# How to calculate self-discharge time of capacitors given the leakage current?

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?