Say I have a 1F capacitor that is charged up to 5V. Then say I connect the cap to a circuit that draws 10 mA of current when operating between 3 and 5 V. What equation would I use to calculate the voltage across the capacitor, with respect to time, as it is discharging and powering the circuit?


4 Answers 4


charge on a cap is a linear product of capacitance and voltage, Q=CV. If you plan to drop from 5V to 3V, the charge you remove is 5V*1F - 3V*1F = 2V*1F = 2 Coulombs of charge. One Amp is one Coulomb per second, so 2C can provide 0.01A for 2C / (0.01 C/sec) or 200 seconds. If you actually withdraw charge from the cap at a constant current, the voltage on the cap will decrease from 5V to 3V linearly with time, given by Vcap(t) = 5 - 2*(t/200).

Of course, this assumes you have a load that draws a constant 10mA even while the voltage supplied to it changes. Common simple loads tend to have relatively constant impedance, which means that the current they draw will decrease as the cap voltage decreases, leading to the usual non-linear, decaying exponential voltage on the cap. That equation has the form of V(t) = V0 * exp(-t/RC).

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    \$\begingroup\$ "Common simple loads tend to have relatively constant impedance" -- these days switching power converters are common, and they tend to draw constant power, not constant current (meaning the current goes up w/ increasing voltage) \$\endgroup\$
    – Jason S
    Commented Oct 7, 2010 at 12:51
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    \$\begingroup\$ with constant power, wouldn't current go up with decreasing voltage? \$\endgroup\$
    – JustJeff
    Commented Oct 7, 2010 at 22:00

The general equation for the voltage across the capacitor is

\$ V = V_0+\dfrac{1}{C} \int {i dt}\$

In the special case where \$I\$ is constant this translates to

\$ V = V_0 + \dfrac{I \times t}{C} \$

We want to find \$t\$, so rearranging gives us

\$ t = \dfrac{C (V - V_0)}{I} = \dfrac{1F (3V - 5V)}{-10mA} = 200s\$ = 3 minutes and 20 seconds.

The more general solution is where \$I\$ is a function of time. I'll assume that the 10mA is the initial current, at \$V_0\$ = 5V. Then the discharge resistor \$R = \dfrac{5V}{10mA} = 500\Omega\$. The time constant \$RC\$ is then 500s. Then

\$ V = V_0 \times e^{\left(\dfrac{-t}{RC}\right)} \$


\$ t = -RC \times ln{\left(\dfrac{V}{V_0}\right)} = -500s \times ln{\left(\dfrac{3V}{5V}\right)} = 255s \$ = 4 minutes and 15 seconds.

This makes sense. Following an exponential discharge will get us at 3V later than with the linear discharge.

  • \$\begingroup\$ Your time calculation doesn't appear to take ESR into account. Where would that fit in? \$\endgroup\$ Commented Aug 22, 2018 at 1:39

\$ \Delta U = \dfrac{I \times T}{C}\$ for DC current only! (I - current, T - time, C - capacitance).

in general:

\$ u(t) = \dfrac{1}{C} \times \int{i dt}\$

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    \$\begingroup\$ So is that the total time? \$\endgroup\$ Commented Oct 7, 2010 at 12:00
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    \$\begingroup\$ wouldn't that be the time to dissipate 63 percent of the voltage? \$\endgroup\$ Commented Oct 7, 2010 at 12:01
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    \$\begingroup\$ @Jordan; T - yes that is total time; the 63% is the value change in one RC constant. In this case you are discharging through current not resistance. \$\endgroup\$ Commented Oct 7, 2010 at 12:08
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    \$\begingroup\$ He is asking about 5v down to 3v. Not asking for the time constant. \$\endgroup\$
    – Kellenjb
    Commented Oct 7, 2010 at 15:02
  • \$\begingroup\$ @Kellenjb, that is why I gave the equations on how the capacitor is discharged using constant current, the RC mention is to explain why he may think that T is 63% \$\endgroup\$ Commented Oct 7, 2010 at 20:22

The answer is already given above but this is the way I think about it:

Assuming a constant current: I=C*dV/dt --> dt=C*dV/I

dv=5V-3V =2V, I=10mA, C=1F --> dt=1F*2V/10mA= 200sec


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