Bleed resistors for series electrolytic caps?

Hello,

we are doing a 330W offline full bridge .

Do we need bleed resistors across our 200V rated electrolytics at the input.?
-its two 200V rated electrolytics in series of value 680uF.

Sometimes the smps is connected to 265VAC , but other times it is connected to 90VAC with a voltage doubler link for the rectifier.

How do we size these resistors.

-i presume its to do with the leakage current that flows through each capacitor....and the fact that its not quite the same in each, and so one of them charges all the way up to the rail?

When high values tank caps are used in series the leakages and high ripple current are sufficient to charge both the caps .
However , when 2200ufd or higher in series ,we generally add 1M ohm x 2 with center point to charge both equally .
Be careful to select the matching selection in pairs

i thought that the leakage current through a capacitor didnt actually charge it?...but just sort of leaks through it?

Yes it does pass thru the top but charge is available for getting stored thru the other . At least a part of it which is biasing the other for build up of capacitance ..
The ripple current passing thru one gets into the second thru the series impedance of the first .

1.

its two 200V rated electrolytics in series of value 680uF.

Click to expand...

Capacitors in series are calculated by (A x B) / (A + B). You don't have 680uF (assuming that's what you wanted). Rather you have half that.

2.

i presume its to do with the leakage current that flows through each capacitor....and the fact that its not quite the same in each, and so one of them charges all the way up to the rail?

Click to expand...

Capacitors in series do not necessarily attain equal charge. One usually acquires more than the other. If its V rating is exceeded then it won't survive.

You really need a single capacitor, rated for the highest crest of your incoming AC voltage.

3.

Do we need bleed resistors across our 200V rated electrolytics at the input.?

Click to expand...

Good idea.

If you wish to bleed off the charge in 5 seconds, then you want a time constant of 1 second. The formula for time constant and capacitors is R x C.

With 680uF, this comes out to 1470 ohms. When you apply 265 V (I'm taking the highest value you mentioned) it will conduct 180mA continuously and will dissipate (waste) 48W continuously while power is applied.

Not what you need.

Say you allow bleed off time of 20 seconds. Then the resistor can be 29.4 K. It will pass 9 mA continuously and waste 2.4 W continuously.

Your bleedoff resistor(s) should be rated for twice the expected wattage. So a 5W rating is good.

Etc. These are just sample calculations.

Note: You need a bleeder resistor across each capacitor, if you keep the caps in series. When caps are in series, one of them may have acquired a higher charge (as stated earlier). Then as is discharges it could drive the other electrolytic past zero into reverse polarity (if you try to use a single resistor). That tends to ruin a 'lytic.

---------- Post added at 13:48 ---------- Previous post was at 13:15 ----------

Vimalkhanna's advice is valid. Putting 1 meg resistors across each cap.

In this application they serve more as equalizing resistors than bleeder resistors.

The two caps must be matched as to uF amount.

See if you can use lower than 1 meg. The idea is to prevent unequal charges on the caps.

If you use 1/2 W resistors then the value can be 281k each. At 132V per resistor that makes 1.9 mA flow. The resistors are dissipating 1/4W each. That's half what they're rated for, which gives you the recommended safety margin.

I see 220K resistors in this place in many ATX power supplies.

Hi,

> How do we size these resistors.

What I write is based on my experience with UL tests:

1) The UL norm that you should read is the UL60950

2) UL60950 defines a "Permanently Connected Power Supply", a power supply that has a screw connector instead of a plug.

3) Permanently connected power supplies DO NOT have to pass the "discharge time after unplug" test.
In this case, you don't need any BLEEDING RESISTOR

4) If the power supply has a plug, than the "discharge time after unplug" test has to be passed.
Read UL60950 or ask an UL safety engineer how the test is done and the limits.

5) If the 2 resistors are SMD, than I suggest you to use an 1206 or 1812 package.

6) The value must be tuned manually.

Enrico Migliore

I think primary goal here is ensure equal voltage across two series resistor. For that I would say 1Meg is bit high. I would choose anywhere between 200K to 500K depending upon how much power loss one can afford. If there are a safety regulation to be followed, then choose "bleed" resistor to ensure capacitor discharge to sufficiently low voltage in time specified by standards. This could be same of different set of resistor, it all depends upon on time allowed to discharge input capacitor in load condition.

These resistors serve ptotection purpose also, to discharge capacitor when power is off. These capacitors can stay in charged state for very long time. People are killed due to charged capacitors in gadgets not in use for last 20 years.