Answer:

The particularity of the power supply for sensors equipped with a 4-20mA (2-wires) amplifier (Option C or C6) is that it depends on the value of the load resistance of the reading instrument (R_L) increased by the resistance of the wiring (R_W).

Indeed, for a correct operation, the power supplyat the terminals of the amplifier (V) must be at least 9 to 28VDC. However V = V_S - I x (R_L + R_W) with I being able to reach 20 mA.

Consequently, Vs must be greater than or equal to 9 + (R_L + R_W) x 0.02.

Examples:

1.  Input resistance of the reading device RL: 33 Ohms Wiring resistance Rw: 4 Ohm. To be able to generate a 20mA signal, the supply voltage must be at least: 9V + ((33+4)*0.02) = 9.74V

    2.  With R_L + R_W = 1 kΩ (abnormally high value), V_S must be at least: 9 + 20 = 29 V to be able to generate a 20mA signal with the specified linearity.

Likewise, for a specified supply voltage, the value of the sum of the load resistors will be limited to (V_S - 9) / 0.02.

Example: With V_S = 24 V, the value of R_L + R_W may not exceed 750 Ω in order to generate a 20mA signal with the specified linearity.