The following is a circuit that wirelessly transfers power via inductive coupling. The circuit is supposed to power a 12 V 0.15 A DC fan and a typical smartphone, which requires 5 V DC across it. The loads are connected in parallel to each other, as seen in the circuit schematic(though there is no capacitor at the output of the voltage regulator). As you can see in the picture below, it powers both just fine, but the phone only charges for a few seconds. And it can only charge both the fan and the smartphone only when the transmitter and receiver coils are very close to each other. The transmitter coil has 9 turns and the receiver coil has 39 turns.

enter image description here enter image description here I then used a digital oscilloscope to measure the voltage induced in the receiver coil at two different loading conditions - at no load and at full load.

The voltage induced in the receiver coil hand an RMS voltage of 28.3 V as shown in its waveform. It also had some ringing, which I assume is because the PCB terminal block acts as a capacitor. Anyways, I assumed that the voltage would remain 28.3 V because the DC fan and the phone are both connected in parallel so the the voltage at the receiver coil shouldn't change, or that's what I thought.

enter image description here

So I then measured the voltage at the receiver coil when both loads were added, and, as you can see in the waveform below, two things changed: the ringing is now gone and the RMS voltage decreases to 13.0 V.

enter image description here

Can someone please explain why the receiver coil voltage decreased even when the loads were added in parallel?

I am sorry for the long post, but I just wanted to make sure every little detail is given.

Edit: Thank you, everyone, for the answer regarding the decrease in receiver coil voltage. I fully understood it.

  • \$\begingroup\$ The receiver coil is acting like the transmitter coil, but generate the reverse field (you pass high-frequency current threw receiver coil. Now it is the second transmitter for 3rd load) then The field is reduced in total \$\endgroup\$ – M KS Nov 24 '18 at 16:49

I think your real problem lies in understanding parallel and series circuits, and what they really mean.

If you have this:


simulate this circuit – Schematic created using CircuitLab

Then you can rightfully expect the voltage to be constant, regardless of whether the switch is opened or closed.

This is what you expect your wireless power transfer to do.

What you don't realize is that the current changes depending on whether the switch is opened or closed. You need more current when the switch is closed - both resistors are drawing current, so your voltage source must deliver more current.

Unfortunately, it can't do that. It can only deliver as much current as it can pull through the magnetic field between the two coils.

If they are close by, or coupled through a piece of magnetic material, then you can draw more current and everything is as you expected it.

As it is, it looks more like this:


simulate this circuit

R3 represents the current limitation of the coils. With that in there, closing the switch causes more current to flow, which causes the voltage across R1 and R2 to change.

How limited the current is depends on how well the coils are coupled. Better coupling (closer together, or using a core) makes the effect smaller. Coils further apart makes the effect larger.

The same thing happen when using different sized batteries.

All batteries have some internal resistance. The smaller the battery, the higher the resistance and the less current you can draw from it.

  • \$\begingroup\$ Just one more question: did the ripple in the receiver coil voltage disappear because the voltage decreased? And thank you for the answer. \$\endgroup\$ – DigiNin Gravy Nov 24 '18 at 18:17
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    \$\begingroup\$ I don't know. The ringing might come from the driving side or the receiving side. If it is on the receiving side, then the additional load might be absorbing the ringing. It wouldn't be because the voltage is lower - the voltage is lower because of the higher load, and the higher load also soaks up the ringing. It might be on the driving side, though, and that would have a different mechanism. \$\endgroup\$ – JRE Nov 24 '18 at 18:44

Can someone please explain why the receiver coil voltage decreased even when the loads were added in parallel?

If instead of the coils in your question you had a regular HF transformer coupling power to the load(s) you would see very little reduction in voltage when adding more parallel loads. This is because the two coils inside a regular HF transformer are very closely coupled and the leakage inductances on primary and secondary are very small. More coupling means less leakage inductance.

Unfortunately, when you have coils that are lightly coupled the first thing to notice is that the Vin:Vout relationship is no longer 9:39 (that's the turns ratio by the way) - it is somewhat less and that totally depends on the coupled flux from transmit coil to receive coil. If the coupled flux falls to 50% then the voltage ratio reduces to half i.e. 9:19.5.

That's under no-load conditions and it's worse when you take a load current (or more load current) because there is leakage inductance to consider. Leakage inductance is the inductive element of both coils that isn't coupled to each other. Less coupling between transmit and receive coils means more leakage inductance.

Leakage inductance is in series with both the transmit coil and receive coil so, as you take more power (more load current), there is a bigger voltage dropped across these leakage indutances and, inevitably, the real output voltage reduces.

  • \$\begingroup\$ Just one more question: did the ripple in the receiver coil voltage disappear because the voltage decreased? And thank you for the answer. \$\endgroup\$ – DigiNin Gravy Nov 24 '18 at 18:41
  • \$\begingroup\$ Most likely it's because with a higher load current (lower resistance load) the Q of the circuit went from underdamped to close to perfect damping. High Q = resonance and ringing; low Q means less resonant and possibly no ringing. \$\endgroup\$ – Andy aka Nov 24 '18 at 18:46

the coil turn ratio is very less and hence most of the magnetic field is disturbed and lost without idnducing in secondary coil to get maximum power. However this can be eliminated if you calculate the T1 and T2 with respect to Vin anad Vout.

Coming to the load, please draw the V-I characteristic of the circuit and draw the Q-point to know where excatly the circuit is operating at. Usually when the loads are connected in parallel the resistance is low and hence voltage need to drive the circuit will be more in terms where the coil supoort this voltage transfer or not.

By this two possible solution you can understand the problem of high current drawn and voltage drop.

Thank you.


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