Inductively coupled circuits
Two coils with no wire between them transfer energy, isolate a circuit from the mains and turn an 8 Ω loudspeaker into a 5 kΩ load. It all comes from a single number: the mutual inductance.
01Mutual inductance
When the magnetic flux of one coil passes through another, a varying current in the first induces a voltage in the second with no electrical connection between them. This relationship is measured by the mutual inductance \( M \), and it adds a term to the equation of each coil:
The sign of the \( M \) term depends on the winding directions, and it is indicated on the schematic with the dot convention: if both currents enter at the ends marked with a dot, the fluxes add and the mutual term takes a positive sign.
| Coupling | Typical \( k \) |
|---|---|
| Power transformer with laminated iron core | 0.998 or more |
| Ferrite transformer of a switched-mode power supply | 0.99 |
| Qi wireless charger, phone well seated | 0.5 to 0.7 |
| RFID coils a few centimeters apart | 0.01 to 0.1 |
02The ideal transformer
If the coupling is perfect (\( k=1 \)), the losses are zero and the inductances are so large that the current needed to magnetize the core is negligible, the equations reduce to purely algebraic relationships between the two coils, with \( n = N_2/N_1 = \sqrt{L_2/L_1} \):
A power tube performs well with a load of about 5 kΩ, and a loudspeaker is 8 Ω. A transformer with ratio \( n = \sqrt{8/5000} = 1/25 \) makes the loudspeaker look like 5 kΩ from the tube. It is the reason for being of the heaviest part of any tube amplifier.
03From the real transformer to the ideal one
A real transformer departs from the ideal in two ways, and each one rules at one end of the frequency range. At low frequency, the reactance \( \omega L_1 \) of the primary stops being large compared with the reflected load, and the source has to deliver a considerable magnetizing current. At high frequency, the flux that does not couple —the leakage inductance, \( (1-k^2)L_2 \) seen from the secondary— ends up in series with the load and steals voltage from it. The transformer behaves as ideal only in the band that lies between those two limits.
\( L_1 = 10 \) mH, \( L_2 = 2.5 \) mH (ratio 2:1, \( n = 0.5 \)) and a 100 Ω load: the ideal would promise 400 Ω at the input. You move the coupling and the frequency; the curves show how far the real pair departs from the ideal (1 = ideal).
04Where it shows up in electronics
| Device | What it takes advantage of in the coupling |
|---|---|
| Transformer of a linear charger | Stepping 220 V down to a low voltage and isolating the user from the mains |
| Transformer of a switched-mode power supply | The same, at tens of kHz: that is why it is much smaller than a 50 Hz one |
| Current transformer of a meter | Measuring hundreds of amperes without interrupting the conductor |
| Data isolation in Ethernet | Each port carries transformers that pass the signal and block DC and differing ground potentials |
| Wireless charging | Transferring energy with loose coupling, compensated with resonance on both sides |
The relationship between size and frequency in the table is no coincidence: to transfer the same power, the required core falls roughly in inverse proportion to the frequency. Going from 50 Hz to 100 kHz is what allowed a phone charger to fit in the plug.
05In the lab
With two coupled coils, a generator and a two-channel oscilloscope, determine which ends carry the dot by observing whether the voltages are in phase or in antiphase.
Measure \( L_1 \) and \( L_2 \) with an inductance meter; then measure the inductance of the two coils in series one way and the other. With \( L_a - L_b = 4M \), calculate \( M \) and \( k \).
Using this page’s lab, find between which frequencies the input impedance stays within 10 % of 400 Ω for k = 0.99 and for k = 0.9.
06Common mistakes
- Ignoring the dot convention. A wrong sign on \( M \) turns the sum of fluxes into a subtraction.
- Applying the ideal transformer to DC. A transformer does not transfer DC: with \( \omega = 0 \) there is no induction.
- Reflecting the impedance with \( n \) and not with \( n^2 \). Impedances are reflected with the square of the turns ratio.
- Believing that \( k = 1 \) is enough to be ideal. You also need a large magnetizing inductance compared with the reflected load.
- Forgetting galvanic isolation. It is often the main reason for using a transformer.
07Self-assessment
What does the coupling coefficient measure?
The fraction of the flux of one coil that passes through the other: \( k = M/\sqrt{L_1L_2} \).
How is a load reflected to the primary of an ideal transformer?
Divided by \( n^2 \), with \( n = N_2/N_1 \).
What limits a real transformer at low frequency?
The inductance of the primary: its reactance stops being large and the magnetizing current becomes considerable.
And at high frequency?
The leakage inductance, which ends up in series with the load.
Why is a switched-mode power supply transformer so small?
Because it works at tens of kHz, and the core required for the same power falls roughly with the inverse of the frequency.
08Further reading
- Charles K. Alexander and Matthew N. O. Sadiku. Fundamentals of Electric Circuits. 7th ed., McGraw-Hill, 2021. The chapter on magnetically coupled circuits, with the dot convention and the ideal transformer.
- Stephen J. Chapman. Electric Machinery Fundamentals. 5th ed., McGraw-Hill, 2011 (Spanish edition, “Máquinas eléctricas”). The real transformer, its equivalent circuit and the open-circuit and short-circuit tests.
- Colonel Wm. T. McLyman. Transformer and Inductor Design Handbook. 4th ed., CRC Press, 2011. Real transformer design for switched-mode power supplies, with the relationship between frequency and core size.