An ideal transformer should have insignificant power loss, meaning, input VI equals to the output VI.

Question... To achieve this is it true the transformer must have equal copper volume in its primary and secondary windings?

I recently unwound the whole secondary from a 1000VA EI core transformer. This is done without cutting or disassembling of the core. On top of the original primary, I wound 10 turns of 12mm^2 copper wire which gives me about 10 volt which should (theoretically) be able to deliver 100 Amps.

Not knowing anything about primary copper volume, I am wondering how efficient this transformer will be.

Any advice is appreciated.

  • \$\begingroup\$ You are right. Fill the window with copper as much you can. The transformer builders have tables for detemining the cross section of wires with fill factor. \$\endgroup\$ Jul 16, 2016 at 6:40

1 Answer 1


The volumes of copper wire are not required to be equal in the primary and secondary, for theoretical or practical transformer operation.

However, copper is lossy, and any space in the 'iron window', the space in the core available for winding copper that isn't copper (so air, insulation, core former, tape) is making the copper thinner than it could be, so losses higher, and the transformer less efficient.

In a well designed transformer, the copper area in primary and secondary will both tend to be the same at around 25% to 33% of the iron window, the 33% to 50% balance being wire insulation and wasted space. However, if one winding needs very thin wire (so relatively thicker insulation), or very high insulation levels, or multiple secondaries, or space for a winding shuttle to fit through, then less copper area would be used.

The thing determining current in a transformer is heating, the allowable operating temperature for the insulation. While 12mm^2 wire is just about OK for 100A in a single strand in free air, this winding will be packed into a small space, together with heating from the primary and core losses. You will be able to draw 100A from it for a while, maybe a minute, certainly seconds, but not for 10s of minutes, before it gets too hot.

I recently built what sounds like a similar transformer, and used 32mm^2 (8 by 4mm^2 in parallel) for my 100A continuously rated secondary. It was threaded through a 600VA toroidal core after secondary removal, so 80% of the winding length was in fresh air.

I'm going to go out on a limb and guess your 1000VA transformer is from a microwave oven? If so, be aware that the core and primary losses are terrible, and forced air cooling is mandatory. If it is indeed a 'proper' core, better still a toroid, then your secondary will benefit from the lower heating and better cooling, and you'll be able to use 100A for longer.

Don't worry about the primary volume. If a manufacturer put it there as a transformer primary, it's going to be far better than anything you could modify.

For the highest efficiency use the thickest copper with the thinnest insulation, and fill the remaining space in the window. In practice, this means magnet wire, not plastic insulated 'wiring' wire. It will run at a higher temperature as well. Plastic works, it's a lot easier to wind by hand without damaging the insulation, but is less efficient.

  • \$\begingroup\$ Thank you. From your text what I conclude is more copper, higher the efficiency of the unit. Since the secondary works on very low voltage, ideally the insulation can be very thin as long as it is thermally stable. So I'm thinking of using the standard 12mm^2 PVC insulated 'wiring' wire without the PVC but insulated by a layer of kapton tape. This should give more space for copper and thermal stability. Is this okay? \$\endgroup\$ Jul 16, 2016 at 12:40
  • \$\begingroup\$ That's almost exactly what I did in another transformer rewind, used heat-shrink tubing to insulate a short length of heavy wire. Kapton should also be fine. The only purpose of insulation below a few hundred volts is to stay physically intact while being wound, and to keep the wires apart. It's only well above that voltage that it needs to be a continuous jacket. \$\endgroup\$
    – Neil_UK
    Jul 16, 2016 at 13:22
  • \$\begingroup\$ Just for your information, the transformer is not from a microwave oven but from a american made KNS brand gold wire bonder. Its almost double the weight of standard microwave oven transformers. It has about 10 terminals on its primary for various input voltage configuration from 100V to 240V. Original secondaries are 6V, 12V and 110V. No where it's mentioned about its outputs current rating or the VA rating. I assumed it must be at least 1000VA merely based on its size. How accurate my estimation is? \$\endgroup\$ Jul 16, 2016 at 16:15
  • \$\begingroup\$ A standard MOT delivers about 1kVA, with forced air cooling and deep into saturation. Something double the weight ought to be comfortably 1kVA. \$\endgroup\$
    – Neil_UK
    Jul 16, 2016 at 16:28
  • \$\begingroup\$ Another interesting factor is the wire bonder machine is made for universal ac frequency, meaning this transformer must be specified for both 50 and 60 Hz. I'm currently operates it at 240V 50Hz. \$\endgroup\$ Jul 16, 2016 at 16:32

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