacitors

My buddy Mark One dropped off a pair of Maxwell PC10 10 farad Ultracapacitors. We both recall our respective professors saying that a farad is an impractical unit, there’d never be such a thing as a 1 F capacitor, and it would be the size of a barn anyway…

These are 25x30x3 mm.

The downside, of course, is that they’re rated at 2.5 V DC with an absolute maximum of 2.7 V.

On the other paw, they have a maximum current of 2.5 A and a whopping 19 A short-circuit current. Serious risk of fire & personal injury there…

Charged one up from an AA NiMH cell I had lying around on the desk, which took a while, then let it discharge all by itself while taking notes. The results look like this:

10 uF Ultracap - Self DischargeTime    Voltage – mV
13:02    1353
13:08    1350
13:36    1338
13:56    1333
14:09    1329
14:21    1326
14:54    1318
15:13    1314
15:49    1308
16:06    1305
17:42    1291
18:49    1283
19:03    1282

Now, maybe that’s not exactly the extreme top left end of an exponential drop, but it looks close enough:

V(t) = V0 * exp (-t/τ)

Pick any two points on the curve to find τ, the time constant:

V(t1) / V(t2) = exp (-t1/τ) / exp (-t2/τ)

Take the log of both sides and remember that the log of a ratio is the difference of the logs:

log V(t1) – log V(t2) = (-t1 + t2) / τ

Plug in the first and last data points to get:

0.02341 = 21.6 ks / τ

Reshuffle and τ = 923 ks. Close enough to a megasecond for my purposes.

How to find the capacitance? Charge the cap up fram a pair of NiMH cells, discharge it at a constant current using a battery tester, thusly:

10 uF Ultracap - 100 mA Load

10 uF Ultracap - 100 mA Load

That curve isn’t exactly linear, but it’s close enough that we can use the familiar capacitor equation:


Reshuffle to get capacitance over there on the left side:

C = I * ΔT / ΔV

The lower axis is minutes, not seconds, with truly poor grid values. Eyeballometrically, call it 4 min * 60 = 240 seconds.

Plug in the appropriate numbers and find that

C = 0.1 A * 240 s / 2.5 V = 9.6 F.

Close enough.

Knowing τ and C, find the self-discharge resistance R = τ/C = 96 kΩ. That seems pretty low, but at 2 V it amounts to 25 µA. The cap’s self-discharge current is rated at 40 µA, so that’s well within spec.

Now, admittedly, the cap doesn’t hold much energy:

  • NiMH 2 x AA = 1 Ah @ 2.4 V = 2.4 Wh = 8600 Ws = 8600 J
  • Ultracap 10 F @ 2.4 V = 1/2 * C * V^2 = 29 J

But, heck, it’s pretty slick anyway… it’ll make a dandy backup power source for a clock I’m thinking of making.

Memo to Self: Datasheet says to add balancing resistors that carry 10x the self-discharge current when stacking in series. That’d be 10 kΩ, more or less, which seems scary-low.