Mutual induction

Question

I need to find the mutual induction between two coils.

^{simulate this circuit – Schematic created using CircuitLab}

The circuit look more or less like the one I have linked. I don't know how to calculate the mutual induction between the two coils. Does anyone have any experience they would like to share? I would appriciate it very much!

Answer 1

Does anyone have any experience they would like to share. I would appreciate it very much!

_{For simplicity, I'm keeping this answer based around a 1:1 transformer where the primary and secondary inductances are both 13.5 μH (as per the question).}

100% coupling, no-load

If the coils are 100% coupled, the mutual inductance is \$\sqrt{L1\cdot L2}\$ and, the output voltage is the same as the input. However, if the output voltage is only 50% of the input voltage, then you can be sure that the coupling coefficient is half. This is because only half the flux produced by the primary couples with the secondary and, as per Faraday's law of induction, the induced voltage will be half. So, the unloaded transfer voltage ratio is a good measure of "\$k\$" when the transformer is a 1:1 type.

Adding a load, coupling less than 100%

When we factor in \$k\$, the mutual inductance formula becomes \$k\sqrt{L1\cdot L2}\$: -

When you add a load of high value you can usually assume that the transfer ratio defines the coupling coefficient \$k\$ but, be careful. With (say) an apparent 50% transfer ratio and a load that varies you can become mistaken. For instance, here's a 50% coupled transformer with a load that varies between 1 ohm and 100 ohm: -

As you can see, if you had a 1 ohm load resistor and your measured the transfer ratio at 100 kHz you'd see an attenuation of nearly 24 dB. At low frequencies (say 1 kHz) the attenuation is 6 dB (50%) so there's no problem measuring at 1 kHz but at higher frequencies you can misdirect yourself into assuming the wrong coupling coefficient.

Equivalent circuit

The reason why this happens because when the coupling is only 50%, leakage inductances become present and these, along with the load resistor, form a potential divider like this: -

In the above circuit, I've halved the coupled inductances to 6.75 μH and made k = 1 then I've added the leakage inductances as new inductors L3 and L4. This produces exactly the same result as my original circuit up above.

Now, the mutual inductance could be taken as \$\sqrt{6.75 \text{ μH}\cdot 6.75\text{ μH}}\$ which equals 6.75 uH or, could be taken as this: -

$$0.5\cdot\sqrt{13.5\text{ μH}\cdot 13.5\text{ μH}} = 6.75\text{ μH}$$

And, finally, the equivalent circuit not using a transformer is this: -