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When talking about how wireless power transfer is done, the standards (e.g Qi) distinguish between inductive and resonant mode. (Link)

They claim that when the units are in very close proximity it is more efficient to oscillate at the inductive side of the resonance frequency, but when further apart it is better to ride on top of the resonance peak.

Im not quite sure as why this is the case.

I know that oscillating on the inductive side have some pros, like ZVS to increase efficiency in the switching, but they also claim higher power output by using this at close proximity. why is this not the case when the units move further away from each other?

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In most inductive coupling products, the primary and the secondary are both tuned. When the two tuned circuits physically come-together they detune each other: -

enter image description here

When coupling is small (k < 0.01) there is quite a tight central peak (about 10 MHz) but as coupling gets larger, the central peak splits into two peaks that start to move away from each other as coupling improves so, you could choose to drive the primary at a slightly lower or a slightly higher frequency than 10 MHz and get an improvement in coupling. Clearly as coupling approaches 0.1 the peaks are quite seperate so this technique sounds like a good idea.

They claim that when the units are in very close proximity it is more efficient to oscillate at the inductive side of the resonance frequency, but when further apart it is better to ride on top of the resonance peak.

As far as I can tell it will work on either side of resonance. See this answer that shows the same happens with capacitively coupled tuned circuits.

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  • \$\begingroup\$ Thanks for your comment! So what i gather from your comment is that driving the tank circuit directly at resonance is the best case in terms of efficiency? So why would you drive it a lower or higher frequency? Is it because of the very high voltage gain at the peak frequency(high voltage requirements for the components)? \$\endgroup\$ – Linkyyy Sep 26 '18 at 21:37
  • \$\begingroup\$ @Linkyyy I’m not sure who your comment was intended for. \$\endgroup\$ – Andy aka Sep 27 '18 at 6:35
  • \$\begingroup\$ @Andyaka, when you connect a load to Vout, the frequency splitting disappears at a higher coupling. the critical coupling point (maximum coupling at which frequency splitting disappears) is related with 1/Q1Q2. \$\endgroup\$ – Pojj Sep 27 '18 at 7:41
  • \$\begingroup\$ @Pojj yes if the Q lowers the two individual peaks merge. \$\endgroup\$ – Andy aka Sep 27 '18 at 10:23
  • \$\begingroup\$ @Andyaka: The comment was intended for you, but i guess John Birchhead kind of answered the question i had :) \$\endgroup\$ – Linkyyy Sep 27 '18 at 17:38
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The principle of inductive coupling is to have two coils with good "mutual inductance" so that the magnetic field generated from the charging coil passes through the core of the other, in the manner of a transformer. A charger's magnetic coupling with the device to be charged depends on this mutual inductance, and the magnetic field is weaker as the devices are moved apart. So the charged device is placed in the flux path of the charging device, and orientation and distance are important. When a non-resonant charger is well positioned, leakage inductance is minimized, resulting in good efficiency.

In resonant charging, the charging and receiving devices operate at their resonant point. The receiving device oscillates sympathetically with only small excitation from the charger, as can be the case when the devices are farther apart or poorly aligned. The result is that the primary magnetic field is "in sync" with the receiver's secondary magnetic field (from resonance) and that the magnetic coupling between the two coils is improved. However, the efficiency is not as good when both types are ideally located due to leakage inductance (at least in early implementations of resonance charging). But in many cases, the charging device's power is not limited, so this not a major drawback.

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  • \$\begingroup\$ Thanks for you comment! So what i understand from your comment is that when you move the devices further from each other and they have very low mutual inductance, you kinda have to treat it as an RF system with the coils being the antennas? (far-field?) So to get maximum power transfer you drive it at the resonance(assuming both the transmitter and reciever is tuned to the same frequency) \$\endgroup\$ – Linkyyy Sep 26 '18 at 21:43
  • \$\begingroup\$ Good analogy. My own analogy is the example of an opera singer breaking a glass by hitting its resonant point. A glass that's a couple of feet away can be broken in this manner if it resonates at the singer's frequency, while a second glass, six inches away and with a different resonant frequency, would not. \$\endgroup\$ – John Birckhead Sep 27 '18 at 13:37
  • \$\begingroup\$ Hehe yes. But no matter what the coupling is, driving it at the resonance frequency would always be best in terms of efficiency right? \$\endgroup\$ – Linkyyy Sep 27 '18 at 17:36
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The terminology of resonance is not consistent in different articles. The resonance of the complete wireless power system and the resonance of individual coils are sometimes different. When it comes to the resonance frequency of the complete system, it will identical to individual resonance when the coupling is small. when the coupling is higher, two additional resonances will appear. the critical coupling is defined as $$ k_{critical}=1/Q_{1}Q_{2} $$ where Q1 and Q2 are loaded quality factors of the primary and the secondary.

One important feature is that the efficiency and power profile show different frequency characteristics when the coupling is higher than critical coupling. Efficiency is always maximum at the self-resonance frequency of individual coils, however, maximum power transfer occurs at higher and lower resonances. For example, see the below simulation and the efficiency and voltage ratio. enter image description here enter image description here

It is important to note that the source is assumed to be a an ideal sinusoidal source for the efficiency calculation. If you consider the losses of your converter, the efficiency will be slightly lower at the resonance frequency.

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