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Hello I have a circuit that takes a 4-20mA reading from a pressure transmitter and converts it to analog 0-5V then to I2C. But it doesn't seem to be reading I just see about 0.5V on the serial monitor regardless of if the transmitter is connected to my PCB or not.

So I believe the circuit has an error somewhere, but because it is relatively complicated I am not sure where to begin. The transmitter is connected to the board via screw terminals and has 12V+,12V-, GND connections its current signal is converted to 0-5V using an IC RCV420JP next the analog signal is taken to an ADS1115 ADC which converts the reading to I2C. Because my MCU's logic levels are at 3.3V though I translate the 5V digital readings to 3.3V using a voltage translator TXS0108EPWR. The I2C signals are then sent to the MCU ESP32-WROOM-32UE

I have looked at this a little bit and not been able to spot the error. Does anyone have a clue as to what may be the issue?

circuit1 circuit2 circuit3 circuit4

Edit Some have suggested that I should connect to both the In- and In+ as I am currently connected to only the In+. I used figure1 and 8 in the RCV420JP datasheet and there doesn't appear to be a connection on In-. After some time, I inquired to Texas Instruments and received a very thorough answer and reference circuit: TI answer figure1 from datasheet figure8 from datasheet

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  • \$\begingroup\$ have you measured anything with a digital multimeter? \$\endgroup\$
    – vicatcu
    Oct 24, 2021 at 1:44
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    \$\begingroup\$ I just checked and all of the ICs have supply voltages, and noticed that the analog voltage was about 8V for some reason, which is too high. My transmitter goes from 4-20mA at 0 psig to 30 psig and my system is around 0 psig. So it seems like there is a problem with the RCV420JP circuit \$\endgroup\$
    – Feynman137
    Oct 24, 2021 at 2:09
  • \$\begingroup\$ The input is a current loop and you don't appear to have one. Look at the data sheet for the RCV420Jp. \$\endgroup\$
    – Jack Creasey
    Oct 24, 2021 at 3:36
  • \$\begingroup\$ @Jack Creasy see the screw terminal in the first circuit (J301) the 12V is the supply voltage to the pressure transmitter. The “PressureA420” is the current output from the transmitter. This is the current loop. \$\endgroup\$
    – Feynman137
    Oct 24, 2021 at 3:54
  • \$\begingroup\$ @Feynman137 ....I see the 12V and GND that supply the sensor ....but see only one signal wire being used. The 4-20mA signal should be able to float anywhere between the supply and ground ...and on the RX IC you need to use +in and -in. You have no current loop. \$\endgroup\$
    – Jack Creasey
    Oct 24, 2021 at 4:10

1 Answer 1

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After skimming through the datasheet I am under the impression that the RCV420JP requires a bipolar supply, that is, V- should be at least -5 V, not ground.

I think you might be able to replace the entire converter part of the circuit with a load resistor between GND and PressureA420, and measure the voltage drop across it with the ADC. This should work if your sensor is the only thing on the 4-20 mA loop and is located close to the ADC in a reasonably noise free environment. You might even be able to do away with the level shifting and run the ADC at 3.3V.

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  • \$\begingroup\$ See figure 8 in the datasheet, which was one of the reference circuits I used for my board. It shows the V- being connected to 12V-. Which on my board is connected to GND. As for the comment about replacing the IC with a resistor this does seem possible but I'd like to make the IC work if possible and keep the output at 5V which gives me a good analog range. \$\endgroup\$
    – Feynman137
    Oct 25, 2021 at 1:25
  • \$\begingroup\$ @Feynman137 precisely, I see no indication that the chip can work with a single 12V supply. As for the range, what makes you think 3.3V would be bad? \$\endgroup\$
    – DamienD
    Oct 25, 2021 at 7:17
  • \$\begingroup\$ Just to be clear, what I read on Figure 8 is that there is 24 V between V+ and V-, and 12V between CT and V-. Your circuit has 12V and 0V, respectively. \$\endgroup\$
    – DamienD
    Oct 25, 2021 at 7:26
  • \$\begingroup\$ It's not 12V-, it really is minus twelve volts. \$\endgroup\$
    – DamienD
    Oct 25, 2021 at 7:48
  • \$\begingroup\$ I believe you were correct about the bipolar supply. I inquired to TI today on their forum and I got an answer. \$\endgroup\$
    – Feynman137
    Oct 26, 2021 at 2:40

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