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Software to calculate the inductance of a printed multilayer coil

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Bou

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Hello,

I m looking for a 2D software or calculator to determine the inductance of a printed 2 layers inductor

Usually, I work with ADS but I need another software to compare results

I tried SOnnet but the evaluation version it s limited to 30 M which is not enough in my case

If you know other softwares, would you please share theme with me ;)

10x
 
Last edited:

You can use free FastHenry solver. Design entry is not so convenient but no limitations in complexity.
 

10x FvM for your reply

But i seems diffucult to you use!!!
 

I have used it for simulation of RFID coupler coils and PCB trace self and mutual inductance as well as skin effect losses.

Yes, you need some time to become familiar with the description language. You can also visualize your geometry file with the FastModel tool.
 

Do you know any 2D (2.5D) softawre to simulate 2 layers inductors other than ADS Momentum, SONNET, PEDASOFT?
 

Bou said:
I m looking for a 2D software or calculator to determine the inductance of a printed 2 layers inductor
Click to expand...

I have worked on planar inductor calculation for many years. From my experience, textbook equations are accurate within +/-5% for the low frequency inductance, so you can use that for comparison:
https://www.circuits.dk/calculator_planar_coil_inductor.htm

For a more accurate simulation (with loss and self resonance from parasitic capacitance) your ADS Momentum solver is fine.
 

10x volker@muehlhaus,
For the online calculators, I have tested some but I noticed that they are all used for single layer inductors, I didn't find an online calculator for 2 or multilayer
 

Bou said:
For the online calculators, I have tested some but I noticed that they are all used for single layer inductors, I didn't find an online calculator for 2 or multilayer
Click to expand...

I'm sorry! You need to use the underlying equations where you can use "N" for number of turns. This is the total number of turns, e.g. N=10 if you have 5 turns on each of the two layers.

"Modified Wheeler" formula or "Expression Based on Current Sheet Approximation" both work fine.
http://smirc.stanford.edu/papers/JSSC99OCT-mohan.pdf

 

volker@muehlhaus said:
I'm sorry! You need to use the underlying equations where you can use "N" for number of turns. This is the total number of turns, e.g. N=10 if you have 5 turns on each of the two layers.]


But an single lauer inductor with 10 turns is not equivalent to 2 layers inductor with 5 turns in each one
The mutual inductance between the different segments are not the same
Click to expand...
 

Bou said:
But an single lauer inductor with 10 turns is not equivalent to 2 layers inductor with 5 turns in each one The mutual inductance between the different segments are not the same
Click to expand...

If inner and outer diameter are identical, the inductance for single layer 10 turns and dual layer 2x5 turns is indeed identical. ;-)
 

volker@muehlhaus said:
If inner and outer diameter are identical, the inductance for single layer 10 turns and dual layer 2x5 turns is indeed identical. ;-)
Click to expand...

Can you send me the reference, because I remember that I saw some articles that include a demonstration for the inductance of double layer coil: with some conditions in the thickness of the substrate the Total inductance is 4 times the inductance of single layer
 

You apparently misunderstood the example.

If you have two coils with perfect coupling, the total inductance (series connected) will be the fourfold of a single one.

If you have a flat coil of neglibible thickness with 5 windings, duplicate it with infinitesimal distance between both coils, you also get perfect coupling and thus fourfold inductance of the series circuit. The interesting point is, how fast does the coupling factor decay with windings distance so that the total inductance is somewhere between coupled (fourfold) and uncoupled (doubled) value?

Mathematically, the total inductance of the series circuit is the sum of self and mutual inductance.

Ltot = 2*Lsingle + 2*k*Lsingle, with k varying from 0 to 1.
 

FvM said:
You apparently misunderstood the example.

If you have two coils with perfect coupling, the total inductance (series connected) will be the fourfold of a single one.

If you have a flat coil of neglibible thickness with 5 windings, duplicate it with infinitesimal distance between both coils, you also get perfect coupling and thus fourfold inductance of the series circuit. The interesting point is, how fast does the coupling factor decay with windings distance so that the total inductance is somewhere between coupled (fourfold) and uncoupled (doubled) value?

Mathematically, the total inductance of the series circuit is the sum of self and mutual inductance.

Ltot = 2*Lsingle + 2*k*Lsingle, with k varying from 0 to 1.
Click to expand...

Thank you very much

Now it's more clear for me :razz::razz::razz:
 

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