*3.4. Efficiency*

The power-transfer efficiency *η* or power gain of the CPT system is defined as

$$
\eta = \frac{P\_{out}}{P\_{in}} \,\tag{18}
$$

with the expressions for *Pin* and *Pout* given by Equations (15) and (17). Hence, *η* is written as a function of the parameters of the circuit and the port voltages.

#### **4. Maximum-Efficiency Solution**

In the previous section, the general expressions for input power, output power and efficiency were found. Now, the configuration at maximum power-transfer efficiency will be considered. The notation *ηmax* is applied in this section to indicate that the power-transfer efficiency equals its maximum attainable value.

#### *4.1. First-Order Necessary Condition*

To find the output voltages at the maximum system efficiency configuration, the first-order necessary condition is applied to generate a system of 2*N* equations [20,26]:

$$\frac{\partial \eta}{\partial v\_n^{r\epsilon}} = 0,\tag{19}$$

$$\frac{\partial \eta}{\partial v\_{nl}^{im}} = 0.\tag{20}$$

The optimal input voltages *vn* are the solution of the above system. Unfortunately, solving the system directly is not straightforward. First, the quotient rule for derivatives is applied. With Equation (18), the system becomes

$$P\_{\rm in} \frac{\partial P\_{\rm out}}{\partial v\_n^{rc}} - P\_{\rm out} \frac{\partial P\_{\rm in}}{\partial v\_n^{rc}} = 0,\tag{21}$$

$$P\_{in} \frac{\partial P\_{out}}{\partial v\_{n}^{im}} - P\_{out} \frac{\partial P\_{in}}{\partial v\_{n}^{im}} = 0. \tag{22}$$

Substituting the derivatives of Equations (15) and (17) to *vren* and *vimn* into the system Equations (21) and (22), and taking into account Equation (18), the solution for the optimal normalized input voltages *vop<sup>t</sup> n* = *vre*,*op<sup>t</sup> n* + *jvim*,*op<sup>t</sup> n* at each input port is found.

$$v\_n^{rc,opt} = 0,\tag{23}$$

$$v\_n^{im,opt} = \frac{k\_{0n}Q\_n(\eta\_{\max} - 1)}{2\eta\_{\max}}v\_0. \tag{24}$$

If the values of the normalized input port voltages *vn* (*n* = 1, ... , *N*) equal Equations (23) and (24), the maximum attainable efficiency *ηmax* is reached. It is important to note that the voltages here are not only expressed as a function of the parameters of the circuit network (which are known and fixed) and the reference output voltage *v*0, but also of the maximum efficiency *ηmax*, which is (for now) an unknown value. In Section 4.3, the value of *ηmax* will be determined.

#### *4.2. Optimal Input and Output Power*

By substituting Equations (23) and (24) into Equations (15) and (17), the input power *Pop<sup>t</sup> in* and output power *Pop<sup>t</sup> out* at the maximum efficiency configuration are determined:

$$P\_{in}^{opt} = \frac{v\_0^2 (1 - \eta\_{max}^2) a\_N^2}{8 Q\_0 \eta\_{max}^2},\tag{25}$$

$$P\_{out}^{opt} = -\frac{v\_0^2}{2Q\_0} \left[ 1 + \frac{(\eta\_{max} - 1)a\_N^2}{2\eta\_{max}} \right],\tag{26}$$

where the following notation is introduced

$$a\_N^2 = \sum\_{n=1}^N a\_{n\prime}^2\tag{27}$$

with

$$
\alpha\_n = k\_{0n} \sqrt{Q\_0 Q\_n} \,. \tag{28}
$$

The parameter *αn* is named *the extended kQ-product* of the link between the *n*-th transmitter and the receiver, analogous to [27–29]. The variable *αN* is called the *system kQ-product*, a naming borrowed from [14,15,29]. The introduction of these variables seems artificial at this point, but will be further discussed in Section 5.

The value of *ηmax* is still unknown and will be determined in the following subsection.

#### *4.3. Maximum Efficiency*

A quadratic equation in *ηmax* is found by substituting Equations (25) and (26) in Equation (18):

$$
\eta\_{\max}^2 - \left(2 + \frac{4}{a\_N^2}\right) \eta\_{\max} + 1 = 0. \tag{29}
$$

In order to alleviate the notation, the symbol *γ* is introduced:

$$
\gamma = \sqrt{1 + a\_N^2}.\tag{30}
$$

The quadratic Equation (29) results in two solutions:

$$
\eta\_{\max,1} = \frac{\gamma - 1}{\gamma + 1},
\tag{31}
$$

and

$$
\eta\_{\text{max},2} = \frac{\gamma + 1}{\gamma - 1}. \tag{32}
$$

Equation (32) is physically not possible since 0 ≤ *ηmax* ≤ 1. The maximum attainable power-transfer efficiency *ηmax* is therefore expressed by Equation (31).

The maximum efficiency *ηmax* is now determined as a function of the characteristics of the circuit parameters only, which implies that the optimal input voltages Equations (23) and (24), input Equation (25), and also output power Equation (26) are expressed as functions of the circuit characteristics only.

For example, the optimal output and input power are given by

$$P\_{out}^{opt} = \frac{\gamma}{2Q\_0} v\_{0^\prime}^2 \tag{33}$$

$$P\_{in}^{opt} = \frac{P\_{out}^{opt}}{\eta\_{max}} = \frac{\gamma}{2Q\_0} \frac{\gamma + 1}{\gamma - 1} v\_0^2. \tag{34}$$

#### *4.4. Optimal Input Voltages, Currents, and Admittances*

Combining Equations (23), (24), and (31), the optimal normalized input voltages *vop<sup>t</sup> n* = *vre*,*op<sup>t</sup> n* + *jvim*,*op<sup>t</sup> n* are found:

$$
\sigma\_n^{rc,opt} = 0,\tag{35}
$$

$$
\upsilon\_{\rm in}^{im,opt} = \frac{k\_{0u}Q\_{\rm u}}{1 - \gamma} \upsilon\_0. \tag{36}
$$

From Equations (11), the optimal normalized input currents *iop<sup>t</sup> n* = *<sup>i</sup>re*,*op<sup>t</sup> n* + *jiim*,*op<sup>t</sup> n* follow

$$\dot{q}\_{n}^{ref,opt} = \frac{v\_0}{1 - \gamma} \sum\_{i=1}^{N} k\_{0i} k\_{in} Q\_{i\prime} \tag{37}$$

$$i\_n^{im,opt} = \frac{\gamma k\_{0n}}{1 - \gamma} v\_0. \tag{38}$$

The optimal normalized input admittance *yin*,*op<sup>t</sup> n* at port *n* thus equals

$$y\_n^{in,opt} = \frac{I\_n^{opt}}{v\_n^{opt}} = \frac{\gamma}{Q\_n} - j\frac{1}{k\_{0n}Q\_n} \sum\_{i=1}^{N} k\_{0i}k\_{in}Q\_i. \tag{39}$$

From this equation, it can be concluded that the cross-coupling between the transmitters can be compensated by a normalized shunt inductance *bSn* equal to

$$b\_n^S = \frac{1}{k\_{0n}Q\_{nn}} \sum\_{i=1}^N k\_{0i}k\_{in}Q\_{ii} \tag{40}$$

or unnormalized

$$B\_n^S = \frac{\mathcal{G}\_{nn}}{b\_{0n}} \sum\_{i=1}^N \frac{b\_{0i} b\_{in}}{\mathcal{G}\_{ii}}.\tag{41}$$

Equation (39) implies that the optimal input conditions can be obtained by a set of *N* independent current generators operating in maximum power-transfer conditions. The internal normalized shunt admittances of these generators are

$$(y\_n^S = (y\_n^{in,opt})^\* = \frac{\gamma}{Q\_n} + j\frac{1}{k\_{0u}Q\_u} \sum\_{i=1}^N k\_{0i}k\_{in}Q\_{i\cdot} \tag{42}$$

and their normalized currents are

$$\mathbf{i}\_n^S = \mathbf{i}\_n^{opt} + y\_n^S \mathbf{v}\_n^{opt} = 2jk\_{0n} \frac{\gamma}{1 - \gamma} v\_0. \tag{43}$$

In this way, maximum power transfer from the generators to the network is achieved. At this point, the maximum-efficiency solution also becomes the one that provides the maximum output power.

The corresponding unnormalized values of the shunt admittances and currents of these generators are (Figure 2a)

$$Y\_n^S = \gamma g\_{nn} + j \frac{g\_{nn}}{b\_{0n}} \sum\_{i=1}^N \frac{b\_{0i} b\_{in}}{g\_{ii}},\tag{44}$$

$$I\_n^S = 2j b\_{0n} \frac{\gamma}{1 - \gamma} V\_0. \tag{45}$$

Since *v*0 was chosen as reference phasor, the condition for the input current sources can be practically achieved by imposing that the ratios of the input currents must satisfy

$$\frac{i\_G^S}{k\_{01}} = \frac{i\_2^S}{k\_{02}} = \dots = \frac{i\_N^S}{k\_{0N}}\tag{46}$$

or

$$\frac{I\_1^S}{b\_{01}} = \frac{I\_2^S}{b\_{02}} = \dots = \frac{I\_N^S}{b\_{0N}}.\tag{47}$$

(a) transmitter *n*

$$I\_n^\mathcal{S} \xleftarrow[\underbrace{Y\_n^\mathcal{S} \prod\_{\text{input}} \stackrel{\text{input}}{\text{port}}}\_{\text{port}} \quad \underbrace{\text{output}}\_{\text{port}} \underbrace{\prod\_{\text{output}} \prod\_{\text{output}} \gamma\_{\mathcal{S}|\text{output}}}\_{\text{port}}$$

(b) receiver

**Figure 2.** Maximum-efficiency solution for (**a**) the transmitter ports with generator *ISn* and internal shunt admittance *YSn* , and (**b**) the receiver port with the optimal load susceptance and load conductance.

#### *4.5. Optimal Load Admittance*

Finally, it is possible to determine the optimal load that realizes the maximum-efficiency solution. The optimal normalized load admittance is given by

$$y\_0^{L,opt} = g\_0^{L,opt} + jb\_0^{L,opt} = -\frac{i\_0^{opt}}{v\_0^{opt}} \tag{48}$$

where *g<sup>L</sup>*,*op<sup>t</sup>* 0 and *<sup>b</sup>L*,*op<sup>t</sup>* 0 are the optimal normalized load conductance and susceptance, respectively. FromEquation(11),itfollowsthat

$$d\_0^{\sigma\varepsilon\rho pt} = \frac{\upsilon\_0}{Q\_0} + \sum\_{n=1}^{N} k\_{0n} v\_n^{im\,\rho pt} \,\,\,\,\,\tag{49}$$

$$d\_0^{im,opt} = -k\_{00}v\_0 - \sum\_{n=1}^{N} k\_{0n} v\_n^{rc,opt}.\tag{50}$$

Substituting Equations (35) and (36) results into

$$
\dot{u}\_0^{rc,opt} = -\frac{\gamma}{Q\_0} v\_{0\prime} \tag{51}
$$

$$
\dot{u}\_0^{im,opt} = -k\_{00}v\_{0\prime} \tag{52}
$$

which leads to the optimal normalized load conductance *g<sup>L</sup>*,*op<sup>t</sup>* 0 and load susceptance *<sup>b</sup>L*,*op<sup>t</sup>* 0 as functions of the parameters of the network:

$$\mathbf{g}\_0^{L,opt} = \frac{\gamma}{\mathbb{Q}\_0},\tag{53}$$

$$b\_0^{L\rho pt} = k\_{00}.\tag{54}$$

The corresponding unnormalized values are (Figure 2b)

$$\mathbf{G}\_0^{\mathsf{L}, \rho pt} = \gamma \mathbf{g}\_{00\prime} \tag{55}$$

$$B\_0^{L,opt} = b\_{00}.\tag{56}$$

## **5. Discussion**

In order to practically achieve the maximum attainable efficiency *ηmax*, three conditions must be met simultaneously:


Additionally, if the internal shunt admittances of the generators equal Equation (44), the maximum-efficiency solution also becomes the one that maximizes the output power.

The first condition indicates that, the higher the coupling between transmitter *n* and the receiver, the lower the necessary value of the current source for that transmitter. From Equations (35), (36), and (45), it follows that the phasors of the optimal current sources and optimal voltages of all input ports are orthogonal to the reference output voltage *V*0.

Instead of applying current sources at each transmitter, one could also apply voltage sources for which the ratios must satisfy

$$V\_1: V\_2: \ldots: V\_n = \frac{b\_{01}}{\mathcal{G}\_{11}}: \frac{b\_{02}}{\mathcal{G}\_{22}}: \ldots: \frac{b\_{0n}}{\mathcal{G}\_{nn}}.\tag{57}$$

The optimal input voltages are in other words determined by the ratios between the transmitter–receiver coupling strength and the resistive losses of the transmitter.

The second condition, the insertion of shunt susceptances at the input port, is necessary to compensate for the cross-coupling between the transmitters. If no cross-coupling is present, no shunt susceptances have to be inserted, as can be seen by Equation (41).

Under the optimal conditions, the maximum efficiency *ηmax* given by Equation (31) is reached. Notice that *ηmax* is independent on the cross-coupling between the transmitters; the optimization of input voltages *Vop<sup>t</sup> n* and load admittance *<sup>Y</sup>L*,*op<sup>t</sup>* 0 eliminates the influence of the cross-coupling. Nevertheless, the presence of the shunt susceptances at the input ports is necessary for achieving *ηmax* and to ensure that the optimal voltages *Vop<sup>t</sup> n* are reached from the current sources *IS n* that supply the power for the CPT system.

The third condition refers to the terminating load. The optimal load conductance *<sup>G</sup>L*,*op<sup>t</sup>* 0 of the receiver is proportionate to its parasitic conductance *g*00. For high coupling (*<sup>α</sup>N* >> 1) between the transmitters and receiver, the optimal conductance can be approximated by *<sup>G</sup>L*,*op<sup>t</sup>* 0 = *g*00*αN*. The output load susceptance must equal Equation (56) and thus cancels out the self-susceptance of the receiver resonator.

The optimal terminating load admittance corresponds to the value found in scientific literature for a CPT system with a single transmitter ( *N* = 1) coupled to a single receiver [30–32].

Notice that not only the maximum efficiency *ηmax*, but also the optimal load and input current ratios are independent on the cross-coupling between the transmitters. For an uncoupled system, the maximum efficiency *ηmax* and optimal load are the same as for a coupled system, since the shunt susceptances at the input ports compensate for the cross-coupling. This does not imply that the *efficiency* is not influenced by cross-coupling for a *general* CPT system; it is the *maximum* efficiency that is invariant for cross-coupling for an *optimized* system towards efficiency.

The efficiency rises with higher couplings between transmitter and receiver, and lower conductances *g*00, *g*11,..., *gNN* of the system.

It is not surprising that the maximum efficiency *ηmax* is expressed as a function of a single variable; it is a general property of *any* reciprocal power transfer system that the efficiency can be stated as a function of a single scalar [28]. In the context of WPT with multiple transmitters and/or receivers, this variable is often called the system kQ-product [14,15,20,23]

From Equation (27), it can be concluded that the square of the system kQ-product equals the sum of the squares of the kQ products of each individual transmitter–receiver link. Determining the kQ product for each single transmitter–receiver pair thus results in a prediction for the entire system's power-transfer efficiency.

The higher the system kQ-product *αN*, the higher the efficiency of the system, as can be seen by Equations (27), (28), and (31). The maximum efficiency *ηmax* of the CPT system can thus be increased by adding more transmitters to the system, even if the transmitters themselves are coupled. Indeed, the cross-coupling between the transmitters can be compensated by the shunt susceptances *B<sup>S</sup> n* at the input ports. There is no optimal number of transmitters. The more transmitters, the higher the maximum attainable efficiency *ηmax*.

#### **6. Numerical Verification**

In order to validate the theory, an example of CPT system with three transmitters ( *N* = 3) and a single receiver is considered; it is assumed that there is a cross-coupling present between the transmitters themselves (Figure 3a). The parameters within the dashed rectangle are assumed to be given and fixed, including the coupling strengths. They can be represented by the admittance matrix *Y*. In order to optimize the CPT system towards efficiency, it is possible to act on the value of the load *Y<sup>L</sup>* 0= *G<sup>L</sup>* 0+ *jB<sup>L</sup>* 0, the supply current sources *IS G*, and the input shunt susceptances *B<sup>S</sup> n*.

**Figure 3.** (**a**) Equivalent circuit of a capacitive wireless power transfer system with 3 transmitters (left) and a single receiver (right). The (desired) electric couplings between transmitters and receiver are depicted by the full arrows. The (undesired) cross-couplings between transmitters themselves are indicated by the dashed arrows. (**b**) Applied equivalent circuit for the simulation of the capacitive coupling.

The numerical values indicated in Table 1 are considered, the operating frequency is *f*0=10 MHz. No specific design consideration is assumed; a range of different desired and undesired coupling factors (Table 2) were chosen to verify the analytical derivation. Further, it is assumed that the receiver and the first transmitter have a self-susceptance *C*00 and *C*11, respectively.


**Table 1.** Given network simulation parameters for the analyzed numerical example.



Electric coupling is realized by the coupled capacitors *Cj* (*j* = 0, 1, 2, 3). In each transmitter and in the receiver, a resonant circuit is constructed by adding a shunt inductor *Lj* with value

$$L\_j = \frac{1}{\omega\_0^2 \mathcal{C}\_j}.\tag{58}$$

The corresponding values are given in Table 3.

In order to verify the analytical formulas, the numerical example has been simulated in AWR NI. Figure 3b depicts the applied equivalent circuit for the simulation of the capacitive coupling [25].

First of all, the admittance matrix of the link has been calculated, obtaining the following values:

$$\mathbf{Y} = \begin{bmatrix} 1 + 31.42j & -1.26j & -3.14j & -3.14j \\ -1.26j & 1.5 & -0.628j & -9.42j \\ -3.14j & -0.628j & 0.5 & -6.28j \\ -3.14j & -9.42j & -6.28j & 1 + 31.42j \end{bmatrix} \cdot 10^{-3} \cdot \tag{59}$$

By using the values reported in Equation (59), the extended *kQ*-product *αn*, the system *kQ*-product *αN*, and the inductors *Lj*, can be calculated from Equations (28), (27), and (58), respectively, and are listed in Table 3.

**Table 3.** Calculated network simulation parameters for the example capacitive power transfer (CPT) system.


As per the maximum efficiency, a value of 84.9% is attainable, according to Equation (31). The parameters at which the maximum efficiency configuration is reached are listed in Table 4. A current source *IS*1 of the first transmitter of 100 mA was chosen, resulting in the optimal current sources of the other transmitters, according to Equation (47).

The optimal load admittance is calculated by Equations (55) and (56). The optimal load susceptance is negative, i.e., it corresponds to a shunt inductor *<sup>L</sup>L*,*op<sup>t</sup>* 0.

Additionally, according to Equation (41) a shunt susceptance at each transmitter side is necessary to compensate the transmitter's cross-coupling. At the first transmitter, the shunt susceptance is an inductor *L<sup>S</sup>* 1 . At the second and third transmitter, it is found that the shunt susceptances are capacitors *C<sup>S</sup>* 2and *C<sup>S</sup>* 3. All the values calculated from theoretical formulas are summarized in Table 4.


**Table 4.** Calculated values for the maximum efficiency solution.

First, the simulation is executed at the maximum-efficiency configuration of Table 4. The simulation program returns a power-transfer efficiency *η* of 84.9% and confirms that the output voltage *V*0 is orthogonal to the input voltages and currents.

Next, the load conductance and load susceptance are varied, respectively, while keeping the other parameters fixed at their optimal value given in Table 4. The simulation results are depicted in Figures 4 and 5.

**Figure 4.** The simulated efficiency *η* as a function of varying load conductance for the given system of three transmitters and a single receiver. The load conductance is varied, while keeping the other system parameters at their optimal value.

**Figure 5.** The simulated efficiency *η* as a function of varying shunt load inductance for the given system of three transmitters and a single receiver. The load inductance is varied, while keeping the other system parameters at their optimal value.

Regarding effect on the efficiency of the compensating shunt susceptances *L<sup>S</sup>* 1 , *C<sup>S</sup>* 2 , and *C<sup>S</sup>* 3 , this is investigated in Figures 6 and 7. The simulated efficiency confirms that maximum efficiency is obtained by using the compensating shunt susceptances calculated by using the theoretical formulas.

**Figure 6.** The simulated efficiency *η* as a function of the compensating susceptance *LS*1 . *LS*1 is varied, while keeping the other system parameters at their optimal value.

**Figure 7.** The simulated efficiency *η* as a function of the compensating susceptance *CS*2 and *CS*3 . *CS*2 (black curve) or *CS*3 (red curve) is varied, while keeping the other system parameters at their optimal value.

Additionally, the simulation confirms that the shunt susceptances at the input ports eliminate the cross-coupling. For the circuit without the compensating shunt susceptances and no cross-coupling, the same maximum efficiency of 84.9% is reached.

Finally, the effect of the amplitude of the input currents has been analyzed. The achieved results are summarized in Figure 8. It is observed that the efficiency is always lower than the optimal current distribution from Table 4 (i.e., *I*2/*I*1 = 3 and *I*3/*I*1 = 2). For example, if all input current sources are equal to 100 mA, an efficiency of 76.2% is attained.

The case of voltage sources has been also analyzed; in this case, by using Equation (57) and the values of Equation (59), the optimal voltage ratios can be determined, e.g.,

$$\frac{V\_2}{V\_1} = \frac{\frac{h\_{02}}{\mathcal{S}\_{22}}}{\frac{h\_{01}}{\mathcal{S}\_{11}}} = 2.\tag{60}$$

Analogously, the optimal ration *V*3/*V*1 = 4 is found. This is confirmed by circuital simulation results summarized in Figure 9.

**Figure 8.** The simulated efficiency *η* as a function of the ratio of the input currents. The efficiency is always lower than the optimal current distribution *I*2/*I*1 = 3 and *I*3/*I*1 = 2. For example, if all input current sources are equal (i.e., *I*2/*I*1 = 1 and *I*3/*I*1 = 1), an efficiency of 76.2% is attained.

 **Figure 9.** The simulated efficiency *η* as a function of the ratio of the input voltages.

In conclusion, circuital simulations confirm the data provided by the theory for the analyzed example: a maximum power-transfer efficiency is achieved for the optimal values of Table 4, calculated according to analytical derivation.
