Does the Arduino clock frequency affect analogRead()?

Projects Programming
1

I keep getting the wrong voltage. It should be about 2.11 volts but it shows .42 volts. I am using the 8MHz internal clock on the atmega328 non-picopower model. What is going on?

EDIT:

Here's my code:

// Configuration:
// 5v------10k-------Sensor------GND
//               |
//            Analog

void setup()
{
  Serial.begin(9600*2);  //Start the serial connection with the computer
}

float getTemp() {
  float aread = analogRead(0) / 1024.0;           // calculate voltage
  Serial.println(aread);                                   // print voltage
  float resistance = 330.0 / (5.0 / aread - 1.0); // calculate resistance of sensor
  Serial.println(resistance);                             // print resistance
  float temp = 46 / pow(resistance, .40984);     // calculate temperature with resistance
  Serial.println(temp);                                    // print temp
  return temp;
}
 
void loop()
{
  getTemp();
  delay(1000);
}
2

Code ? Schematics ?

3

Why not just tell the compiler you're using a clock rate of 8MHz? Check out the tools->board menu and select the pro with 8MHz

4

@WizenedEE
That doesn't work. It just gives the same numbers.

5 
float getTemp() {
  return temp;
}
 
void loop()
{
  getTemp();

The getTemp() function returns a value that you just throw away. Why?

6

@PaulS - It would be used to change the speed of a motor via PWM, but I'm just trying to get the correct temperature right now.

7

Hi,
are you sure the equation for calculating the resistance is correct? From the schematics in your code:

// Configuration:
// 5v------10k-------Sensor------GND
//               |
//            Analog

it should be:

float aread = 5.0 * (analogRead(0) / 1024.0) // voltage in Volts
float resistance = (10000 * aread) / (5.0 - aread); // resistance in ohm
8

@ea123 - I figured this out already, but thanks anyway.

By the way:

10000 / (5 / aread - 1) = 10000 * aread / (5 - aread)