Transformer modelling for harmonic studies

[waqasakbar323]

Why transformer core is modeled as resistance even though it is frequency dependent.With increasing frequency current rises. Doesn't it seem capacitive reactance behaviour? Which decrease with increasing omega?

 

[Baluncore]

Why transformer core is modeled as resistance even though it is frequency dependent.

A power transformer is specified for operation at a fixed frequency.
The real power loss is in phase with the voltage, not in quadrature as VAR.
Core loss is when flux change is greatest, which is in phase with the voltage, so must be resistive.

 

[waqasakbar323]

A power transformer is specified for operation at a fixed frequency.
The real power loss is in phase with the voltage, not in quadrature as VAR.
Core loss is when flux change is greatest, which is in phase with the voltage, so must be resistive.

Correct, it make sense. But because in reality harmonics exist and are growing with increasing power electronics. So if someone want to model transformer, specially for harmonic studies. Than the generic model is not enough. For example capacitance between windings should also be considered. But i am confused abit about core. But as you said even we consider harmonics it would still be resistive due to inphase V and I.

 

[Baluncore]

Are you considering the core used for 50/60 Hz power,
or the small high frequency isolation transformer in a switching voltage regulator ?
There should be some low-pass network to prevent switching noise entering the AC supply.
There should be power-factor control of the rectifier input to a switching converter.

 

[Babadag]

Core losses depend on frequency. See attached articles.

 

[Delta2]

Core losses depend on frequency. See attached articles.

I took a closer look only to the 1st paper, I agree with the analysis there.

However though the voltage in a loop of eddy current has 90 degrees phase difference with the current that drives the coil whose core we study, this voltage is in phase with the eddy current, and though the eddy losses are proportional to the square of the frequency you still can't model it as a capacitance or inductive reactance, for the simple reason that in a capacitor or inductor the energy balance in 1 cycle is zero, while the eddy loss in 1 cycle is not zero.

I believe the correct modeling is as ohmic resistance whose value R depends on frequency. This modeling is happening in skin effect for example, where increasing frequency makes almost all of the current to flow in a thinner outer layer of the conductor, thus reducing the effective cross section of the conductor and thus increasing ohmic resistance.

 

[DaveE]

The real power loss is in phase with the voltage, not in quadrature as VAR.

Yes! @Baluncore nailed it with his reference to the phase of the impedance in your modelling of the losses.

Ohm's law is a really good simple model for many things, but it's also often not true. Resistance can vary with frequency, with excitation level, etc. This is common in RF circuits where the losses in a capacitor dielectric or and inductor core are usually frequency dependent (nearly everything is, if you look closely enough). It's also true in things like light bulbs, thermistors, or semiconductors where resistance is a strong function of temperature because temperature often depends on the power loss.

The key point is that losses are modeled with real valued impedance (i.e. resistance) that has current that is in-phase with the voltage. An impedance can be modeled with a real part (lossy) and an imaginary part (energy storage, not loss). When you do the power calculation (integration of voltage times current) the imaginary, out of phase, part sums to zero. This is ultimately a definition IMO, if there's power dissipation, it has to be the resistive part, that's how the math works out.

 

[anorlunda]

When you do the power calculation (integration of voltage times current) the imaginary, out of phase, part sums to zero. This is ultimately a definition IMO, if there's power dissipation, it has to be the resistive part, that's how the math works out.

@Baluncore and @DaveE both gave good advice. I would like to add that the imaginary portion of the power also cause real power losses via the resistance in the wiring that brings current to the device in question.