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TLDR: given a sensing device that produces a varying resistance, how does a "sensor" change this into a 4-20 mA current flow from a 24 V supply?

Various sensors (flow, fluid depth, ...) use a 4-20 mA output to convey their measurements: you put +24V into one of the two sensor wires, feed the other into, say, a 500 Ω resistor, and then connect that to ground:

+24 --- SENSOR --- R500 --- GND
                |        |
                |        |
                A        B

and then you can measure the voltage between A and B. If the current is 4 mA, you get 2 V; if the current is 20 mA, you get 10 V. You can then use this voltage to drive further circuits, etc.

I'd like to understand what's in the "SENSOR" part of this diagram. Because whatever it is has to be powered by a voltage-drop somewhere between 22 and 14 V, and I'm not used to thinking about circuits that don't have some sort of fixed supply voltage.

Can someone sketch out how this is done, assuming for the sake of illustration that the actual sensing device in this situation acts as a variable resistor, or point me to a reference?** [My knowledge-level is about "midway through a first semester electronics course", except that I know a lot more about vacuum tubes than most students, because of my age.]

** Yes, I've googled. I get a million hits for places that I can buy 4-20 mA sensors, but not much on their internals.

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Typically, a 4-20 mA transmitter that is connected to a variable resistance sensor has a voltage regulator with very low quiescent current. The regulated output of this voltage regulator is used to power the variable resistance sensor. Usually, the current through the sensor itself is not enough to sink 4 to 20 mA, so usually there is a dependent current sink in the transmitter that will sink 4 to 20 mA depending upon the current through the variable resistance sensor. This current sink provides the (the bulk of) the 4-20 mA signal.

Here is an example of a 4-20 mA current loop transmitter circuit. I provide it for illustration purposes. There are many more such ICs that are provided by Texas Instruments and other manufacturers.

A rough schematic for the chip within a typical application may be useful for understanding the description that I gave above.

enter image description here

The schematic shows the regulator, whose output is Vreg. Vreg might be connected to a voltage divider consisting of a fixed resistor and the variable resistance sensor. The output of that voltage divider might be fed into the terminals of "Vin". The amplifier, together with the transistors (both internal and external) plus the various resistors form a dependent current sink, which is dependent on the voltage across the "Vin".

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    \$\begingroup\$ Ah...lovely. So a circuit like this would work with a V+ (on pin 7) that doesn't really need to be 24V --- just enough to provide the rails for the op-amp and to make the 5V regulator work, although with a lower-than-24V input, the voltage across R_L would need to be measured relative to that supply voltage. Thanks very much! \$\endgroup\$
    – John
    Jul 16, 2023 at 15:09
  • \$\begingroup\$ The voltage drop between pin 7 and pin 4 does not need to be 24V -- just enough to provide the rails for the op-am and to make the 5V regulator work. But you still need headroom voltage in your supply to cover the voltage drop across the receiver's sense resistor, typically 250 ohms. Of course, you can use a smaller resistor, but 250 ohms is quite common. \$\endgroup\$ Jul 16, 2023 at 15:16
  • \$\begingroup\$ Thanks ... good point. I've got an actual pressure sensor I'm working with, and it works beautifully at 24V; I'm going to try it out at some lower voltages today. :) \$\endgroup\$
    – John
    Jul 16, 2023 at 15:18

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