## 3 Interference of Extra Data Transfer Channel on Power Transfer  
### Interference of Extra Data Transfer Channel on Power Transfer  
When considering the influence on power transfer due to the extra data transfer channel, two issues should be of concern. The first is the interference of data carrier transfer with power transfer; the second is the power transfer loss due to the data transfer circuit. However, since data is transferred when the switch \(S\) is off, so the interference of data carrier transfer on power transfer can be ignored. As for the power transfer loss due to the data transfer circuit, it can be monitored by the attenuation of output voltage \(U_{L}\) whether the data transfer circuit is added or not. The simplified power transfer circuit without and with the data receiver circuit are shown in Figure 6a,b respectively, where \(i_{i}\) is the equivalent input current source.  
Figure 5: Circuit of the two working modes: (**a**) \(S\) is on and \(S_{d}\) is off; (**b**) \(S_{d}\) is on and \(S\) is off.  
For the original system shown in Figure 6a, the reflected resistance to the primary side is given by:  
\[Z_{r}=\frac{\omega^{2}M^{2}}{Z_{s}} \tag{3}\]  
where \(Z_{s}\) is the secondary side loop impedance, given by \(Z_{s}=j\omega L_{s}+1/j\omega C_{s}+R_{eq}\). \(M\) is the mutual inductance.  
The primary inductance current \(i_{p1}\) can be derived as:  
\[i_{p1}=i_{i}\frac{(1/j\omega C_{p})}{Z_{p1}} \tag{4}\]  
where \(Z_{p1}\) is the impedance given by \(Z_{p1}=j\omega L_{p}+1/j\omega C_{p}+Z_{r}\).  
The output voltage can be given by:  
\[u_{eq1}=j\omega Mi_{p1}\frac{R_{eq}}{Z_{s}} \tag{5}\]  
The transfer function from the input current \(i_{i}\) to the output voltage \(u_{eq1}\) is:  
\[G_{pp1}(\omega)=\frac{u_{eq1}}{i_{i}}=\frac{MR_{eq}}{C_{p}\big{(}j\omega L_{p} +1/j\omega C_{p}+Z_{r}\big{)}Z_{s}} \tag{6}\]  
As mentioned before, the data carrier is injected to the secondary side only when switch \(S\) is off (i.e., power transmission is blocked), the data transmitter circuit does not affect the power transfer. As for the data receiver circuit added in the primary side shown in Figure 6b, the influence can be monitored. The primary inductance current \(i_{p2}\) can be rewritten as:  
\[i_{p2}=i_{i}\frac{(1/j\omega C_{p})}{Z_{p2}} \tag{7}\]  
Figure 6: The simplified power transfer circuit: (**a**) without data receiver circuit; (**b**) with data receiver circuit.  
where \(Z_{p2}\) is the impedance given by:  
\[Z_{p2}=j\omega L_{p}+\frac{1}{j\omega C_{p}}+Z_{r}+Z_{d} \tag{8}\]  
and \(Z_{d}\) is the impedance of the data receiver circuit, given by:  
\[Z_{d}=\frac{1}{j\omega C_{d}+1/j\omega L_{d}+1/R_{d}} \tag{9}\]  
\(Z_{p2}=j\omega L_{p}+1/j\omega C_{p}+Z_{r}+Z_{d}\), \(Z_{d}=1/(j\omega C_{d}+1/j\omega L_{d}+1/R_{d})\).  
So the output voltage can be given by:  
\[u_{eq2}=j\omega Mi_{p2}\frac{R_{eq}}{Z_{s}} \tag{10}\]  
The transfer function from the current input \(i_{i}\) to the output voltage \(u_{eq2}\) can be calculated as:  
\[G_{p\
eq 2}(\omega)=\frac{u_{eq2}}{i_{i}}=\frac{MR_{eq}}{C_{p}\big{(}j\omega L_{ p}+1/j\omega C_{p}+Z_{r}+Z_{d}\big{)}Z_{s}}. \tag{11}\]  
By comparing the Bode plots of Equations (6) and (11), the interference on power transfer due to the addition of the data transfer channel can be monitored, and this will be presented in the following section.