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I am trying to use a pressure sensor which is MPX2100DP. Datasheet is; https://www.nxp.com/docs/en/data-sheet/MPX2100.pdf

My goal is to measure the air pressure inside a pipe. First of all, I tried it alone. 1st = GND, 2nd = A1, 3rd = +12V. (I connected GND to power supply and arduino.) Whether I close the pipe or not, the result is 1023.. Also, I tried plug the hose into the other port. There arent any difference. Then, I tried to use a differential amplifier to compare the normal air with the air in the pipe and increase the value, but the values ​​​​reflected unchanged at the opamp output. I used the instrumentation amplifier AD620AN, thinking that it is lost because of the low signal, but I still could not get any results. Where is my fault? Someone could advice me? I will convert voltage to pressure but firstly ı want to solve the problem of value.

enter image description here

Arduino code is ;

Serial.begin(9600);
}
void loop() {
 float sensorValue = analogRead(A1);
 Serial.print("A/D is:    ");
 Serial.println(sensorValue);
 delay(750);
}

With AD620AN; enter image description here

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    \$\begingroup\$ Please, post a proper schematic. If you don't have a schematic in some CAD program, you can even draw it in paint (try to make it readable). Also, how many volts do you actually have on chip output? Measure with multimeter please. \$\endgroup\$
    – Ilya
    Feb 15, 2022 at 14:35
  • 2
    \$\begingroup\$ The ADC result of 1023 indicates that you are reading a very high voltage. Try measuring the sensor output voltage with a multimeter first. \$\endgroup\$
    – Klas-Kenny
    Feb 15, 2022 at 15:54
  • \$\begingroup\$ How did you connect the AD620? Power supplies? \$\endgroup\$
    – Antonio51
    Feb 15, 2022 at 18:58
  • \$\begingroup\$ I hope my drawing was self explanatory. I connected the output directly to the arduino and got 1023, that is 5V, as an analog value. I will measure with a multimeter and reply again,thank you. \$\endgroup\$
    – zehra
    Feb 15, 2022 at 20:12
  • \$\begingroup\$ I added a new drawing circuit with the AD620AN. I was using +12V power supply. I used 330 ohm and my gain was 150, thank you for your reply. \$\endgroup\$
    – zehra
    Feb 15, 2022 at 20:38

2 Answers 2

Reset to default
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Did you use the AD620 correctly?
Note that "common-mode" voltage is eliminated for output.

Here is enter image description herewhat you should get ...

Be careful if Gain and Vref are not well chosen.
The output voltage will be higher than ADC Arduino full range.
Don't forget decoupling capacitor of AD620, nearest supply pins.

enter image description here

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  • \$\begingroup\$ Oh, I think ı used the AD620AN wrong.. Ok, ı will search your reply thank you sir. \$\endgroup\$
    – zehra
    Feb 15, 2022 at 20:41
  • \$\begingroup\$ I edited my circuit. i designed a voltage reducer for vref=2.5V and the gain is 50 with 1k ohms. My results are: Vout is 3.1V on the multimeter. While the value is 616 on the arduino when the hose is open, the value is 400 when i turn it off. i think these are the correct values. Now i am going to convert the voltage to pressure, thank you for your reply and helps. My circuit diagram is; link \$\endgroup\$
    – zehra
    Feb 17, 2022 at 9:10
  • \$\begingroup\$ is the link i just added? \$\endgroup\$
    – zehra
    Feb 17, 2022 at 10:41
  • \$\begingroup\$ Yes. Norton gives a warning in red... No worries, make an answer at your own question and add the picture ... \$\endgroup\$
    – Antonio51
    Feb 17, 2022 at 10:43
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This is my last circuit and i get correct values. (batteries represent power supply.)

Note: Actually I didn't understand where I should place the decoupling capacitors. I tried to put it on pin 2 and pin 3 of AD620AN but got the same values.

enter image description here

And it's my arduino code. There is a conversion of voltage to pressure.

int offset =0; 
int pressure; 

void setup() {
Serial.begin(9600);
}

void loop() {
  int Value = 0;
  float  Vout = analogRead(A1);
  Value = Value + analogRead(A1);

 
  float Voltage = Vout /1023 *5;
  pressure = ( (Vout - offset) / 0.0004 ) / pow(10,3) ; 

 
  Serial.print("Vout: ");
  Serial.print(Vout);
 
  Serial.print("     Voltage: ");
  Serial.print(Voltage);
 
  Serial.print("     Pressure " );
  Serial.print(pressure,1);
  Serial.println(" kPa");
  delay(750);
}




//When hose is open; Vout: 620.00     Voltage: 3.03     Pressure 1550 kPa
//When hose is closed; Vout: 420.00     Voltage: 2.05     Pressure 1050 kPa
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  • \$\begingroup\$ Decoupling is a precaution against some noise generated by supplies and characteristics of sensible opamps are "stabilized" ... Much less trouble. \$\endgroup\$
    – Antonio51
    Feb 18, 2022 at 7:42
  • \$\begingroup\$ ok sir ı got it, thanks again.. \$\endgroup\$
    – zehra
    Feb 18, 2022 at 10:07

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