inductance of transformer windings

Inductance of magnetically coupled turns do not add in proportion to the turns count. If they are perfectly coupled, they add up in proportion to the square of the turns count. For small coils, the coupling is certainly not perfect (all the flux created by any turn does not surround all the other turns). But the squared approximation is still a first guess. So lets see what the inductance per turn squared is for this coil and then calculate the total inductance if all the turns are put in series aiding (all creating flux in the same direction with the same current passing through all turns).

Coil 2 471 uH / (2 turns squared) = 118 uH/T^2 Coil 1 706 uH / (3 turns squared) = 78 uH/T^2 Coil 3 941 uH / (4 turns squared) = 58 uH/T^2 Coil 4 1880 uH / (8 turns squared) = 29 uH/T^2

As the number of turns goes up, the inductance per turn squared is going down, a sure sign of poor coupling.

If we take the inductance factor of the 8 turn coil and use it as an upper limit for all 17 turns in series, the estimate would be 29 uH * (17 turns squared) = 8.4 mH.

You say the inductance of the whole 17 turns is 4 mHy. Is this a measured value, with all turns in series? It is not impossible, but it would indicate that there is essentially no magnetic coupling between the individual sections, and that would not make it a very effective transformer.

Well, I screwed that up.

I was taking the 4 coil inductances as measured values, but think I see now, that the only known inductance is the total of the 17 turns, and you are trying to calculate the section inductances.

So the full coupled assumption would be that inductance factor per turns squared would be:

So, by the turns squared assumption: Coil 2 ((2*T)^2)*13.8 uH / T^2 = 55 uH Coil 1 ((3*T)^2)*13.8 uH / T^2 = 124 uH Coil 3 ((4*T)^2)*13.8 uH / T^2 = 221 uH Coil 4 ((8*T)^2)*13.8 uH / T^2 = 883 uH

Since the actual flux coupling will be less than perfect, these are minimum possible values. At zero coupling your figures are what the section inductances would converge to.

The actual value will be somewhere in between. >

Thanks John, I got those same alternative numbers too using this turns squared formula I made

coil_turns(inductance) = (1 / (17 / coil_turns(turns))^2) * 4mH

I guess that the magnetic coupling of the transformer is the important thing to know now! What would you estimate it would be for a modern planar transformer design?

cheers, Jamie

I put 0.98 in ltspice as a first guess, but I guess there is a way to measure the coupling if you have the physical transformer..

cheers, Jamie

The secondary has 2 turns + 5 turns + 5 turns + 2 turns, (14 turns) can I calculate its total inductance by: secondary turns/primary turns * primary inductance = 14/17*4mH = 3.294mH

And then use the same method above to get the inductance of each winding I guess?

Also does 4mH seem high for only a 17turn primary? This is the number they told me but it seems quite high.

cheers, Jamie

You don't say how you are measuring inductance. You will get very different results for an inductor with a core, depending on things like measurement frequency and excitation level. Paul Mathews

(snip)

Depending on what the transformer will be used for, it might be very important. I would not estimate it at all, but measure it.

Jamie Morken wrote: (snip)

The coupling factor in LTspice is a representation of the voltage coupling efficiency between inductors. That means that if you apply 1 volt per turn to one winding and you get out only .98 volts per turn from another winding, then the coupling factor between that pair of windings is .98.

In your case, there will be a different coupling factor between each pair of sections, though the values may be very close to each other.

(snip)

I would estimate its total inductance by: ((14*T)*13.8 uH/T^2=2.7 mH

If the turns are all tightly coupled (assuming a coupling factor of 1) then this turns squared formula will work for any number of turns. Why do you need to calculate the inductance of various windings?

It implies a very high permeability core with no air gap. That also means that it will have a high tolerance, since it depends so strongly on the material properties and the accuracy of the ground contact surfaces and their proper assembly. But it also implies a coupling factor very close to 1 and good accuracy of the turns squared model of the inductance of any number of turns.