Posted on December 08, 2016 at 11:25To add info on:
1)The antenna circuit connected between TX1 and TX2 (including the tuning circuit + RX circuit + the antenna) receives a square voltage between VPS_TX* Zload/(Zload+Zout) and – VPS_TX*Zload/’Zload+Zout).
This signal is AC and no more DC. So when looking at voltage and current values from antenna circuit side, all calculation must be done considering AC signals.
At VPS_TX=3V, Zout = 27ohm which leads to a square voltage between +1.5V and -1.5V.
This signal is a square signal at 13.56MHz. The magnitude of the sinusoidal 13.56MHz component of this signal is given by 1.5V*4/pi=1.9V.
As a consequence, the AC current at 13.56MHz flowing into Zload is ITX1-TX2 = 1.9V/27=70mA. For RF calculations on antenna side, only signal part at 13.56MHz are considered.
2)
VPS_TX, DC current consumption includes all currents sink on TX1/TX2 (at 13.56MHz and all other harmonics) in addition to all additional current consumption due to the analog circuit itself. Table 31 of the datasheet shows that the typical measured DC current consumption on VPS_TX is 70mA while we guarantee that it will never exceed 100 mA (taking measurement variation and a guard band).
3)
The current driven to antenna depends on VPS_TX, the input impedance of the antenna matching circuit (“antenna tuning�) and the antenna impedance itself.
Current in antenna is obtained by dividing the antenna voltage by the antenna impedance. Since the antenna impedance is inductive, current calculated in 1) and current in antenna are not in phase and completely different.
The CR95HF EMI filter calculation sheet (
CR95HF EMI FILTER CALCULATION.xlsm)
provides a voltage graph showing the voltages calculated at different location of the antenna matching circuit. Antenna voltage is shown. Current in antenna can be calculated by dividing this voltage by the antenna impedance.
For instance, if you enter the CR95HF demo board component values and antenna parameters, voltage on antenna is around 8V peak, which lead to a current Iantenna= 8/(424nH*2*pi*13.56e6)= 221mA.
Regards