# 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. Nov 8, 2016 at 8:56
• Thank you. It helps. But my question becomes should I use 1:1 or sqrt(L1):sqrt(L2)? Nov 8, 2016 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. Nov 8, 2016 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. Nov 8, 2016 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. Nov 8, 2016 at 9:57

• 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? Nov 14, 2016 at 1:55
• 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. Nov 14, 2016 at 8:46