Frequently Asked Question

4-20 mA current output Sensor
Last Updated 3 years ago

The sensor draws current from its power source in direct proportion to the angle/acceleration it measures.
Take QG40N-KIXv-360-AI-CM-UL as an example. At -180°, the sensor draws 4 mA from its power source. At +180° the sensor draws 20 mA. At 0° the sensor draws 12 mA and so on.


The relationship of angle measurement to current output is almost always linear, allowing the resulting current loop data to be scaled with a simple Iout = 12 + 8*(α/180) [mA] formula.

you'll make sure your supply is possible to cover at least:
1. maximum current consumption of the sensor: 30mA.
2. current loop: 4-20mA.
3. voltage drop on the wire.
Usually with 10-30V dc, you should have no problem at all.


Output load
No instruments measure current directly. They all do it indirectly by measuring the voltage dropped across a resistor of known value, and then they use Ohm’s Law to calculate actual current. The resistor is referred to as a “Output load”.


You also need to ensure that it doesn’t burden the loop, so lower values are better than higher. Because we are working with current, we can take the load all the way to zero Ohms for a current source without consequence. Our industry has standardized on a value of 250 Ohms.

3-wire sensors
Our current output sensors are 3-wire sensors, which means you should connect the load between the output and ground. 3-wire sensors with a process current output have a separate wire for ground, signal (4-20 mA), and the power supply. This configuration is the easiest for current loop beginners to grasp, one input for power and a second for the current loop with a common ground(at the same potential voltage).


The primary advantage of a 3-wire sensor over its 2-wire counterpart is its ability to drive higher resistive loads. Resistors drop voltage for any given current in direct proportion to their resistance value. Holding current constant, higher resistances drop more voltage. Turning back to the 2-wire sensor and holding current constant, as the shunt resistance increases the voltage drop across it also increases. You might reach a point where the voltage dropped by the shunt lowers the voltage drop across the sensor below the minimum required for it to operate properly.


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