0
\$\begingroup\$

I am trying to model a wireless energy transmitter as a transformer equivalent circuit with a large air gap. Because of the large air gap I’m assuming that for a realistic model fringing fluxes would have to be taken into account, I can’t find an easy to understand equivalent circuit that allows me to model the effect of a large air gap (or other mediums e.g concrete+air) by some formula for the coupling coefficient (I think?) that allows for a large air gap because most seem to assume the air gap is small.

Any help would be greatly appreciated.

\$\endgroup\$
1
  • \$\begingroup\$ Note only: As per Andy's advice - resonance of both tx and rx sides essentially essential for substantial power transfer at any sort of efficiency. Higher range will require higher Q (see Andy's comments about coupling factor) \$\endgroup\$
    – Russell McMahon
    Feb 10, 2014 at 17:58

2 Answers 2

2
\$\begingroup\$

A large air gap will give you a large leakage inductance, which appears as series inductance in your model.

enter image description here

Normally a transformer designer tries to minimize leakage inductance since it negatively affects regulation, but sometimes it's necessary (as in your case) or even an advantage (as in the case of discharge tube transformers).

\$\endgroup\$
2
  • \$\begingroup\$ Leakage inductance is also purposely made higher than the minimum in some power transformers, to limit the maximum current that can flow during a fault. See electronics.stackexchange.com/questions/96646/… . \$\endgroup\$ Feb 10, 2014 at 13:28
  • 1
    \$\begingroup\$ @Li-aungYip Yes, as well as that high power stuff, small "impedance protected" UL Class 2 transformers. Also old-school non-inverter welders which have adjustable leakage inductance. \$\endgroup\$ Feb 10, 2014 at 13:38
0
\$\begingroup\$

You can use the model of a transformer as suggested in Spehro's answer but it's a little complex. This is because at a large distance between sending and receiving coils, the leakage inductance gets bigger BUT the magnetizing inductance gets smaller (shown as kL\$_P\$ on Spehro's diagram). It is shown this way correctly in Spehro's diagram (also note the "1-k" term) but, when ever I simulate these scenarios I use coupled inductors of fixed value with a variable coupling coefficient to account for air gap.

This means that L1 (driving inductor) is a simple inductor and small series resistor and ditto for L2 (receiving inductor). Why do it my way? Almost without doubt, if you are to achieve maximum power throughput on large gaps you'll need to parallel tune L1 with a capacitor and the same applies with L2 - this means you can easily set the tuning capacitors and their respective inductors to maximize power throughput at a large distance then forget about those values - at short distances (coupling approaching 100%) they will naturally de-tune each other but that is what you'd expect for this type of circuit: -

enter image description here

Note how when coupled at a factor of 0.5 there are two distinct peaks in the spectrum. As coupling falls the peaks gradually come together eventually becoming one.

Regarding the difference between concrete and air and the difference it may make - it won't - only ferromagnetic materials or conducting materials will alter the relationship - ferromagnetic materials will enhance the coupling and conducting materials (due to eddy currents being induced in them) will lower the coupling factor.

\$\endgroup\$
4
  • \$\begingroup\$ Note "concrete" parts of a building usually have steel reinforcement. \$\endgroup\$ Feb 10, 2014 at 13:29
  • \$\begingroup\$ @Li-aungYip good call - iron/steel reinforcement of the sort of size used in concrete (and at the likely frequencies of excitation) will have both ferromagnetic effects and eddy current effects - these can nearly cancel-out but there will be losses i.e. the ironwork will remove energy from the mag field and produce small amounts of heat - remember, at the sort of frequencies (between 50kHz and maybe 2MHz) iron will be totally conducting in a very thin skin at its surface whereas some steels will conduct better because they may have zero permeability (stainless e.g.) \$\endgroup\$
    – Andy aka
    Feb 10, 2014 at 13:35
  • 1
    \$\begingroup\$ Andy's advice is good. A magnetically large gap might be 0.005" (0.13mm) which some might not consider all that significant. \$\endgroup\$ Feb 10, 2014 at 13:58
  • \$\begingroup\$ @SpehroPefhany that's very kind of you to say so \$\endgroup\$
    – Andy aka
    Feb 10, 2014 at 14:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.