0

I am trying to display the battery level as it is being used to power the arduino.

I have tried powering the arduino through the USB and the battery voltage is read accurately, when compared to the multimeter reading. But when the battery is powering the arduino and being read, it keeps displaying 5.00V, when it should not be.

Just wondering what might be happening and how I can go about resolving this issue.

I have been trying to research some answers, and came across something about an 1.1V reference voltage, but not entirely sure what they mean.

Would appreciate any answers! Thanks!

Here is my code for reference.

#include <SoftwareSerial.h>
SoftwareSerial BTserial(10,11);

int battPin = A1;

void setup() {
  pinMode(battPin, INPUT);
  BTserial.begin(9600);
  Serial.begin(9600);
}
void loop() 
{
  int voltReading = analogRead(battPin);
  float volts = (voltReading/204.6);
  BTserial.print(volts);
  BTserial.print(";"); 
  Serial.println(volts);
}

I am using a 3.7V Lipo battery. It is connected to A1 and also Vin.

2

AFAIK, Arduinos are set up to use their operating voltage as the analogRead reference voltage by default.

So, what I think is happening is that you are trying to measure the battery voltage with the battery itself as a reference voltage.

This will, of course, always show "full scale".

You could use a voltage divider to bring the measured battery voltage down to about 1V (max) and use the internal reference of 1.1V for measuring.

Note that the 1.1V internal reference is stable, but not necessarily exactly 1.1V.

| improve this answer | |
0

You need to divide down the battery voltage into the 1.1V max range, and use the Internal bandgap voltage reference for the ADC conversions.

Connect the Battery to VCC, not Vin, you don't want it going thru the regulator.

| improve this answer | |
  • By "divide down the battery voltage" do you mean with circuitry, such as a voltage divider, or through the code? And what is the internal bandgap voltage reference? – acho Dec 23 '19 at 17:44
  • Yes, two resistors, in series between VCC and GND. Connect Ax to the junction. See the datasheet for more info on the internal source. – CrossRoads Dec 23 '19 at 18:23
  • connecting the battery to the VCC, means connect to the 5V pin? Is that okay? I read that if you connect the USB while battery is connected to the 5V pin, it could damage stuff. – acho Dec 23 '19 at 21:17
  • Maybe. 3.7V into the barrel jack will be overcome by the USB input and the power switching circuit, and USB will be used. – CrossRoads Dec 23 '19 at 23:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.