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bridge rectifier question

Discussion in 'General Electronics Discussion' started by EvilBunny, Feb 1, 2011.

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  1. EvilBunny

    EvilBunny

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    Feb 1, 2011
    We were troubleshooting a power supply today and the faulty component turned out to be a 0.1 ohm resistor (rated @ 20 watts).

    This resistor was the only thing between chassis ground and the negative dc side of the bridge...

    This is a standard full wave bridge rectifier connected by two wires to a xformer on the ac side and a positive and negative terminal on the dc side. It's a 28VDC power supply 120VAC input.

    any ideas on what the purpose of this 0.1 ohm resistor is?

    One guy said it was part of the filter but there were smoothing caps on the other (positive terminal) side that serve that purpose (along with the bleeder resistor), so I'm not buying it... Another guy said something about eliminating noise from high frequencies, but I'm not sure about that either... although the application does happen to be an RF transmitter operating in the 100-ish megahertz range.
     
  2. shrtrnd

    shrtrnd

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    Jan 15, 2010
    Resistance value is too small to be a bleeder resistor for the caps.
    Hewlett Packard used a lot of 'Zero Ohm Resistors' in their old gear, basically as fuses to protect circuits between boards in their test and measurement gear.
    I may be wrong (I'm sure somebody will tell us that when they read this if I am), but
    my GUESS would be that it IS intended to act like a fuse. Either to protect the circuit, or people touching the chassis during a fault. There's probably a good technical reason why they'd use an extremely low ohm resistor, rather than an actual fuse, but I don't know what that would be.
    Anyway, that's my two cents worth after working a lot of HP test and measurement gear.
    Maybe somebody will respond to this post, with the exact reason for your resistor.
    Since it shut-down your circuit. I would think the power supply shared ground with another board in your device, THROUGH the .1 Ohm resistor. And to protect the other board, when a fault developed in your insturment, the .1 ohm resistor opened, to protect your power supply, or other boards in the device, by disconnecting them.
     
  3. davenn

    davenn Moderator

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    Sep 5, 2009
    A bleeder resistor is going to be across the caps between positive and negative rails and several k Ohms in value, not from negative rail to ground.
    Nor is it going to do any RF filtering. Anyway small value (0.01 - 0.1uF) capacitors are used for that and they would also be between positive and negative rails.
    Nothing can go THROUGH the resistor to another board IF the resistor is from negative rail to ground.

    Are you totally sure its from negative to grn ?? post a clear pic :)

    Dave
     
  4. Resqueline

    Resqueline

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    Jul 31, 2009
    A resistor between the bridge and the filter cap's will reduce the peak charging currents, protecting the bridge. It comes at the expense of extra power dissipation though.
    It may be used to measure current consumption/delivery too.
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Resqueline beat me to it. I was thinking essentially the same thing.

    A resistor *can* be used as part of a filtering network. R before the C can reduce the slope of the charging side of the 100Hz ripple, but this resistor is not a value I'd expect to see in that application.

    Given the rating, I'd tend to lean more towards limiting peak current into the filter capacitors at switch-on.
     
  6. davenn

    davenn Moderator

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    Sep 5, 2009
    but its NOT between the bridge and the filter caps
    read again what he said

    Dave
     
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    I presumed the negative side of the filter capacitors was connected to chassis ground. And that the resistor was the only thing connected to the -ve output of the bridge.

    That;s how I visualised it anyway. You may well be right. It may be a connection between electrical ground and chassis ground. Why such a high wattage for that though I could not imagine. Unless they're utilising the inductance of a wire-wound resistor :eek:
     
  8. EvilBunny

    EvilBunny

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    Feb 1, 2011
    It is absolutely tied to the negative pole of the bridge. The power supply didn't work without it... It failed open and there was no return path for the circuit. The only thing attached to the negative lead of the bridge was that resistor and the other side of the resistor was attached to the chassis... The schematic confirms this as well.

    I don't really understand what purpose it serves. As has been discussed, it is obviously not being used for a bleeder for the filter. I just don't understand why it's there. A jumper across the resistor had the unit working perfectly.

    Still confused as to what it's there for...
     
  9. davenn

    davenn Moderator

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    Sep 5, 2009
    ok thanks for the confirmation :)
    looks like its just being used as a very slow blow fuse
    But really strange that the rest of the negative rail is coming from the chassis rather than the negative rail of the rectifier

    Ummmm... it means that you should be seeing a wire from the negative terminal/s of the big smoothing capacitor/s going to the chassis as well ?
    the cct should show where the rest of the negative rail is connected to the chassis

    Dave
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    What current is the power supply rated at?

    20W 0.1R implies something around 10A (since 14A would cause it to dissipate 20W and I assume the designer left some margin to allow for elevated temperatures etc).
     
  11. EvilBunny

    EvilBunny

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    Feb 1, 2011
    Yes, this is an old power supply (circa 1970's). Apparently it was common to use the chassis as a return path back then???

    It's a drawer in a transmitter rack, and yes the neg side of the caps were connected to chassis as well... along with various other components in the circuit. I would post the schematic but I'm not sure I'm allowed to... as funny as that may sound.

    The thing I don't really understand about it being an overcurrent protector is that the whole circuit is protected by a 10 amp slow blow fuse. That resistor would effectively be wired in series with the entire circuit since everything would have to flow through it to get to the negative post of the bridge rectifier.

    The system has a battery backup (without a low voltage disconnect circuit, which most of our other equipment has), and the batteries were dead. Not sure if we lost ac power, drained the batteries then lost the resistor due to overcurrent upon recharging the batteries, or if we lost the resistor first which caused the batteries to drain. The chicken or the egg... so to speak.

    I don't know... I just thought it was strange to have such a low value resistor there. Still don't really understand it.

    Hope that all made sense... it's been a long day.

    Thanks for the replies!
     
  12. davenn

    davenn Moderator

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    Sep 5, 2009
    where in the cct is that fuse ?
    AC Primary?, or AC secondary (between the transformer and the rectifier)?
    or in the +V rail from the rectifier to the smoothing caps ?

    Dave
     
  13. Resqueline

    Resqueline

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    Jul 31, 2009
    I hate to repeat myself but the only reason for that resistor to be there is to limit the peak repetitive current pulses from the bridge. Turn-on current benefits too of course.
    Bridges often (or used to) have a specified maximum value of the smoothing capacitor. Going above that would require measures to be taken.
    I can imagine old transmitting gear having a large cap to reduce hum.
    So, what's the spec's on that bridge and how big is that cap?
     
  14. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    I hate to agree with Resqueline again :)p), but I agree.

    The resistor seems rated fairly conservatively for normal operation, and almost certainly is called on to dissipate *well* over 20W at switch on.

    This will reduce stress on the rectifier and will probably only be called on to handle this load for under 1/2 a cycle on switch on.

    It will also reduce the peak current drawn at the top of each half cycle as the bridge rectifier tries to charge the capacitors back up to the peak voltage of the AC input (less forward drop, etc...)

    Without knowing the specs on the rectifier (and indeed on the capacitors, and possibly even the transformer) it is hard to be definitive. You might find that replacing the bridge rectifier with a more modern high current device, and possibly the capacitors with something capable of handling higher ripple current would reduce the need for the resistor, however you may then find that the fuse ends up failing for a similar reason as the resistor.

    If this hypothesis is correct, the resistor was under significant stress due to peak loads exceeding its continuous rating and it probably died of fatigue. This is also a method of killing fuses, although slow blow fuses are designed to minimise this.

    Replacing the resistor with a link may lead to premature rectifier failure (which will typically be followed up milliseconds later by a fuse failure).
     
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