How do I calculate the resistor size to bleed a capacitor?

Question

I was wondering how to calculate the proper size for both resistance and wattage to bleed out a capacitor. The capacitor if for smoothing a DC motor and will be around 10,000 to 15,000 uf at 48 volts.

Answer 1

The time constant for a resistor and capacitor is RC where R is in Ohms and C is in Farads. The result is seconds. When discharging a capacitor the capacitor will reduce its voltage to 37% of its value every time constant.

For power use, P = V^2/R where P is watts, V is volts and R is Ohms.

So assume that you want to use a 1/2 watt resistor, with V = 48 then R = 4608. The nearest 5% value would be 4.7K. The time constant for C = 10000uf and R=4.7k would be 47 seconds. So the capacitor would be discharged in about 3 time constants or about 150 seconds.

Answer 2

It's basically dealers choice. Pick a time you want to discarge the capacitor in. Bleeders are there for human safety so usually a second or so's more than adequate. Calculate the resistor from the time constant and then using that value caculate the power dissipation. Double that value for safety margins.

Answer 3

reliable rule is to easily use an particularly great resistor (not actual length) like 50,000 ohms or extra. this might save the amps very low and for this reason save the warmth low. i'm unsure how long the potential would be off to allow bleeding. If it desires to bleed rapidly, you will might desire to calculate the time and hit upon a compromise between that and the warmth produced.