2
\$\begingroup\$

I am hoping somebody could help me diagnose the cause of failing electrolytic capacitors in a circuit. The context is a charger circuit, more specifically its power supply unit:

Given a diode bridge rectifier that is connected on one set of terminals to the mains via a filter circuit and on the other side to two electrolytic capacitors connected in series. Parallel to each capacitor are two discharge-resistors. (And of course the actual consumer circuit.)

The problem is, one of the two capacitors in series, the one connected to the - terminal of the diode bridge, explodes. This fault occurs since the device was subjected to intermittent AC-supply: unstable frequency, probably over- and under-voltages.

I am wondering where to search for the problem:

  1. diode bridge? - I played around a bit with spice and the only way I could get reversed polarity on one of the capacitors was to assume two diodes short-circuited and the other two became isolators. This does not seem a very likely failure mode...
  2. circuit fault after the power supply unit? - Any hint on what to look for? The consumer circuit is non trivial...

The capacitors in question were 200Vdc, 1000uF for a device to be plugged into 230Vac. I replaced both after the first failure, plugged it in and immediately the same capacitor blew again.

Unfortunately I do not have access to an oscilloscope at the moment but only a digital multimeter.

Many thanks!

Edit 1

I took a closer look at the PCB and got the following partial schematic. The main transformer is in the top left hand corner. The PWM IC chip etc. are connected to the hanging MOSFET the the bottom of the schematic.

The TRIAC is part of a STR 81145, which automatically switches to voltage doubling mode if input is 115V. (It takes it's supply from the positive rail as well.)

After removing the capacitors in question, C7 & C8, the resistance to the positive and negative rails are equal to the 50kOhm set by the voltage divider.

Partial Schematic

\$\endgroup\$
  • \$\begingroup\$ Y'know, some electrolytics just go whenever they want. \$\endgroup\$ – tyblu Mar 9 '12 at 6:28
  • \$\begingroup\$ Yes, but right after they were replaced is far to coincidential... IMHO this is quite clearly a circuit fault. \$\endgroup\$ – ARF Mar 9 '12 at 17:20
  • \$\begingroup\$ 230 VAC or 110 VAC? I have seen the doubler try to work on 230 VAC :-) !!!! Try!. \$\endgroup\$ – Russell McMahon Mar 9 '12 at 20:22
  • \$\begingroup\$ @RussellMcMahon If the doubler engages, I should be able to measure 460 Vrms between the positive and the negative rail, right? If so, this should be easy enough to check... \$\endgroup\$ – ARF Mar 10 '12 at 9:40
3
\$\begingroup\$

230V AC swings between +230V and -230V RMS, but that means it reaches +/- 325 V peak-to-peak. With a rectifier, that's between 0 and 325V, which split on two 200V capacitors would be just about enough -- except the capacitors have to be perfectly matched for that to be true.

If there are differences in manufacture, then the ESR of the capacitors will be slightly different, and one will see more voltage than the other.

Also, there may be a max amperage for the capacitor, and given that a capacitor is akin to a short when it's empty, you may get too much surge current (check the ripple current rating of the capacitor data sheet.)

\$\endgroup\$
  • \$\begingroup\$ Thanks Jon, the capacitors were originally identical. I replaced both of them with the same model when one blew. Also, given the voltage divider connected in parallel to the capacitors, the voltage on each capacitor should only be 325V/2, no? \$\endgroup\$ – ARF Mar 9 '12 at 17:16
  • 2
    \$\begingroup\$ The effect of the voltage divider depends on the effective ESR of the capacitors. When just charging up, the capacitors have very low impedance, and basically shorts out the voltage divider. I know 400V capacitors are bigger and more expensive, but you should try with a pair of those. Also, you might want to have an oscilloscope hooked up across each of the capacitors when powering up the circuit, and capture/measure the actual voltage it develops. \$\endgroup\$ – Jon Watte Mar 11 '12 at 5:15
  • \$\begingroup\$ The difference in ESR seems to have indeed been the problem. It appears the device did not have a huge engineering margin. Many thanks. \$\endgroup\$ – ARF Mar 28 '12 at 20:25
2
\$\begingroup\$

The two capacitors that were connected in series are probably rated for lower voltage than the one that the power supply actually produces. In this case, if one of the capacitors gets shorted or something, the other sees the full supply voltage and goes explodes.

Replace both capacitors with ones that are rated for more the full power supply voltage, assuming there is enough space in the device.

\$\endgroup\$
  • \$\begingroup\$ The capacitors in question are part of the power supply. They were 200Vdc 1000uF for a device to be plugged into 230Vac. I replaced both after the first failure, plugged it in and immediately the same capacitor blew again. I do not think they are rated too low: after all the device worked well for years. \$\endgroup\$ – ARF Mar 8 '12 at 12:45
  • \$\begingroup\$ @ArikRaffaelFunke It could be that one of the resistors is bad (shorted or open), that an also cause a capacitor to see the full voltage (which is about 320V for 220V mains). Is the "center tap" connected anywhere? If it is, one half of the rest of the circuit can be shorted. If it is not, check the resistors. \$\endgroup\$ – Pentium100 Mar 9 '12 at 5:57
  • \$\begingroup\$ Pentium100, the resistors appear to be good. I measure 100k rail to rail and 50k between the centre tap and both rail. - The centre tap has a number of connections. Please see the updated question with a partial schematic. \$\endgroup\$ – ARF Mar 9 '12 at 17:18
  • \$\begingroup\$ I agree with @Pentium100. If you replace both capacitors with parts rated for 450V, you will at least get a chance to make some measurements before they explode! \$\endgroup\$ – Adam Lawrence Mar 9 '12 at 19:15
2
\$\begingroup\$

It seems likely that the centre point of the two capacitors connects to the equipment concerned and that this centre point is connected to the positive rail by a fault - possibly due to a shorted switching transistor or a shorted diode - see below.

You do not say what is connected to the capacitors, but usually such an arrangement is a bridge circuit with the capacitors forming one centre tapped "leg" of the bridge and two transistors forming the other "leg" with the load connected between the two bridge legs. Look for two largish transistors possibly MOSFETS, probably on a heatsink and probably the same model / part number. The one which connects to positive will quite possibly be short circuit. Note also the reverse diode possibility as described below.

The diagram below shows the general arrangement diagrammatically - here if Q1 is short-circuited it will destroy C2.

enter image description here

The diagram below is functionally the same but more complicaed and more like what you'd see in practice. Again, Q1 is short-circuited it will probably destroy C2 BUT the diagram is harder to follow - trace the path from +ve via Q1 then on to C2. Shorting the reverse diode across Q1 will achieve the same result.

enter image description here

Both diagrams are from here

\$\endgroup\$
  • \$\begingroup\$ Thanks Russell, this really helped. The design I have is slightly more complicated but seems to follow the same principle. (See the partial schematic in the updated question.) If if (in your schematics) Q1 shorted out, should I be able to measure a short with a multimeter? I only get the value of one leg of the voltage divider connected in parallel to the capacitors. \$\endgroup\$ – ARF Mar 9 '12 at 17:10
  • \$\begingroup\$ Complex. Could be something that drices a driver transistor hard on. Could need some voltage to make it conduct. May be somewhere else :-). It's a guide but alas faults can lurk almost anywhere. It does show approx where to look if it is the same cap that is always stressed. So comparison of the two driver halves may help. They will not be 100% symmetrical but are usually close. \$\endgroup\$ – Russell McMahon Mar 9 '12 at 20:19
  • \$\begingroup\$ Hi Russel, thanks for all your help so far. I think I am slowly getting somewhere... One last question regarding your initial answer: If Q1 or D1 short out, how is C2 connected to the positive rail? The centre-tap of the transistors is isolated with a capacitor (in series with the transformer) from the centre-tap of the capacitors, no? - Maybe I am a bit slow on the uptake... \$\endgroup\$ – ARF Mar 13 '12 at 14:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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