LTspice Tip 3 – Modeling a transformer
Using transformers in LTspice by Mike Englehardt is found LTMag-V16N3-23-LTspice_Transformers-MikeEngelhardt.pdf.
A transformer can be simulated using transformer_example.asc (right click to save)
The negligible resistor, 1 nano ohm, is inserted to ensure that the source voltage, Vs, doesn’t drive L1 directly.
The spice directive (K1 L1 L2 1) is used to specify the coupling between the primary, L1, and the second L2 (below 1 or 100%)
Note that this is NOT a practical transformer for operation at 60 Hz (the inductances are too small, see comment at the bottom).
Right clicking on the inductors enables/disables showing the phase dots (above the secondary is in phase with the primary).
For quick reference if Vp and Ip are the voltage across, and current through, the primary (L1 above) and Vs and Is are the voltage
across, and current through, the secondary (L2 above) then
Vp/Vs = Is/Ip = Np/Ns (the easy way to remember this is power, Vp*Ip = Vs*Is)
where Np and Ns are the number of turns of wire used in the primary and secondary wraps around the magnetic core. Also,
Np/Ns = root(L1/L2) = Vp/Vs
So, above, since L1 is 100 times larger than L2 we expect Vp to be 10 times larger than Vs (noting we can flip the transformer
around if we want the output to be larger than the input). Then Ip is 10 times smaller than Is.
(Vp*Is)/(Vs*Ip) = Zp/Zs = (Np/Ns)^2 = L1/L2
Where Zp is the impedance of the 100k (= Zs) reflected to primary (so Zp = 10MEG here).
Okay, one last (important) comment related to this simulation, while the reflected impedance the primary sees is 10 MEG, and
the current from this impedance is 174/10MEG or 17.4 uA, there is still an AC current from the source driving the 1 mH inductor.
This current, at 60 Hz, has a peak value of 174/(2*pi*60*10^-3) = Vp/(2*pi*f*L) = 461 A! This current is significantly larger
than the current due to the transformer action and so this is not a practical model for a transformer operating at 60 Hz! To make
the model more practical (significantly) increase the inductances so this later component of current is much smaller than the current
supplied to the secondary load.