if you have as in your exsample a permited voltage drop of 7 volt by max sensor output but your voltage drop is max 5V over the cable so there's still 2V left open
you have to adjust this on your power supply or put another resistor so that there is a voltage drop over the "compensation" resistor from 2V
do i understand that good?
Robert-Jan
hi,
No, another resistor is not required.
Consider the 'sensor' has a quoted voltage operating specification of 12Vdc thru 24Vdc.
This means the 'sensor' has an inbuilt voltage regulator that can handle 12Vdc thru 24Vdc.
So any 'excess' voltage appearing at the 'sensor' is taken care of by the internal regulator.
Think of the 'sensor' as a LM317 connected as a Constant Current source,say set for 20mA.
If you now connected the LM317 to a variable psu, the constant current would remain
the same over a wide range of input voltages.
Consider now that the resistor thats sets the constant current in the LM317 is able to change, acting as the 'sensor element' then the constant current would vary.
You could measure this change in current by adding a series resistor, say 250R and measuring the voltage across the resistor.
Another important point about 4 to 20mA loops, is the current will still remain the same even if the 250R is now 0R.
If its open circuit, the sensor will not work.