The minimum drive requirement follows from the minimum receiver noise.
In case of an integrating receiver we minimally need:
\begin{equation} A_{R}=0.002412 \end{equation}The system gain $A_v$ is obtained as:
\begin{equation} Gain=- \frac{0.001135}{B_{T}} \end{equation}Its required value follows from the specifications:
\begin{equation} Gain=0.25 \end{equation}The coefficient $B_T$ of the transmittion matrix thus needs to be:
From this we can obtain the maximum value $B_{Tmax}$ of $B_T$:
\begin{equation} B_{Tmax}=0.004542\,\left[ \mathrm{\Omega}\right] \end{equation}From the above, we find the minimum peak value $I_{oMin}$ of the drive current:
\begin{equation} Io_{min}=220.2 \end{equation}The voltage drive requirement of the transmitter is found by multiplying the peak voltage of the source with the voltage gain from the source to the voltage across the transmit coil, at the largest full-power frequency and add it to the input voltage,
\begin{equation} V_{TRcoil}=\frac{1.414 V_{i} \left|{L_{r}^{0.5} L_{s}^{0.5} k_{c}}\right|}{\left|{A_{R}}\right| \left|{B_{T}}\right|} \end{equation} \begin{equation} V_{TRcoil}=0.3536 \end{equation}At the highest frequency the voltage across the coil has 90 degrees phase shift with respect to the input voltage. The total voltage $V_{TRp}$ at the tramnsmitter output thus equals:
\begin{equation} V_{TRp}=1.061 \end{equation}The voltage slew rate $SR_{vT}$ of the transadmittance transmittance amplifier should minimally be:
\begin{equation} SR_{vT}=2.121 \pi f_{fpl} \end{equation}After substitution of all the known variables, we obtain:
\begin{equation} SR_{vT}=3.332 \cdot 10^{4} \end{equation}The peak input voltage $V_{rec}$ of the receiver is found by taking the voltage at node 'o' as detector voltage.
\begin{equation} V_{rec}=\frac{2.828 \pi V_{i} \left|{L_{r}^{0.5} L_{s}^{0.5} f_{fpl} k_{c}}\right|}{\left|{B_{T}}\right|} \end{equation} \begin{equation} V_{rec}=26.79 \end{equation}The voltage drive capability of the receiver is specified in the application description ($V_o$).
The voltage slew-rate of the receiver amplifier should minimally be:
\begin{equation} SR_{vR}=2 \pi V_{o} f_{fpl} \end{equation}After substitution of all the known variables, we obtain:
\begin{equation} SR_{vR}=7854 \end{equation}The current-drive capability $I_L$ of the receiver should minimally be:
\begin{equation} I_{L}=\frac{V_{o}}{R_{\ell}} \end{equation}After substitution of all the known variables, we obtain:
\begin{equation} I_{L}=0.000125 \end{equation}Go to gainDistribution_index
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Last project update: 2023-08-04 17:10:25