The split-type LVDT outputs 4~20mA – is a load resistor required?
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I’m seeking help. I currently have a domestic LVDT sensor, paired with a separate transmitter; it is powered by 24VDC and outputs a signal of 4–20mA, using a 4-wire connection. Connect the 24VDC power supply, and connect the signal output to a multimeter; at zero point, the multimeter should indicate a current of around 4.7mA, while the normal value should be 4mA. I consulted the manufacturer, who said that a 250Ω resistor needs to be connected in series with the signal + cable so that the output can function properly. What is the principle behind this?
Logically, it’s not necessary; after all, when you use a multimeter to measure current, the multimeter itself creates a current circuit. Since the manufacturer has given their opinion, why not try connecting a resistor in series?
I tried it by connecting resistors, and indeed the output worked normally, but I really don’t understand the principle behind it!
That proves that the loop resistance is indeed too low. Many of the output and input circuits are gate circuits; if the loop resistance is too low, it’s not possible to trigger the output. Of course, this doesn’t mean there is no output at all
A series resistor is used because the resistance of your transmitter is too low; since it falls within the lowest range of the output range available for that transmitter, it is necessary to increase the resistance. If there are multiple output levels to choose from, then a series resistor is not required! High school physics knowledge. . . .
The instrument has a load capacity indicator; it will deliver a proper output only within its specified load range.
Generally, the board connected to the LVDT comes with a potentiometer; a multimeter can be connected in series to it, and the potentiometer can be adjusted accordingly! If it’s Xinhua’s VCC card, electrical zeroing can also be done directly using the VCC software. I hope this helps!