Coupling Inductors

Is there a way to couple 2 inductors with a variable coupling coefficient?

I'm trying to model an Armstrong regenerative AM radio detector and I can't figure out how to lightly couple the tickler coil to the tank coil.

I'm a complete novice at simulating circuits and CircuitLab has been a wonderful education tool. Thank you!

DanM

by DanM
August 23, 2012

signality - thanks for the reply. I'll stay with CL on my Mac and look for a different type regenerative circuit to try. My computer literacy is about equal to my simulation literacy - ie. zilch. So I really appreciate CL's lack of computer jargon and it's "handholding". A great learning tool for an old retired guy.

by DanM
August 23, 2012

Two thoughts:

  1. You may be able to use the Transformer model to get the coupling you want, as these are essentially (perfectly) coupled inductors.
  2. You could use a pair of voltage-controlled-voltage-sources (VCVS) in series with inductors. The control voltage of the "primary" VCVS would be the voltage across the "secondary" inductor, and vice versa. You'd have to do some math to get the gains exactly right (inductance ratios, coupling coefficient, etc.), but that's basically how coupled inductors are modeled anyway.

Do you have some numbers (L1, L2, K) or a public circuit we can look at?

Hope that helps!

by mrobbins
August 23, 2012

Based on some things I made earlier ...

Here are two ways to make a pair of inductors with an adjustable coupling coefficient (I've referred to it as coupling factor or k):

I think you can extend the lower version using the CL transformer symbol, to use the CL centre tapped transformer.

The upper version using a behavioural model can also be extended to an arbitrary number of windings but I'm not exactly sure how to do it based on this model.

Here is a multi-winding transformer - or multiple coupled inductors - model using a different set of behavioural sources but it is very messy to use and has not yet been properly tested with the leakage inductance/coupling factor expressions added.

by signality
August 24, 2012

mrobbins & signality: Thank you for your replies! I need to take some time to think about what you've said and to read some inductor theory. Teaching old dogs new tricks etc. :-)

I was able to to get the transformer to resonate like an isolated inductor, but I need to think through the affect of N on signal levels and bandwidth that are simulated. Thank you again. This really is fun stuff. DanM

by DanM
August 24, 2012

A few memory jogs:

For a 1:N turns ratio;

Lsec = Lpri*N^2

For a coupling coefficient of k;

The primary mutual inductance Lmpri = Lpri*k

The secondary mutual inductance Lmsec = Lsec*k

The primary leakage inductance Leakpri = Lpri*(1-k)

The secondary leakage inductance Leaksec = Lsec*(1-k)

Sanity check:

If k=1 (perfect coupling) the leakages go to zero and the mutual inductances equal the respective winding inductances.

Similarly, if k=0 (uncoupled inductors) the leakage inductances form the total inductance of each respective inductor and the mutual inductance fall to zero.

You can rearrange these formulas to express the inductances in slightly different ways but I find this the easiest way to remember it all.

Assuming no core and copper losses (and no other parasitics) Pout = Pin

Vout = Vin*N

Iout=Iin/N

For impedance matching: Zout=Zin*N^2

So for the same resonant frequency at the primary and the secondary:

Lpri*Cpri = Lsec*N^2 * Csec/N^2

Exactly what happens to the resonances when k<1, I am less clear about - it's a long time since I had to think about that - but that's what the simulator is for!

Have fun!!

:)

p.s. to see how the formulas are formatted in this post (and more) have a look at:

https://www.circuitlab.com/comments/syntax_overview/

by signality
August 24, 2012

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