# Calculate charge time of a capacitor through a current limiting power supply?

Is there a formula or a way to calculate Charge time of a capacitor given that it is charged by a current limiting power supply. This would lengthen the charge time but is there a way to calculate this due to the power supply limitations. Let's say that we are charging a 1uF Capacitor to 5,000VDC through 20kOhms of resistance in series.

• I realize it is just an example, but I'd be a tiny little bit careful when charging capacitors to 5kV on your work bench. Otherwise good question. Mar 8, 2013 at 21:48

As it is not a piecewise continuous system, you'll need to calculate, in two steps:

First: you'll need to calculate the time of charging the capacitor until it reaches

$$(Vb-Vc)/R = Imax$$

with constant current of Imax. if the current is constant that capacitance does not change this is a simple straight ramp curve upto the point where the current is no longer limited by the constant current.

Second: you can use the equations for power supply without limited current power supply.

Which are the exponential curves from that point onward to the asymptote.

– user17592
Mar 8, 2013 at 15:02
• OK. My fault... but I can't comment yet. Mar 8, 2013 at 15:07
• Jean this is a perfectly good answer, and you're welcome to create your own. Mar 8, 2013 at 15:21
• @rawbrawb: Camil's complaint was about a comment I made to him (in my answer, since i could not comment others commentaries yet) that there was a current limitation, then Imax could not be Vb/R. I edited the answer then. Mar 8, 2013 at 15:25
• @JeanWaghetti I edited your answer, if you don't like please rollback, just some feedback. Mar 8, 2013 at 15:25

No, there is no formula to calculate charge time of a capacitor. Or yes, there is: $t_C=\infty$: the capacitor will never get 100% charged. What you can do is calculating the charge on a given moment, as shown here: Specifically interesting is the formula for Q as shown in the graph:

$$Q=CV_b(1-e^{-\frac{t}{RC}})$$

We have $C=0.001F$, $V_b=5000V$ and $R=20000\Omega$. We can calculate our time constant, $RC=R\cdot{}C=0.02s$. Our formula now is:

$$Q=0.001\cdot5000\cdot(1-e^{-\frac{t}{0.02}})=5\cdot(1-e^{-\frac{t}{0.02}})$$

We can rewrite this so that t becomes a function of Q:

$$t\approx-0.02 log(0.2 (5\cdot-Q))$$

Now, just fill in the charge you want to reach and you'll get the time needed.

As rawbrawb mentions in the comments, we generally use as a rule of thumb that the capacitor is loaded after 5 to 6 time constants. So, in your case, that would be 0.1 to 0.12 seconds.

• In a purely mathematical sense, yes charge time is $\infty$ but the general rule of thumb is 5 to 6 time constants. Because in reality there are noise floors or even the precision of the system. Mar 8, 2013 at 15:13

If it is a normal PSU with a current limiter then your cap will charge linearly. Think of it as charging a cap with a current source.

$$Q = C\cdot{}U = 1\mu{}\textrm{F} \cdot{} 5000 \textrm{V} = 5\textrm{mC}$$ $$t_{charge} = \frac{Q}{I_{lim}}$$ Assuming 250mA current limit ($5\textrm{kV}/20\textrm{k}\Omega{}$) your cap will charge in approximately 20ms.

• 20ms is only true when there is only current limiting and no series resistance. Mar 8, 2013 at 21:52