I am working on a circuit to convert 4-20 mA signal to 0-5 V to finnaly read with an analog input of an MCU (PIC16F886). The IC I'm using to convert the signal is the RCV420K which works fine. On the documentation of the IC it says that it needs to have -1.25 V on the output at 0 mA so at 4 mA can be 0 V. I am reading the voltage signal with no problems but when I unplug the signal from the IC the voltage goes negative and I suspect it is going to be harmful for the MCU.

The thing is the IC outputs very little current (0.10 mA on my readings) but when the signal source is removed it reads -1.0 mA. Is this safe for my MCU? If not, Is there any way to ground the negative signal so it doesn't affect the MCU?

Thanks in advance!! I'll attach an schematic.

Wiring of RCV420, at VOUT5 its the 0-5 V. If there's no input signal, this will be -1.25V.

  • \$\begingroup\$ I'm not familiar with the device but converting from 4 - 20 mA to 1 to 5 V can be done with a 250 Ω resistor. Usually the 1 V offset can be removed in software if you're feeding an ADC or microcontroller. \$\endgroup\$ – Transistor Aug 7 at 21:47
  • \$\begingroup\$ I considered that but voltages which I'm going to read current from are on ranges from 12, 24, and/or -24 so I can't rely on making a conversion like that. \$\endgroup\$ – Joel Duran Aug 10 at 14:40

It's probably safe enough but datasheet says the current could be as much as -13mA which is less than the absolute maximum of 20mA for chip survival but could cause odd things to happen with the MCU.

You can simply add a Schottky diode such as BAT54 from the ADC input to ground, which solves the problem related to your question here.

You might also consider an RC filter and clamping the receiver output on the positive side, but that's outside the scope of this answer.

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  • \$\begingroup\$ A shunt diode is actually a good idea for maintaining acuracy, I'll see what happens. \$\endgroup\$ – Joel Duran Aug 10 at 14:43

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