# Full bridge rectifier and resonance for wireless power transfer in the receiver circuit

I was wondering which is the most suitable method to set up the receiver circuit for wireless power transfer circuit. Currently I have a circuit with the tesonance tank, i.e., the receiver coil and the equivalent capacitors in parallel to provide the resonance at a specific frequency. Followed by this are my full bridge rectifier diodes and a finally a load at the end. This based off the circuit given in thid application note ANP032g from Würth page 11. There is also another application note from Würth ANP070b, which shows the resonant capacitors in series with the receiver coil and is followed by the full bridge rectifier. In another website, they have the full bridge rectifier circuit connected to the Rx coil, and then have the resonant capacitors given at Theorycircuit Which is the most suitable circuit layout for the receiver circuit?

I feel like I have understood the concept well enough, but the more I read and see different approaches to wireless power transfer, the more confused I am.

• The circuit with alleged tuning capacitors after the rectifier is rubbish . Even if for no other reason than that the large filter capacitor will swamp the effect of the small capacitors. Nov 10, 2022 at 6:49

Which is the most suitable circuit layout for the receiver circuit?

A receiver with a series resonant capacitor does not help at all when the coupling between transmit coil and receive coil is weak. This is because a series resonant receiver coil does not provide any voltage amplification.

Hence, those applications that are unable to have a strong coupling will use a parallel resonant receive coil. Parallel resonance (with weak coupling) can provide substantial voltage amplification and, is the preferred method for the wireless power circuits that I have designed (several).

Of course, if the "transformer" coupling between transmit coil and receive coil is good, using series resonance on the receive coil will prevent a massive overload of voltage that arises when a parallel resonant receive coil is used and placed up-close to the transmit coil.

But, if the coupling is expected to work from short to long distances, there are alternative ways of restricting (regulating) the received voltage when parallel resonant receive coils are used. I've used switching converters to regulate the receive voltage and, I've used data transmission back to the transmit coil so that it reduces its driving amplitude when coupling is close.

I would strongly urge you to use a simulator and see the effects for yourself: -  With coupling at a low value (0.01) the parallel receive tuning slaughters the performance of series receive tuning. If coupling was increased to 0.1 then parallel tuning is still excellent: - However, the received voltage is now approaching quite a high level and may need some form of regulator to prevent it over-powering the receiver circuits.

Before I start, I'd like to say that I personally didn't like the last circuit taken from an amateur website. So I'm just ignoring it and I won't make any comments.

We can model the series and parallel receivers as following: simulate this circuit – Schematic created using CircuitLab

Note that the voltage couples to the receiver coil is shown as a series-connected voltage source. And the following analysis is based on an equal voltage for both, $$\v\$$. The currents $$\i_x\$$ and $$\i_y\$$ indicate the load currents for both models. And also the effective series resistances (or DC resistances) of the coils are neglected for simplification.

We can calculate the tank currents for both models now.

Series: $$i_s=\frac{v}{R_L+Z_{Lr}+Z_{Cr}}=\frac{v}{R_L+j\omega L_{r}+\frac{1}{j\omega C_r}}$$

Parallel: $$i_p=\frac{v}{Z_{Lr}+Z_{Cr}||R_L}$$

To make the life easier and the comparison more understandable, we can transform the parallel model to series model using $$\Q\$$: simulate this circuit

$$Q=\omega_0 C_r R_L \\ \Rightarrow C_{rx}=\Big(1+\frac{1}{Q^2}\Big)\ C_r \ \ \ and \ \ \ R_{Lx}=\Big(\frac{1}{1+Q^2}\Big)R_L$$

As can be seen from the conversion, effective $$\C_r\$$ is higher, and effective $$\R_L\$$ is lower, on converted model.

So after the conversion, the tank current for parallel model is

$$i_{px}=\frac{v}{R_Lx+Z_{Lr}+Z_{Crx}}=\frac{v}{R_L+j\omega L_{rx}+\frac{1}{j\omega C_{rx}}}$$

This will lead to a result of that the power delivered to the load in parallel tank is slightly higher than that in series tank.

But...

For a very high parallel tank $$\Q\$$ factor i.e. for $$\Q=\omega_0 C_r R_L>>1\$$ the effective component values will be equal. Therefore the power delivered to the load will be equal on both configurations.