How a transformer works: mutual induction, turns ratio and losses

Mutual induction

Two coils wound on the same iron core share the same magnetic flux. When an alternating voltage V1 is applied to the primary winding, it drives a current that sets up an alternating flux Φ in the core. By Faraday's law the changing flux induces an EMF in every turn of both windings. The voltage per turn is the same for both because the same flux passes through each:

e = dΦ/dt per turn, so V1 / N1 = V2 / N2

The primary voltage and the flux are tied together. For a sinusoidal supply, the peak flux is

Φ_max = V1 / (4.44 × f × N1)

and the peak flux density is B_max = Φ_max / A, where A is the core cross-section. This formula explains why a transformer designed for 50 Hz overheats on 20 Hz: the flux goes up as the frequency goes down, until the iron saturates. The flux lags the applied voltage by 90°.

Turns ratio, voltage and current

For an ideal transformer, with no losses and perfect coupling:

V1 / V2 = N1 / N2

I1 / I2 = N2 / N1

V1 × I1 = V2 × I2

A step-down transformer has fewer turns on the secondary: lower voltage, higher current. A step-up transformer is the opposite. The transformer does not make power; it trades voltage against current. That is why power lines use very high voltage: for the same power the current is small, so the I²R loss in the wires is small, and transformers at each end convert to and from a safe voltage.

A load resistance R on the secondary appears at the primary as R × (N1 / N2)². This impedance transformation is used for matching in audio and radio circuits.

The real transformer

A real transformer departs from the ideal in several ways.

• Magnetizing current: even with no load the primary draws a small current to set up the flux in the core. It is typically a few per cent of the rated current and lags the voltage by nearly 90°.
• Copper losses: the winding resistances cause I²R heating, proportional to the load current squared.
• Core losses: the changing flux produces hysteresis loss and eddy-current loss in the iron. They depend on the voltage and frequency, not on the load. Cores are made of thin insulated laminations to cut eddy currents.
• Leakage flux: a part of the flux links only one winding. It acts like a small series inductance, which causes a voltage drop under load. The coupling coefficient k, between 0 and 1, describes how well the windings share flux; well-built power transformers have k very close to 1.

Because of these, the secondary voltage falls a little as load increases, and the output power is less than the input.

Why a transformer needs AC

Faraday's law needs a changing flux. If you connect a transformer to a DC source, there is a short pulse at the moment of switching on, then a steady flux induces nothing in the secondary. Worse, the primary resistance is small, so the DC current is limited only by the winding resistance and burns out the winding. A transformer therefore works only with AC (or pulsed DC), and a higher frequency allows a smaller core for the same voltage, which is why switch-mode power supplies work at tens of kilohertz and aircraft use 400 Hz.

Efficiency and rating

Efficiency is the output power divided by the input power:

η = P_out / (P_out + P_core + P_copper)

Large power transformers reach more than 98–99%. They are rated in volt-amperes (VA or kVA), not watts, because the heating depends on the voltage and current, whatever their phase. Maximum efficiency occurs when the copper loss equals the core loss, which is usually at about half to three quarters of the rated load. At light load the fixed core loss dominates, so efficiency drops.

Worked example: a transformer has a 2400 V primary and a 240 V secondary and supplies 10 kVA to a resistive load.

• Turns ratio N1 / N2 = 2400 / 240 = 10.
• Secondary current I2 = 10 000 / 240 = 41.7 A.
• Primary current I1 = I2 / 10 = 4.17 A.

If the losses are 200 W (core 100 W, copper 100 W), the input is 10.2 kW and the efficiency is 10 / 10.2 = 98.0%.

How the simulator models the transformer

Curvo Studio's single-phase transformer simulator is a sinusoidal steady-state teaching model with a fixed core cross-section of 25 cm² and a saturation limit of 1.6 T. It is not a full equivalent circuit.

• The no-load secondary voltage is V1 × (N2 / N1) × k, where k is the coupling coefficient.
• The peak flux density is B_max = V1 / (4.44 × f × N1 × A). If it exceeds 1.6 T, the core is saturated and the secondary voltage is scaled down accordingly.
• The load is a resistance. The secondary current is V2 / R, limited so that the output power does not exceed 94% of the rated VA.
• The efficiency is a simple function of load, rising from 88% at light load to 95% at full load and capped at 96%. The primary current used internally is the output power divided by that efficiency and the primary voltage.
• Magnetizing current, hysteresis, leakage reactance and temperature are not modelled separately.

The default is 500 VA, 230 V, 50 Hz, 500 : 50 turns, k = 0.98 and a 10 Ω load. The simulator shows V₂ = 22.5 V, I₂ = 2.25 A and a peak flux density of 0.83 T. Compare this with the ideal calculation: 230 / 10 = 23 V and 2.3 A. Internally the model also computes an output power of about 50.8 W and an efficiency of about 88.7%, but these are not shown as readouts.

Try it in the transformer simulator

Open the single-phase transformer simulator and try the following.

• Check the turns ratio: with 500 primary and 50 secondary turns, the Turns ratio is 0.1 and V₂ is about 22.5 V at 230 V. Set the Coupling coefficient to 1.00 and V₂ becomes exactly 23 V.
• Raise Secondary turns to 1000 (and the Load resistance to 100 Ω). The Secondary voltage becomes about 451 V, so the transformer steps the voltage up. The Secondary current is held to about 1 A by the model's output power limit of 94% of the rated VA, far below the 4.5 A that 451 V across 100 Ω would give.
• Lower the Load resistance to 1 Ω. I₂ rises only to about 21 A, because the output power is capped at 470 W (94% of 500 VA); without the cap it would be 22.5 A.
• Lower the Supply frequency from 50 Hz to 20 Hz. Bmax (the Core flux density) climbs above the 1.6 T saturation limit (to about 2.1 T), the core is flagged as saturated and V₂ falls. This is the frequency lesson in action.
• Set the Primary voltage to 0 to see that with no supply, no flux and no secondary voltage exist.

The Voltage and magnetic flux waveforms charts show the primary and secondary voltages in phase and the core flux density lagging the voltage by about 90°.

Try it in the simulator