MC-A5 · Efficiency & voltage regulation

Efficiency and Voltage Regulation

A transformer has two losses of different character. Core loss is fixed and independent of load; copper loss grows with the square of the load. Vary the load and find for yourself the condition where efficiency peaks.

At what load is efficiency greatest?

Core loss is a flat line independent of load; copper loss is a curve that grows with the square of load. Push the load to find where the two losses meet. At exactly that load, efficiency reaches its peak.

Load fraction mm = 0.90

Slide for continuous load. Where the two losses meet is the peak.

Efficiency and the two losses

η = 95.5% · Pcu=3.24 / Pfe=1.00 kW

The load is far from the peak. Copper and core losses differ widely, and one of them drags the efficiency down.

Misaligned

Fixed loss and variable loss

Core loss arises in the shunt exciting branch. Once voltage and frequency are set, the flux is set, and the hysteresis and eddy losses it fixes flow even at zero load. So it is a fixed loss, independent of load. Copper loss arises in the winding resistance of the series branch as the square of the load current. At load fraction m, copper loss is m² times the rated copper loss.

ObservePcu(m) = m² Pcu

Copper loss grows with the square of load — variable.

The condition for peak efficiency

Efficiency is output divided by output plus losses. At load fraction m, η = mScosφ / (mScosφ + Pfe + m²Pcu). Setting its derivative with respect to m to zero gives Pfe = m²Pcu, the condition that fixed and variable losses are equal. Solving, the load fraction for peak efficiency is m = √(Pfe / Pcu).

Choosemax η: Pfe = ?

Peak efficiency is where variable loss equals fixed loss.

Fill inm = √(?)

Solve for the load: the root of the core-to-copper ratio.

Voltage regulation comes from the series drop

When the load current passes the series impedance, the secondary voltage falls below its no-load value. The ratio of this change is the voltage regulation, approximated by ε ≈ p cosφ + q sinφ, where p is the percent resistance drop and q the percent reactance drop. A lagging (inductive) power factor pulls the voltage down so the regulation is positive; a leading (capacitive) factor lifts it and can make the regulation negative.

On your ownε = p cosφ + ?

Regulation is the sum of the resistance and reactance drops.

Back to the first screen

Efficiency peaked at the load where the two loss curves met. Core loss is flat and load-independent, while copper loss grows with the square of load and somewhere equals the core loss. At that point loss per output is least and efficiency is greatest. In load fraction it is m = √(Pfe / Pcu), the load where variable and fixed losses are equal. Meanwhile the voltage drop across that same series impedance shows up as the regulation ε ≈ p cosφ + q sinφ.

The efficiency: η = output / (output + core loss + copper loss). Copper loss goes as the square of load (m²Pcu); core loss is fixed. Peak efficiency occurs at m²Pcu = Pfe, i.e. variable loss = fixed loss, at load fraction m = √(Pfe / Pcu). The voltage regulation from the series drop is ε ≈ p cosφ + q sinφ, positive for a lagging factor and possibly negative for a leading one.

Closing out group A

That completes magnetic circuits and transformers. We read flux with the magnetic Ohm’s law (A1), saw the energy pooled in it (A2), swapped voltage and current by the turns ratio (A3), captured the non-idealities in an equivalent circuit (A4), and computed performance from the losses (A5). The next group B is the DC machine, which makes rotation on the same magnetic circuit. It begins with how the commutator turns AC into DC.