I have a simple 12V 10 A power supply with just a transformer and a rectifier. After doing some research and simulations, I've added 3 10 mF capacitors in parallel to smooth out the output.

My problem is that after turning the supply off, capacitors remain charged for quite some time. I can get small sparks after shorting the output even 5 minutes after turning the supply off. Right now I only have a single LED connected to the capacitors and it takes more than 10 minutes for it to turn off completely after powering the supply down and the capacitors still aren't fully discharged when it turns off.

The most obvious way to solve the problem would be to put a resistor and a switch on the output and connect the resistor to the capacitors after turning off the supply by hand, but I'm hoping to get something a bit smarter and a bit safer.

Another point is that I want to use the supply's original case which has very little free volume now that I've added the capacitors, so just putting a ceramic 11 W resistor could be a problem because there would be very little free space around it for safe cooling.

  • \$\begingroup\$ 3 mF or 3 µF​​? \$\endgroup\$
    – endolith
    Commented Apr 6, 2011 at 19:14
  • 1
    \$\begingroup\$ @endolith 30 mF or 30000 µF . \$\endgroup\$
    – AndrejaKo
    Commented Apr 6, 2011 at 19:22

3 Answers 3


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.

  • \$\begingroup\$ How would I determine correct value? \$\endgroup\$
    – AndrejaKo
    Commented Apr 6, 2011 at 18:07
  • 2
    \$\begingroup\$ I used 1\$k \Omega\$ resistor and it's not on fire yet, so I guess this works. \$\endgroup\$
    – AndrejaKo
    Commented Apr 6, 2011 at 20:40
  • \$\begingroup\$ 30mF and 1k ohm at 12 V would mean it takes ~69 seconds to discharge to 1.2 volts. And, of course, the leakage current is ~12 mA. stevenvh's MOSFET idea hit me as well and seems reasonable to me, though I'm sure that (unlike me) he's actually seen it used IRL :-) \$\endgroup\$
    – exscape
    Commented Jun 19, 2012 at 11:51

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.

  • \$\begingroup\$ hi Steven, I have a 3KW strobe light LED array that has 81mF capacitance so that it can sustain 120Amp discharge for 5ms without dropping more than 4-5V on the rail. This all works nicely, but when I turn off the system power switch just like the OP of this question, the capacitor banks stay charged. I have actually already tested in my design the use of 4 infineon BSS159N depletion mode N channel MOSFETS, quickly shown here dropbox.com/s/1jv57olhhryco12/example.JPG . My circuit doesnt quite work though, as my 5V pull up (to turn them 'off' on power up) is powered by the 24v rail.. \$\endgroup\$
    – KyranF
    Commented Jun 26, 2014 at 7:13
  • \$\begingroup\$ So my question is, what could I do to convert my circuit so that I can somehow separate my power supply (which is 24V input, goes to cap banks and into a 5V DCDC module, which gives me my 5V rail..). As the entire system input is turned off, it takes quite a while (only a few LEDs) to discharge, as my MOSFETs manage to keep their voltage at their pins because I unfortunately forgot that my 5V supply is fed by the 24V and will not drop-out until the 24V rail dips to ~5.2V. One option i have is my controller gets Power over Ethernet which is independant of the 24V input! perhaps I could connect \$\endgroup\$
    – KyranF
    Commented Jun 26, 2014 at 7:15
  • \$\begingroup\$ something to the external power switch which indicates to my PoE powered controller to pull the MOSFET gates to 0V? \$\endgroup\$
    – KyranF
    Commented Jun 26, 2014 at 7:16

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.

  • \$\begingroup\$ There's no regulation and there's nothing to tune inside. The circuit is basically what I showed here plus a circuit beaker and LED on the output and fuse and a power switch on the input. I've had around 8 V of ripple on the output when the supply is unloaded, now I have around 0.1 V. +1 for mentioning high frequency noise. I totally forgot to measure the actual frequencies. Also, where would I add the inductor? \$\endgroup\$
    – AndrejaKo
    Commented Apr 7, 2011 at 9:40
  • \$\begingroup\$ You just add DCDC converter, which will guarantee 0.01V ripple with 100uF output cap and will be much much better that this. They does not cost much. \$\endgroup\$ Commented Apr 7, 2011 at 12:27
  • \$\begingroup\$ That's my plan. I'll eventually have one at the output and solve the problems. I was concerned that horrible rectifier and transformer side will cause problems to the regulator, so I added some extra capacitors. \$\endgroup\$
    – AndrejaKo
    Commented Apr 7, 2011 at 12:31

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