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

Question

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.

Answer 1

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.

Answer 2

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.

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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.

Answer 3

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.