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Take for example the schematic below of a inductive receiving circuit in RFID, where we want to maximize the voltage over the load resistor R4 (R3 is the resistance of the inductive coil, and the voltage source represents the signal that is received and induced in the coil)

schematic

simulate this circuit – Schematic created using CircuitLab

The voltage is now V2*R4/(R3+sL1+R4), which in my opinion can easily be maximized by adding a series resonant capacitance, such that the impedance is real at resonance (and further maximized by increasing R4). However, I read that voltage maximization for circuits as this one is done by adding a parallel capacitance. When doing the calculations, I always obtain a voltage that is lower than the one for the simple series resonant circuit I present here.

EDIT: when considering the measuring device to have an infinite input impedance, the measured voltage for the series resonant circuit equals the applied voltage V2. For the parallel resonant circuit, it equals 1/(sRC) times the applied voltage V2. Is it possible that the eventual best solution is depending on the size of R and C and the operating frequency (whether their product is larger or smaller than 1)?

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2 Answers 2

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That's because the model you are assuming does not well replicate an rfid receiver.

The EM wave that couples with the coil produces some varying voltage at the coil's terminals, so the voltage generator, with a series resistor if you want, should be in parallel with the coil. Now the best you can do is adding a parallel capacitor so that the LC parallel resonates at the right frequency provoking a very high voltage gain (not power gain).

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  • \$\begingroup\$ Although intuitively it is weird, I thought that the effect of mutual coupling between two coils can be modelled as a series voltage source at the receiver, and a series impedance at the transmitter? \$\endgroup\$
    – BNJMNDDNN
    May 20, 2015 at 14:44
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The RFID receive coil is induced with a voltage from the transmit coil. This voltage is generated in series with the receive coil and, when you add a "so-called" parallel capacitor you are actually making a highly resonant LC low pass filter: -

schematic

simulate this circuit – Schematic created using CircuitLab

It's still a series circuit but you are taking the voltage across the capacitor i.e. there can be very significant voltage amplification. We call it parallel thru laziness I suspect and I am as guilty of this as anyone.

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  • \$\begingroup\$ If you add a series resistor to 'Coupled V' my parallel configuration is as good as yours, or not? \$\endgroup\$ May 20, 2015 at 14:40
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    \$\begingroup\$ @VladimirCravero The voltage generator you mention can never be in parallel with the inductor with or without a series resistor. That bit is incorrect as I see it. I would also say that resonating can get more power - i've seen tuned LC coils producing a magnetic field totally "loaded" when a tuned coil is brought near them. If the coil were (say) just shorted the effect hardly could be seen until the coils got much closer. \$\endgroup\$
    – Andy aka
    May 20, 2015 at 15:00
  • \$\begingroup\$ can you explain why? \$\endgroup\$ May 20, 2015 at 15:01
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    \$\begingroup\$ @VladimirCravero The voltage induced is in the coil and the amount of induced voltage is due to the coupling factor. The coupling factor (say 10%) is like having 10% of the windings bypassed and directly connected to the transmit voltage source whilst 90% of the receiver inductance left as uncoupled and therefore in series with the induced voltage. \$\endgroup\$
    – Andy aka
    May 20, 2015 at 15:05
  • \$\begingroup\$ As I wrote in the edit of my question: Is it possible that the eventual best solution is depending on the size of R and C and the operating frequency (whether their product is larger or smaller than 1)? I suppose that for strong overdamped LC-circuits, a "parallel" resonant capacitance is actually disadvantageous compared to a series resonant capacitance (although in that case the series capacitance may as well be left out)? \$\endgroup\$
    – BNJMNDDNN
    May 21, 2015 at 14:53

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