I am having some trouble understanding the behaviour of my toroidal transformer. Currently, I am using a 400V to 12V transformer in reverse, meaning I supply 12V and around 1.5 A to the secondary side of the transformator and try to get 400V output voltage.

My problem is, when I connect anything to the primary side, my voltage breaks down, meaning I only get around 2V of input voltage and 60V of output voltage.

I am supplying the secondary side with a OPA548, which should be able to easily supply my 12V 1.5A, but no matter how and if I limit the current, the voltage always breaks down.

Can someone please help me understand what I am missing about my circuit?

EDIT: Could it be that because of the low input impedance of the secondary windings, I nearly short my circuit and therefore the voltage drops?

  • 1
    \$\begingroup\$ What sort of signal is the OPA548 supplying to the winding and how is it connected. Draw a schematic please. \$\endgroup\$
    – Andy aka
    Commented Jul 14, 2020 at 9:12
  • \$\begingroup\$ The OPA548 is supplying a sine wave at 50Hz and 16Vpp. The schematic is just the output and ground of the OPA548 connected to the secondary winding of the transformer, so pretty much nothing. \$\endgroup\$ Commented Jul 14, 2020 at 9:15
  • \$\begingroup\$ The primary voltage should be 12V~50Hz. In other words, it should be 34Vpp and not 16Vpp as stated by you. \$\endgroup\$
    – vu2nan
    Commented Jul 14, 2020 at 9:50

1 Answer 1


The OPA548 is supplying a sine wave at 50 Hz and 16 Vpp.

This won't work. You have not factored in the magnetization requirements when driving the secondary.

Ignoring the primary (high voltage output winding), the secondary acts like an inductor and the inductance will be too low to adequately magnetize the core of the transformer without high reactive currents that the OPA548 cannot be expected to supply. The primary winding might have a magnetization inductance of 10 henries and, it has this to avoid massive current draw from the 400 volt AC supply when connected conventionally.

The transformer ratio is 400:12 hence, the inductance of the secondary will be (400:12)² times smaller. So 10 H might reduce to 9 mH. That's the effective magnetization inductance of the secondary.

At 50 Hz, that has an impedance of 2.82 ohms. Basically the OPA548 has to drive that impedance adequately at 12 volts RMS. That's a reactive current of over 4 amps RMS and beyond what the OPA548 can deliver.

You should also consider that the OPA548 output voltage is limited compared to its power rails. If your supply (for instance) is 20 volts then you cannot deliver more than about 13 volts p-p at the output under load conditions.

  • \$\begingroup\$ Thanks a lot! I already knew that the OPA548 would not be able to realistically drive that much current given these conditions. I connected a 5 ohms 25W resistor between the output of the OPA548 and the transformer and now the voltage does not drop. Just one thing I don't understand, how do you get the value of 10 henries? I searched all possible information I could get for my transformer and could not find any information on the winding parameters. \$\endgroup\$ Commented Jul 14, 2020 at 10:16
  • \$\begingroup\$ @RobertReichel yes, the value of 10 henries is based on experience. Transformer manufacturers are pretty tight on delivering this figure in their data sheets. I base it on a figure of 10 uH per turn and 1000 turns. Now to convert inductance of one turn to 1000 turns you square the turns to get a million and a million x 10 uH = 10 henries. You should probably also add a schematic of the op-amp and its power rails and how the secondary is connected. \$\endgroup\$
    – Andy aka
    Commented Jul 14, 2020 at 11:40

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