Transformer EMF Equation, Losses and Efficiency with Worked Examples
The transformer is the reason AC won over DC for power distribution. It steps voltage up for transmission and down for use, with efficiencies above 95% even in small units and above 99% in large power transformers. It has no moving parts, yet its design depends on a few precise relationships between voltage, frequency, turns and core flux.
In this article you will derive the EMF equation, use it to calculate winding turns for a 25 kVA single-phase transformer, and then work out losses, efficiency at different loads, the load for maximum efficiency, and voltage regulation.
Working Principle and the EMF Equation
A transformer has two windings on a laminated silicon steel core. The primary, connected to AC, sets up an alternating flux in the core. The same flux links the secondary, inducing a voltage by Faraday's law. Because both windings see the same flux, the voltage per turn is the same on both sides, and the voltage ratio equals the turns ratio.
For a sinusoidal flux φ = Φm sin ωt, the flux changes from +Φm to −Φm in half a cycle. The average EMF per turn over a quarter cycle is 4fΦm. Multiplying by the form factor of 1.11 gives the RMS value.
E = 4.44 f N Φm (volts RMS) Φm = Bm × Ai (Wb; Bm in tesla, Ai = net core area in m²) E1/E2 = N1/N2 = K⁻¹ (K = N2/N1, transformation ratio) I1/I2 ≈ N2/N1 (ideal transformer, ampere-turns balance) V1 I1 ≈ V2 I2 (VA in = VA out, ideal)
Worked example: turns for a 25 kVA, 2200/220 V transformer
Assume maximum flux density Bm = 1.2 T, net core area 0.0225 m², and supply frequency 50 Hz.
- Φm = 1.2 × 0.0225 = 0.027 Wb.
- Volts per turn = 4.44 × 50 × 0.027 = 5.994 V.
- Primary turns N1 = 2200/5.994 = 367 turns.
- Secondary turns N2 = 220/5.994 = 36.7, rounded up to 37 turns.
- Full-load primary current I1 = 25000/2200 = 11.36 A.
- Full-load secondary current I2 = 25000/220 = 113.6 A.
Rounding N2 up raises the secondary voltage slightly, which helps offset the voltage drop at load. The secondary conductor must carry ten times the primary current, so it uses a much thicker conductor with fewer turns.
Transformer Losses
| Loss | Cause | Depends on | How it is reduced |
|---|---|---|---|
| Hysteresis loss | Repeated magnetisation of core steel | f and Bm (about Bm^1.6, Steinmetz) | Grain-oriented silicon steel, amorphous cores |
| Eddy current loss | Circulating currents in the core | f² and Bm² and lamination thickness² | Thin insulated laminations |
| Copper loss | I²R heating of windings | Square of load current | Thicker conductors, lower current density |
| Stray and dielectric losses | Leakage flux in tank, insulation | Load and voltage | Design and shielding |
Hysteresis and eddy losses together form the iron or core loss Pi. With constant voltage and frequency, it is practically constant from no load to full load. It is measured by the open-circuit test, done on the low-voltage side at rated voltage with the high-voltage side open. Copper loss varies with load and is measured at full-load current by the short-circuit test, done on the high-voltage side with the low-voltage side shorted and a reduced voltage applied.
Efficiency and Maximum Efficiency
Output at load fraction x = x × S × cos φ Copper loss at fraction x = x² × Pcu,FL η = x S cos φ / (x S cos φ + Pi + x² Pcu,FL) × 100% Maximum efficiency when copper loss = iron loss: x = √(Pi / Pcu,FL)
For the 25 kVA transformer, assume the open-circuit test gave Pi = 300 W and the short-circuit test gave full-load copper loss Pcu = 500 W. These are example values chosen for the calculation.
| Load and PF | Output | Copper loss | Total loss | Efficiency |
|---|---|---|---|---|
| Full load, 0.8 PF | 20000 W | 500 W | 800 W | 96.15% |
| Half load, 0.8 PF | 10000 W | 125 W | 425 W | 95.92% |
| Full load, 1.0 PF | 25000 W | 500 W | 800 W | 96.90% |
| 77.5% load, 0.8 PF | 15492 W | 300 W | 600 W | 96.27% |
Maximum efficiency occurs at x = √(300/500) = 0.775, or 77.5% of full load, where copper loss equals iron loss at 300 W each. Distribution transformers stay energised round the clock but are lightly loaded at night, so they are designed with low iron loss and reach maximum efficiency below full load. All-day efficiency compares energy output to energy input over 24 hours for this reason.
Voltage Regulation
Regulation is the change in secondary voltage from no load to full load, as a percentage of full-load voltage. It is caused by the resistance and leakage reactance of the windings. A good approximation uses per-unit resistance and reactance.
% Regulation ≈ x (εr cos φ ± εx sin φ) (+ for lagging PF, − for leading PF) Example: εr = 2%, εx = 4%, full load, 0.8 PF lagging % Reg = 2 × 0.8 + 4 × 0.6 = 1.6 + 2.4 = 4.0% Secondary falls from 220 V to about 211.5 V
Summary
- E = 4.44 f N Φm links voltage, turns and core flux; volts per turn is the key design figure.
- Iron loss is constant and comes from the OC test; copper loss varies with load squared and comes from the SC test.
- Maximum efficiency occurs when copper loss equals iron loss, at x = √(Pi/Pcu).
- Regulation depends on winding resistance, leakage reactance and load power factor.