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source: https://circuitdigest.com/electronic-circuits/simple-wireless-power-transmission-circuit-diagram

Above circuit is the very simple wireless power transfer circuit, but I could not find any explanation online about how the circuit actually works. How does the circuit on the left work as an oscillator? What is the purpose of the center tap, and how does it convert DC source into an oscillating current? Please help understand the circuit operation!

Also, how could this circuit be extended to a three phase one?

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    \$\begingroup\$ Precisely what do you mean by 3 phase? The present circuit runs from a 1.5 volt battery and is pretty low quality and inefficient. If you might be thinking about powering from 3 phase then you will need a fire extinguisher handy. \$\endgroup\$ – Andy aka Sep 2 at 19:06
  • \$\begingroup\$ The link you posted explains it. Read the entire page. \$\endgroup\$ – David Sep 2 at 19:29
  • \$\begingroup\$ @David What I'm looking for is a more detailed explanation. Of course I read the entire page. \$\endgroup\$ – user207787 Sep 2 at 19:39
  • \$\begingroup\$ I don't think that circuit works as the article describes either. Like you said, I don't understand how that circuit would oscillate. \$\endgroup\$ – David Sep 2 at 19:45
  • \$\begingroup\$ FYI, I found out that this circuit configuration is called Joule Thief. \$\endgroup\$ – user207787 Sep 3 at 13:20

It seems to work just fine (with some guessed values):


Note the polarity of the inductances. When power is turned on, there's no current through L1, but there is through L2+R2 due to the BE junction being directly polarized. The transistor opens and allows current to flow through L1. When the transistor is saturated, there is no more variation in the current, so the induced L1 is now transferring back to L2, and the cycle continues. Since the transistor can work with relatively high dI/dt, the values of the voltage can get very high, so some protection might be needed (or limit dI/dt). This schematic is a simple way of having a self-oscillating circuit.

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  • \$\begingroup\$ Thank you so much for this! \$\endgroup\$ – user207787 Sep 3 at 11:59
  • \$\begingroup\$ @user207787 You should know two things: 1) the NPN needs an anti-parallel diode, to avoid damaging the transistor during reverse operation, and 2) as noted in the comments and in my conclusion, this circuit is very crude and inefficient; at best it has an educational value, to prove a concept, rather than to prove the schematic works. If you intend to use it, choose another. Not lastly, if any of the answers solve your problem, mark one of them (the check mark) so that future searches will show an accepted answer for this question. \$\endgroup\$ – a concerned citizen Sep 3 at 12:08
  • \$\begingroup\$ Done! Also, I was asked to extend this to a three phase one, and I have no idea what that means. Can you give me some suggestion? \$\endgroup\$ – user207787 Sep 3 at 13:21
  • \$\begingroup\$ @user207787 That makes no sense to me. Can't you ask for clarifications from whomever asked you this in ther first place? Was it your teacher? It would great if you could add that clarification to your question, I am always glad to learn something new. \$\endgroup\$ – a concerned citizen Sep 3 at 13:48
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    \$\begingroup\$ @Giga-Byte Yes, that came out weird. I meant that, since L1 and L2 are in anti-phase, they behave like a transformer, so the induced voltage appears in L2 after the cycle in L1 finishes -- when the transistor is saturated there is no more variation in the current through L1. Did my translating neuron made it this time? \$\endgroup\$ – a concerned citizen Sep 4 at 13:00

Firstly the coil wounds are same, each with 17 turns. So the induced EMF is neither decreasing nor increasing but I don't think it's possible for it to convert DC as low as 1.5V into an oscillating current.

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  • \$\begingroup\$ The number of turns is not the key here, it's the polarity of the inductances, an effect of the center tap, which causes the two inductances to work alternatively, and which makes the whole circuit a self-oscillating one. I may have exaggerated with the coupling factor in my answer, but it works, even at k=0.3, even at 1.5 V. \$\endgroup\$ – a concerned citizen Sep 3 at 7:17

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