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3 phase rectifier with cap filter w/overheating wire.

  1. Feb 24, 2010 #1
    First post here. Found this place while looking for a solution to my problem. My background is Navy trained electronic tech. (FC to be precise) So an engineer I'm not, but I do fix the stuff.

    Anyway, at work, I have a 3 phase full wave rectifier with a capacitor ripple filter. Here's a very basic, and badly modified, stolen schematic from wikipedia of it.

    http://www.rjsoap.com/work/rectifier.jpg [Broken]

    They're actually SCR's and a firing board for it, along with all kinds of monitoring circuits, bleed down resistor for the filter cap when the powers turned off, etc... but I'm taking it down to the basics.

    The rectifier output is rated at 28.5vdc, 600amp continuous with 2000 amp peak. The capacitor filter is actually an 8 capacitor bank in parallel, 50v 100000uF each, so in my book that's a .8F cap sitting there. Ripple voltage is 1.8v p-p

    Before I get to the problem, I want to preface this with all readings are nominal. Output Vdc is 28.5, ripple voltage is 1.8v p-p.

    The problem is, the physical wires that connect the capacitor bank to the bus bars on the rectifier are getting very hot.

    My thought is the wires are too small. They are 6 awg.

    We put a 300 amp load on the rectifier and ran it for 5 minutes or so. We took the temp of the wire, and it's sitting around 220 degrees f.

    I then put a clamp meter on the wire, and it's reading around 280-290 amps AC with a 300 amp load on the rectifier. Way too much for a 6 gauge, I know. If I put a 2000 amp load on it, the wire reads about 400 amps.

    Now, here's the deal. The "engineer" thinks something is wrong with one of the SCR's in the rectifier OR something's wrong with one of the caps to pull that much current. My thought on a failed scr is, if something was wrong with one of the SCR's I would never get 28.5 vdc out. I've had these fail shorted before, and that's a spectacular fail! I've also had them fail open, and that's a boring failure, but it would never reach 28.5vdc out.

    I've had caps fail shorted, that's a spectacular fail too. Although I've never had large caps like these fail. I'd imagine the little white pressure relief plug would come flying out. If one failed open, no big deal, but I think the ripple voltage might be a little higher. (Could be wrong on that one.) And if the capacitance in one is going down, I'd never know, because today I realized out of all the meters we have to measure capacitance, we don't have one that can measure that high. So I ordered a new one... Should be in next week.

    Anyway, I'm sticking with my "The wire is too small theory" as all other readings are nominal, and with all the other monitoring circuits, if something was off it would shut down.

    The reason I mentioned the temperature of the wire is the engineer says he ran a 600 amp load on the prototype rectifier for 12 hours straight, and he didn't fry the wire. I asked him if he physically touched the wire to see if it was hot, and he didn't. I looked up the insulation rating for that wire and it's rated for 200 degrees f. Which tells me, and I can't prove it so this is just a guess on my part, but it probably would need to be around 400-500 degrees before the insulation would start melting.

    We have a new rectifier that should be coming off the line tomorrow or Friday, which I will be looking at to settle this matter. But was wondering if anyone had any other thoughts on this.

    Thanks!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 24, 2010 #2

    dlgoff

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    Welcome to PF Kias.
    If I picture this right; the 6 gauge wires are going to the capacitors don't carry the load? If so, do the wires get hot without a load? Could you have a capacitor becoming resistive? Are any caps getting hot?
     
  4. Feb 24, 2010 #3
    Without a load, the wires are fine, and there's no detectable current in the wire with the clamp meter. I also noticed the capacitors are not even getting warm with a full 2000 amp load. Yet another reason I think there's nothing wrong with the caps.

    All 8 caps were replaced before we even plugged it in for testing, as the overheating wires was the initial complaint. ...and I'll be checking those old caps next week when I get the new meter in.

    The only thing I can find wrong is those wires are overheating, which really makes me think they're just too small to begin with.
     
  5. Feb 24, 2010 #4
    Ok, wait... I think I see what you're asking. The wires to the capacitors do not carry the load.

    The load is taken from two bus bars on the rectifier on 1/0 cables. The caps are wired to the bus bar with 6 gauge wires.
     
  6. Feb 24, 2010 #5
    Your suggestion that the wires are too small is probably correct. With such a large cap bank, the caps are charging during a small fraction of the ac cycle. This means that the average value of I2R (where R is the cable resistance) is much larger than if there were no cap bank. For example, if the caps were charging over 10% of the charging cycle, the peak current would be 10 x higher, the peak I2 would be 100 x higher, and the average I2R would be 10 x higher than with no cap bank.

    Bob S
     
  7. Feb 25, 2010 #6
    Are you saying you want to run this rectifier at less than 300 amps or that you want your circuit to handle the high amp output? To lower your amperage you could modify your SCR driver. (You could use the SCR's to rectify less of the AC waveform). Otherwise it seems you simply undersized your wire. Not sure what you want to do.
     
  8. Feb 25, 2010 #7
    No, the whole circuit was designed to handle a 600 amp load, with a 2000 amp peak. The only problem was the wires connecting the filter caps to the rectifier bus bars were overheating.

    A new unit came off the line today and they rolled it over for me to test. These wires are overheating also. (Which means a hundred or so of them are also overheating out in the wild, but that's another story. ...and best of all not my department that'll get yelled at.)

    So, my thought was correct.

    Now I'm trying to figure out the math to determine the amps through the wire. I already know what the answer is, as I put a meter on it. It's around 290 amps with a 300 amp load on the rectifier.

    I don't ever remember doing anything like this in school 20 some years ago. So somebody slap me if I'm wrong on my assumptions.

    Even though the caps are sitting on the DC side of the rectifier, I'm assuming this is irrelevant as once the caps are charged to 28.5vdc (the rectifier output) and there's no load placed on the rectifier, there's no current through the caps.

    So I need to look for the the amps with AC response of the cap using the ripple as the AC.

    I=V/Xc

    So I need to find Xc

    Xc = 1/2πfC

    I'm having a problem with finding what the frequency is. The clampmeter says about 200hz. The oscope's built in measuring function is all over the place from 100 to 300 hz. The waveform of the ripple is all over the place too, as would be expected from the cap filtering the DC, so I can't manually figure out the frequency either. 100 to 300hz doesn't work out anyway, as it would come out as 400 amps and up. This of course is using the ripple voltage of .8v

    Interestingly though, if I use 60 hz as the freq (The rectifier input freq), and .8v as the voltage (The ripple voltage), the answer comes real close to the amps we're reading on the meter.

    So this makes me wonder if 60hz should be the frequency I'd plug into the formula. It is a three phase input, so my brain should explode any moment now.

    If anyone could shed some light, or even light a match, I'd appreciate it much!

    Thanks!
     
    Last edited: Feb 25, 2010
  9. Feb 25, 2010 #8

    dlgoff

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    Wouldn't the frequency be 360 Hz? i.e. 6 pulses per 360° of "rotation"

    http://www.allaboutcircuits.com/vol_3/chpt_3/4.html" [Broken]
     
    Last edited by a moderator: May 4, 2017
  10. Feb 25, 2010 #9
    The ripple without capacitors would be ~30 volts x [1-cos(30 degrees)] peak-to-peak (p-p), or ~4 volts p-p. If your ripple is ~1.8 volts p-p, then you are charging the capacitors for less than 10 degrees out of the 30 degrees, so if your average current is 300 amps, then the peak current has to be over 900 amps. So the peak I2R is ~9 times the average, but this is for a ~33% duty cycle. Thus the averaged I2R heating is ~3 times what it would be for a constant 300 amp dc current.

    Bob S

    [update] because the cables are not inline with the load, the cables carry only the currents for charging and discharging the capacitors. The capacitors charge only for about 20 degrees of every 60 degree period (even less if the SCRs are limiting voltage). So the cables are carrying about +600 amps for 20 degrees, and -300 amps for 40 degrees. So the averaged I2R heating is equivalent to 600 amps continuous dc. A more exact I2R(t) can be calculated by time integration over the 60 degree period, including both the charging and discharging portion of the cycle.
     
    Last edited: Feb 26, 2010
  11. Jun 20, 2010 #10
    hi, can any one give me the formula for calculation dc-link capacitor for three phase diode rectifier or any reference to it?
     
  12. Jun 20, 2010 #11
    Years ago designed high current rectifier using circuit shown in first post.
    Input was 60 hertz and had excessive output ripple.
    Problem was physical layout. The capacitors were wired as shown.
    The capacitors should have been individually connected across the bus.

    If you connect the capacitors directly across the bus, you should check:
    The RMS current rating of the capacitors against the actual current.
    The peak and RMS current rating of the SCR's should also should be checked.

    In summary if there is significant resistance in series with the capacitors, the capacitors are not operating as efficiently as they could be.
     
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