How to discharge smoothing capacitors?

Appropriate bleeder resistors are the usual solution. They aren't usually switched, although they can be.

The value depends on the time you require to discharge the capacitors. The formula is

$$ V_{t} = V_{0} \, e^{ -t / RC } $$

where \$V_{t}\$ is the voltage at time t and \$V_{0}\$ is the initial voltage at time 0. It's an exponential function, so I'd just assume 1/10 of the initial voltage.

It isn't a power function, as someone edited it!

You should find that the power taken by the bleeder resistors is negligible compared to the 120W capability of the supply.


What you want is a switch which is open when the circuit is powered, and closed when it is switched off. When closed it should discharge the capacitor over a resistor. You don't want to short the capacitor; they don't like that. Two approaches I can think of (from the top of my head):

  1. Use a depletion MOSFET as the switch. Depletion MOSFETs conduct when there's no voltage applied to the gate. Apply a voltage to switch it off. This voltage can not be derived from the capacitor you want to discharge! Otherwise the MOSFET would never be switched off. (You think about this, if you don't get it tell me, and I'll try to explain.)

  2. Use an ordinary NPN transitor which you drive from the capacitor's voltage. As long as there's a voltage present, it will discharge. Pull the transistor's base to ground if the circuit is switched on. Again, the voltage to do this is from a separate power supply.


Such huge caps seems to be an overkill... If it's regulated (linear/pulse) you would need to tune it till ripple would be acceptable with much less output capacitor. If you have alot of high-freq noise - you would need to add several ceramic caps. Also, make sure that your inductor at the output is calculated correctly.