# What is the formula for charging a capacitor with constant current?

I read that the formula for calculating the time for a capacitor to charge with constant voltage is 5·τ = 5·(R·C) which is derived from the natural logarithm.

In another book I read that if you charged a capacitor with a constant current, the voltage would increase linear with time. Is this true, and if it is, what is the formula used for calculating this? Would a complete voltage charge be possible with a constant current?

• To achieve a constant current through a capacitor implies that the voltage across the capacitor increases without limit. In reality, "without limit" is limited by the capacitor exploding. 5 tau is generally taken to be "good enough" at 99.3% charged. Commented Nov 20, 2016 at 18:28
• A real constant current source such as a LM334 will "drop out" at its lower compliance limit and tail the charge current off as a result when the cap's fully charged up, provided the cap is rated well enough to not go bang first. Commented Nov 20, 2016 at 18:31

Normally I would let you go and look as this is not a hard question to solve, but as I am feeling generous here is how we get there:

From fundamentals, we know that $Q =CV$

If we take the derivative with respect to time (remembering that $I = \frac {Q} {T}$) we yield

$i = C\frac {dv} {dt}$

Rearranging, we find that $\frac {i} {C} = \frac {dv} {dt}$

Therefore charging a capacitor from a constant current yields a linear ramp (up to the compliance of the current source).

I will leave finding the solution in terms of time versus some voltage to you.

• charge time = C x V/I Commented Jan 27, 2023 at 17:32

Here is the equation:

charge time = C x V/I

• This should be the accepted answer. Rather than the pompous reply which stops short of actually answering the question. Commented Apr 21 at 4:31