# SPICE model of coupled inductor I want to write a spice model of a coupled inductor. I’m wondering how to find N1:N2 in the figure below. I have the same windings but different inductances in my two coils. Is N1:N2 equal to 1:1 or $\sqrt{L1}$:$\sqrt{L2}$?

I also found the SPICE syntax of a coupled choke. For example,
L1 1 2 1u
L2 3 4 5u
K1 L1 L2 0.6

Can this example get the same result to the model below?

• N1:N2 is probably the transformer ratio so it could be 1:1 but it does not have to be. Maybe this article will help: cds.linear.com/docs/en/lt-journal/… Yes it's about LTSpice but the same principles will apply to Spice. – Bimpelrekkie Nov 8 '16 at 8:56
• Thank you. It helps. But my question becomes should I use 1:1 or sqrt(L1):sqrt(L2)? – sinlin612 Nov 8 '16 at 9:17
• see here: linear.com/solutions/5092 from that I conclude that N1/N2 = sqrt(L1/L2) so it does not matter what you use. – Bimpelrekkie Nov 8 '16 at 9:31
• Thank you. Do you think the meaning of N1 and N2 in the LTspice for transformer is equal to the meaning of N1 and N2 in the above equivalent circuit? I'm not sure. – sinlin612 Nov 8 '16 at 9:51
• No I do not think it is as in (LT)Spice you model a transformer with 2 inductors and (to actually make it a transformer) add mutual coupling between those inductors as is shown in your Spice syntax example. The coupling factor K must be between -1 and 1, see ltwiki.org/?title=Mutual_Inductance In your case I think making K = 1 will do the job, that would then simulate an ideal (not-lossy) transformer. – Bimpelrekkie Nov 8 '16 at 9:57

You model a coupled inductor in Spice and other simulators by using a component called "k". "k" needs to be setup (quite simply) by stating which inductors it applies to and, it also needs a coupling factor i.e. by how much the magnetic field of one inductor couples to the other inductor. A negative value for k is the same as the output inductor being wound the opposite way i.e. it produces an inverted output waveform.

So, if one inductor is 1 henry and the other inductor is 10 milli henry (a 100:1 inductance ratio) there is an implied turns ratio of 10:1 when k = 1. If k = 0.5 there is still an implied turns ratio of 10:1. Both the below scenarios produce identical results: - This is how I set up coupled inductors but there's no reason why you can't split an inductor into its leakage element and its coupled part. But why bother when k does that for you and it looks neater as a schematic symbol.

• Thank you. So you think the three line ways is equal to the figure above if I know how to calculate all values in it? I'm very curious how to calculate every components in above figure in the currect way. – sinlin612 Nov 8 '16 at 10:55
• Yes I do and I'll add some more detail to my answer as a demonstration. – Andy aka Nov 8 '16 at 11:06
• I got very different result from the above model I memtioned. I have a six-line netlist like this:<br/> Lk1 1 5 0.011u<br/> Lk2 2 6 0.006u<br/> Lm 5 3 0.277u<br/> L1 5 3 100<br/> L2 6 4 100<br/> k L1 L2 1.00<br/> These values are from:<br/> $Lm=\frac{N1}{N2}\sqrt{L11L22-(1-k^{2})*L11L22}$<br/> $Lk1=L11-Lm$<br/> $Lk2=L22-Lm{(\frac{N2}{N1})}^{2}$<br/> L11 and L22 are the inductance measured in one side by opening other sides. Do I have any mistakes? – sinlin612 Nov 14 '16 at 1:55
• It's been decades since I used netlists to describe a circuit so I'm not able to understand what circuit you are describing. – Andy aka Nov 14 '16 at 8:16
• My netlist is describing the figure I post above in the beginning. I used N1:N2=$\sqrt{L1}$:$\sqrt{L2}$ to find all inductance, however I can't get the same results to the model with only three lines L1, L2, and k. – sinlin612 Nov 14 '16 at 8:46