Mutual Inductance

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I think I may have found a possible solution to this problem. Here is a link to the type I originally used:

http://uk.farnell.com/webapp/wcs/stores/servlet/ProductDisplay?catalogId=10001&langId=44&urlRequestType=Base&partNumber=1749245&storeId=10151

These are Panasonic ELC-18B series.  Looking at the datasheet fro the ELC series I see there is an E version which looks like this:

http://uk.farnell.com/panasonic-electronic-components/elc18e682l/inductor-6800uh-0-39a-10-radial/dp/2423581?categoryId=700000005459

Would I be right in thinking that this form of enclosed construction would have far less of an extraneous field and hence far less mutual inductance, perhaps even to the point where its effect is second order?

Cheers

Ian
 
ruffrecords said:
Thanks Abbey. I will replace all the inductors on the board with this new type and retest.

Cheers

Ian
This will make a difference, how much I can"t tell, but I would think it makes center frequencies close enough, probably within +/-5%, which is what you can expect in regard with tolerances (+/-10%).
 
abbey road d enfer said:
This will make a difference, how much I can"t tell, but I would think it makes center frequencies close enough, probably within +/-5%, which is what you can expect in regard with tolerances (+/-10%).

I measured more than 10 inductors of each value and they are all consistently higher than the specified value (as you might expect for an inductor intended for use in s SMPS) . I used the mean to calculate the required capacitance and selected caps to better than 5%. This is one reason I noticed the frequencies were wrong and then that  the inductances were wrong.

Thanks for your help.

Cheers

Ian
 
  Hi, how are you people, not so often around but here we are...

  Ian, I think what you are seeing is not mutual inductance but flux  path modifications, as you are talking about a single inductor value. You would get a similar effect if the inductors where close to any ferromagnetic material, not just inductors. Even one of this inductors close to a pot one would see the effect, while probably the pot one wouldn't or the effect is really small.

  Mutual inductance would be if you have two of those inductors in series and the value is not the value of the sum of both. That is, mutual inductance can be added or subtracted depending on the turns direction between the two inductors. This could also be happening on your board as I guess you do have series connected inductors there, but it wasn't the case for your first post.

  To solve it, yeah, keep the inductors apart or use closed(ish) cores, with this is not only the value that behaves strangely, once you build the whole thing you will start having mutual inductance between inductors with different signals (freq, phase and amplitude changes between them) and you will never know what is going on once you switch in or out one of them. That could be a much funnier troubleshoot than a 10% freq response shift! 

JS
 
Curious to see your signal tests/plot of your 'amusement machine' with the E replacements.  :)  Are the shields made of ferrite? 


http://www.farnell.com/datasheets/2059688.pdf?_ga=2.164009919.1090840084.1526520888-1365115998.1523225744
 
M>

https://www.electronics-tutorials.ws/inductor/mutual-inductance.html


do not know what use knowing the value of mutual inductance would serve , not really used to design transformers,

easy to work though>
 

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> not really used to design transformers,

It "is"; but instead of puny 75% or 33% coupling we normally strive for 99.9% coupling. The practical implications become different. The main inductance does not change (0.1%, NBD). The 0.1% not-coupled inductance, "leakage inductance", determines bandwidth (HF response).
 
simple definition;

"Mutual Inductance is the interaction of one coils magnetic field on another coil as it induces a voltage in the adjacent coil"

for two coils with the same turns like a 600;600 with 100% coupling;

"When the coefficient of coupling, k is equal to 1, (unity) such that all the lines of flux of one coil cuts all of the turns of the second coil, that is the two coils are tightly coupled together, the resulting mutual inductance will be equal to the geometric mean of the two individual inductances of the coils.

Also when the inductances of the two coils are the same and equal, L 1 is equal to L 2, the mutual inductance that exists between the two coils will equal the value of one single coil as the square root of two equal values is the same as one single value as shown."


so for equal coils: M = (L1 x L2) ^.5 = L
 
CJ said:
...
so for equal coils: M = (L1 x L2) ^.5 = L
That's for k=1 (total coupling), not needed for the coils to be equal, just that you are just calling L=L1=L2. Then, if coupling factor isn't 1 the mutual inductance is:
M =k* (L1 x L2) ^.5

Then when you connect them in series, the coupling inductance is added twice, with the sign depending of the direction of the turns.
Lt = L1 + L2 ± 2*M

If you want to find out the mutual inductance of two coils, one way is to connect them in series, measure the total inductance of the series and apply use the formula above. The other is to short one of the coils and measure the inductance in the other which will be Ls (parasitic inductance) and get
k = Sqrt( 1 - Ls / L1 )

  It is a useful value for both, series inductors and transformers, as it will affect the freq. response and efficiency of transformers and the value of the inductors.
 

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