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I'm interested in how the efficiency of a transformer changes with the power factor of the load.

I know that in principle you can increase efficiency through power factor correction (i.e. making your load net capacitive).

I'm not sure how to derive the efficiency analytically in the case of a leading power factor from the standard set of equations. It seems like I'd have to decompose the load impedance (which is a huge pain if you're interested in the efficiency at different %s of the rated apparent)

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The basic formula for transformer efficiency is below, where \$\cos(\phi)\$ is the power factor (i.e. the ratio of real to apparent power consumed by the load. \$\phi\$ is essentially the impedance angle (so 45° would give a PF = 0.707).

I'm confused about V2 - I'm 90% sure this should be the voltage across the load, but some sources seem to use the rated transformer voltage. Is this a valid approximation?

$$\eta = \frac{\text{output power}}{\text{input power}} = \frac{\text{output power}}{\text{output power + losses}}$$

$$\eta = \frac{\text{output power}}{\text{output power + iron losses + copper losses}}$$

$$\eta = \frac{V_2 I_2 \cos(\phi_2)}{V_2 I_2 \cos(\phi_2) + P_i + P_c}$$

If V2 is in fact the voltage across the load, how do we calculate this if we have a leading power factor? It's trivial to simply subtract the voltage drop across the series resistance and inductance in the case of a unity/lagging PF, but I get the feeling that this wouldn't be valid in the case of a leading PF (because for a fixed load resistance, increasing the capacitor value can actually reduce the total circuit impedance up to the resonance point, which would increase the voltage drop across the resistor).

If we have a load power factor of say 0.5 leading versus unity PF, wouldn't the total circuit impedance be lower (meaning the resistive component of the load would need to increase to keep load current constant)? If this is the case, surely the efficiency of the transformer would increase, since the power dissipated across the load is increasing while every other parameter is staying constant?

How exactly do I go about solving for the efficiency without explicitly finding the load resistance and calculating the real power dissipated, or is there no other way?

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  • \$\begingroup\$ Have you tried to simulate it? \$\endgroup\$
    – winny
    Jul 2, 2022 at 9:27

1 Answer 1

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A transformer is properly rated in terms of Volt-Amperes (VA), and not watts (W), and this is largely determined by winding resistance and safe temperature rise based on insulation class. At unity power factor, VA=W, but at a power factor of 0.707, only 70% of the VA is applied to the load. So the effective efficiency is about 70%, and reduced further by the actual transformer efficiency. Thus a 100 VA transformer can only supply 70 watts to a load to stay within its rating. The impedance of the fully loaded transformer will stay the same but will contain resistive and reactive components.

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  • \$\begingroup\$ This is thoroughly confused. Power factor does not have anything to do with efficiency. VA ratings are useful for determining the current rating for fuses and leads, but not for energy efficiency. Transformers are pretty efficient actually when feeding a real load. The worst efficiency is with a capacitive load. \$\endgroup\$
    – user107063
    Jun 29, 2023 at 1:16
  • \$\begingroup\$ What I was trying to say is that the efficiency of the whole system is reduced if the load power factor is low. If the load has low PF, it will draw considerable reactive current, so the resistive heating losses will increase, and efficiency will be poor. But you may be right, in that the efficiency of the transformer will not change. It will just see more heating due to the higher current needed to get the required power. \$\endgroup\$
    – PStechPaul
    Jun 29, 2023 at 1:40

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