I'm not looking for critique on what I'm doing that has me asking this question, but I'm slightly confused by the definition of the VA rating on control transformers, specifically a 2kVA transformer for stepping 600 volts down to 240/120 volts I'm messing with.

What exactly happens when you exceed the VA rating in real power drawn from a transformer designed for this purpose? Will the voltage drop proportionally to limit the current outputted by the transformer so the total real power does not exceed the VA rating plus some error/safety factor? If not, does that mean despite the voltage dropping and no longer behaving as an ideal transformer, can the total real power output actually exceed the VA rating of the transformer when it's overloaded, just with degraded performance and excessive heating?


  • \$\begingroup\$ Is this of help in answering your question? electronics.stackexchange.com/questions/80561/… \$\endgroup\$
    – vu2nan
    Commented Aug 12, 2023 at 5:22
  • \$\begingroup\$ @vu2nan no I'm asking what happens when you overload a transformer \$\endgroup\$
    – LAX1DUDE
    Commented Aug 12, 2023 at 5:47
  • 1
    \$\begingroup\$ For a short time generally nothing but it will start to get hot and the voltage will drop. \$\endgroup\$
    – Gil
    Commented Aug 12, 2023 at 5:52
  • 1
    \$\begingroup\$ You almost certainly mean average power, not rms power -- rms is used for specifying voltages and currents, because their squares are proportional to power (through a resistive load). \$\endgroup\$ Commented Aug 12, 2023 at 11:53
  • \$\begingroup\$ "RMS power" is a completely meaningless term. Or rather, it is well defined at least, but it has no useful meaning. \$\endgroup\$
    – Hearth
    Commented Aug 12, 2023 at 17:36

4 Answers 4


The 2kVA nameplate rating of your mains frequency transformer would be for a certain safe temp rise with continuous loading. Such a temp rise spec depends on the insulation materials used in the transformer construction. The rise could be say 30 degrees kelvin. If you pull say 4 kVA the copperlosses will quadruple which will in time cook the transformer windings. The large thermal mass of the heavy 2kVA transformer should give you many minutes. The voltage drop will be greater in approx proportion. So if at 2kVA the drop is specified at say 3% then expect 6% at 4 kVA. Some reputable transformer suppliers state the overload performance, example 50% overload for 1 hour. If you have any doubts then place a overtemp switch on your transformer.

  • \$\begingroup\$ So basically the kVA rating at least on some transformers is limited by thermal dissipation when the transformer is being used at its intended voltage? I am wondering if it would be worth covering the outside of the core of the transformer with peltier devices and computer CPU heat sinks just because it looks so funny. \$\endgroup\$
    – LAX1DUDE
    Commented Aug 12, 2023 at 23:55
  • \$\begingroup\$ @LAX1DUDE cooling a transformer by whatever means will increase its cont rating .This is usually done on large transformers and usually not on small transformers \$\endgroup\$
    – Autistic
    Commented Aug 13, 2023 at 4:32

"The usual" way to overload a transformer is to draw more than rated current.

The good news is increasing RMS current will not increase core losses.
Temperature rise will increase, which may affect insulation, and does increase resistance leading to even higher resistive losses and reduced output voltage.
The transformer has non-trivial thermal capacity:
As long as nothing is overheated, it's fine.

Causing more than rated flux (applying higher voltage/frequency (Vs)) drives the core farther into saturation, resulting in higher magnetising current and associated losses.
Core losses rise with voltage.

  • \$\begingroup\$ I'm using the transformer at its intended voltage so this is just a matter of RMS current. If I understand correctly, you're saying putting the wrong voltage/frequency is what will cause core losses and that maximum current the transformer can output at the correct voltage/frequency is simply based on the resistance of the internal components of the transformer, or is there still more inductive effects going on that will also additionally limit the current along some curve even with the transformer running at the right voltage/frequency? \$\endgroup\$
    – LAX1DUDE
    Commented Aug 12, 2023 at 8:32
  • \$\begingroup\$ The transformer operating on AC, the reactances are not to be ignored. Primary&secondary resistance being somewhat easy to measure, a short-circuit measurement of secondary current can be used to compute the sum of primary and "transformed"/"referred" secondary reactance. \$\endgroup\$
    – greybeard
    Commented Aug 12, 2023 at 9:01
  • \$\begingroup\$ Would you suggest the best way to answer this question would be to hook the transformer up in a short circuit with an amp clamp and voltage meter for a few seconds and just draw my own conclusions on the power draw? \$\endgroup\$
    – LAX1DUDE
    Commented Aug 12, 2023 at 23:49
  • \$\begingroup\$ There seem to be two "official" questions here: Title: Can a transformer output more than rated VA without instant permanent ill effects? A: Yes. Body: What exactly happens when you exceed the VA rating in W at rated V? A: Excessive W at rated V implies excessive A…. That said, just power the 600 V winding for a less stressful short circuit experiment. This allows to compute inrush current in your application - FWIW. Energy lost in the transformer per charge would be most useful. The capacitors gets charged with average current, the transformer gets heated with effective current. \$\endgroup\$
    – greybeard
    Commented Aug 13, 2023 at 5:05
  • \$\begingroup\$ yeah you're right, it would make more sense to use a large resistor to overload it by a known amount so more useful details about the transformer can be determined from the test \$\endgroup\$
    – LAX1DUDE
    Commented Aug 13, 2023 at 22:42

Overloading a transformer falls into several categories: the coils have internal resistance, and carrying current heats them according to that resistance. The coil packets are comparatively compact and the ability for heat to escape is constrained. If the packet overheats, the insulation of the wires may get damaged and windings short out, causing progressive damage.

Then there is is damage related to magnetic overload, such as remagnetisation losses (heat again, but in a different place). The core is usually laminated with individual plates isolated with paint to minimize eddy currents (which are actually electric heating again). That paint may get damaged, adding to the heat sources. Heated by remagnetisation and eddy currents, the core may get hot enough to start melting stuff: in extreme cases the core itself, but things like plastic carriers of coil packets are also a possible target.

Then there is magnetic overload: if the core magnetisation saturates, the coils stop offering inductive resistance to current changes, becoming pure resistive heating coils. The resulting large currents will cause thermal damage very fast.

Transformers tend to put out something like the tenfold nominal current as peak but that does not mean that it can tolerate this for brief or extended amounts of time.

Charging capacitors with 2kV to 850V will require careful design of your current control circuits. It likely makes more sense to charge them with a flyback transformer circuit that tracks the capacitor voltage than with a continuously running transformer.

  • \$\begingroup\$ I asked for answers not intended to critique the validity what I am applying this information to, I just want to know electrically how the transformer's output responds to being overloaded. And I have the transformer stepping 120v or 240v to 600v, that's why I said 850v, I just feed the 600v output directly into a bridge rectifier without any voltage regulation. \$\endgroup\$
    – LAX1DUDE
    Commented Aug 12, 2023 at 8:14
  • 1
    \$\begingroup\$ Drawing extra current through a transformer does not cause magnetic flux to increase, but to decrease. If volt-seconds was not sufficient to cause saturation before drawing extra current, the core will not saturate when drawing overload current. Downvoting. \$\endgroup\$ Commented Aug 12, 2023 at 10:52
  • \$\begingroup\$ @MathKeepsMeBusy Primary current is not independent from secondary current (or we wouldn't need a transformer in the first place) and magnetic coupling is not complete, so the (winding-number adjusted) drawn current will be offset with more than its equivalent on the primary side. But yes: for transformers not dimensioned for constant primary voltage (like tube amp transformers), the worst case is the load-less one. \$\endgroup\$
    – user107063
    Commented Aug 13, 2023 at 16:12
  • \$\begingroup\$ Actually, it is the leakage inductance on the primary side which causes the magnetic flux to decrease when a load is drawn on the secondary. Consider the standard model of a real transformer. The secondary load and leakage inductance when referred to the primary, are in parallel with the magnitization inductance. The primary leakage inductance is in series with these. The primary leakage inductance and the magnetization inductance form a voltage divider that is loaded by the 2ndary leakage inductance and load referred to primary. Load secondary, decrease effective volts on I_mag. \$\endgroup\$ Commented Aug 13, 2023 at 18:04

I just want to know electrically how the transformer's output responds to being overloaded.

Here is a partial explanation of what will NOT happen if you overload a transformer by placing a low resistance across its secondary.

Overloading a transformer by placing a resistive load across the secondary which draws more amps than the transformer is rated for, does not increase the magnetic flux in the core or lead to core magnetic saturation. Rather, it leads to a reduction in magnetic flux.

A standard way to model a non-ideal transformer is this:


simulate this circuit – Schematic created using CircuitLab

In this model, \$R'_{s}\$, \$L'_{s(leakage)}\$ and \$R'_{load}\$ are the secondary coil resistance, the secondary coil leakage inductance, and the load on the secondary, all referred the primary.

\$R_{core}\$ represents core losses, and \$L_{mag}\$ represents the magnetization inductance.

In this model, it should be clear that decreasing \$R'_{load}\$ will decrease the voltage across \$L_{mag}\$, and consequently decrease the flux generated in the core. Furthermore, the greater the primary leakage inductance, the more pronounced will be this effect.


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