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So I've been working on fixing this window AC unit for a while, and when I started I knew next to nothing about circuits. At present today, I know a bit about circuits -- enough that i at least know what the various components are for, plus how to use the multimeter, but I am faced with a mystery I was wondering if anyone else might be able to put an educated guess on, because this puzzling event happened back when I DIDN'T know what I was doing, so the details meant nothing at the time. Now that i'm more knowledgable, I am unable to reproduce the same result.

Some background: Our apartment suffered an overvoltage event, which fried all of our surge protectors as well as the two appliances that were plugged directly into the wall: the AC unit being the more valuable. Upon taking the thing apart, i noticed that the board had a burn mark on the back side which corresponded to where an MOV varistor was mounted on the front.

Here is the point where I wish i knew at the time more about what I was doing, because when I first saw this I pulled out my cheap dollar store multimeter, which I had no clue how to use, and put it on SOME SETTING and put the leads to each side of the varistor, which suddenly caused the AC unit to jump to life and begin working! Without knowing what the purpose of the varistor was, I immediately ordered a set of new ones with the goal of soldering a new one on. I used wire cutters to remove the old varistor off the board.

Now about a month later, I understand what a varistor does, and so am at a loss as to how bridging its leads on ANY SETTING on the multimeter would have caused the entire unit to suddenly turn on. I also can no longer reproduce whatever it was that I did, and any attempts to bridge the leads now results either with A) nothing or B) it immediately blowing the 3.15A fuse placed in series next to it.

So here is the puzzle. What possible reason would it, when i original bridged the two leads of the varistor with the dollar store multimeter, have caused the AC unit to turn on? From my understanding of varistors now, it seems like the only possible results of putting something other than a varistor in between those two points in a protected circuit would either A) short it or B) do nothing, passively. I don't understand how the circuit would be turned on by doing something to the protected side in parallel.

Anyone have any thoughtful guesses? I am at a loss. enter image description here

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    \$\begingroup\$ Do you know the difference between a voltage surge suppressor that goes in parallel and a current surge suppressor that might be used in series for soft starts and now your current meter fuse is blown. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 23 '20 at 1:59
  • \$\begingroup\$ "any attempts to bridge the leads now results either with A) nothing" - Are you saying the AC unit still doesn't work? \$\endgroup\$ – Bruce Abbott Jun 23 '20 at 3:52
  • \$\begingroup\$ Tony I think you may be right—I had been assuming the varistor surge protection was of the type that is in parallel, in other words a totally passive component that could be removed without any effect on the circuit (Er.. well no effect other than the circuit now being vulnerable to surge event damage, that is). \$\endgroup\$ – SubtleHyperbole Jun 28 '20 at 2:22
  • \$\begingroup\$ Oops, hit send accidentally, what I meant to say: Tony I think you may be right! I had been assuming the varistor surge protection was of the type that is in parallel, in other words a totally passive component that could be removed without any effect on the circuit (Er.. no effect other than the circuit being vulnerable to surge events that is). This is why I couldn’t understand how bridging it in any way shape or form would cause the device to turn on and function — it would be like a truck with failed brakes veering onto the runaway truck ramp in order to reach its intended destination. \$\endgroup\$ – SubtleHyperbole Jun 28 '20 at 2:32
  • \$\begingroup\$ The RV1 is a shunt Varistor clamp. Don't worry about figuring out a doubtful trigger. Just hope it does not reoccur. Check Mains voltage for nominal value within 10% Vac and no need to use AMP setting. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 28 '20 at 15:47
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A varistor behaves electrically as a resistor. Your overvoltage event undoubtedly burned out the varistor so it appeared as an open circuit. This prevented the AC from turning on. When you bridged the varistor with your multimeter, you inserted some resistance in the circuit which was enough to allow the AC to turn on. Since you don't know what setting you used it is difficult to give more details. Note, however, that a multimeter will appear as a resistance to the circuit being measured for almost all of its settings whether current, voltage or resistance. The exact value of resistance will vary with the settings and can range from an ohm or two up to many megohms. Again, without knowing what you did, no more can be said. As to not being able to reproduce the original scenario, you probably haven't tried the original multimeter setting (since you are blowing fuses, I assume you have not run through all of the possible multimeter settings).

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  • \$\begingroup\$ thank you, I think yours and the other answer are both excellent. I really appreciate the speculation, especially given the lack of available info. Also yes, because of the fuses being blown, I haven't run through all possible settings. Each time a fuse blows its rather startling, which after the third time created new burn marks on the board, so I stopped. How much of an issue could these burns present? I have a ton of fuses of the right rating, so running out of them is not the issue that is limiting me -- what I am concerned about is damage caused to the board each time one burns out. \$\endgroup\$ – SubtleHyperbole Jun 24 '20 at 21:27
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Now about a month later, I understand what a varistor does, and so am at a loss as to how bridging its leads on ANY SETTING on the multimeter would have caused the entire unit to suddenly turn on.

My best initial guess is that the varistor you bridged with the meter was a series inline NTC varistor used for reducing inrush current into your circuit. This is quite common practice.

You possibly bridged the varistor with your meter in mA or current mode and, that would have acted like a short circuit AND, if the said varistor had been damaged (gone open circuit due to the "overvoltage event"), then bridging it with a low impedance would have brought the circuit back into operation.

Some time later you repeat the same experiment but pick the wrong varistor (one that is actually across the incoming supply lines to prevent overvoltage surges) and bang, you blow the fuse.

Purely speculation.

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  • \$\begingroup\$ I too was thinking that the varistor possibly was there to somehow limit inrush current -- however the varistor spot is definitely the same one the entire time (or did you mean, I may have ordered the wrong type of varistor as a replacement?) The latter is possible, the former is not. \$\endgroup\$ – SubtleHyperbole Jun 24 '20 at 21:34

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