resistor in a wall wart

I was using a 12V 1A wall wart yesterday to power a project when I heard so mething in it go "pop". My project draws approximately 0.5A. Postmortem rev ealed 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 mo ney on a fuse? or does it actually serve some purpose other than as an easy point of failure?

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

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


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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.

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Mark Storkamp

Probably. It's too small to be an inrush limiter.

or does it actually serve some purpose other than as an easy point of failure?

Measure true, DC coupled RMS voltage or current!

Probably something downstream failed first. Mosfet or electrolytic cap most likely.

John Larkin         Highland Technology, Inc 

jlarkin at highlandtechnology dot com 
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John Larkin

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.

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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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Spehro Pefhany

Some use the wire gauge in the transformer as a fuse.


Reply to
Frank Galikanokus

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