Can a control transformer output more real power than its VA rating when overloaded?

Question

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?

Thanks.

Answer 1

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.

Answer 2

"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.

Answer 3

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.

Answer 4

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.