Capacitors have a “leakage current”, i.e. when you charge them up fully, the leakage current will cause them to slowly lose that charge. I was wondering about this in the context of an ultra-low power JeeNode, which has a 10 µF buffer cap right after the voltage regulator. Does its leakage affect battery lifetimes?
Time to do a quick test – I used the 47 µF 25V cap included with JeeNode kits these days:
So how do you measure leakage currents, which are bound to be very small at 3.3V? Well, you could charge up the cap and then insert a multimeter in series in its most sensitive range. This multimeter goes down to 0.1 µA resolution, although its accuracy is specified as 1.6 % + 5 digits, so the really low values aren’t very precise.
A simpler way is to use the RC time constant as a basis. The idea is that a real-world cap can be treated like a perfect cap (which would keep its charge forever) plus a resistor in parallel. That resistor merely “happens” to be situated inside the cap.
What I did was charge the cap from a 3x AA battery pack which was just about 4.0V, then disconnect the battery and watch the discharge on the oscilloscope:
As you can see, it took 500 seconds for the charge in the capacitor to drop by some 2.5V – note the exponential decay, which matches the assumption that the leakage comes from a fixed resistance.
Can we derive the leakage from this? Sure we can!
The formula for RC discharge is:
T = R x C
Where T (in seconds) is the time for the cap to discharge by 63.2 percent, R is the discharge resistor (in ohms), and C is the capacitor size (in farads).
Above, it took 500 seconds to drop from 3.98 V to 1.48 V, which by pure accident is almost exactly 63.2 %, so T = 500 and C = 0.000,047 – giving us all the info needed to calculate R = 500 / 0.000,046 = 10638298 ≈ 10.6 MΩ.
Using ohm’s law (E = I x R), that means the leakage current at the start is 4 V / 10.6 MΩ = 0.376 µA.
The good news is that such a result would not be of any concern with ultra-low power JeeNodes – the regulator + ATmega + RFM12B use an order of magnitude more than that, even when powered down.
But the bad news is that this result is in fact completely bogus: to measure the charge, I placed the oscilloscope probe over the cap, and it happens to have 10 MΩ internal resistance itself. So basically the entire discharge behavior shown above was caused by the probe i.s.o. the capacitor’s own leakage!
So it looks like I’ll need a different setup to measure real leakage, which is probably in the nanoamp range…