Inductive coupling is when there is no capacitor to tune the circuit.I am using inductive coupling method, to place a cap in parallel with the Tx and Rx coils. This means I can only get resonance at a frequency within the input frequency range. Therefore, Q values at different input frequency shoudl be different.
Depending on how you have defined the Q factor (it can be Q factor of coil, or the Q factor = Q_tx * Q_rx or some Q related to your wireless link) it is dependent on the values of the inductors. The coils geometry influences the "k" factor as you said. Coils geometries influences the inductance. I conclude then, the Q factor is related to "k" somehow.If we assume the Vpp at Rx can be determined by L_rx, L_tx, and the coupling factor, can I conclude that the coupling factor is a weak function of Q?
I only know the coupling factor is related to the distance between the coils and the coils' diameter. However, I did not find the equation that relates the coils' coupling factor and the Q.
Inductive coupling is when there is no capacitor to tune the circuit.
If any capacitor is there in order to tune the circuit to 1 frequency, then it is called resonant coupling, not inductive coupling as the Qi standard uses.
Depending on how you have defined the Q factor (it can be Q factor of coil, or the Q factor = Q_tx * Q_rx or some Q related to your wireless link) it is dependent on the values of the inductors. The coils geometry influences the "k" factor as you said. Coils geometries influences the inductance. I conclude then, the Q factor is related to "k" somehow.
"k" is related to coil geometry --> Coil geometry is related to the Inductance --> Inductance is related to Q factor
====> "k" is related to Q factor.
Experimentally: Do a Bode plot of the magnitude of your link.How can I explain that I get similar Vpp at Rx when input frequency is 1MHz, 3MHz and 5MHz on bench?
The Q of the coils and any components connected to them, is to a very large degree independent of the coupling between coils, when an adjacent coil pulls power out of a driven coil - it may look like the Q is reducing - indeed the Q of the overall system is - as you are introducing a load or losses - but the Q of the individual coils is independent of coupling.
Experimentally: Do a Bode plot of the magnitude of your link.
Theoretically: knowing "k" and all other parameters of your design, find V_rx / Vin and do a Bode representation. If there is no improvement at the resonant frequency, I guess your circuit is too tightly coupled and hence detuned already due to the frequency split phenomena. But as I said, that is just a wild guess.
Why 10 pF and not 521.2 pF ?(2) both caps are 10pF to resonant with coils at 3MHz.
Why 10 pF and not 521.2 pF ?
I guess "testing the data transfer between two wireless coils" involves operating it without a receiver coil load. That's different from the intended operation of these coils for wireless power transfer, where a coil Q of about 100 is almost irrelevant because the real load dominates all circuit losses.
Secondly you are operating the coils in a higher frequency range where ferrite losses are probably dominant.
Finally "coupling factor" in your post seems to refer to tx/rx voltage ratio of a resonant circuit, which is not the same as "k" in usual coupled inductor models. Your definition of Q is neither quite clear. It's usually described as frequency dependent parameter of a single inductor, Q = jωL/ESR. How do you determine it?
Easy P. is quite correct.
What happens at great distance is that coupling is weak and the two tuned circuits can be peaked independently, but energy coupling is poor because of the distance.
As the two tuned circuits are moved closer the amplitude increases and you reach a condition known as "critical coupling".
If you move the tuned circuits even closer together a very strange thing happens. You get two new frequency peaks either side of the original resonant point, and a dip in the middle.
Its called "over coupling" and is actually quite useful for increasing the bandwidth of some kinds of inductively coupled filters.
https://www.ee.bgu.ac.il/~intrlab/lab_number_7/Two inductively coupled RLC circuits.pdf
In actual practice there are so many secondary effects such as the Q of the two circuits and coil geometry, its pretty difficult to predict a result. It really needs to be tested.
Not sure about a Bode plot, but the amplitude versus frequency effect of over coupling is certainly very readily visible on a spectrum analyser.
The main thing is that with very loose coupling right up to critical, both tuned circuits resonate quite independently of each other.
With tighter coupling you are then dealing with a "tightly coupled system" with a very high degree of interaction. You loose the independence, and each resonant circuit heavily loads the other.
Phase through a coupled radio system is not really relevant where there is only one signal path.
Phase only becomes important when there is something else to compare it to.
Why 10 pF and not 521.2 pF ?
Resonance frequency shifts because the inductance increases with reduced coil distance.
I'm not sure if the in circuit Q observed in your measurements has much to do with coil losses. I guess it's mostly determined by the generator impedance, not yet mentioned in the thread.
Yes. Actually, if tuned perfectly i.e. very little parasitics, on that range of frequencies shows a flat band.do you think it makes sense that I do not have too much variation for the waveforms between 1MHz to 5MHz?
Do you mean Mutual inductance, right ?Resonance frequency shifts because the inductance increases with reduced coil distance.
Yes. Actually, if tuned perfectly i.e. very little parasitics, on that range of frequencies shows a flat band.
As the coupling factor is increased (magnetic coupling factor, not Rx/Tx), as FvM said, the peak shifts. Maybe parasitics play an important role on your circuit which shows a peak at 6 MHz.
Do you mean Mutual inductance, right ?
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Now that you know that you placed the wrong cap, could you do as well the bode plot around the 21 MHz freq and see what shows there?
No because the right hand side of the bode plot decreases very fast. Moreover, in the parallel-parallel resonant topology is more difficult to see or to meet the condition for the 2 peaks, in contrast of the series-series topology.is it possible that frequency splits already happened when I have the distance 1cm? This means all I saw in the bode plot are the lower frequency component of the splitted frequency, while the upper frequency component is higher than 21MHz? Thank you!
No because the right hand side of the bode plot decreases very fast. Moreover, in the parallel-parallel resonant topology is more difficult to see or to meet the condition for the 2 peaks, in contrast of the series-series topology.
I think somehow you have reached the ~130 pF with parasitics... otherwise, I do not know why the resonance is at ~6 MHz. In that circuit, at first sight you already have 10 pF + 6.4 pF (parasitic of the coil), 7.5 pF for Tx and Rx respectively.
To make sure that the true resonance of the circuit is around 6 MHz, do a Bode plot at a very large distance (i.e. very loosely coupled where the true resonance is not influenced) e.g. 7 cm or so. Then, the true resonance of the circuit is at that frequency. Make sure this time you are able to see up to 15 MHz because it will shift from upper frequency to lower.
Hi CataM, thank you so much for your help! Sorry I still have some questions regarding your answers, would you teach me again?
(1) I model the parasitic resistance (DCR and trace impedance of the pcb) to be in series with the L, and calculate the peaking.
In this plot, the left hand side and right hand side of the peak are pretty symmetric. I did not see the "flat band" on the left hand side, nor "the right hand side of the bode plot decreases very fast". Would you tell me what is wrong in my calculation or derivation?
(2) regarding the equivalent capacitance of the Rx side, there is another DC block cap connected to the output of the receiver, and then feed as the input of a logic inverter. This is the picture that showed in an earlier post in this tread. I am using this inverter (https://www.ti.com/lit/ds/symlink/sn74lvc1g97.pdf).
My guess is, if we assume the Cgs of the input MOSFET of the inverter is still in pF level, Cgs will connect in series with this 10nF DC blocking cap, then stay in parallel with that 10nF cap C_SEC. So the overall load of the secondary data coil is still several pF level, like what you have pointed out, around 130pF. Is this analysis correct?
Thank you!
Not only mutual inductance. Also the coil main inductance. If you look at the core geometry, it's obvious why.Do you mean Mutual inductance, right ?
Not only mutual inductance. Also the coil main inductance. If you look at the core geometry, it's obvious why.
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