One of the main challenges of near-field WPT, in particulary IPT, is the sensitivity of the system towards changes in distance between transmitter and receiver, misalignment, load variation and changes in frequency. As a result, practial WPT systems often experience unstable behaviour.

A way to increase the stability of the WPT system is applying parity-time symmetric WPT (PT-symmetric WPT).

The name “parity-symmetric” refers to the fact that the circuit does not change when both parity and time are changed:

• Changing parity: the position is changed, i.e., the spatial coordinate is flipped, e.g., $x$ becomes $-x$. In the context of WPT (due to the fact that the wavelengths are long compared to the circuit), it corresponds to switching the labels of transmitter (1) and receiver (2), e.g., a coil $L_1$ at the transmitter side becomes $L_2$, the coil at receiver side, and vice versa. In other words, there has to be symmetry present in Kirchoff's laws.
• Changing time: the direction of time is flipped from $t$ to $-t$. Since current is the time derivative of charge, it results in flipping the current in the context of WPT. If we apply this on Ohm's law, this results in changing the sign of a resistor: we get negative resistors! Instead of a linear current decrease when the voltages lowers, we now have a lineary increasing current for a voltage reduction. Energy is now not dissipated any more in the resistor, but the negative resistor generates energy. The negative resistance is not any more a passive component, but an active component, supplying energy to the system. In other words, the supply of the system is now represented by a negative resistance. For inductors and capacitors, nothing changes when we apply time reversal. Applying Ohm's law also inverts the current, but the time derivative in the voltage-current relationship flips the sign again, resulting in a cancellation of the minus sign.

A WPT circuit is PT-symmetric if it remains exactly the same when (1) you first switch the indices from transmitter and receiver and then (2) flip the sign of all resistors. Do not change the sign of the reactances (inductors and capacitors).

There are two main ways to realize a negative resistor.

• The first one uses an operational amplifier (opamp). However, this method is only suitable for low power (mW) levels, and as such, it is not very useful for WPT applications.
• As a result, a controlled power electronic inverter is typically used. It measures the output voltage and/or current of the electronic inverter and sends it to a controller module. This controller adjusts the driver to realize a fixed phase difference between the output voltage and current of the inverter. As a result, a negative resitor (= 0° or 180° phase difference between voltage and current) can be realized.

The negative resistor is only realized within a certain voltage or current region.

• In the passive sign convention, with R<0, the AC current and voltage are in opposite phase (180° phase difference).
• In the active sign convention, with R<0, the AC current and voltage are in phase (0° phase difference)

Consider the equivalent circuit of a series-series (SS) compensated IPT circuit, with $R_1$ and $R_2$ the losses in the system (mainly the series resistance of the inductors), $R_L$ the load and $-R_N$ the supply, represented by a negative resistor. $V_1 (I_1)$ and $V_2 (I_2)$ represent the voltage (current) at the input port (over the supply $-R_N$) and output port ($R_L$), respectively.

From Kirchoff's laws, we obtain: $$\begin{bmatrix} R_1+j(\omega L_1-\frac{1}{\omega C_1}) & j\omega L_{12}\\ j\omega L_{12} & R_2+j(\omega L_2-\frac{1}{\omega C_2}) \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \end{bmatrix} = \begin{bmatrix} V_1 \\ V_2 \end{bmatrix} $$

Since $V_1=-(-R_N)I_1$ and $V_2=-R_LI_2$, we obtain:

$$\begin{bmatrix} -R_N+R_1+j(\omega L_1-\frac{1}{\omega C_1}) & j\omega L_{12}\\ j\omega L_{12} & R_L+R_2+j(\omega L_2-\frac{1}{\omega C_2}) \end{bmatrix} \begin{bmatrix} I_1 \\ I_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}~~~~~~\text{(eq. 1)} $$

In order to obtain a PT-symmetric circuit, Kirchoff's laws have to be symmetric. As a result, the following conditions must be met: $$L_1=L_2$$ $$C_1=C_2$$ $$-R_N+R_1=R_L+R_2$$

Note that the first two conditions imply that the self LC-resonance frequency of the transmitter resonator must equals the self resonance of the receiver circuit: $\omega_1=\omega_2$ since $1/\sqrt{L_1C_1}=1/\sqrt{L_2C_2}$

The same derivation can be made for a parallel-parallel (PP) compensated WPT circuit. However, other PT-symmetric conditions are obtained: $$L_1=L_2$$ $$C_1=C_2$$ $$R_1=R_2=0$$ $$R_N=R_L$$ Note in particulary that a PP topology is only exact PT-symmetric if no internal losses are present. In well-designed WPT systems, the resistive losses should be small, resulting in a good approximate PT-symmetric PP system.

Whereas it is possible to obtain an exact PP-symmetric circuit for SS and PP compenation topologies, this is not the case for SP and PS topologies. However, approximations are possible: if the coupling factor is low, and the coil quality factors are large (i.e. the resistances $R_1$ and $R_2$ are small), then SP and PS are in a good approximation also PT-symmetric, even though the circuits are obviously not parity symmetric (e.g., SP: serie in transmitter vs parallel compensation at receiver side == not symmetric).

Solving equation (1) for frequency results in equating the determinant to zero:

$$ [-R_N+R_1+j(\omega L_1-\frac{1}{\omega C_1})].[R_L+R_2+j(\omega L_2-\frac{1}{\omega C_2})]+\omega^2L_{12}^2=0 ~~~~~~\text{(eq. 2)} $$

In order to obtain a PT-symmetric circuit, Kirchoff's laws have to be symmetric. As a result, the following conditions must be met, as stated earlier: $$L_1=L_2$$ $$C_1=C_2$$ $$-R_N+R_1=R_L+R_2~~~~~~\text{(eq. 3)} $$

Solving equation (2) with respect to $\omega$ results in two solutions: $$\omega_{o1}=\frac{\omega_0}{\sqrt{2(1-k^2)}}\sqrt{2-(\frac{R_2+R_L}{\omega_0 L_2})^2+\sqrt{[2-(\frac{R_2+R_L}{\omega_0 L_2})^2]^2-4(1-k^2)}}$$ $$\omega_{o2}=\frac{\omega_0}{\sqrt{2(1-k^2)}}\sqrt{2-(\frac{R_2+R_L}{\omega_0 L_2})^2-\sqrt{[2-(\frac{R_2+R_L}{\omega_0 L_2})^2]^2-4(1-k^2)}}$$

where we introduced the coupling factor $k$ and the self resonance of each resonator $\omega_0$

$$ k = \frac{L_{12}}{\sqrt{L_1L_2}}$$ $$ \omega_0=\frac{1}{\sqrt{L_1C_1}}=\frac{1}{\sqrt{L_2C_2}}$$

Note that the first two conditions for exact PT-symmetry imply that the self LC-resonance frequency $\omega_1$ of the transmitter resonator equals the self resonance $\omega_2$ of the receiver circuit: $\omega_1=\omega_2=\omega_0$.

The solutions $\omega_{o1}$ and $\omega_{o2}$ are only real when the following condition is met:

$$ k \geq k_c=\sqrt{1-\frac{1}{4}[2-(\frac{R_2+R_L}{\omega_0 L_2})^2]^2}$$

We call $k_c$ the critical coupling coefficient.

In other words, we can only achieve a PT-symmetrical system is the coupling factor $k$ is higher than $k_c$.

If $k<k_c$, we do not achieve a PT-symmetrical system. We call this the “broken PT-symmetric state”.

From the imaginary part of equation (2), we find the only solution:

$$\omega_{o3}=\omega_0$$

From the real part of equation (2), we find that the solution $\omega_{o3}=\omega_0$ only holds when the value of the negative resistor is given by:

$$R_N=R_1+\frac{\omega_0 k^2L_1L_2}{R_2+R_L}~~~~~~\text{(eq. 4)} $$

When $k \geq k_c$, there are two resonance frequencies, $\omega_{o1}$ and $\omega_{o2}$. In this case, from equation (3), the negative resistor automatically compensates the other resistors in the network $R_1, R_2$ and $R_L$.

When $k=k_c$, both resonance frequencies coincide into $\omega_{o3}=\omega_0$

For $k<k_c$, the self resonant frequency, $\omega_{o3}=\omega_0$, remains the only resonance frequency in the system. From equation (4), it can be seen that the negative resistor is dependent on the coupling factor $k$ and does not automatically compensates the resistances in the network.

The same conclusions can be drawn for the PP compensated circuit.

From equation (1), both the current ratio $I_1/I_2$ and the voltage ratio $V_1/V_2$ can be determined, with $V_{1}$ the voltage over the negative resistor, and $V_2$ the voltage over the load.

Substituting the conditions for the exact PT-symmetric state, equation (3), and the corresponding resonance frequencies $\omega_{o1}$ and $\omega_{o2}$ into the current and voltage ratio expression, results in:

$$ \frac{I_1}{I_2}=1$$ $$ \frac{V_{1}}{V_{2}}=\frac{R_1+R_2+R_L}{R_L}$$

We notice that, in the exact PT-symmetric region, the current and voltage ratios are constant: they do NOT depend on the value of the coupling factor $k$.

Substituting the value for the negative resistor in the broken PT-symmetric region, equation (4), and the self-resonance frequency $\omega_{o3}=\omega_0$ into the current and voltage expressions, results in expressions that are dependent on the coupling factor $k$. In other words, a varying coupling between transmitter and receivers changes the current and voltage ratios in the broken PT-symmetric region.

Calculating the output power $P_L$ in the load, taken into account the conditions and resonance frequencies in the PT-symmetric region, result into:

$$ P_L=\frac{L_1L_2R_LU_{in}^2}{L_1^2(R_2+R_L)^2+2L_1L_2R_1(R_2+R_L)+L_2^2R_1^2}$$

Analogously, we get for the power transfer efficiency $\eta$:

$$ \eta=\frac{L_1R_L}{L_2R_1+L_1(R_2+R_L)} $$

These two equations indicitate the important aspect of parity-time symmetric WPT: both the output power and efficiency are constant, i.e., independent on the coupling factor $k$, in the exact symmetric PT-region ($k \geq k_c$).

The underlying explanation is that in a PT-symmetric region, the eigenvalues of the system are pure imaginary. As a result, even when losses are present, there is no change in the eigenstates of the system.

This is not the case in the broken PT-symmetric region, when $k<k_c$. Taken into account the value for the negative resistor in the broken PT-symmetric region, equation (4), and the self-resonance frequency $\omega_{o3}=\omega_0$, it can be calculated that

$$ P_L=\frac{\omega_0^2k^2L_1L_2R_LV_{1}^2}{[\omega_0^2k^2L_1L_2+R_1(R_2+R_L)]^2}$$

$$ \eta=\frac{\omega_0^2k^2L_1L_2R_L}{R_1(R_2+R_L)^2+\omega_0^2k^2L_1L_2(R_2+R_L)}$$

We can see that now both the output power and efficiency are NOT constant; they change with varying coupling factor $k$.

In the broken PT-symmetric region, the eigenstate amplitude increases or decays exponentially, i.e., no stable region is obtained.

References

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