Say I have a Maxwell BCAP0005 supercap (2.7V, 5F), which has a leakage current of 0.015mA. I'd like to estimate the time it takes to discharge to a certain voltage.

I've tried applying a formula for constant current discharge,

$$ t = \frac{C}{V_\text{initial}-V_\text{discharge}}I $$

So, for $$V_\text{initial}=2.7V, V_\text{discharge} = 0V, C = 5F, I = 0.000015A$$ $$t = 900,000\text{ sec (10.4 days)}$$

And if $$V_\text{initial}=2.7V, V_\text{discharge} = 2.0V, C = 5F, I = 0.000015A$$ $$t = 233,333\text{ sec (9.7 days)}$$

But this seems like an oversimplification. For example, is leakage current constant? Does the ESR affect the discharge time? What other assumptions need to be clarified?


In practice the leakage current specs in the datasheet are only an approximation. Hopefully they will give you an upper bound, but that's about all. The real self-discharge time will vary greatly depending on just about everything. It will of course vary from cap to cap, but also by temperature, age, and lots of other things. Also, your circuit may be the biggest "leaker" of them all.

Your circuit will also have a "useful lower voltage limit", which will also vary depending on everything. Basically, some of your PCB's might stop functioning when the cap reaches 2.0v, but other PCB's might work down to 1.6v.

Simply put: the only way to be half-way sure about the self-discharge rates will be to build up a bunch of prototypes and test them. While that will be the most accurate way to figure that out, there will still be a lot of variations and future batches of your PCB's might discharge faster or slower than what you measured initially.


More importantly, your formula does not have units of time! The formula should be C*(Vi-Vd)/I. Always check units!

That said, there is a more accurate version of the formula which takes the exponential decay rate into account (constant leak resistance, not constant current): t = -RClog(Vi/Vd)


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