Within various chapters of my book, I have spoken to voltage drop on a cable. I provided examples of voltage drop from the feedback of an actuator. In that example, I gave the voltage drop expected across the cable to understand the difference between the actuator position and what the controller registered as the actuator position. Elsewhere in the book, I discussed how to calculate the voltage drop across a cable using Ohms law. What was missing was an example bringing these two examples together for a complete example on how to calculate the input to the controller based on the voltage drop across a cable, by calculating everything from first principles. Here is that example.
A differential pressure transmitter is powered by 24Vdc, providing an output of 0-10Vdc for a range of 0-4Bar. The sensor is connected through a run of cable 400 meters long, with a resistance of 30.5 Ω/km, to an input on a controller with an input impedance of 5kΩ. To start let me draw out the circuit:
This circuit can be updated to a traditional electronic circuit as follows:
For a single core of the cable, the resistance is calculated as:
Now, we know the cable resistance and the resistance of the input, the circuit resistance can now be added:
The worst-case scenario for voltage drop will exist at the high end of the sensor range of 10Vdc. Our example will be based on this but you could equally base it on the maximum voltage you expect the transmitter to output. Using Ohms Law we can equate voltage, resistance and current.
We know the voltage of the circuit shall be 10Vdc and the circuit resistance is:
Current through the circuit, from Ohms Law, shall be:
Now that we have the current and we know the cable resistance we can calculate the voltage drop on the cable:
Therefore, using a cable with a resistance of 30.5 Ω/km to connect a water differential pressure transmitter providing 0-10Vdc output for a range of 0-4Bar connected to an input, 400 meters away, gives a voltage drop of 0.048Vdc at full range.
BMS Explained 