I have been working a little bit on a hobby project of mine involving inductive power distribution. I have never done anything like this before and before I hook things together I would like to avoid blowing stuff up.

I plan on powering between 1 and 3 inductive loads (i.e. hand-wound coils about 4-6cm in diameter shaped like a ring with probably around 50-100 turns) in parallel from an H-Bridge. The coils will probably be wound with 28-30awg wire and my target is to transmit up to 2A between coils placed about 2-3mm apart. The 2A figure is the amount of current consumed by the "transmitting" coil and not the current induced in the "receiving" coil. Any of the coils being driven by the H-Bridge may or may not have a corrisponding coil lined up (in other words, some of them may have no "load" and others may have a load). This whole circuit may or may not be powered by an inductive method itself as well which further complicates matters of power.

I plan on running everything somewhere in the neighborhood of 100Khz, but that's really up for debate. I just know that it has to be a higher frequency since the coils are coupled inductively without any sort of core.

I understand that at the point of resonance for the inductor (which has to do with the capacitance between turns as I understand it), the inductor will stop being there in terms of impedance and the only limiting factor for current would be the resistance of the wire. I would imagine this would blow up my H-Bridge rather quickly unless I seriously oversized things, so I know I need to stay away from the resonant frequency of my inductor.

Here are my questions:

  • How does the inductor look from the viewpoint of the H-Bridge when nothing is inside its magnetic field other than itself? I imagine it looks just like an inductor

  • How does the inductor look from the viewpoint of the H-Bridge when another coil is placed in close proximity to it with a load attached?

  • What would happen if of 3 inductors in parallel, only one (or two) was coupled magnetically to a "receiving" coil?

Something I plan on trying out is the possibility of using the same coil for either sending or receiving power depending on sensor inputs (current and voltage on a coil, basically). What would happen if I were to drive an inductor which was inside the magnetic field of another inductor that was also being driven? What would I see from the perspective of either side in terms of impedance?

Thanks very much in advance for any insights.

  • \$\begingroup\$ Resonance is your very great friend in such applications. An ideal parallel resonant circuit will have infinite impedance at resonance. An ideal series resonant circuit will have zero impedance at resonance. A resistor loaded secondary resonant circuit will appear to the tx as a resistive load. Multiple resonant tx coils on the same drive cct but not coupled to each other magnetically won't affect the loading that occurs on other members. | If you magnetically couple 3 or more inductors and drive two of them you will get coupling and energy transfer much as you'd expect for a coupled system. \$\endgroup\$
    – Russell McMahon
    Dec 30, 2013 at 5:04
  • \$\begingroup\$ I agree with the above but I'd also add a couple of things. If you use a tuning cap across your receive coil your receive coil circuit acts like a series tuned circuit and if up close to your transmit coil it can sap too much energy and degrade the field generated and you might find you get stupidly high excessive voltages that damage what you are trying to power. At 100k, I'd be looking at 5 to 10 turns and 10n to 100n tuning. Get an oscilloscope and experiment. Also use a simulator and specify inductor coupling in the range 0.01 to 0.5 to see what happens. LTspice is free. \$\endgroup\$
    – Andy aka
    Dec 30, 2013 at 11:47

1 Answer 1


The basic reason for using a parallel tuned circuit as the exciter is one of efficiency. If your inductor and tuned capacitor are 33uH and 75nF, resonant frequency will be about 101kHz. If you do the math you'll see that a lossless tuned circuit like this exhibits infinite impedance but still circulates high current between cap and inductor.

Lossless circuit are of course impossible but making the losses as low as possible means that if your inductance is 33uH and your applied voltage is (say) 40VRMS at 100kHz, the current in the inductor is: -

\$\dfrac{40V}{2\cdot\pi\cdot F\cdot L}\$ = 1.93 Amps

Your H bridge won't even be breaking into a sweat because it won't be supplying anything like this current. This current is due to the voltage across the inductor but the capacitor has the effect of performing "power-factor" correction and because the losses are low maybe the H bridge will be supplying in the order of 50mA to a couple of hundred mA.

However, your H bridge is exciting the coil/cap with a square wave and there will be losses due to the harmonics within the square wave. Because of this it makes sense to feed the coil/cap via an inductor too - somewhat smaller than the coil (maybe 1 quarter). You will also need to retune the capacitance to compensate for this. Some experimentation in this is required to get best results but, you should aim to reduce the H bridge's current to avoid it overheating.

I'd also say make one larger coil suitable for all three inductive loads. The larger coil can be any regular shape that suits your requirements for placement of the receiving coils.

Optimum performance is when the receiving coils are also tuned with a capacitor but, because the induced voltage is in series with a receive coil, the tuned circuit behaves like a series tuned circuit and, if the coupling is too great it will heavily detune the transmit coil when it is close by. You should aim for a minimum gap or incorporate circuits in the H bridge that current limit.

I strongly advise you to use something like LTSpice for simulating this - you'll learn a lot about the various interactions. I'd also recommend you read a bit about tesla coils because that is what are are intending to build (when tuned as per my thoughts).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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