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resistor in a wall wart

Discussion in 'Electronic Design' started by [email protected], Jul 24, 2013.

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  1. Guest

    I was using a 12V 1A wall wart yesterday to power a project when I heard something in it go "pop". My project draws approximately 0.5A. Postmortem revealed a dead resistor on the 120V input to the wall wart, which appears to be a 1 ohm 1/2 watt resistor. The wallwart is a swithmode power supply.

    1) Am I correct in assuming this 1 ohm resistor was some attempt to save money on a fuse? or does it actually serve some purpose other than as an easypoint of failure?

    2) Is there a way to compute how much power was being dissipated across theresistor? I suppose I could replace it and measure the voltage drop acrossit?

    3) Was it a bad resistor that popped, or a bad wall wart design? or both?
     
  2. ehsjr

    ehsjr Guest

    1) Google fusible resistor

    2) Power dissipated in the resistor is negligible. Power in
    = Power out, so ignoring switcher efficiency and assuming
    1 amp at 12V out, pout (and therefore Pin) = 12 watts. If mains
    voltage is 120, then current is 12watts/120volts or .1 amps.
    That means power dissipated by the resistor (I^2R) is .01 watts
    at maximum rated load. At your .5 amp load, it's half that.

    3) Don't know. Probably not a bad fusible resistor. Could be
    a short (perhaps the leads touched?) on whatever the wall wart
    was feeding. BTW - wall wart designs can be horrible in any
    event, whether there is a failure or not. At least this thing
    had a fusible resistor.

    Ed
     
  3. You would think if it was rated for 1A, 0.5A would not be a problem. I
    recently read an article posted somewhere out in the vast interweb(1)
    where someone did a detailed test of a number of USB chargers. Basically
    all just 5V wall warts. Very few, and only the major name brands that
    had not been counterfeited, could actually source their rated current.
    Many couldn't even do half. Voltage regulation was iffy at best, and
    noise incredibly bad. Based on that, and experience I've had in
    rebuilding chargers for cordless drills, I would say it was a bad wall
    wart design that you were expecting far too much from by trying to get
    ..5A out of it. Having seen far too many literally melt, (and the recent
    electrocution of the woman in China) I never go for low price in power
    supplies, which unfortunately does not save you from the counterfeits.

    (1) Meaning I've lost the link and leave it as an exercise to anyone who
    wishes to find it.
     
  4. Guest

    Seems I was a little off in my load estimation. I've had it hooked up to my bench supply with the current limiter set to 850ma and I just saw it trip. So, I think it's entirely possible that I did exceed the capabilities of the wall wart.

    Also, I tried replacing the resistor and the wall wart is dead, so something else did indeed go kablooey in it.
     
  5. On Wed, 24 Jul 2013 09:35:12 -0700 (PDT), the renowned
    Maybe to reduce inrush current and, assuming it's a fusible resistor,
    it could act as a fuse.
    You'd have to measure the voltage with a true-RMS meter or you'll get
    the wrong answer using that method. The input current will be spikey
    and quite far from sinusoidal.

    You could make a rough but fairly accurate calculation simply if you
    knew the input circuit (filter cap and any series resistance). Or
    simulate it.
    If it popped at power-up, _maybe_ just the resistor. Otherwise, I'd
    wager the power transistor and maybe a diode or two (and perhaps more)
    are toast, so hardly worth playing with.

    Try the diode check range on your multimeter.. input rectifier and,
    the output transistor. If any are not junctiony then much evil has
    visited.



    Best regards,
    Spehro Pefhany
     
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