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Electrical Technology II · 144 h · Topic 5 of 6

Transformers

It is the simplest electrical machine—it has not a single moving part—and the one that made power distribution as we know it possible. It steps the voltage up for transmission and down for use, with efficiencies that no other machine can match.

Electrical machines Mutual induction Turns ratio Efficiency

01The principle

Everything follows from what we saw in electrodynamics and induction: a varying current produces a varying magnetic flux, and that flux, passing through another winding, induces a voltage in it. The iron core is there so that almost all of the primary's flux passes through the secondary.

flux Φ N1 = 1100 N2 = 60 220 V 12 V 0.273 A 5 A primary secondary same power on both sides: 220 × 0.273 = 12 × 5 = 60 W
Figure 1. Step-down transformer, animated. The alternating flux circulates through the core and links the two windings. With fewer turns on the secondary, the voltage drops in the same proportion and the current rises: the power going in is the power coming out, minus the losses.
V1V2=N1N2=I2I1=a The turns ratio a. Watch the order: the currents are inverted with respect to the voltages, and that is the key to everything that follows.
It only works with AC

What induces a voltage is the change in flux. With DC the flux is constant and nothing is induced: the only current is whatever the wire resistance allows, and the winding burns out. Connecting a mains transformer to a battery destroys it, and it is a mistake you only make once.

Example · A 220 V to 12 V transformer

Ratio a = 220/12 = 18.33. If the primary has 1100 turns, the secondary has 1100/18.33 = 60 turns.

With a load drawing 5 A from the secondary, the primary current is 5/18.33 = 0.273 A. The power: 12 × 5 = 60 W on the load side, and 220 × 0.273 = 60 W on the mains side. The voltage goes down and the current goes up in the same proportion.

02The real transformer: where energy is lost

An ideal transformer would transfer 100 % of the power. A real one loses power in two different places, and telling them apart is what lets you diagnose a fault or calculate an efficiency.

Core (iron) losses · PFe
  • Hysteresis: the energy it takes to reorient the magnetic domains on every cycle. It is reduced with grain-oriented silicon-steel laminations.
  • Eddy currents: currents induced inside the core itself. They are reduced by laminating the core and insulating the laminations from one another: that is why cores are a stack of thin sheets and not a solid block.
  • They depend on the voltage and frequency, not on the load: they exist even with the secondary open. They are what makes a transformer draw power—and hum—with nothing connected.
Copper losses · PCu
  • Joule heating in the winding resistance: P = I²·R.
  • They depend on the square of the load: at half load they are a quarter of what they are at full load.
  • They set the thermal limit of the transformer, and they are what you feel when you touch it after it has been working for a while.
50 % 75 % 100 % 25 % 50 % 75 % 100 % 125 % load relative to rated load maximum: P_Cu = P_Fe load 25 % η = 87.0 % iron: 2.0 W (constant) copper: 0.25 W (∝ I²) output: 15.0 W load 50 % η = 90.9 % iron: 2.0 W (constant) copper: 1.00 W (∝ I²) output: 30.0 W load 71 % η = 91.4 % iron: 2.0 W (constant) copper: 2.02 W (∝ I²) output: 42.6 W load 100 % η = 90.9 % iron: 2.0 W (constant) copper: 4.00 W (∝ I²) output: 60.0 W load 125 % η = 90.1 % iron: 2.0 W (constant) copper: 6.25 W (∝ I²) output: 75.0 W 60 VA transformer with 2 W of iron loss and 4 W of copper loss at full load. At no load the efficiency is zero: it delivers nothing and still dissipates the 2 W of iron loss.
Figure 2. Efficiency versus load, animated. The iron losses are constant and the copper losses grow with the square of the current. The efficiency is highest exactly where the two are equal, which is why distribution transformers are sized to operate near that point.
η=P2P2+PFe+PCureg=V0−VcVc·100 The regulation compares the no-load voltage with the full-load voltage: how much the transformer “sags” when loaded. In a small one it can be 10 to 20 %; in a distribution transformer, 2 to 4 %.
The two classic tests
  • Open-circuit test: the primary is supplied at rated voltage with the secondary open. Almost no useful current flows, so the wattmeter reads practically just the iron losses. It also gives the magnetizing current.
  • Short-circuit test: the secondary is short-circuited and the primary voltage is raised until rated current flows—only 3 to 6 % of the rated voltage is needed—. Since the flux is small, the iron hardly loses anything: the wattmeter reads the copper losses, and the short-circuit impedance follows from that.

With those two tests, carried out with very little energy, a transformer of any power rating is fully characterized. It is an elegant method and it turns up in any exam on the subject.

Example · Efficiency of a 60 VA transformer

Open-circuit test: PFe = 2 W. Short-circuit test at rated current: PCu = 4 W.

  • At full load (60 W output): η = 60/(60 + 2 + 4) = 60/66 = 90.9 %
  • At half load (30 W): the copper losses drop to a quarter, 1 W. η = 30/(30 + 2 + 1) = 30/33 = 90.9 %
  • At 25 % (15 W): PCu = 0.25 W. η = 15/(15 + 2 + 0.25) = 86.9 %

The maximum efficiency is where PCu = PFe, that is, at a load of √(2/4) = 71 %. And the other thing shows up too: at no load the efficiency is zero, because it delivers nothing and yet dissipates 2 W. That is the argument against leaving chargers plugged in when not in use, although with modern switching power supplies the figure is considerably lower.

03Designing a small transformer

It is an approximate calculation, but good enough to wind a laboratory transformer, and the procedure is still used in the workshop.

S=1.2PNV=45Sd=2I/Jπ S is the core cross-section in cm² and P the power in VA; the 45/S relation holds for 50 Hz with a flux density of about 1 T. J is the allowable current density, 3 A/mm² in small transformers.
Worked example · 220 V to 12 V, 60 VA transformer
  1. Core cross-section: S = 1.2 × √60 = 1.2 × 7.75 = 9.3 cm². Choose a set of E-I laminations whose center leg times the stack thickness gives approximately that value.
  2. Turns per volt: N/V = 45 / 9.3 = 4.84.
  3. Primary: 220 × 4.84 = 1065 turns.
  4. Secondary: 12 × 4.84 = 58 turns, plus 5 % to compensate for the drop under load → 61 turns.
  5. Currents: I₂ = 60/12 = 5 A; I₁ = 60/220 = 0.273 A.
  6. Secondary wire: cross-section = 5/3 = 1.67 mm² → d = 1.46 mm (AWG 15).
  7. Primary wire: cross-section = 0.273/3 = 0.091 mm² → d = 0.34 mm (AWG 28).

Then you have to check that the winding fits in the window of the lamination: if it does not, increase the core cross-section (fewer turns) or accept a higher current density at the cost of a higher temperature.

Why a transformer hums

Because of magnetostriction: the laminations change dimension slightly with the flux, a hundred times per second. If they are loose, they vibrate and the hum is amplified. A hum that suddenly gets louder usually points to loose laminations, overvoltage—the core goes into saturation—or a shorted turn, which will also heat it up quickly.

04Impedance transformation

This is the transformer's third function, the least obvious and the most used in electronics: seen from the primary, whatever is on the secondary appears multiplied by a².

Z1=a2·Z2a=Z1Z2 It is used to match a source to its load and transfer the maximum power, which is the central problem of output stages and transmission lines.
Example · Matching a loudspeaker to an output stage

A vacuum-tube stage needs to see 5000 Ω at its plate and the loudspeaker is 8 Ω:

  • a = √(5000/8) = √625 = 25
  • With 2500 turns on the primary, the secondary has 100.

The same idea explains the 70 V line transformer of public-address systems, the balun of an antenna and the reason why in RF people always talk about 50 Ω: if the load is not what the line expects, part of the energy is reflected back, as seen in impedance measurements.

05Types you need to know

TypeWhat is special about itWhere it appears
PowerTwo separate windings, laminated E-I or core-type (legged) corePower supplies, distribution, substations
1:1 isolationDoes not change the voltage: provides galvanic isolationEssential for measuring mains-connected circuits with an oscilloscope
AutotransformerA single tapped winding: smaller and cheaper110/220 V step-up converters, motor starting. Does not isolate: primary and secondary share a conductor
ToroidalRing-shaped core: less leakage, less hum, less stray fieldHigh-quality audio, instrumentation
Current transformer (CT)The primary is the conductor itself; it delivers a proportional current (5 A typical)Metering and protection in switchboards; the same principle as the clamp meter
Voltage transformer (VT)Steps the voltage down to a measurable value (110 V typical)Metering at medium and high voltage
High-frequencyFerrite core, not laminated steelSwitching power supplies: at 100 kHz the core can be hundreds of times smaller
Never open the secondary of a current transformer

With the primary in service, if the secondary of a CT is opened, the current that was compensating the flux disappears: the core saturates and voltages of thousands of volts appear across the open terminals. It is one of the most dangerous operations in a switchboard. Before disconnecting an instrument fed by a CT you must short-circuit the secondary using the terminal block provided for that purpose.

Why switching power supplies are so small

Because the required core cross-section is inversely proportional to frequency. At 50 Hz a 100 W transformer weighs more than a kilogram; at 100 kHz—two thousand times faster—the same job is done by a ferrite core that fits in the palm of your hand. That is why almost all power supplies today are switching ones, and also why they produce interference and need filters (filters).

06In the lab

Lab 1 · Turns ratio

With a multi-tap laboratory transformer, measure the input voltage and the voltage at each output with no load, and calculate the ratios. Compare them with the turns ratio if the manufacturer states it. Then wind 10 turns of wire around the core and measure the voltage that appears: dividing gives you the volts per turn value experimentally.

Lab 2 · The two tests

Open-circuit test: supply at rated voltage with the secondary open and measure V, I and P. Short-circuit test: with the secondary short-circuited, raise the primary voltage gradually with a variac until rated current is reached, and measure. With those data, calculate the efficiency at full load and at half load, and plot the curve.

Lab 3 · Regulation

Measure the secondary voltage with no load and with rated load (lamps or resistors). Calculate the regulation. Repeat with a higher-power transformer and compare: the larger one regulates better. That is why an oversized transformer gives more stable voltages.

Lab 4 · Impedance matching

With an audio output transformer, measure the resistance seen by the generator with the secondary loaded with 8 Ω and with the secondary open. Compare the measured value with a²·Z₂. Then drive a loudspeaker with and without the transformer from a high-impedance stage and compare the volume: the difference is the matching.

Working with an isolation transformer

Any oscilloscope measurement on a mains-connected circuit must be made with an isolation transformer. The oscilloscope ground is tied to the earth of the wall outlet: if you connect it to a live point, you create a dead short through the instrument. An autotransformer is not suitable for this, because it does not separate the circuits.

07Common mistakes

SymptomUsual cause
Gets very hot with no loadShorted turn, poorly insulated laminations or a supply voltage above the rated value (saturation).
The output voltage drops a lot when the load is connectedPoor regulation: undersized transformer or wire that is too thin.
Hums more than usualLoose laminations, overvoltage or an incipient winding fault.
The primary fuse blows when you switch it onInrush current: when energized at the worst instant, the flux adds to the remanent flux and saturates the core. Solved with a slow-blow fuse or an inrush-limiting NTC thermistor.
Gives the correct voltage but delivers no currentYou are measuring with no load. A transformer with shorted turns or insufficient power reads fine with no load and collapses under load.
The 12 V secondary measures 13.5 VThis is normal: small transformers are wound with 5 to 10 % extra turns to compensate for the drop under load.
Burned out when connected to DCWithout a change in flux there is no inductive reactance: only the wire resistance remains, and it is very low.
An autotransformer “does not isolate” and there was an accidentCorrect: it shares a conductor between input and output. To isolate you need one with two separate windings.
Spark when disconnecting an instrument from a CTThe secondary of the current transformer was opened. It must be short-circuited first.

08Self-assessment

A 220 V to 24 V transformer has 1100 turns on the primary: how many does the secondary have?

a = 220/24 = 9.17. N₂ = 1100/9.17 = 120 turns.

If the secondary delivers 4 A, what current does the primary of the previous example draw?

I₁ = I₂/a = 4/9.17 = 0.436 A. Check by power: 24 × 4 = 96 W and 220 × 0.436 = 96 W.

How do iron losses differ from copper losses?

The iron losses (hysteresis and eddy currents) depend on the voltage and frequency and are practically constant: they exist even at no load. The copper losses are I²R and grow with the square of the load.

At what load is the efficiency highest?

When the copper losses equal the iron losses. With PFe = 2 W and PCu = 4 W at full load, that happens at √(2/4) = 71 % of the rated load.

What is the short-circuit test for, and why does it not destroy the transformer?

To measure the copper losses and the short-circuit impedance. It does not destroy it because only 3 to 6 % of the rated voltage is applied: just what is needed for rated current to flow.

Why is the core made of laminations and not solid iron?

To cut down the eddy currents. In a solid block, the varying flux induces internal currents that heat it uselessly. Laminating and insulating the sheets greatly reduces that loss.

You want to match a 4 Ω load to a 900 Ω source: what turns ratio is needed?

a = √(900/4) = √225 = 15. That is, 15 turns on the primary for every one on the secondary.

What is the difference between an autotransformer and an isolation transformer?

The autotransformer has a single winding with a tap: input and output share a conductor and there is no galvanic isolation. The isolation transformer has two independent windings, which is why it is the only one suitable for working safely on mains circuits.

Why is the transformer in a switching power supply so small?

Because it operates at tens or hundreds of kHz instead of 50 Hz. The required core cross-section is inversely proportional to frequency, so at 100 kHz a tiny ferrite core is enough.

A 100 VA transformer is supplied with 240 V instead of 220 V. What happens?

The flux increases proportionally and the core approaches saturation: the no-load current grows much more than the voltage, and the transformer hums and heats up. This is the typical fault when the mains voltage is high or when a 220 V transformer is connected to a 240 V supply without a suitable tap.

Development of the topic “Transformers” of Electrical Technology II (Year 5), based on the “Curriculum Proposal – Second Cycle of the Technical-Vocational Track, Secondary Education – Electronics,” Ministry of Education of the Province of Córdoba, DGETyFP. Back to the Topic Map · catto.ar