Burning the arduino

Official Hardware Nano Family Classic Nano
1

Hello,
I am trying to read the battery voltage level with an easy partition resistance method.

I am using this schematic:
Burn

It's very easy.
But every time I connect the battery (at this moment about 16V) the Arduino nano gets burning.

Code is very easy:

int sensorPin = A6;   // select the input pin for the potentiometer
int relayPIN = 1;      // select the pin for the LED
int VoltsValue = 0;  // variable to store the value coming from the sensor

void setup() {
  //pinMode(sensorPin, INPUT) ;  //this should be just optional: A6 in nano is only ADC
  pinMode(relayPIN, OUTPUT);
  Serial.begin(9600);
}

void loop() {
  // read the value from the sensor:
  int anaRead = analogRead(sensorPin);
  VoltsValue = anaRead/1024 * 3.3;
  Serial.print("A:");
  Serial.println(anaRead);
  Serial.print("V:");
  Serial.println(VoltsValue);
  delay(1000);
}

Can anyone explain why is nano burning even I read a voltage level lower than 3.3V before I connect the arduino??

2

No, what you described is not the cause of the Nano burning. There must be some other reason. Post a more complete schematic showing the Nano and the power supply/supplies.

3

Check the voltage divider with a voltmeter.

4

It's burned without anything else connected.... and not only 1 pcs, I burned 5 pcs! I could not believe it could be caused by the circuit, and after checking again everything I tried again and again. After 5 pcs... I asking for help here.

5

I did, without arduino connected I read 2.83V

6

I hope you did connect a DC voltage to voltage divider?

7

what do you mean? I connected as I posted the schematic. where do I make mistakes?

9

You know the difference between AC and DC?

An Arduino does not survive AC.

10

18V came from a battery. it's DC of course.

11

@AlessandroLecce
You shouldn't connect any voltages to the I/O pins when the Nano is off, it could damage the Nano.
First turn on the Nano then connect the battery.
Then disconnect the battery before turning the Nano off

12

Before you do what @paulpaulson asked in Post #2, I will go ahead and tell you that anaRead will always be zero here because analogRead returns a value between 0 and 1023 and the your division there is performed with integer arithmetic, yielding 0. Always.

Besides that, the formula itself is wrong, you will not calculate the voltage correctly.

Finally, VoltsValue is declared as an integer, and can only store …integer values.

And do not use Pins 1 for the relay if you plan to use Serial. Nor 0.

But, let’s first see how you are connecting the Nano

2 Likes
13

With signals less than rated maximum (5.5v for Atmega328) the chance to burn something is about to zero

14

@AlessandroLecce
Could you show a clear photos of your setup?

15

It's not the voltage but the current. 18V/18K = 1mA

16

I was thinking about it too, but I read until 40mA I shouldn't care...because Arduino can manage up to 40mA on ADC.

17

it's exactly the schematic ... and nothing else.

18

If you need a help - please provide the photos

19

Not while the Nano has no power.

20

I don't want to argue with you, but my experience contradicts what you say. I have a device that consists of a powerful consumer (LED panel) and a control unit with microcontroller. The panel and MCU both has its own power supplies.
I always turn on the screen first, and then the control unit - and turn it off in the reverse order.
I’ve been doing this for 5-6 years now - and so far not a single MCU has burned out.

1 Like
21

how much should I higher the resistors?? to be safe when off and able to read when turned on?