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Can wall wart's be damaged by diodes?

chopnhack

Apr 28, 2014
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I was using my w.w. to power a breadboard and didn't notice that one of my jumper wires was in the wrong location. The LED did power up, but the light gradually diminished. I disconnected and found the misplaced jumper, but instead of having 16.4v I now only had about 2v. Any ideas of how that happened? There goes my power supply, LOL :D:eek: :oops:
 

davenn

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Sep 5, 2009
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without seeing your circuit and where you mis-wired ... very difficult to tell what mite have happened

show a schematic or some photos of the correct and incorrect wiring you did
I would have to assume you somehow shorted out the plugpack

Dave
 

chopnhack

Apr 28, 2014
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No schematic as it was a simple circuit from the barrel connector through a resistor to the LED, I had already broken down the breadboard to create a different circuit to test the LED - which is working fine, that is what confused me! I would have to agree, somehow the w.w. got shorted or something? I believe that the misplaced jumper was on the positive side (should have been the negative) such that the LED was getting positive from both anode and cathode, odd that it would dim slowly.

hmm.. I just checked it again after leaving it unplugged for awhile. Its back up to 16.4v! Is it possible a PTC kicked in? As a test I think I will leave it plugged in with no load for awhile and see if it still provides 16v otherwise it will be of limited value for breadboarding.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Is it possible a PTC kicked in?

It's possible. It's also possible that there's a chip in there which has some sort of thermal shutdown.

Probably a good indication that you should carefully check your wiring. :)
 

chopnhack

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Very good point Steve! The kids and I were playing with snap circuits - childrens electrical activity center and I was just telling them that this morning! Check your picture to make sure you have everything connected first!! Do as I say, right?

I left the transformer powering a LED for about 30 minutes and came back to find the circuit out. The transformer was only putting out about 2V. I unplugged it and let it rest and when I checked it again, it was back at 16v. The resistor circuit is, 680Ω 1/4w resistor delivering 20mA across a single LED - the power dissipation across the resistor is enough to get it pretty hot after awhile - 0.272W if my calculation is right (P=I*I*R) - as the resistor gets hotter doesn't its resistance increase? Could this possibly cause the transformer to shut down? If the resistance increases, I would have thought that the corresponding draw in mA would drop. Little stumped as to why the transformer would react this way...

I am going to use two resistors in series to spread the heat dissipation and check how long the transformer can keep the LED lit.
 

Gryd3

Jun 25, 2014
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Perhaps the initial burst of tripping the thermal shutdown has caused some damage in the transformer to prevent it from dissipating heat as efficiently as it once had.
 

chopnhack

Apr 28, 2014
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Perhaps the initial burst of tripping the thermal shutdown has caused some damage in the transformer to prevent it from dissipating heat as efficiently as it once had.
Its very possible. I have had this one LED running now for nearly 2 hours and the voltage has held steady to within 0.02V. Its such a small load though, I am looking at wiring a group of LED's in parallel, IIRC that uses more current at the same potential. I will have to review whether the current setting resistors will be able to handle it!

Thanks for the input, you may very well be correct.
 
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