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9.1 — Transformers and Why AC Won
In the 1880s two systems competed to electrify cities. Edison's ran on direct current at 110 V. Westinghouse and Tesla's ran on alternating current. Edison lost, and he lost because of one component that only works on AC.
Chapter 1.1 gave the reason in one equation. Power lost in a cable is I^2R, so moving power at low loss means low current, and since P=VI fixes the product, low current means high voltage. A DC system in 1890 had no way to change voltage. An AC system had the transformer.
1. How it works
Two coils wound on a shared magnetic core. Nothing connects them electrically.
Current in the primary produces a magnetic flux in the core. Because the current alternates, the flux alternates, and by Faraday's law of induction a changing flux through a coil induces a voltage in it:
V=-N\frac{d\Phi}{dt}
The minus sign is Lenz's law — the induced voltage opposes the change that produced it, which is what makes the whole thing energy-conserving rather than a perpetual motion machine.
Both coils share the same core and therefore the same flux. So:
V_1=N_1\frac{d\Phi}{dt}, \qquad V_2=N_2\frac{d\Phi}{dt}
Divide:
\boxed{\frac{V_2}{V_1}=\frac{N_2}{N_1}}
The voltage ratio is the turns ratio. Nothing else enters it.
And since an ideal transformer neither creates nor destroys power:
V_1I_1=V_2I_2 \;\Longrightarrow\; \frac{I_2}{I_1}=\frac{N_1}{N_2}
Step the voltage up and the current steps down by the same factor. That is the whole device.
Why it cannot work on DC. With a constant current the flux is constant, d\Phi/dt is zero, and no voltage is induced at all. The secondary sits dead while the primary, seeing only the resistance of its winding, draws enormous current and burns out. That is the entire reason Edison lost.
2. Impedance transformation
Connect a load Z_L to the secondary. What does the source see?
Z_{in}=\frac{V_1}{I_1}=\frac{V_2(N_1/N_2)}{I_2(N_2/N_1)}=\frac{V_2}{I_2}\left(\frac{N_1}{N_2}\right)^2
\boxed{Z_{in}=\left(\frac{N_1}{N_2}\right)^2Z_L}
Impedance transforms as the square of the turns ratio.
Worked example. An 8 Ω loudspeaker to be driven by a valve amplifier that needs a 5 kΩ load:
\frac{N_1}{N_2}=\sqrt{\frac{5000}{8}}=\sqrt{625}=25
A 25:1 transformer. This is the output transformer in every valve amplifier, and it exists precisely because Chapter 1.4's maximum power transfer requires a matched load and a valve's plate impedance is nothing like 8 Ω.
The same principle appears in radio matching networks (Chapter 7.7), in the balun that converts between a balanced dipole and unbalanced coaxial cable, and in every switching power supply.
3. Where the ideal model fails
Real transformers deviate in five ways, and each one is a design constraint.
Winding resistance. The copper has resistance, so I^2R is lost in both windings. This is the copper loss, and it rises with load.
Core loss, which has two parts:
Hysteresis loss — the core's magnetic domains resist reversing, and the energy to flip them is lost each cycle:
P_h=k_hfB_{max}^n, \qquad n\approx1.6\text{ to }2.0
Eddy current loss — the core is a conductor sitting in a changing magnetic field, so currents circulate within it:
P_e=k_ef^2B_{max}^2t^2
The t^2 is the lamination thickness, and it is why cores are made of thin sheets insulated from each other rather than solid iron. Halving the lamination thickness quarters the eddy loss, and a typical mains transformer uses 0.35 mm laminations.
Above a few kilohertz, laminations are not enough and cores are made of ferrite — a ceramic with magnetic properties and very high resistivity, so eddy currents essentially cannot flow.
Leakage inductance. Not all the flux from the primary links the secondary; some escapes into the air. That escaped flux behaves as an inductance in series with each winding, causing the output voltage to sag under load.
Sometimes it is deliberate. A microwave oven transformer has a magnetic shunt deliberately built in to increase leakage, which limits the short-circuit current — so the magnetron cannot draw unlimited power.
Magnetising current. Even with no load, the primary draws current to establish the flux. The transformer is, from the source's point of view, an inductor with the load in parallel.
Winding capacitance. Between turns and between windings, which sets an upper frequency limit and provides a path for high-frequency noise to cross the isolation barrier — which is why sensitive designs add an electrostatic shield between the windings.
4. The equivalent circuit
All five effects fit into one diagram that is the working model for every transformer calculation.
Voltage regulation measures the sag:
\text{regulation}=\frac{V_{no\ load}-V_{full\ load}}{V_{full\ load}}\times100\%
Small transformers: 5 to 15%. Large power transformers: 1 to 3%, because the copper is proportionally cheaper at scale.
Efficiency:
\eta=\frac{P_{out}}{P_{out}+P_{copper}+P_{core}}
And the maximum occurs where copper loss equals core loss. That is not a coincidence: core loss is fixed and copper loss rises with the square of load, so differentiating the efficiency expression and setting it to zero gives exactly that condition.
Distribution transformers are therefore designed so the two are equal at about 50 to 70% of rated load, since that is where they spend most of their life. Large power transformers reach 99.5% efficiency, which is higher than almost any other energy conversion device ever built.
5. Cores and materials
| Material | Frequency | Where |
|---|---|---|
| Silicon steel laminations | 50/60 Hz | mains, distribution |
| Ferrite | 20 kHz–1 MHz | switching supplies |
| Powdered iron | 100 kHz–100 MHz | RF, filters |
| Amorphous metal | 50 Hz–20 kHz | high-efficiency distribution |
| Air (no core) | above 1 MHz | RF |
The core exists to guide the flux, raising the coupling between windings from a few percent to over 99%. Its permeability \mu_r is typically 2,000 to 100,000, which is how much better it conducts flux than air.
Saturation is the hard limit. Beyond a certain flux density the core's domains are all aligned and it cannot carry more, so the inductance collapses and the current rises without limit.
B_{max}=\frac{V}{4.44fNA}
Read the frequency in the denominator. At a fixed voltage, halving the frequency doubles the flux density — which is why a transformer designed for 60 Hz can saturate on 50 Hz, and why equipment rated for both is designed for the 50 Hz case.
And it is why switching supplies are small. A 50 Hz transformer delivering 100 W needs a core weighing a kilogram. Raise the frequency to 100 kHz and the same power needs a core 2000 times smaller in the flux calculation, giving a device a few grams in weight. That single relationship is why a modern phone charger fits in a plug and a 1980s one weighed half a kilogram.
6. Three-phase
Generation, transmission and distribution are all three-phase, and the reasons are worth stating because they are not obvious.
Three voltages, each 120° apart:
v_a=V\sin\omega t, \qquad v_b=V\sin(\omega t-120°), \qquad v_c=V\sin(\omega t-240°)
Why three:
1 — Constant power. Single-phase power pulsates at twice the supply frequency, going to zero twice per cycle. The sum of three phases is constant:
p_a+p_b+p_c=3\frac{V_mI_m}{2}\cos\theta
with no time-varying term at all. A three-phase motor therefore produces smooth torque, where a single-phase motor vibrates at 100 or 120 Hz.
2 — A rotating magnetic field for free. Three windings 120° apart in space, fed by three currents 120° apart in time, produce a magnetic field that rotates smoothly. This is what makes the induction motor of Chapter 9.2 possible, and it needs no commutator, no brushes and no electronics.
3 — Less copper. For the same power, a three-phase system needs about 75% of the conductor material of a single-phase one, because the return currents partly cancel.
Star and delta.
Star (wye): three windings joined at a common neutral.
V_{line}=\sqrt3\,V_{phase}, \qquad I_{line}=I_{phase}
Delta: three windings in a closed triangle.
V_{line}=V_{phase}, \qquad I_{line}=\sqrt3\,I_{phase}
The \sqrt3 appears in both, and it comes from the 120° geometry: two phasors 120° apart, subtracted, give a resultant \sqrt3 times as long.
Worked example. European distribution: 230 V phase to neutral in star gives
230\times\sqrt3=398\approx400\ \text{V line to line}
Which is exactly what a three-phase socket provides, and why industrial equipment is rated at 400 V while domestic equipment is rated at 230 V — the same supply, measured between different pairs of terminals.
The star-delta starter exploits the difference: start a motor in star, where each winding sees 1/\sqrt3 of the line voltage and therefore draws a third of the current, then switch to delta once it is up to speed. A three-to-one reduction in starting current, from a switch.
7. The grid
Generation at 11 to 25 kV — limited by the insulation possible inside a rotating machine.
Step up to 400 kV or more for transmission.
Why so high. For a given power, I = P/(\sqrt3 V\cos\theta), so:
P_{loss}=3I^2R=\frac{P^2R}{V^2\cos^2\theta}
Loss is inversely proportional to the square of the voltage. Going from 132 kV to 400 kV — a factor of 3 — reduces the loss by a factor of nine.
Worked example. Transmitting 500 MW over 200 km of line with 0.05 Ω/km per phase:
At 400 kV, with a power factor of 0.95:
I=\frac{500\times10^6}{\sqrt3\times400\times10^3\times0.95}=760\ \text{A}
P_{loss}=3\times760^2\times10=17.3\ \text{MW}=3.5\%
At 132 kV the current would be 2303 A and the loss 159 MW — 32% of the power, which is unusable.
Step down in stages: 400 kV → 132 kV → 33 kV → 11 kV → 400/230 V.
Each step is a transformer, and there are perhaps five between a power station and a light bulb.
Why HVDC exists despite everything above. For very long distances or undersea cables, DC wins because:
- No reactive power. An AC cable's capacitance draws charging current whether or not power flows, and beyond about 80 km of submarine cable the charging current consumes the entire capacity.
- No skin effect, so the whole conductor carries current.
- Two conductors instead of three.
- It can link grids that are not synchronised, which is why the UK-France and Scandinavian links are DC.
The cost is the converter stations, tens of millions each. The break-even is around 600 km overhead or 50 km submarine, and beyond that DC wins outright — which is a striking reversal of the 1890s argument, made possible entirely by power semiconductors that Edison did not have.
8. Practical transformer types
Autotransformer. One winding with a tap, so part of it is shared between primary and secondary. Smaller and cheaper for ratios near 1:1, and it provides no isolation, since the two circuits share a conductor. Used in variable transformers (Variacs) and in grid interconnections between similar voltages.
Isolation transformer. 1:1 ratio, existing solely to break the electrical connection. Used in medical equipment and test benches, where the point is that touching either output terminal and earth simultaneously does not complete a circuit.
Current transformer. A single conductor passing through a toroidal core with many turns wound on it. The primary is one turn, so a 1000-turn secondary gives a 1000:1 current reduction — 500 A becomes 0.5 A, which an ordinary meter can read.
And the compulsory warning: never open-circuit a current transformer's secondary while current flows in the primary. The primary current is fixed by the load it is measuring, so the core drives whatever voltage it takes to push the secondary current — and with an open circuit, that is thousands of volts. Current transformers have shorting links for exactly this reason.
A clamp meter is a current transformer with a hinged core, which is why it measures current without breaking the circuit.
Pulse transformer. Designed for fast edges rather than sine waves, with very low leakage inductance and capacitance. Used to drive the gates of high-side power transistors, where the gate must float at hundreds of volts above ground.
Chapter 9.2 takes the rotating magnetic field that three-phase supply produces for free, and uses it to turn electricity into motion.
Every formula above, built from scratch
None of the results in this chapter are worth memorising, because each one can be rebuilt in under a minute from something simpler. What follows is that rebuilding, one result at a time, so the formula and the reason for it sit on the same page as the explanation that needed them.
Transformers
V=-N\frac{d\Phi}{dt} \qquad\text{(Faraday, with Lenz's minus sign)}
Both windings share the same core flux, so dividing their induced voltages gives
\frac{V_2}{V_1}=\frac{N_2}{N_1}, \qquad \frac{I_2}{I_1}=\frac{N_1}{N_2}
Impedance transformation:
Z_{in}=\frac{V_1}{I_1}=\frac{V_2(N_1/N_2)}{I_2(N_2/N_1)}=\left(\frac{N_1}{N_2}\right)^2Z_L
As the square of the turns ratio — for a valve amplifier driving 8 Ω from a 5 kΩ plate load, the ratio is \sqrt{5000/8}=25.
EMF equation:
V_{rms}=4.44\,fNA\,B_{max}
Where the 4.44 comes from: for sinusoidal flux \Phi = \Phi_m\sin\omega t, the induced voltage is N\omega\Phi_m\cos\omega t, whose rms value is N\omega\Phi_m/\sqrt2 = 2\pi fN\Phi_m/\sqrt2 = 4.44fN\Phi_m.
B_{max}=\frac{V}{4.44fNA}
Halving the frequency doubles the flux density at fixed voltage, which is why 60 Hz equipment can saturate on 50 Hz.
Core losses:
P_{hysteresis}=k_hfB_{max}^n, \quad n\approx1.6\text{–}2.0
P_{eddy}=k_ef^2B_{max}^2t^2
The t^2 on lamination thickness is why cores are laminated — halving the thickness quarters the eddy loss.
Efficiency and regulation:
\eta=\frac{P_{out}}{P_{out}+P_{cu}+P_{core}}, \qquad \text{regulation}=\frac{V_{NL}-V_{FL}}{V_{FL}}
Maximum efficiency occurs where copper loss equals core loss, because core loss is fixed and copper loss rises as load squared — differentiate and set to zero.
Three-phase
\text{Star: } V_L=\sqrt3\,V_{ph}, \quad I_L=I_{ph}
\text{Delta: } V_L=V_{ph}, \quad I_L=\sqrt3\,I_{ph}
Where \sqrt3 comes from: two phasors 120° apart, subtracted, give a resultant of 2V\sin60°=\sqrt3 V.
P=\sqrt3\,V_LI_L\cos\theta
Instantaneous three-phase power is constant, with no double-frequency term — unlike single phase, which pulsates to zero twice per cycle.
Transmission loss:
P_{loss}=3I^2R=\frac{P^2R}{V^2\cos^2\theta}
Inversely proportional to voltage squared, which is the entire case for high-voltage transmission.
What the next chapter fixes
A transformer moves energy between two coils that never move. Chapter 9.2 covers the machines where something does move — where electrical energy becomes rotation and rotation becomes electrical energy, using the same law of induction running in both directions.