Wireless power transfer efficiency calculation

So I'm looking into retrofitting a Qi transmit eval and receive eval kit with coils of a larger diameter than the stock ones, then looking at what playing with the separation will afford me for my theoretical use-case (I would be exceeding the recommended Qi 5mm max sep). To clarify, I understand I may be taking them out of the Qi specs for now and that's fine for the purpose of my study. Starting with this stack exchange discussions:

Is the QI norm about the components or the coil?

and

How big does my wireless charging coil need to be?

I thought I could in theory, take the field generated by the Tx'er Bt(z=0), then that received by the Rx'er Br(z), solve them for current, then take the ratio of received to transmitted current multiplied by the required voltage ratio and get the required power ratio to calculate efficiency. This would give me the variables of the separation of the coils 'z' and their associated radii directly influencing the efficiency (power ratio), but the numbers don't make sense. Have I made a gross oversimplification or is my logic not valid outright? Thanks in advance for any insight or any pointers on a direction I should go from here, b

See this official WPC link for example. You can't know for sure unless you measure the Tx and Rx powers, not fields -- directly calculate the voltage and current, cos$\phi$ included (you're transmitting active power). As an informative number, readily found kits work with an efficiency of 50%-70%, due to the air core coupling and relative distance, but it can be made better with less distance and, perhaps, an opened magnetic core -- but don't forget that the frequency lies between 105kHz-210kHz (IIRC). For more details, it's always a good idea to see what the WPC has to say about it, it's a bunch of links with information, very useful.

• Thanks, I saw this link, but clearly they're using modeling or maybe even experimentation to generate those curves on the log-plot. Agreed, that direct measurements of the constituent parameters would be best, but I just thought I could get at least a scalar of sorts to scope efficiency, which is apparently more squirrely than anticipated. – benny d Jun 18 '18 at 17:48
• @bennyd Alas, no free lunch, but the links could be of a help in ordering the meal. – a concerned citizen Jun 19 '18 at 5:52

Wireless power transfer efficiency calculation

Theoretically there is 100% power efficiency. Whatever you drive your perfect transmit coil with, if there is nothing leeching-off power, then you are not putting any power into the coil. So inefficiencies come from resistive losses in your coil, driver losses in your oscillator/amplifier and eddy current losses in any localized metals that surround your transmit coil.

It's a coil of wire and you can, with care, make it very efficient.

And, with a really good coil design for both transmit and receive, the only other inefficiencies are in converting the received signal to DC (rectifiers) and maintaining a stable voltage (regulators are of course pretty good but not 100%.

So, in short, what you get out in terms of power is what you put in minus all the inefficiences of the oscillator, coil-resistance, eddy currents, rectifiers and regulators.

I thought I could in theory, take the field generated by the Tx'er Bt(z=0), then that received by the Rx'er Br(z), solve them for current, then take the ratio of received to transmitted current multiplied by the required voltage ratio and get the required power ratio to calculate efficiency.

No.

• Thank you for the break down, DC coil resistance is something I'm considering, I'd need a bit more info from the manufacturer sheets than I'm seeing to get the remainder of the info needed, however. – benny d Jun 18 '18 at 17:53
• But, if one were to vary the separation as an exercise, wouldn't you have to add a factor of how the field strength would fall off with distance as ~1/z^3? – benny d Jun 18 '18 at 18:00
• No, absolutely not. The receiver would take less power and, as such, the transmit circuit would not be taxed so heavily because of the reduced power taken in the receiver. – Andy aka Jun 18 '18 at 21:53