inductance of transformer windings

[Jamie Morken]

Hi,

I have a transformer primary of 17 turns and 4mH inductance.

There are 4 coils in series (3 taps / 5 leads)

I am trying to figure out the inductance of the 4 coils.

coil1 3turns 706uH

coil2 2turns 471uH

coil3 4turns 941uH

coil4 8turns 1.88mH

that adds up to 4mH but I read before that when you
double the turns the inductance quadruples, so I am
not sure if the above calculations are correct.

cheers,
Jamie

 

[John Popelish]

Hi,

I have a transformer primary of 17 turns and 4mH inductance.

There are 4 coils in series (3 taps / 5 leads)

I am trying to figure out the inductance of the 4 coils.

coil1 3turns 706uH

coil2 2turns 471uH

coil3 4turns 941uH

coil4 8turns 1.88mH

that adds up to 4mH but I read before that when you
double the turns the inductance quadruples, so I am
not sure if the above calculations are correct.

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.

 

[John Popelish]

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. (snip)

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:
4000 uH / (17 turns squared) = 13.8 uH / T^2

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.

 

[Jamie Morken]

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:
4000 uH / (17 turns squared) = 13.8 uH / T^2

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.

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

 

[Jamie Morken]

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?

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

 

[Jamie Morken]

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:
4000 uH / (17 turns squared) = 13.8 uH / T^2

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 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

 

[neon]

Gee he measured inductance at what frequency nobody knows I don't know either interesting

 

[Paul Mathews]

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






- Show quoted text -- Hide quoted text -

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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

 

[John Popelish]

John Popelish wrote:

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?

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

 

[John Popelish]

Jamie Morken wrote:
(snip)

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..

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.

 

[John Popelish]

John Popelish wrote: (snip)

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

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

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

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?

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

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.