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We want to measure the information from three sensors that generate current (4mA for the minimum value and 20 mA for the maximum).

Obviously, a simple solution could be to use the classic current-to-voltage converter with an OP.

enter image description here

But if I want to do something a little more professional, I imagine that there is some dedicated integrated circuit that does this conversion.

Am I right?

Any comments or suggestions are welcome.

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    \$\begingroup\$ What sensors? Depending on that the solution might be as simple as placing a shunt resistor. What is "professional"? Perhaps you mean an off-the-shelf solution, but it is definitely does not guarantee any better quality of the solution. \$\endgroup\$ – Eugene Sh. Apr 19 at 16:41
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    \$\begingroup\$ Google 4-20ma current loop receiver maybe? \$\endgroup\$ – Dejvid_no1 Apr 19 at 16:43
  • \$\begingroup\$ Do the sensors need to take power from the signal loop? (It's a very common configuration in industrial sensors.) Do you have a specific voltage range that the 4-20mA signal needs to be translated to? In either case, the shunt resistor solution is still very common in industrial applications. \$\endgroup\$ – Theodore Apr 19 at 17:16
  • \$\begingroup\$ Whole books written on this topic. The most professional solution is the one that solves your problem best. For mA current and no specified bandwidth, that is probably a resistor. \$\endgroup\$ – user1850479 Apr 19 at 18:23
  • \$\begingroup\$ Don't forget the reason for 4 mA being the minimum. Its so you can detect broken wire faults. Example, if the sensor is controlling the temperature of an industrial furnace, you don't want a broken sensor wire to cause the heaters to run continuously, overheat, and burn down the building. Be sure to check for faults and shut down. \$\endgroup\$ – Mattman944 Apr 19 at 21:19
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I recommend the RCV420 chip.

enter image description here

And read the data sheet so that you understand what you can do with it.

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    \$\begingroup\$ There's really not enough information in the question to say that this level of complexity is necessary to make an adequate solution. \$\endgroup\$ – Theodore Apr 19 at 17:06
  • \$\begingroup\$ It is only a recommendation, I'm not saying that it is the solution. \$\endgroup\$ – user283383 Apr 19 at 18:52
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The standard way to convert 4-20 mA current loop to a voltage is with the precision current to voltage converter below:

schematic

simulate this circuit – Schematic created using CircuitLab

If you expect current to be drawn from the \$V_{out}\$ terminal, you can add an op-amp configured as a voltage follower. Use an op-amp with low offset voltage and low input current.

schematic

simulate this circuit

However, if you are feeding \$V_{out}\$ into an ADC (which is very common) and the ADC does not draw too much current, then the op-amp voltage follower may be both un-necessary, and un-necessarily degrade the accuracy of your measurement.

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  • \$\begingroup\$ It may work; but doesn't look professional enough; IMHO. \$\endgroup\$ – AJN Apr 19 at 17:28
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    \$\begingroup\$ 250.000000 Ohm. \$\endgroup\$ – tobalt Apr 19 at 17:57
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    \$\begingroup\$ @tobalt with gold leads. \$\endgroup\$ – Math Keeps Me Busy Apr 19 at 18:17
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    \$\begingroup\$ This idea I think would work much better with a differential amplifier measuring the voltage across the resistor \$\endgroup\$ – user3249244 Apr 19 at 20:14
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    \$\begingroup\$ @user3249244 Yes, you can add an op-amp configured as a voltage follower to the output. Use an op-amp with low offset voltage and input current for accuracy. \$\endgroup\$ – Math Keeps Me Busy Apr 19 at 20:15
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It may work; but doesn't look professional enough; IMHO. – AJN

It does to me.

This idea I think would work much better with a differential amplifier measuring the voltage across the resistor – user3249244

Don't think so. A diff amp will introduce an input offset voltage error. And, its output still will be referenced to the ground reference potential of the input. Both of these are true with the voltage follower. Other than making the negative feedback path a 250 ohm resistor to reduce bias current error, the follower should work fine with careful component selection and attention to wiring details.

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