[SOLVED] Basic capacitor question

Its been too long since I learned this stuff. A capacitor rated to 900v & 600uF is accross a 500v 1amp power supply. The questions I have are
1. Whats the max charge in micro farads will capacitor obtain from supply?
2. How long to receive that charge?

I know the formula for charge in a capacitor is Q=C*Vsupply [1-e^-t/R*C]; were R=cap esr = 3mOhms & C = 600uF but I am uncertain because the cap rating is above power supply volts.

With supply 500V @ 1A, implies a resistance of 500 ohms.

Therefore RC time constant is 500 x 600 x 1 millionth, or 0.3 second.

Convention says a capacitor is fully charged after 5 time constants.

Will it actually charge to 600uF? 500v is far below its rated voltage.

900 V defines the maximum capacity of the capacitor. If you are charging using 500 V, it will get charged uptill 500 V.

That is Q = 600uF x 500 V.

The cap rating needs to be greater than maximum voltage which you expect it to charge to. This means the 900V rating is extra safe. You could get by with a 550 V rating.

Unit of charge is As, not Farad (or As/V).

Capacitance of film capacitors is completely voltage independent, capacitance of electrolytic capacitors almost.

Since I don't trust my math skills I rely on the online calculators. For capacitor charging I go to
"Capacitor charge and discharge - Must Calculate". When I plug the OPs values in I get a charge time of approximately 1.4 seconds at 1 amp. But brads number gives a charge time of 0.3 seconds.
I'm wondering why the online calculator is so much longer. The only thing I can figure is that i am not using the calculator properly. These are the numbers I plugged in.

From voltage- 0
To voltage - 500
Supply voltage- 500
Capacitance - 600u
Resistance - 500
Time - 1.5 seconds

- - - Updated - - -

I think I see my mistake. The 0.3 represents 1 time constant and not complete charge time.

Hi,

Brad´s math calculates tau. Tau = R x C.
It is a non_constant_current charging of a cpaacitor.

After one tau the voltage of the capacitor rises from 0 to about 63% of the supply voltage. In other words the error is about 37% or 0.37
After n x tau the error will be 0.37 ^n
exactly the remaining error is 1/euler which is: 0.36788...
1 tau => error of 0.37^1 = 0.37, charged to 63%
2 tau => error of 0.37^2 = 0.135, charged to 86,5%
3 tau => error of 0.37^3 = 0.0498, charged to 95%
...
This is true for every (constant) R and (constant) C.

The error will never reach 0, but it becomes very close to 0. The capacitor voltage will never reach supply voltage.

*****
Brad calculated correctly tau to be R x C = 500 Ohms x 600uF = 0.3s.

Klaus

Yes, since one time constant calculates as 0.3 sec, then 5 time constants make 1.5 sec which is close to your result for overall charge time.