I have readed in many articles that for high coupled inductors (high k) you get a better efficiency using non-resonant inductive link. On the other side, for low coupled inductors (low k) you get a better efficiency using a resonant inductive link.
So I wanted to check this information through a simulation on LTSpice. Below you have two designs, one resonant and one non-resonant designed for 100kHZ.
I have simulated the efficiency for high k=0.8, blue curve=resonant, yellow=non-resonant
For low k=0.1
As expected at low k, you get a better efficiency for non-resonant. But for high k, non-resonant has almost the same efficiency as resonant.
Is it normal or am I missing something in my model? Why do we use only non-resonant inductive link for high coupled inductors?
Sorry I wanted to write: As expected at low k, you get a better efficiency for resonant. This is what you can see on the blue curve for k=0.1, you get an efficiency of 96% @ 100khz.
To be more specific, resonance refers specifically to nulling out the leakage inductance(s) which would normally inhibit power transfer. When k is near 1, the leakage inductance is negligible, so there's basically nothing to resonate with.
On the resonant circuit(second circuit), with high K, change the 240 nF capacitors by 0.1266515 uF capacitors and let me know what is the result please.
Not with series resonance, change the 2nd circuit to parallel resonance. place 10 uH || 0.1266515 uF in the Tx and in the Rx.
Ok If understand well, if you take a resonant circuit and you bring it in high coupled mode, it will work as a non-resonant circuit. (this is why the blue and the yellow curve are almost the same in the simulation at k=0.8)
But then I don't understand why all those alliances (AirFuel and WPC) distinguish resonant/non-resonant and tells you the benefit of one or the other? I mean you could just build a resonant circuit and it will become non-resonant when high coupled.
But then I don't understand why all those alliances (AirFuel and WPC) distinguish resonant/non-resonant and tells you the benefit of one or the other? I mean you could just build a resonant circuit and it will become non-resonant when high coupled.
Hard to say without reading the original claims you're referring to. But in principle resonant drive should always be better (assuming the resonant capacitors have no loss), since it minimizes reactive power flow. But when K is high then the advantage likely becomes negligible.
Well, when I add your capacitor of 0.1266515µ, the curves become the same, resonant and non-resonant at k=1. Is this what you wanted me to see? How did you find this value?
At K=1, the 2 inductive coils behave like one, so I wanted to see if there is resonance. I calculated the Capacitor to place in parallel by equaling the Capacitor reactance Xc(f) with the inductor reactance XL(f), and i got C=1/(4Π^2*f^2*L)=2.533*10^-(7), but when k=1, as they are one, the coils will have both capacitors in parallel, which capacitance is summed, so I divided 2.533*10^(-7) by 2 and get the capacitance 0.1266515 uF.
Ok I think I have understood. I was confused about the wireless power consortium who was promotting the non-resonant link for his Qi standard. I thought the choice of resonance/non-resonance was based on performance, but in fact it's only a choice of application.
Ok I think I have understood. I was confused about the wireless power consortium who was promotting the non-resonant link for his Qi standard. I thought the choice of resonance/non-resonance was based on performance, but in fact it's only a choice of application.
Extreme close coupling, as in a conventional transformer is always non resonant and the results are much more predictable.
Way below critical coupling where considerable relative distances and very loose coupling are involved, resonant operation will be better where both tuned circuits can resonate and be tuned to a peak independently without mutual interference.