I've connected my ADS1115 to an Arduino Uno as in the diagram (Vdd - 5V, Gnd - Gnd, SDA/SCL - SDA/SCL), all powered from the USB on a PC, with A0-3 unconnected, and am getting readings of around 3000 (i.e. 3000/65535 on 5V supply = 0.23V) from all four analog pins. Then if I connect them to 3.3V output on the arduino they give me only 17625 (i.e. 1.34V) i.e. about 40% of the full range I expected. If I try with a seperate 5V supply, it's still the same problem. Why is this?

P.S. sorry the code format below isn't very good, but basically it reads the four pins by
int16_t V1 = ads.readADC_SingleEnded(PIN).

enter image description here

#define SHELLPTPIN 2
#define SPAREPIN 3
#include < Wire.h >
#include < Adafruit_ADS1015.h >

Adafruit_ADS1115 ads(0x48);  //16 bit ADC add-on for arduino for better temperature resolution
float Voltage = 0.0;

void setup(void)

void loop(void)
    int16_t V1;
    int16_t V2;
    int16_t V3;
    int16_t V4;
    long cumV1;
    long cumV2;
    long cumV3;
    long cumV4;
    float avgV1;
    float avgV2;
    float avgV3;
    float avgV4;

    //take readings of voltages between resistor and thermsitor
    int i=0;

    while(i<6) {
        V1 = ads.readADC_SingleEnded(THERMISTORPIN1); //5 readings to allow the arduino capacitors to equalize with the pin voltage, to increase accuracy
        V2 = ads.readADC_SingleEnded(THERMISTORPIN2);
        V3 = ads.readADC_SingleEnded(SHELLPTPIN);
        V4 = ads.readADC_SingleEnded(SPAREPIN);

    while(6<=i && i<16) {
        cumV1+= ads.readADC_SingleEnded(THERMISTORPIN1);  //multiple measurements from which to take an average, to increase precision
        cumV2+= ads.readADC_SingleEnded(THERMISTORPIN2);
        cumV3+= ads.readADC_SingleEnded(SHELLPTPIN);
        cumV4+= ads.readADC_SingleEnded(SPAREPIN);


    //print out information to serial monitor on laptop
    Serial.print(" ");
    Serial.print(" ");
    Serial.print(" ");
    Serial.print(" ");
    Serial.print(" ");
    Serial.print(" ");
    Serial.print(" ");
  • first off, single ended readings are only 15 bit. secondly, providing 3.3v will limit the measurement to 3.3v; 66% of 5v.
    – dandavis
    Feb 15, 2018 at 12:34

1 Answer 1


Your readings are correct because you have the default gain set. You have incorrectly assumed that you can convert the ADC reading from your 16-bit ADC to Volts by dividing by 2^16 - 1. Your ADC has gain settings to have more accurate readings for certain voltage intervals. If we look at the example sketch, we see

 void setup(void) 

  Serial.println("Getting single-ended readings from AIN0..3");
  Serial.println("ADC Range: +/- 6.144V (1 bit = 3mV/ADS1015, 0.1875mV/ADS1115)");

  // The ADC input range (or gain) can be changed via the following
  // functions, but be careful never to exceed VDD +0.3V max, or to
  // exceed the upper and lower limits if you adjust the input range!
  // Setting these values incorrectly may destroy your ADC!
  //                                                                ADS1015  ADS1115
  //                                                                -------  -------
  // ads.setGain(GAIN_TWOTHIRDS);  // 2/3x gain +/- 6.144V  1 bit = 3mV      0.1875mV (default)
  // ads.setGain(GAIN_ONE);        // 1x gain   +/- 4.096V  1 bit = 2mV      0.125mV
  // ads.setGain(GAIN_TWO);        // 2x gain   +/- 2.048V  1 bit = 1mV      0.0625mV
  // ads.setGain(GAIN_FOUR);       // 4x gain   +/- 1.024V  1 bit = 0.5mV    0.03125mV
  // ads.setGain(GAIN_EIGHT);      // 8x gain   +/- 0.512V  1 bit = 0.25mV   0.015625mV
  // ads.setGain(GAIN_SIXTEEN);    // 16x gain  +/- 0.256V  1 bit = 0.125mV  0.0078125mV


Your ADC has a default gain of 2/3, or 0.1875mV per count. You say when you connect 3.3V to the pin, you get a reading of 17625. Thus we have

17625 * 0.1875 / 1000.0 [V]
= 3.3046875 [V]

Which is a correct reading. A reading on a floating pin (nothing attached) is irrelevant. Use the correct factors from above to convert your read ADC value to voltage.

Side note: In line


You have the float avgV1 on the left and cumV1 (a long) divided by 10 on the right; this will do a integer division. Make sure to divide by 10.0f if you want a floating point average.

Side note 2:

These factors do make sense. If you apply positive 6.144V on the 2/3rd gain setting, you will read the maximum positive value of a int16_t, i.e. 2^15-1 = 32767 (solve above formula for the count). If you apply -6.144V, you get -32768; thus the full 16-bit number space is used. On differnt gain settings it analogous.

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