LTSpice calculates the turns ratio of a transformer with SQRT(Lp/Ls). This is however only true if both windings have the same geometry.

I have an air-core transformer with an Lp of 10μH, an Ls of 30μH, a turns ratio of 10 (and a coupling coefficient of 0.5). How can I simulate this? I thought about splitting up Ls into 29.9μH and 0.1μH and only coupling the second part. This would give the correct turns ratio, but I think it wouldn't be equivalent.


1 Answer 1


I have an air-core transformer with an Lp of 10μH, an Ls of 30μH, a turns ratio of 10 (and a coupling coefficient of 0.5). How can I simulate this?

LTSpice uses a linear component called "k" to couple coils. The value of k is the same as the value of 0.5 in your question. Use "k" and everything will turn out just right without having to make any calculations of turns ratio (unless of course you want to).

The way it works is like this: -

  • If \$L_P\$ is 10 μH and \$k\$ is (say) 0.4 then 4 μH couples 100% to 40% of \$L_S\$.

  • If \$L_S\$ is 30 μH then the inductance that 100% couples to the primary is 12 μH.

  • So, 4 μH of \$L_P\$ 100% couples to 12 μH giving a true voltage magnification of \$1:\sqrt3\$.

  • The primary has a leakage inductance of 10 μH - 4 μH = 6 μH and, this drops voltage.

  • So, when the secondary is unloaded only 40% of the full primary voltage is "transformed"

  • This means that if the primary voltage is (say) 12 volts RMS, then only 4.8 volts get magnified by \$\sqrt{3}\$ and this appears as 8.31 volts on the unloaded secondary output.

Comparison between what I talked through above and using "k" directly: -

enter image description here

Ignore the microscopic phase angles of sub nano-degrees; that's just the simulator trying to be "too-clever". But, do note that the output voltage for both cases is 8.31 volts. For the record, the current values are with a source frequency of 100 kHz.

And, of course, if you draw current from the secondary, extra current is drawn into the primary and a bigger voltage is "lost" across the primary leakage inductance thus reducing the output (in addition to the volt drop (lost voltage) produced by the secondary leakage inductance of 18 μH).

But, you don't necessarily need to know any of this if you are using a simulator with the component called "k" that links primary and secondary inductances.

  • \$\begingroup\$ When the coupling constant is 1, the relationship must meet \$\frac{L_{_\text{PRIMARY}}}{L_{_\text{SECONDARY}}}=\left(\frac{N_{_\text{PRIMARY}}}{N_{_\text{SECONDARY}}}\right)^2\$. The OP states that this isn't a correct statement of the situation. I believe that LTspice also uses something approximating \$L_{_\text{LEAKAGE}}=L\cdot\left(1-K^2\right)\$. I think the OP wants to know how to properly frame their figures (which I feel are questionable, regardless) into LTspice. I haven't tried. But does applying the OPs figures work out okay? Are you sure? \$\endgroup\$
    – jonk
    Mar 7, 2022 at 20:21
  • \$\begingroup\$ @jonk - the OP's figures are miles out. That's why I gave a worked example (which hopefully I haven't screwed up on). I wasn't about to try and figure out what the OP was thinking. \$\endgroup\$
    – Andy aka
    Mar 7, 2022 at 20:38
  • \$\begingroup\$ @Andyaka Thanks for this awesome explanation. I'm still trying to fully understand the part with the leakage inductance and confirm for myself how this all works out. \$\endgroup\$
    – ismxy
    Mar 7, 2022 at 21:05
  • \$\begingroup\$ @ismxy once you find the "k" component and link the two inductors that make the transformer with a coupling of 0.5, you can parallel an alternate solution using the numbers I derived above with a "k" factor of 1.00 to see how it works. \$\endgroup\$
    – Andy aka
    Mar 7, 2022 at 21:15
  • 3
    \$\begingroup\$ @ismxy Just to be clear: k is not particular to LTspice, but to SPICE, in general, and ever since 50 years ago. \$\endgroup\$ Mar 7, 2022 at 21:24

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