A transformer is two coils and a lump of iron, and it does three separate jobs that no other single component can do: it changes a voltage, it isolates one circuit from another completely, and it makes one impedance look like a different one. Every mains power supply on your bench contains one, most valve and many solid-state output stages contain one, and a balun at the feedpoint of your dipole is one. The exam asks about it more than any other passive component.
The principle: mutual inductance
Wind two separate coils on the same iron core. Feed alternating current into the first — the primary. Its current is continually changing, so its magnetic flux is continually changing, and the core carries that changing flux through the second coil — the secondary. A changing flux through a coil induces an EMF in it. That is Faraday's law, and one coil inducing an EMF in another is mutual inductance, exactly as the inductors lesson defined it.
Note what identifies each winding. The primary is the winding connected to the source and the secondary is the one connected to the load — not the one with more turns. A step-up transformer has fewer turns on the primary and a step-down transformer has more, so counting turns tells you nothing about which is which.
The turns ratio
Each turn of wire on the core has the same changing flux through it, so each turn has the same EMF induced in it. Put four times as many turns on the secondary and you get four times the voltage. The voltages are therefore in the same ratio as the turns:
Vs ÷ Vp = Ns ÷ Np
secondary volts ÷ primary volts = secondary turns ÷ primary turns
- Step-up: Ns is greater than Np, so the secondary voltage is higher than the primary.
- Step-down: Ns is less than Np, so the secondary voltage is lower. The 230 V mains adaptor on your bench is a step-down transformer: many turns on the mains side, few on the low-voltage side.
Worked example 1 — secondary voltage
A transformer has 100 turns on the primary and 10 turns on the secondary, and is connected to 230 V AC mains. What is the secondary voltage?
Vs = Vp × (Ns ÷ Np) = 230 × (10 ÷ 100) = 230 × 0.1 = 23 V
Ten times fewer turns gives one tenth of the voltage. Multiplying by the ratio instead of dividing produces 2300 V, which is the wrong answer the paper always offers.
Current runs the other way
A transformer is not an amplifier. It has no power source of its own, so whatever power the secondary delivers must have come in through the primary. Ignoring losses:
Vp × Ip = Vs × Is
watts in = watts out
Rearrange that and the current ratio is the inverse of the turns ratio. Step the voltage down and the current available goes up in the same proportion; step the voltage up and the current you can draw falls. A step-up transformer does not create power out of nothing — that is the intuition the exam is probing.
Ip ÷ Is = Ns ÷ Np
Worked example 2 — primary current
An ideal transformer delivers 2 A at 12 V from a 230 V mains primary. What current does it draw from the mains?
The secondary is delivering P = V × I = 12 × 2 = 24 W. The primary must
therefore take 24 W from the mains:
Ip = P ÷ Vp = 24 ÷ 230 = 0.104 A, about 0.1 A
Nineteen times the voltage, roughly a nineteenth of the current.
Type numbers into any two boxes below and watch all three relations move together — voltage with the ratio, current against it, impedance with its square.
Transformer calculator enter any two
Turns and voltage
Current — runs the opposite way
Impedance — follows the square of the ratio
Enter any two values that fix a ratio.
Why transformers are rated in volt-amperes
Two separate things limit a transformer, and neither of them is watts. The voltage is limited by what the insulation and the core will stand, and the current is limited by the size of wire in the winding, which heats as I²R. Multiply those two limits together and you get volt-amperes, which is what the nameplate states.
How many of those volt-amperes turn into real watts depends on the phase angle of a load the transformer knows nothing about. A load that draws current out of phase with the voltage — a motor, a choke — still heats the winding to the full extent of that current, while consuming fewer watts. Rating the part in watts would therefore be a lie for every load except a purely resistive one. The full story is power factor, in Phase, Reactance, Impedance and Power Factor.
The four losses
No transformer is 100 % efficient. The syllabus wants all four losses by name, and the second one is the most-asked transformer question in the whole paper.
| Loss | Where | Cause | Cure |
|---|---|---|---|
| Copper | Windings | I²R heating of the wire as load current flows | Thicker wire; rises with the square of the load |
| Eddy current | Core | The changing flux induces currents in the core metal itself, which loop within it and heat it | Laminate the core |
| Hysteresis | Core | Energy spent dragging the core round its magnetisation loop, twice every cycle | Choose the material — silicon steel or ferrite |
| Flux leakage | Between windings | Flux that escapes the core and misses the secondary altogether | Closed core, windings placed over one another |
Laminations, and why
The core is iron, and iron conducts. The changing flux does not distinguish between the secondary winding and the core, so it induces EMFs in the solid metal too, driving currents that swirl round inside it — eddy currents. They do no useful work and turn straight into heat.
The fix is to build the core from thin sheets — laminations — each varnished or oxide-coated so it is insulated from its neighbours, and stacked in the plane the flux runs along. The magnetic path is unaffected, but the electrical path an eddy current would take is chopped into many short, high-resistance loops, so far less current can flow round. That is the sole purpose of laminating a core: to reduce eddy-current loss. It does nothing for hysteresis.
Hysteresis loss is attacked differently. Every cycle the core is magnetised one way, demagnetised, magnetised the other way and demagnetised again, and the material's reluctance to follow costs energy each time round. Silicon steel is chosen because adding a few per cent of silicon to iron makes it magnetically softer — it follows the reversals with much less lost energy — and, incidentally, raises its electrical resistivity, which helps with eddy currents as well. At radio frequencies iron is hopeless and ferrite or an air core is used instead.
η = (power out ÷ power in) × 100 %
A well-designed mains transformer reaches 90–98 %. Note that core losses are present whenever the transformer is energised, loaded or not — an unloaded mains transformer left switched on still hums and still gets warm — while copper loss appears only when current is drawn.
Isolation and the autotransformer
Primary and secondary are linked only by the magnetic field. No conductor runs between them. That is what an isolation transformer sells: a 1:1 transformer that changes no voltage at all, but gives the secondary circuit no direct electrical connection to the mains. Touch one side of an isolated secondary while standing on the floor and no current flows through you, because there is no return path to the supply. Bench work on live mains-derived equipment is done through one.
An autotransformer throws that away deliberately. It has a single tapped winding, part of which serves as both primary and secondary, so it is smaller and cheaper for a small change of voltage. The Variac-style variable voltage supply is one. The price is that there is a direct connection between input and output, so an autotransformer offers no isolation whatever and must never be used where isolation is the point.
The transformer as an impedance matcher
This is a General-grade syllabus item, but it explains the output transformer in an audio stage and the balun on your antenna, so learn it now.
Voltage scales with the turns ratio; current scales inversely with it. Impedance is voltage divided by current, so it scales with the square of the turns ratio:
Zp ÷ Zs = (Np ÷ Ns)²
Worked example 3 — matching a loudspeaker
An output stage needs to work into 1600 Ω, and the loudspeaker is 4 Ω. What turns ratio is required?
Zp ÷ Zs = 1600 ÷ 4 = 400
The turns ratio is the square root of that: √400 = 20. A
20 : 1 step-down transformer — say 2000 turns on the primary and 100 on
the secondary — makes the 4 Ω speaker look like 1600 Ω to the valve or transistor
driving it.
Why bother? Maximum power is transferred from a source to a load when the load impedance matches the source impedance. Connect that 4 Ω speaker straight to a 1600 Ω stage and almost all of the available power stays in the source as heat; only a trickle reaches the cone. The transformer does not amplify anything — it just presents each side with the impedance it wants to see.
Practice
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