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Is there a practical reason that the roles of the primary and secondary of a simple single-phase transformer cannot be swapped?

Say you have a transformer (from the old vacuum tube days) that has rated for primary 120 VAC 1 A and secondary 12.6 VAC 10 A (so there is about 0.120 kVA for this transformer).

Is there any reason that one cannot take that transformer and drive the secondary with 12.6 VAC and up to 10 A and get from the primary power at 120 VAC up to 1 A?

I cannot imagine a reason that, as long as the specs on the windings are kept that the roles of the primary and secondary windings of a simple power transformer cannot be swapped. But I would be happy to learn differently.

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    \$\begingroup\$ The transformers of a lot of UPSs do exactly that \$\endgroup\$
    – PlasmaHH
    Commented Apr 17, 2017 at 9:02
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    \$\begingroup\$ Knowing that the transformer will be used reversibly, as in a UPS, will affect its design slightly, so it won't need to be de-rated. \$\endgroup\$
    – user16324
    Commented Apr 17, 2017 at 9:30
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    \$\begingroup\$ I've done exactly that when building a vacuum tube preamp - one transformer to bring 240VAC down to 12VAC (to power the tube heaters), and then another transformer to bring 12VAC back up to 240VAC (then rectified and filtered) to provide the high voltage for tube plates with isolation from the mains. As has been mentioned by some of the answers, I didn't quite get 240VAC after the conversion back up, due to transformer losses. \$\endgroup\$
    – Andrew Guy
    Commented Apr 18, 2017 at 2:14
  • \$\begingroup\$ I've done something similar to @AndrewGuy to test small power supplies on ~120V (we use 230V here) as well as for isolation. \$\endgroup\$
    – Chris H
    Commented Apr 18, 2017 at 14:02

3 Answers 3

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A simple answer is not always right. The first problem you have is your quoted ratings. 120 V @ 1 A - 120 VA. If the load is entirely resistive, ie, there is no reactive load, then the maximum you can get out of the secondary will be about 12.6 V @ 9 A, if the efficiency is 95% - not 10 A. Brian Drummond's answer is correct. I have designed transformers and been a production manager in a transformer manufacturing plant. I know what he and I are talking about - and he's correct. There will always be losses that the primary is designed to handle. So, if you reverse feed your transformer, the previously dedicated secondary will have to cover the magnetising, copper and eddy losses. So, you will get less power out of the 120 Vac winding. However, there is one more point: the primary winding will generally have a higher insulation from the core of the transformer, than the secondary. So, you need to consider carefully the source of the power you feed into the previous secondary.

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  • \$\begingroup\$ everyone has said that this transformer will not work as efficiently in reverse. that i get. i should expect about 10% less (maximum) power to be delivered to the load when swapping the roles of primary and secondary windings. $$ $$ thank you, Brian, for sharing your experience (as a manufacturer). i could change my selected answer, but i am reluctant to. \$\endgroup\$ Commented Apr 19, 2017 at 6:10
  • \$\begingroup\$ so i did change the selected answer. \$\endgroup\$ Commented Apr 21, 2017 at 1:36
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You can do exactly that.

Respect the maximum voltage on each winding. Transformers cannot be over-volted and still behave themselves.

Due to the various small non-idealities, winding resistance and leakage inductance being the major ones, you will notice that the voltage ratio one way when under load is slightly less than the reciprocal of the ratio the other way.

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Yes, that'll work - with one caveat.

A carefully designed transformer will take into account the internal losses, to minimise them while reducing cost. So the primary may be wound with slightly thicker wire for lower resistance than in an "ideal" transformer, or the secondary with slightly thinner wire because it only has to handle the output power, minus the losses.

Driving a transformer backwards, these changes work against you, so I would de-rate it to 90% - or maybe 80% - of its rated power to be on the safe side - that is, 120V at 0.8 or 0.9A rather than 1A - and as Neil says you may also observe a minor discrepancy in the voltage ratios.

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  • \$\begingroup\$ thanks. i had to choose one of these answers as the "accepted answer". i chose Neil only because he answered earlier and has fewer rep points than you, Brian. sorry. \$\endgroup\$ Commented Apr 17, 2017 at 9:36

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