**4. Electric field coupling**

Compared with magnetic field coupling, the electric field coupling without resonance uses a displacement current instead of an electromagnetic induction (**Figure 7**), as Eq. (37). Where ε, **E**, and **D** are permittivity, electric field intensity, electric flux density, respectively [17–19].

$$\mathbf{I} = \frac{\partial \mathbf{D}}{\partial \mathbf{t}} \bullet \mathbf{S} = \varepsilon \frac{\partial \mathbf{E}}{\partial \mathbf{t}} \bullet \mathbf{S} \tag{37}$$

In **Figure 8**, the mutual inductance of Cm between two capacitances on the distance x between the two coils can be written as Cm <sup>¼</sup> <sup>k</sup> ffiffiffiffiffiffiffiffiffiffi C1C2 <sup>p</sup> .

The electrical field coupling schematic diagram can be equivalent to π-type coupling, as shown in **Figure 9**. Based on the mesh-analysis method, the basic mathematical expression can be described as a matrix, according to Eq. (38).

**Figure 7.** *Diagram of displacement current.*

*Wireless Power Transmission on Biomedical Applications DOI: http://dx.doi.org/10.5772/intechopen.103029*

#### **Figure 8.**

*Circuit topology of electric field coupling with resonance.*

**Figure 9.**

*π-type equivalent circuit for electric field coupling with resonance.*

$$\mathbf{Z\_{\dot{m}2}} = \frac{\mathbf{1}}{\mathbf{j}\mathbf{o}\mathbf{C}\_2 + \frac{1}{\mathbf{R}\_2 + \mathbf{j}\mathbf{o}\mathbf{L}\_2 + \mathbf{R}\_L}}\tag{39}$$

$$\mathbf{Z}\_2 = \frac{1}{\mathrm{o}^2 \mathbf{C}\_\mathrm{m}^2 \mathbf{Z}\_{\mathrm{in}2}} \tag{40}$$

$$\mathbf{Z\_{in1}} = \frac{\mathbf{1}}{\mathbf{j}\mathbf{o}\mathbf{C\_1} + \frac{1}{\mathbf{Z\_2}}} + \mathbf{j}\mathbf{o}\mathbf{L\_1} + \mathbf{R\_1} \tag{41}$$

To analyze the input impedance, the extended π-type circuit with a separate capacitance is shown in **Figure 10**. The input impedance of each stage can be expressed as Eqs. (38), (39), and (40).

When operating frequency is equal to the resonant frequency on the secondary side, the imaginary part of the Zin2 becomes zero. Then, the resonant frequency can be acquired as Eq. (42).

$$\rho\_2 = \sqrt{\frac{\mathbf{1}}{\mathbf{L}\_2 \mathbf{C}\_2} - \left(\frac{\mathbf{R}\_2 + \mathbf{R}\_L}{\mathbf{L}\_2}\right)^2} \tag{42}$$

**Figure 10.** *π-type circuit with a separate capacitance for electric field coupling with resonance.*

In the same way, when the operating frequency is equal to the resonant frequency on the primary side, the imaginary part of the Zin1 becomes zero. The L1 can be obtained as Eq. (43).

$$\mathbf{L}\_{1} = \frac{\mathbf{C}\_{1} \left(\mathbf{R}\_{2} + \mathbf{R}\_{\mathrm{L}}\right)^{2} - \alpha\_{1}^{2} \left\{ \mathbf{C}\_{2} \left[ \left(\mathbf{R}\_{2} + \mathbf{R}\_{\mathrm{L}}\right)^{2} + \left(\alpha \mathrm{L}\_{2}\right)^{2} \right] - \mathbf{L}\_{2} \right\} \left\{ \mathbf{L}\_{2} \mathbf{C}\_{1} - \left(\mathbf{C}\_{1} \mathbf{C}\_{2} - \mathbf{C}\_{\mathrm{m}}^{2}\right) \left[ \left(\mathbf{R}\_{2} + \mathbf{R}\_{\mathrm{L}}\right)^{2} + \left(\alpha \mathrm{L}\_{2}\right)^{2} \right] \right\}}{\alpha\_{1}^{4} \left\{ \mathbf{L}\_{2} \mathbf{C}\_{1} - \left(\mathbf{C}\_{1} \mathbf{C}\_{2} - \mathbf{C}\_{\mathrm{m}}^{2}\right) \left[ \left(\mathbf{R}\_{2} + \mathbf{R}\_{\mathrm{L}}\right)^{2} + \left(\alpha \mathrm{L}\_{2}\right)^{2} \right] \right\}^{2} + \alpha\_{1}^{2} \mathbf{C}\_{1}^{2} \left(\mathbf{R}\_{2} + \mathbf{R}\_{\mathrm{L}}\right)^{2} \tag{43}$$

To understand the power efficiency, the ratio of energy loss for the primary-side internal resistance, secondary-side internal resistance, and load can be denoted by PR1, PR2, and PRL as Eq. (44).

$$\mathbf{P\_{R\_1}} : \mathbf{P\_{R\_2}} : \mathbf{P\_{R\_L}} = |\mathbf{I\_1}|^2 \mathbf{R\_1} : |\mathbf{I\_2}|^2 \mathbf{R\_2} : |\mathbf{I\_2}|^2 \mathbf{R\_L} \tag{44}$$

The square ratio of I2 and I1 can be calculated as Eq. (45), according to Eq. (38) [16].

$$\left|\frac{\mathbf{I\_1}}{\mathbf{I\_2}}\right|^2 = \frac{\alpha^2 \mathbf{C\_1} \mathbf{C\_2} \left[ \left( \mathbf{R\_2} + \mathbf{R\_L} \right)^2 + \left( \alpha \mathbf{L\_2} \right)^2 \right] + \mathbf{C\_1^2} (\mathbf{1} - \alpha^2 \mathbf{L\_2} \mathbf{C\_2})}{\mathbf{C\_m^2}} \tag{45}$$

The power ratio can be rewritten as Eq. (46), according to Eqs. (44) and (45).

$$\mathbf{P\_{R\_1}:P\_{R\_2}:P\_L} = \left\{ \mathbf{u}^2 \mathbf{C\_1} \mathbf{C\_2} \left[ (\mathbf{R\_2} + \mathbf{R\_L})^2 + (\mathbf{u} \mathbf{L\_2})^2 \right] + \mathbf{C\_1^2} (\mathbf{1} - \mathbf{u}^2 \mathbf{L\_2} \mathbf{C\_2}) \right\} \mathbf{R\_1} : \left\{ \mathbf{C\_m^2} \right\} \mathbf{R\_2} : \left\{ \mathbf{C\_m^2} \right\} \mathbf{R\_L} \tag{46}$$

Finally, the power efficiency can be expressed as Eq. (47).

*Wireless Power Transmission on Biomedical Applications DOI: http://dx.doi.org/10.5772/intechopen.103029*

$$\begin{aligned} \eta(\alpha) &= \frac{\mathbf{P\_L}}{\mathbf{P\_{in}}} = \frac{\mathbf{P\_L}}{\mathbf{P\_{R\_1}} + \mathbf{P\_{R\_2}} + \mathbf{P\_L}} \\ &= \frac{\left\{\mathbf{C\_m^2}\right\} \mathbf{R\_L}}{\left\{\alpha^2 \mathbf{C\_1 C\_2} \left[ \left(\mathbf{R\_2} + \mathbf{R\_L}\right)^2 + \left(\alpha \mathbf{I\_2}\right)^2\right] + \mathbf{C\_1^2 (1 - \alpha^2 \mathbf{L\_2 C\_2})\right\} \mathbf{R\_1} + \left\{\mathbf{C\_m^2}\right\} \mathbf{R\_2} + \left\{\mathbf{C\_m^2}\right\} \mathbf{R\_L}} \end{aligned} \tag{47}$$
