1.11 Transformers: Turns Ratio, Impedance Matching and Eddy Currents

1.11 Transformers: Turns Ratio, Impedance Matching and Eddy Currents

ACMA Exam Focus: Syllabus items 4.24, 4.25 and 4.26 — understand mutual coupling between windings and Faraday's law of induction, apply the relationship between turns ratio, voltage ratio and current ratio, apply the rule that impedance transforms as the square of the turns ratio, and understand eddy currents together with the laminations and ferrite cores used to control them.

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Mutual coupling and Faraday's law (item 4.24)

A transformer is two or more windings sharing a common magnetic path. Alternating current in the primary winding ($N_p$ turns) produces a continuously changing magnetic flux in the core; that flux links the secondary winding ($N_s$ turns) and induces a voltage in it. Energy crosses from one circuit to the other entirely through the magnetic field — the two windings are not electrically connected at all. This linkage is mutual coupling, and the fraction of the primary flux that reaches the secondary is the coefficient of coupling, k, running from 0 for no coupling to 1 for perfect coupling.

The underlying principle is Faraday's law of induction: a voltage is induced in a conductor whenever the magnetic field linking that conductor changes. Nothing moves inside a transformer, so it is the alternation of the current that supplies the necessary change. This is also why a transformer will not work on direct current: a steady current produces steady flux, there is no change, and no secondary voltage appears. Connect DC to a mains transformer primary and the only thing limiting the current is the winding's DC resistance, which is usually the end of both the transformer and the fuse.

Because the windings are electrically separate, a mains transformer also gives galvanic isolation — a safety property that matters in every mains-powered item in the shack.

Voltage, turns and current ratios (item 4.25)

Voltage divides in proportion to turns:

$\frac{V_p}{V_s} = \frac{N_p}{N_s}$

An ideal transformer transfers power without loss, so $V_p I_p = V_s I_s$, and the current therefore goes the other way:

$\frac{I_p}{I_s} = \frac{N_s}{N_p}$

Step the voltage down and you step the current up. A transformer is not a source of free energy: the secondary can supply more amperes than the primary draws only because it supplies them at a lower voltage.

Worked example 1 — a supply transformer. A transformer runs from the 230 V RMS, 50 Hz Australian mains and must deliver 15.0 V RMS to a bridge rectifier. The primary has 1840 turns.

$N_s = N_p \times \frac{V_s}{V_p} = 1840 \times \frac{15}{230} = 120\ \text{turns}$

If the secondary supplies 10.0 A, the primary current is

$I_p = I_s \times \frac{V_s}{V_p} = 10.0 \times \frac{15}{230} \approx 0.65\ \text{A}$

which is why a supply of this size sits happily on an ordinary 10 A power point.

Impedance transforms as the square of the turns ratio (item 4.25)

This is the relationship examined most often, because it is the one that matters on air:

$\frac{Z_p}{Z_s} = \left(\frac{N_p}{N_s}\right)^2$

Turn it around and the turns ratio needed for a given impedance transformation is the square root of the impedance ratio.

Worked example 2. A transformer has 1200 primary turns and 300 secondary turns, and the secondary is loaded with 50 Ω.

1. Turns ratio: $1200 / 300 = 4$.
2. Impedance ratio: $4^2 = 16$.
3. Impedance seen at the primary: $16 \times 50 = 800\ \Omega$.

Worked example 3. A "4:1 balun" is used to feed a folded dipole. The 4:1 describes the impedance ratio, so the turns ratio is $\sqrt{4} = 2$:1. Presented with a 200 Ω feed point, it hands $200 / 4 = 50\ \Omega$ to the coaxial cable.

The point of all this is maximum power transfer: a source delivers its full power only into a matched load. Transformers, baluns and ununs are how an amateur station reconciles a 50 Ω transmitter with antennas and feedlines that are anything but 50 Ω. The word balun itself is a contraction of balanced to unbalanced, describing the other job these transformers do at an antenna feed point.

Eddy currents, laminations and ferrites (item 4.26)

The core carries the alternating flux — and the core is itself a conductor. Faraday's law does not distinguish between a winding and a lump of iron, so the changing flux induces circulating currents inside the core metal. These are eddy currents. They do no useful work at all: they flow through the resistance of the core material, waste power as heat and reduce efficiency. In a solid core, left unchecked, they can make a transformer run hot enough to destroy its own insulation.

Two construction techniques control them:

• Laminations. A 50 Hz mains transformer core is built from thin sheets of silicon steel, each varnished or oxide-coated so that it is electrically insulated from its neighbours, stacked to make up the required cross-section. That insulation breaks one large low-resistance eddy loop into many small high-resistance ones, and the loss falls sharply. The sheets lie parallel to the flux, so the magnetic path itself is unaffected.
• Ferrite and iron-powder cores. At radio and switch-mode frequencies eddy losses grow so fast that even laminations are inadequate. Ferrite is a ceramic magnetic material with very high electrical resistivity, in which eddy currents can barely flow at all. Iron-powder cores achieve the same end by insulating individual particles in a binder. This is why RF chokes, baluns, ferrite beads and switch-mode transformers use ferrite, while the mains transformer sitting beside them uses laminated steel.

The other core loss is hysteresis loss — the energy spent reversing the magnetisation of the core on every cycle. Both losses climb with frequency, which is why a transformer designed for 50 Hz overheats if it is driven at a much higher frequency, and why a ferrite chosen for HF is the wrong material for a mains supply.