michinyon:
The phrase "5 volt rectified AC voltage" is gibberish.
If you have rectified it, it isn't an AC voltage. The electron flow is no longer alternating in direction.
You will expect to see a DC voltage with a voltage ripple. Which is exactly what you are getting.
Depending on the type of rectifier, you will see 65 Hz or 130 Hz ripple. I can't tell from your
sampling rate, exactly which you are getting.
Your DC voltage has a peak of 5 volts and a minimum of about 4.5 volts. That is exactly what you
are getting.
The input to the arduino is 0 to 5volts ac I dont understand why you think its dc. I have attached a print screen of the input to the arduino
It is full wave rectification without a filter capacitor.
Without any load resistor the sample and hold capacitor on the arduino input will act as a smoothing capacitor.
Put a 10K load ( resistor) between the analogue input and ground.
You are supposed to sample at twice the frequency, in this case: 7.65 mS, although because of the full-wave rectification I suspect half that, namely 3.825 mS.
In your other thread on this subject, which I referred to in reply #1, I suggested buffering readings and then printing them out.
Why didn't you do this?
Learning:
Unplug it! It'll go to zero. (or pretty close) 8)
No it won't it will float which is why I suggested you need a load resistor. So why have you not tried this.
Better still put the scope lead on the analogue input and run the code again. Is it different than when the scope is disconnected.
const int analogInPin = A0;
int inputReading [numReadings];
void setup()
{
Serial.begin(115200);
}
void loop()
{
for (int i = 0; i < numReadings; i++) {
inputReading [i]= analogRead(analogInPin); // roughly 9kHz sampling
}
for (int i = 0; i < numReadings; i++) {
float voltage = ((inputReading[i]/1023.0)*5.0);
Serial.println(voltage);
}
}
I tried connecting a 10k resistor as someone suggested. And it WORKED !. I uploaded your code for fun and it also works. Its a bit better than mine....