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I have a transformer with a turns ratio of 6:24 (primary to secondary turns). I measured the primary inductance to be 3.05 microhenries (uH) with the secondary open and 1.725 uH with the secondary shorted. Similarly, the secondary inductance was 38.422 uH with the primary open and 20.79 uH with the primary shorted.

Using these measurements, I calculated the coupling coefficient (k) to be 0.66. This indicates that the magnetic flux is not perfectly coupled between the primary and secondary windings.Coupling Coefficient Formula

In an ideal transformer (k = 1), the voltage ratio would simply be the turns ratio (6/24). However, due to the imperfect coupling (k = 0.66), the actual voltage ratio will be slightly different.

I'm interested in understanding how the coupling coefficient affects the output voltage of the transformer and how does the coupling coefficient factor in while calculation the turns ratio.

Primary With Secondary Open Primary With Secondary Short Secondary with Primary Open Secondary with Primary Short

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  • \$\begingroup\$ Is the low coupling factor an unintended consequence or a wanted parameter in your design? \$\endgroup\$
    – winny
    Apr 18 at 9:23
  • \$\begingroup\$ On my experience, I understand that the coupling coefficient shouldn't be related to the turns ratio, but to the geometry. Under no-load, the turns ratio is ideal. Under some level of load, there will be voltage drop in the resistances (losses) an the inductance (leakage flux) of the equivalent circuit. But your question was presented under some very specific context. \$\endgroup\$ Apr 18 at 15:04
  • \$\begingroup\$ @winny a low coupling factor is an unintended consequence. I wanted to understand if that affects my turn ratio in any way. \$\endgroup\$
    – Neil
    Apr 18 at 15:17
  • \$\begingroup\$ @LuizOliveira I am trying transfer power to an RC. And also trying to identify the resistance of this RC network through my transformer. I know the value of capacitance which is 4pF. \$\endgroup\$
    – Neil
    Apr 18 at 15:20
  • \$\begingroup\$ If you need less leakage/higher coupling factor, fix the problem by changing the transformer structure. \$\endgroup\$
    – winny
    Apr 18 at 15:40

2 Answers 2

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The output voltage of the unloaded (no significant load across the secondary) transformer will be reduced by the coupling factor.

$$ {Vs \over Vp} = k\sqrt{Ls \over Lp} $$

Because of the low coupling factor, I'm guessing that you have an air-core transformer. If you want better coupling, use a core made of ferrite; or some material that suits the frequency range of interest.

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  • \$\begingroup\$ Thanks for the formula. I am using a ferrite based toroid(5967001201) to make the transformer. I am also measuring these values at 8MHz. I am trying to transfer power over 8MHz to a RC load. \$\endgroup\$
    – Neil
    Apr 18 at 4:15
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You forgot the capacitive and resistive value of the inductors

Here is what you get with a generator and internal impedance of 50 Ohm.
kc (2 steps), R1 (4 steps) and C1 (4 steps) as parameters.
Generally, you have a "lowering" of the transfer voltage.

enter image description here

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  • \$\begingroup\$ Thanks! I'm new to transformer design, so I apologize for the basic question. Is this simulation done with LTspice? I'm currently measuring the capacitance & inductance values using a NanoVNA(pictures attached in the question) and the software indicates that both the primary and secondary windings are around 1.4 pF. Are there any more accurate ways to measure this? \$\endgroup\$
    – Neil
    Apr 18 at 13:54
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    \$\begingroup\$ Simulation made with microcap v12 archive.org/details/mc12cd_202110, don't know really good nanoVNA ... \$\endgroup\$
    – Antonio51
    Apr 18 at 14:54

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