Suppose I have a plug-in transformer that reduces 120VAC power to some lower AC voltage (say 12VAC or 24VAC), and suppose I want to do the reverse---increase a low-voltage source back up to 120VAC.
Would there be a problem in connecting the low-voltage power to the plug-in prongs of the wall wart (instead of the normal use: plugging the prongs into the wall outlet) and expecting to get high-voltage power back out the other end?
I guess my question is answered if I know that there are no components other than the transformer inside the wall wart.
My immediate need is to use a 24VCT (center-tapped, 3-wire, 12VAC or
If you attached 12 volts to the line side of the transformer, you'd get 1.2 volts out. You'd want to go the other way to step up , but there are other considerations, such as the wattage you need the transformer to produce and what the wall wart is capable of producing
If it is an AC output and not a DC output wall wart then you can connect 12 volts to it and get out about 120 volts if it is normally a 12 volt output. Just put the 12 volts to the normal 12 volt output and the prongs will have
120 volts on them . You have to check to see how much current or how many watts the transformer is rated for as not to overload it. If your wall wart is good for 12 volts at one amp, then about all you can draw at 120 volts is
1/10 of an amp.
Usuallly transformers do not care if they are connected up either way. Sometimes the output voltage will exceed the insulation of the wiring.
He'd get 120vac because he'd connect the 24VAC to the 24VAC side. The problem is going to be loses and power considerations. He'll be lucky if he ends up with enough power to run a clock.
transformers aren't perfect. there are always loses.
a 24VAC transformer will most likely really be a 30vac transformer with the understanding that there'll be 6vac of loses in the winding when a load is put on it.
Connect 24vac to the secondary of what really is a 30vac transformer, and you'll get 96 volts and only when there is no load whatsoever. Put even a small load on it and you'll be lucky if you get 80 volts.
The voltage drop on any part of the circuit is equal to the resistance times current (E=IR).
It is very likely just a transformer; i.e., does not provide any sort of voltage regulation.
A transformer may be operated in either direction - either winding can be used as the input. Don't exceed the voltage or current ratings of either winding. It is not necessarily true that the center tap on the 24V winding is rated for 700ma, but it probably is.
The ground is probably connected to the transformer core. It might also ground the center tap of the 24V winding. Why not use your ohmmeter or a battery and bulb (sigh, LED) tester to find out?
The ground plug is probably connected to the frame of the transformer or maybe not even connected at all.
As I mentioned , within limits a transformer can be used either way. As yours is rated for 35 watts, you can get 24 volts out at 700 ma or 7/10 of an amp. The 24 volts and 700 ma is only abuot 16.8 watts. If connected in reverse the best you can do is get around 120 volts out at about .14 amps. Not too much you could power with that.
Most wall warts are not too efficiant and the voltage will probably drop a good percentage with a load on them over the no load condition.
It is not likely anything except a transformer and transformers basically work either way. There are some ratio and loss effects as AZ Nomad noted so your actual voltage output will be a little less than what you might expect. If you supply the low voltage winding with 24 VAC you should be able to get about 0.14A from the high voltage winding at a high enough voltage to operate a nominal 110-120V device. Monitor the temperature to be sure it is not overloaded.
The three prong grounded plug does not tell anything relevant about it.
The efficiency might well be the same in both directions, but that efficiency is some number n that is less than 1. If it is the same in both directions, the efficieny of two in a row is n squared, and that will be even lesser than one. So if n is 7/8 (0.875), n**2 is 49/64 or about 0.75.
The power available to the first transformer is for all practical purposes, unlimited, so the current in the primary winding is determined by its impedance and the 110 volts powering it. The current in the secondary is found by I = E/R, where R is the impedance of the secondary winding, and E is 24 volts.
Now that you know what output I is, and rather than calculate it, the transformer probably says on it, that it's 300 ma, or 1500ma, the input current available to the second transformer isn't unlimited like the input to the first. It's maybe 1500 ma max. Raising the voltage from 24 to 110 will lower the available amperage by reciprocal factor. So if the voltage is 110/24, the max amperage is 24/110 * 1500 milliamps. That's about 0.22 * 1500 = 330 ma. And assuming the numbers I assumed, that will be your maximum. Less actually because the transformer is not 100% efficent.
I don't know if you will get 110 volts or not. It would be interesting to do it and see what you get. Wouldn't be at all surprised if it is lower like the AZ said.
I think it tends to indicate that. Of course the voltae will drop when there is a load, because all the current has to go through the secondary winding which has internal resistance (called impedance for alternating current, but when it is only 60 cps, the impedance might not be much greater than the resistance your direct current ohmmeter measures.) When there is no current flow, that resistance has no effect, causes no voltage drop.
Yes and no. Look above at the power rating in the normal direction.
35 watts input, but only 24 volts * 700 ma output. The product is as someone said 16.8 watts, which is less than half of the input power.
When you go the other direction, it will have the same power capability in that it will also put out slightly less than half of the input, or something like 8 watts.
I'm really not sure, but would you say the answer to your question should be yes or no?
Why do you want to do this, btw. Is there no 110 outlet at that location? Maybe you can run one.
Yes you can, but keep in mind they are only something like 80% efficient at best, and your maximum primary current will be the rated output of the device. This means a 1 amp wall wart will put out 100ma of current at 120 volts from a 12 volt supply.
In an AC wall wart there is only a transformer. Don't try it with a DC output unit though!!!!
He builds vacuum tube circuits, and some of them are powered by 2 wall-warts back-to-back, just as you're asking about, like this power supply used for one of his projects:
formatting link
This circuit supplies more than 400 volts of B+ for a small tube amplifier (using a voltage tripler), as well as several other voltages. So yes, it can be done.
Well, you can measure the resistance of the 120volt relay, and from that do a pretty good calculation of the current it will draw, and compare that with what I said would come out of the reversed transformer.
I think I said 300 ma and 8 watts, but you can check. But that was when I guessed the first transformer was 1500 ma, and someone said it was half that.
So that should give you a good idea of whether it would work, and then you can try it and see. It shouldn't take long to try it.
And then if you have to you can get a 24vac relay. I think they are easily found.
Coil current with AC applied will be much lower than the value calculated from DC resistance ... by perhaps a factor of ten or more. Inductive reactance is the dominant contributor to impedance here and it is proportional to frequency and isn't there at DC.
I agree with mm's comment about 24vac relays. Your HVAC thermostat may have a solid-state relay driver that won't be happy with the unexpected load.
HomeOwnersHub website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.