I have found an interesting simultaneous wireless information and power transfer concept where two MOSFETs are used as switches to switch two other resonant capacitors. Because of the switching in the secondary circuit, current amplitude changes can be noticed in the primary circuit. I believe this modulation method is called BASK (Binary Amplitude Shift Keying). Back telemetry is what I am interested in.

What I understand so far is that when the extra capacitors, that are added parallel to the LC circuit, are switched, they will change the resonant frequency of the secondary side, therefore, the current amplitude of the primary side will change.

I tried simulating this circuit and I came up with two questions:

  1. What could be the second MOSFET for (Figure 1)? I do not see any difference in the current amplitude of the primary side after removing it.

  2. Why the amplitude decreases rather than increases when MOSFET is switched (Figure 2)? The only thing I came up with is that perhaps depletion instead of enhancement MOSFETs are used in the offered concept, but also then I am not sure why it would be better in this way if that is the case.

Any ideas/answers would be much appreciated! Thank you.

Offered schematic Figure 1. Principle of data transmission from receiver to transmitter. Offered result Figure 2. Results of data transmission from receiver to transmitter.

  • \$\begingroup\$ when the capacitors are switched in on the secondary side, the current is increased as the voltage rises during each half-cycle and so more of the voltage from the h-bridge ends up across the capacitor in the primary tank, so there’s less voltage across the primary winding. Using a smaller capacitor could increase the data amplitude at the expense of less power being transferred. Notably, the bridge rectifier will only load the transformer at the peaks of each half-cycle but the capacitors will load throughout the cycle; a circuit on the primary side that could detect the change in... \$\endgroup\$ – Frog Apr 18 at 20:14
  • \$\begingroup\$ Phase angle (current is drawn as the signal crosses zero) could be more sensitive if you need fast and reliable data transfer. \$\endgroup\$ – Frog Apr 18 at 20:16

You could use FM and not have to change the power transfer with a Q of 10.

The 2nd FET creates a balanced load to ground with the two series C’s shunting the floating AC signal. Otherwise all it does is clamp one side towards some 0V gnd reference as your load is shown floating.

  • \$\begingroup\$ Thank you, now I see why the second FET is used and why I see no difference when simulating. Only it is not quite clear to me what you meant by not having to change the power transfer with a Q of 10. \$\endgroup\$ – Gabriel Apr 19 at 7:05
  • \$\begingroup\$ Is it because the bit rates are generally limited to 1/10th of the carrier frequency? For me, it is okay since the minimum data rate my system requires is 1kbit/s. \$\endgroup\$ – Gabriel Apr 19 at 7:47
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    \$\begingroup\$ you got it. There are probably 10 thousand papers on WPT. Some may be useful for you. You are adapting a simple carrier modulation over power, but beware of line and load generated noise. \$\endgroup\$ – Tony Stewart EE75 Apr 19 at 12:22
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    \$\begingroup\$ e.g. aip.scitation.org/doi/10.1063/1.5138558 seek and you will find any answer, if you know the right question \$\endgroup\$ – Tony Stewart EE75 Apr 19 at 12:32

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