**2. Magnetic field coupling without resonance**

The magnetic field coupling without resonance type utilizes the principle of electromagnetic induction, as shown in **Figure 3**. According to Ampere's law as Eq. (1), the magnetic pulsing (d*B*/dt) of the primary coil can be generated by time-varying driving current (dI/dt). Where *B*, μ0, J are magnetic field, permeability, and current density. The electric field *E* can be induced on the secondary coil, according to Faraday's law as Eq. (2) [9, 15]. Thus, the charge can be stored *via* a secondary coil.

$$
\nabla \times \mathbf{B} = \mu\_0 \mathbf{J} \tag{1}
$$

$$\mathbf{E} = -\frac{\partial \mathbf{B}}{\partial \mathbf{t}}\tag{2}$$

To understand the principle of magnetic field coupling without resonance, **Figure 3** shows the topology of the circuit analysis. The mutual inductance, Lm <sup>¼</sup> <sup>k</sup> ffiffiffiffiffiffiffiffiffi L1L2 <sup>p</sup> , depends on the distance x between the two coils, where k is the coupling factor (0 < k < 1) between the primary and the secondary coil. Based on Kirchhoff's voltage law, Eqs. (3) and (4) can be obtained as below [16]:

$$\text{joL}\_1\text{I}\_1 + \text{I}\_1\text{R}\_1 + \text{joL}\_m\text{I}\_2 = \text{U}\_1\tag{3}$$

$$\text{joL}\_2\text{I}\_2 + \text{I}\_2\text{R}\_2 + \text{I}\_2\text{R}\_L + \text{joL}\_m\text{I}\_1 = \mathbf{0} \tag{4}$$

I2 can be expressed as Eq. (5), according to Eq. (4).

$$\mathbf{I}\_2 = -\mathbf{I}\_1 \left[ \frac{\alpha \mathbf{L}\_\mathbf{m}}{\alpha \mathbf{L}\_2 - \mathbf{j}(\mathbf{R}\_2 + \mathbf{R}\_\mathbf{L})} \right] \tag{5}$$

The relation between I1 and U1 can be obtained as Eq. (6), according to Eqs. (3) and (5).

$$\mathbf{I}\_1 = \left[ \frac{\mathbf{R}\_2 + \mathbf{R}\_L + \text{joL}\_2}{(\mathbf{R}\_1 + \text{joL}\_1)(\text{joL}\_2 + \mathbf{R}\_2 + \mathbf{R}\_L) + \alpha^2 \text{L}\_m} \right] \mathbf{U}\_1 \tag{6}$$

The relation between I1 and I2 can be expressed as Eq. (7), according to Eqs. (5) and (6).

**Figure 3.** *Circuit topology of electromagnetic induction.*

$$\mathbf{I}\_2 = -\left[\frac{\text{joL}\_{\text{m}}}{(\text{R}\_1 + \text{joL}\_1)(\text{joL}\_2 + \text{R}\_2 + \text{R}\_L) + \text{o}^2 \text{L}\_{\text{m}}^2}\right] \mathbf{U}\_1 \tag{7}$$

The ratio of I1 and I2 can be expressed as Eq. (8), according to Eqs. (6) and (7).

$$\frac{\mathbf{I}\_1}{\mathbf{I}\_2} = -\frac{\mathbf{R}\_2 + \mathbf{R}\_L + \text{joL}\_2}{\text{joL}\_m} \tag{8}$$

To better understand the input impedance and power efficiency of the circuit, a T-type equivalent circuit is used in **Figure 4**.

$$\mathbf{V\_{Lm2}} = \mathbf{j} \mathbf{o} \mathbf{L\_m} \mathbf{I\_1} \tag{9}$$

$$\mathbf{V\_{Lm2}} = \frac{\text{joL}\_{\text{m}}(\text{R}\_{2} + \text{R}\_{\text{L}} + \text{joL}\_{2})}{(\text{R}\_{1} + \text{joL}\_{1})(\text{joL}\_{2} + \text{R}\_{2} + \text{R}\_{\text{L}}) + \text{o}^{2}\text{L}\_{\text{m}}^{2}}\mathbf{U\_{1}}\tag{10}$$

According to Eq. (11), Zin2 can be obtained from Eqs. (8) and (9).

$$\mathbf{Z\_{in2}} = \frac{\mathbf{V\_{Lm2}}}{-\mathbf{I\_2}} = \frac{\mathbf{j} \text{oL}\_m \mathbf{I\_1}}{-\mathbf{I\_2}} = \mathbf{R\_2} + \mathbf{R\_L} + \mathbf{j} \text{oL}\_2 \tag{11}$$

Then, Z2 can be further obtained, according to Eq. (12).

$$\mathbf{Z}\_{2} = \frac{\mathbf{V}\_{\text{Lm}1}}{\mathbf{I}\_{1}} = \frac{\text{joL}\_{\text{m}}\mathbf{I}\_{2}}{\mathbf{I}\_{1}} = \frac{\text{o}^{2}\mathbf{L}\_{\text{m}}\mathbf{I}}{\mathbf{R}\_{2} + \mathbf{R}\_{\text{L}} + \text{joL}\_{2}} \tag{12}$$

Finally, the Zin1 can be acquired, according to Eq. (13).

$$\mathbf{Z\_{in1}} = \mathbf{R\_1} + \mathbf{j} \mathbf{o} \mathbf{L\_1} + \mathbf{Z\_2} = \mathbf{R\_1} + \mathbf{j} \mathbf{o} \mathbf{L\_1} + \frac{\mathbf{o}^2 \mathbf{L\_m}^2}{\mathbf{R\_2} + \mathbf{R\_L} + \mathbf{j} \mathbf{o} \mathbf{L\_2}} \tag{13}$$

To understand the power efficiency, the energy loss ratio for the primary-side internal resistance, secondary resistance, and load can be denoted by PR1, PR2, and PRL as Eq. (14).

**Figure 4.** *T-type equivalent circuit for electromagnetic induction.*

$$\mathbf{P\_{R\_1}} : \mathbf{P\_{R\_2}} : \mathbf{P\_L} = |\mathbf{I\_1}|^2 \mathbf{R\_1} : |\mathbf{I\_2}|^2 \mathbf{R\_2} : |\mathbf{I\_2}|^2 \mathbf{R\_L} \tag{14}$$

The square ratio of I2 and I1 can be expressed as Eq. (15), according to Eq. (8)

$$\left|\frac{\mathbf{I}\_2}{\mathbf{I}\_1}\right|^2 = \frac{\text{o}^2 \mathbf{L}\_{\text{m}}^2}{\left(\mathbf{R}\_2 + \mathbf{R}\_L\right)^2 + \text{o}^2 \mathbf{L}\_{\text{m}}^2} \tag{15}$$

The power ratio can be rewritten as Eq. (16), according to Eqs. (14) and (15).

$$\mathbf{P}\_{\mathbf{R}\_1} : \mathbf{P}\_{\mathbf{R}\_2} : \mathbf{P}\_{\mathbf{L}} = \left\{ (\mathbf{R}\_2 + \mathbf{R}\_{\mathbf{L}})^2 + \mathbf{o}^2 \mathbf{L}\_{\mathbf{m}}^2 \right\} \mathbf{R}\_1 : \left\{ \mathbf{o}^2 \mathbf{L}\_{\mathbf{m}}^2 \right\} \mathbf{R}\_2 : \left\{ \mathbf{o}^2 \mathbf{L}\_{\mathbf{m}}^2 \right\} \mathbf{R}\_{\mathbf{L}} \tag{16}$$

Finally, the power efficiency can be obtained, according to Eq. (17).

$$\eta = \frac{\mathbf{P\_L}}{\mathbf{P\_{in}}} = \frac{\mathbf{P\_L}}{\mathbf{P\_{R\_1}} + \mathbf{P\_{R\_2}} + \mathbf{P\_L}} = \frac{\left\{\mathbf{a^2 L\_{m}^{-2}}\right\} \mathbf{R\_L}}{\left\{\left(\mathbf{R\_2} + \mathbf{R\_L}\right)^2 + \mathbf{a^2 L\_{m}^{-2}}\right\} \mathbf{R\_1} + \left\{\left<\mathbf{a^2 L\_{m}}\right\} \mathbf{R\_2} + \left<\mathbf{a^2 L\_{m}^{-2}}\right>\mathbf{R\_L}}\tag{17}$$
