# Re: What difference between ideal transformer and perfect one

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#16354
MarCatan
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Hi mkthom81,

Consider a simple transformer with a primary and a secondary winding.For the ideal transformer the equations are:
Vp = n* Vs e Ip= Is/n, where n is the ratio turns of secondary and primary winding ns/np or SQRT(Ls/Lp).
As you can see from the screenshot below, the frequency has no influence on the equations, voltages are always in phase (the blue line).
For the perfect transformer the equations become vp = n* Vs e Ip = Is*(1+ ZL/(jw*Ls))/n.
Note the frequency dependence on the green curve (w= 6.28*f). Only under condition that w*Ls>> ZL obtain again the ideal transformer equations.
In the AC screenshot, you can see how at low frequencies Vp and Vs are not in phase and Vs is lower than n*Vs.
in practice there’s a low cutoff frequency.
The example I’ve attached below, shows a transformer model that comes closest to a real transformer, (the red curve) it shows a cutoff frequency at low and at high frequencies.
In the simulation shows a bandwidth of about 10Mhz.

Bye