The transformer

The charger in your pocket

A phone charger plugs into a socket of $230$ V and gives out only $5$ V. Nothing inside it moves. So how does it change the voltage? If you opened one, you would find two coils of wire wound on one piece of iron. This is a transformer.

A transformer works because of mutual induction. When the current in one coil changes, the magnetic flux through a coil close to it changes too. An emf is then induced in the second coil. An alternating current changes all the time, so it keeps the induced emf going without a break.

Two coils on one core

The coil joined to the source is the primary. It has $N_p$ turns. The coil joined to the load is the secondary. It has $N_s$ turns. Both coils are wound on a core of soft iron. The core carries nearly all the flux of the primary through the secondary. Without it, most of the flux would spread into the air and miss the second coil.

The ideal transformer

First we picture a perfect one. It has three properties. The coils have no resistance. All the flux of the primary passes through every turn of both coils. The core wastes no energy.

At any instant the flux $\Phi$ through one turn is the same in both coils. By Faraday's law the emf of a coil is its number of turns times the rate of change of the flux:

The primary has no resistance, so the voltage $V_p$ applied to it is equal in size to its induced emf. The secondary is open at first, so its voltage $V_s$ is equal in size to its emf. Divide one equation by the other. The rate $d\Phi/dt$ cancels:

Each turn has the same voltage across it, whichever coil it sits in. So the coil with more turns has the larger voltage.

• If $N_s > N_p$ then $V_s > V_p$. This is a step-up transformer.
• If $N_s < N_p$ then $V_s < V_p$. This is a step-down transformer.

What happens to the current

A step-up transformer gives a higher voltage. Does it also give more energy? No. An ideal transformer wastes nothing and stores nothing. The power taken from the source is equal to the power given to the load:

Put this together with the voltage relation:

When the voltage goes up, the current goes down by the same factor. The coil with more turns has the larger voltage and the smaller current. All of these are rms values, because only rms values give the power.

How does the primary know how much current to take? With no load, the primary acts as a coil on its own, and the current is small. Now join a load. The current in the secondary makes a flux that opposes the flux of the primary. The net flux starts to fall, and so does the back emf of the primary. More current then flows from the source until the flux is restored. The source gives exactly as much power as the load takes.

A worked example: a doorbell

A doorbell transformer has $1380$ turns in its primary and $72$ turns in its secondary. The primary is joined to the $230$ V mains. The bell takes $1.5$ A from the secondary. Find the secondary voltage and the primary current. Treat the transformer as ideal.

1. The secondary voltage. $V_s = V_p\dfrac{N_s}{N_p} = 230\ \text{V} \times \dfrac{72}{1380} = 12.0$ V.
2. The power. $P = V_sI_s = 12.0\ \text{V} \times 1.5\ \text{A} = 18$ W.
3. The primary current. The primary takes the same power. We want $I_p$, so $I_p = \dfrac{P}{V_p} = \dfrac{18\ \text{W}}{230\ \text{V}} = 0.078$ A.

Check with the turns. $I_p = I_s\dfrac{N_s}{N_p} = 1.5\ \text{A} \times \dfrac{72}{1380} = 0.078$ A. It agrees. The primary current is only about one twentieth of the secondary current.

A step-up example: a sign

A shop sign needs $4600$ V. It is run from the $230$ V mains through a step-up transformer with $120$ turns in the primary. The tube of the sign takes $20$ mA, that is $0.020$ A. Find the number of turns in the secondary and the current in the primary.

1. The turns. We want $N_s$, so $N_s = N_p\dfrac{V_s}{V_p} = 120 \times \dfrac{4600\ \text{V}}{230\ \text{V}} = 2400$ turns.
2. The primary current. $I_p = I_s\dfrac{N_s}{N_p} = 0.020\ \text{A} \times \dfrac{2400}{120} = 0.40$ A.

Check with the power. The sign takes $4600\ \text{V} \times 0.020\ \text{A} = 92$ W. The mains give $230\ \text{V} \times 0.40\ \text{A} = 92$ W. The voltage rose twenty times and the current fell twenty times.

Start with the defaults. What is the ratio of the two voltages, and what is the ratio of the two currents? Now double $N_s$. What happens to the secondary voltage? What happens to the primary current?

Practise with new numbers

Why a transformer needs alternating current

Suppose you join the primary to a battery. The current becomes steady. A steady current makes a steady flux, and a steady flux induces no emf. When you close the switch the flux grows for a moment, so the secondary shows a short pulse. Then the voltage of the secondary falls to zero. A transformer works only while the flux changes, so it needs an alternating source.

There is a second reason. The wire of a primary is thick and has only a small resistance. A steady current is then very large, and the wire may burn. With an alternating source, the inductance of the coil gives the primary a large reactance $X_L$, and this limits the current.

A real transformer wastes some power

A real transformer is close to an ideal one but not equal to it. Some power turns into heat, and it has four common causes.

• Copper loss. The wires have resistance, so the windings heat by $I^2R$. The coil that carries more current is wound with thicker wire.
• Eddy currents. The changing flux induces currents in the iron core itself. They heat the core. We cut them down by building the core from thin sheets, called laminations, that are insulated from each other.
• Hysteresis. The core is magnetised one way, then the other, fifty times a second. Each turn of this cycle costs some energy. A core of soft iron loses little, because its magnetisation is easy to reverse.
• Flux leakage. Some flux of the primary does not pass through the secondary. We keep it small by winding the two coils one over the other on the same core.

A transformer also hums. The core changes its size a little as the flux changes, and the iron vibrates at twice the frequency of the supply.

The efficiency of a transformer is the power given out divided by the power taken in:

It is often given as a percentage. Large power transformers reach $95\ \%$ and more. The losses make the power taken in larger than the power given out, so $\eta$ is always less than $1$.

Example. A step-down transformer takes $540$ W from the mains and gives $513$ W to its load. Its efficiency is $\eta = \dfrac{513\ \text{W}}{540\ \text{W}} = 0.95$, that is $95\ \%$. The loss is $540\ \text{W} - 513\ \text{W} = 27$ W.

Where is the efficiency best?

The iron loss depends on the supply voltage and the core, not on the load. It stays the same from no load to full load. The copper loss is $I^2r$, and it grows with the square of the current. At a small load the iron loss is large compared with the output, so the efficiency is low. At a large load the copper loss grows fast, so the efficiency falls again. In between there is a best load. It comes where the copper loss equals the iron loss.

Find the load where the red point is highest. Compare it with the readout. Then make the iron loss larger and see which way the best load moves.

Try it in Python

The program takes the doorbell transformer and adds an iron loss and a winding resistance. It finds the efficiency and the primary current at four loads.

Common mistakes

• Turning the ratio upside down. The secondary voltage is $V_p\dfrac{N_s}{N_p}$. The coil with more turns has the larger voltage.
• Using the voltage ratio for the currents. The currents are in the inverse ratio: $\dfrac{I_p}{I_s} = \dfrac{N_s}{N_p}$.
• Thinking a step-up transformer gives more power. It gives more voltage and less current. The power is the same, or a little less.
• Putting a transformer on a battery. It needs a changing current, so it needs an alternating source.
• Mixing up the losses. Eddy currents are cut down by laminations, hysteresis by soft iron, and copper loss by thick wire.
• Forgetting that an efficiency is less than $100\ \%$. The power taken in is always more than the power given out.

Remember these

• A transformer works by mutual induction and needs an alternating current.
• Ideal: $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$ and $V_pI_p = V_sI_s$, so $\dfrac{I_p}{I_s} = \dfrac{N_s}{N_p}$.
• Step-up: $N_s > N_p$, larger voltage and smaller current. Step-down: $N_s < N_p$.
• Losses: copper (windings), eddy currents (laminations), hysteresis (soft iron), flux leakage.
• Efficiency $\eta = \dfrac{P_{out}}{P_{in}}$, always below $1$. It is best where the copper loss equals the iron loss.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. An ideal transformer has $500$ turns in its primary and $2000$ turns in its secondary. The primary current is $2.4$ A. What is the secondary current?

2. A student connects the primary of a transformer to a $12$ V battery and reads the voltage of the secondary with a voltmeter. After the first flick the reading is zero. Why?

3. Why is the core of a transformer made of thin insulated sheets, and not of one solid block of iron?

4. A transformer takes $800$ W and gives out $760$ W. What is its efficiency, and how much power is lost?