# Problems with voltage measurement from voltmeter program

Hi! I’ve some problem with voltage measurement…
I’m measuring the output voltage of the first amplifier (circuit attached) with a multimeter and with a voltmeter (code below). I’m looking for a 0.00V, which I think is ok by the fact that is an inverse output and I’m giving positive input voltage to the amplifier.
Using multimeter I find 0.0V; instead with the voltmeter I find 0.01 V…
This is a little difference that changes things a lot, because at the end of the circuit I want to observe millivolts variations.
Could you help me with that?
Thank you!

``````float IRPin=12;
float RedPin=11;
void setup() {

Serial.begin(9600);
pinMode(IRPin, OUTPUT);
pinMode(RedPin, OUTPUT);
digitalWrite(IRPin, HIGH);
digitalWrite(RedPin, LOW);

}

void loop() {

Serial.println(v0);
delay(20);
}
``````

to abserve millivolts you would need to amplify the voltage or better using a 16bit external ADC.

Or you can use an external reduced reference voltage.
With standard internal reference voltage the resoultion of the Arduino-onboard ADC is 10 bit = 2^10 = 1023

5V / 1023 = 0,004778V = 4,778 mV

best regards Stefan

Bear in mind that any digital system only resolves to +- the last digit.

If it reads “0.00” it could be 0.01 or -0.01.

Accuracy and repeatability are further issues.

In your code using float may give you the impression of higher resolution, but it doesn’t and you may get misleading results. ( you only have 1023 steps in the A/D output).

hammy:
( you only have 1023 steps in the A/D output).

No, it was changed to 4096 with 'analogReadResolution(12);'. Notice that I say 4096 not 4095, since accuracy was mentioned. So it should be:

``````  float v0= (analogRead(A2)*3.3)/4096;  //3.3 Vin
``````

What is the ADC result before the math? If > 0, try subtracting that amount before the math.