In this article, is considered an air core transformer. The problem of maximizing the real power on the reciever side is approached through the analysis of the equivalent circuit diagram of the transformer and solving the diagram for the real power in the load. The corresponding function for the power is then maximized by the appropriate choice of the reactive parts of the impedance on both sides.

### Analytical derivation of transmitted power

Fig.1 Air core transformer and its equivalent circuit diagram |

In fig.1 is shown an air core transformer which can be formed by two inductively coupled coils with coupling $k$. Below is shown the equivalent circuit diagram of the transformer. The circuit takes $u_1$ as an input voltage and transmits that to the reciever side as $u_2$. The input current is denoted by $i_1$ and the induced current in the reciever coil is denoted by $i_2$. We assume that the input voltage is a sinusoid, so that we can use AC circuits theory

$$u_1 = u_{1m}\sin{\omega t}$$

Let impedance $z_1$ encorporate the parameters of the first inductor such as the resistance of the coil $R_{coil\,1}$ and its inductance $L_{coil\,1}$. If there is any other real or reactive load on the transmitter side assume that it is encorporated in $z_1$ as well. Thus, for the impedance $z_1$ we have

$$z_1 = R_{coil\,1} + j\omega (L_{coil\,1} - M)$$

$$z_1 = r_1 + j x_1$$

On the reciever side, load impedance $R_{load} + j X_{load}$ is also encorporated in $z_2$.

$$z_2 = R_{coil\,2} + j\omega(L_{coil\,2} - M) + R_{load} + j X_{load}$$

$$z_2 = r_2 + j x_2$$

Impedance $z_3$ only includes the mutual inductance of the two coils and does not have a real part

$$z_3 = j \omega M = j x_3$$

To solve the circuit, we find the input current

$$ i_1 = \frac{u_1}{z_1 + z_2 || z_3}$$

Given that, we can now find the voltage across the reciever coil

$$ u_2 = u_1 - i_1 z_1 = u_1\left(1 - \frac{z_1}{z_1 + z_2 || z_3}\right)$$

Also, we find the induced current

$$i_2 = \frac{u_2}{z_2} = \frac{u_1}{z_2}\left(1 - \frac{z_1}{z_1 + z_2 || z_3}\right)$$

We, then, find the complex power delivered to the load

$$s_2 = u_2 i_2^* = p_2 + jq_2$$

$$s_2 = |u_1|^2 z_2 \left|\frac{z_3}{z_1z_2+z_1z_3+z_2z_3}\right|^2$$

$$p_2 = |u_1|^2 r_2 \left|\frac{z_3}{z_1z_2+z_1z_3+z_2z_3}\right|^2$$

$$q_2 = |u_1|^2 x_2 \left|\frac{z_3}{z_1z_2+z_1z_3+z_2z_3}\right|^2$$

In most applications we would need to maximize the active component of power in the load, therefore

$$p_2 \rightarrow \mathrm{max} \Rightarrow \left|\frac{z_3}{z_1z_2+z_1z_3+z_2z_3}\right|^2 \rightarrow \mathrm{max}$$

$$\left|z_1 + z_2 + \frac{z_1z_2}{z_3}\right|^2 \rightarrow \mathrm{min}$$

$$\left|r_1 + jx_1 + r_2 + jx_2 +\frac{(r_1 + jx_1)( r_2 + jx_2)}{jx_3}\right|^2 \rightarrow \mathrm{min}$$

$$\frac{1}{x_3^2}|(r_1x_3 + r_2x_3 + r_1x_2 + r_2x_1) + j(x_1x_3 + x_2x_3 + x_1x_2 - r_1r_2)|^2 \rightarrow \mathrm{min}$$

For convenince we introduce the following complex function

$$f(x_1,x_2) = (r_1x_3 + r_2x_3 + r_1x_2 + r_2x_1) + j(x_1x_3 + x_2x_3 + x_1x_2 - r_1r_2)$$

And we minimize the magnitude of the function at the cost of varying the imaginary parts of the impedance on both sides. In practice this will mean that we need to include some additional reactive elements in the circuit, perhaps two capacitors, as it is demonstrated below.

$$|f(x_1,x_2)|^2 \rightarrow \mathrm{min}$$

$$\frac{\partial|f|^2}{\partial x_1} = 2\mathbb{Re}(f)r_2 + 2\mathbb{Im}(f)(x_2 + x_3) = 0$$

$$\frac{\partial|f|^2}{\partial x_2} = 2\mathbb{Re}(f)r_1 + 2\mathbb{Im}(f)(x_1 + x_3) = 0$$

If solved simultaneously, these equations have 5 distinct solutions for $x_1$ and $x_2$. Two of them have no physical meaning since they result in either a complex value for $x_1$ and $x_2$ or a negative resistance. After rejecting the unrealistic ones we are left with the following three solutions for $x_1$ and $x_2$.

$$x_1 = -x_3,\quad x_2 = -x_3, \quad p_2 = \frac{|u_1|^2\,x_3^2\,r_2}{\left(r_1r_2 + x_3^2\right)^2}, \quad q_2 = -\frac{|u_1|^2\,x_3^3}{\left(r_1r_2 + x_3^2\right)^2}$$

$$x_1 = \frac{1}{r_2}\left(\sqrt{r_1r_2\left(x_3^2-r_1r_2\right)}-r_2x_3\right), \quad x_2 = \frac{1}{r_1}\left(\sqrt{r_1r_2\left(x_3^2-r_1r_2\right)}-r_1x_3\right), \quad p_2 = \frac{|u_1|^2}{4\,r_1}, \quad q_2 = \frac{|u_1|^2\,x_2}{4\,r_1\,r_2}$$

$$x_1 = -\frac{1}{r_2}\left(\sqrt{r_1r_2\left(x_3^2-r_1r_2\right)}+r_2x_3\right), \quad x_2 = -\frac{1}{r_1}\left(\sqrt{r_1r_2\left(x_3^2-r_1r_2\right)}+r_1x_3\right), \quad p_2 = \frac{|u_1|^2}{4\,r_1}, \quad q_2 = \frac{|u_1|^2\,x_2}{4\,r_1\,r_2}$$

The first solution $x_1 = -x_3,\quad x_2 = -x_3$ seems to be the simpliest and the most practical too. And the corresponding active power is

$$p_2 = \frac{|u_1|^2\,x_3^2\,r_2}{\left(r_1r_2 + x_3^2\right)^2}$$

Note that $x_3$ is defined by the expression

$$x_3 = 2\pi\,f\,k\,\sqrt{L_{coil\,1}L_{coil\,2}}$$

and thus, it depends on coupling $k$, i.e. the separation distance, and the frequency of transmission $f$.

### SPICE model simulation

Fig.2 LTSPICE model of wireless power transmission system |

We conduct simulations utilizing the above results in LTSpice software. Coupling coefficient of $k=0.01$ corresponds to approximately 25 cm of separation between the coils. Both coils have radius of turns of $a_1=a_2=0.1$ m and the number of turns $n_1=n_2=100$. On the transmitter side, a current limiting resistor of 10 Ohm is added. On the reciever side the real load equals also 10 Ohm. The coils are thought to have small resistance. The capacitors are chosen such that

$$\omega L_1 = \frac{1}{\omega C_1}$$

$$\omega L_2 = \frac{1}{\omega C_2}$$

a) Transmitted voltage |
b) Transmitted power |

Fig.3 SPICE simulation of wireless power transmission via coupled inductors |

Note that the transmitted voltage remains the same regardless of the resistance of load. For transmitted power, there exists an optimal value of load resistance.

One can enhance the transmission by

- having more turns in coils
- making coils of bigger size
- increasing the transmission frequency
- reducing coils separation
- introducing a ferromagnetic core, which links the coils