Possible Duplicate:
How to discharge smoothing capacitors?

I have an aluminum capacitor (~600-800uF) smoothing a DC current (~60-70V, ~50-70mA). I have a resistor in parallel to drain the capacitor when powered off. It is important that this capacitor is drained quickly (under 3 sec) when powered off and also that it provides maximum power when powered on.

I've learned that using a lower resistance drains the capacitor quicker but sacrifices running power as wasted heat. On the other hand, a higher resistance provides optimal power and minimal heat but simply drains the capacitor too slowly.

Is there a simple (cheap and small) way to switch a resistor on only when the capacitor needs to be drained? This way maximum power can be achieved and the bleed time can be precisely controlled.


  • 1
    \$\begingroup\$ Here's a concise description of exactly what I'm after ('Dual Bleeder' section) -- en.wikipedia.org/wiki/Bleeder_resistor#Dual_bleeder \$\endgroup\$
    – paulwal222
    Oct 5 '12 at 21:21
  • \$\begingroup\$ Seeing the schematic will make it easier to see what you have available to keep it simple. For example, how is the power switched on and off and is there any intelligence or logic like a power_good signal? You'd need ~750ohms to keep the drain time under 3s which would be a lot of wasted power, indeed. \$\endgroup\$ Oct 5 '12 at 22:34
  • \$\begingroup\$ Assume the worst: power is switched off when a hurricane takes out the transformer down the street. \$\endgroup\$
    – Kaz
    Oct 6 '12 at 4:42

Difficult to answer without a schematic, but if you have at some point a signal that drops to -near- ground when powered down, but is high when power is on, you can use it to drive the gate of a p-mos (or the base of a pnp) which in turn controls whether current flows through your draining resistor or not.

At the gate/base you would have a pull down resistor, so that the transistor defaults to 'on' (draining your cap).

This way negligible power is wasted during normal operation, and the cap is drained when power is off.

  • \$\begingroup\$ Your N-MOSFET should be a P-MOSFET, I presume. \$\endgroup\$
    – stevenvh
    Oct 6 '12 at 18:27

Discharging a cap on loss of power, without the bleeder resistor being in the circuit at all times, can be done with some transistor circuits.

For instance the one in this 1979 patent: http://www.google.com/patents/US7200015

When the power supply is turned off and the load is removed, the transistor is automatically rendered conductive with the removal of the pulsed output of the power supply's power transformer to the transistor, with the energy stored in the capacitors safely and quickly discharged to the output return via the bleed resistor which is placed in circuit by the conducting transistor.

  • \$\begingroup\$ So you have a second small rectifier and smoothing circuit on the basis of a PNP transistor, which controls the bleeder. \$\endgroup\$
    – starblue
    Oct 6 '12 at 7:34

Not the answer you're looking for? Browse other questions tagged or ask your own question.