Wrong analogRead values

I am trying to measure the voltage of an 5volt AC rectified voltage at 65Hz. Using the UNO R3 Below is the current code.

const int numReadings = 25;
int ac_input[numReadings];
const int analogInPin = A0;//Pin for input for AC voltage      
float ac_output = 0;    
int index = 0;

void setup() 

{
  Serial.begin(9600); 
  pinMode(A0, INPUT);
  for (int thisReading = 0; thisReading < numReadings; thisReading ++)
    ac_input[thisReading] = 0;
}

void loop()

{
  ac_input[index]= analogRead(analogInPin);
  ac_output = ((ac_input[index]/1023.0)*5.0);                             
  Serial.println(ac_output);  
  delay(2);
  index = index + 1;

  if (index >=numReadings)
    index =0;

}

The problem is that the output on the serial monitor only shows the peak values of the voltage. That is from 4-5v instead of zero.

5
4.98
5
5
4.95
4.62
4.27
4.26
5
4.93
4.7
4.58
5
4.79
5
5
4.35
4.53
5
5
5
4.91
4.84
4.52
5
5
4.67
4.54
5
4.69
5
5
4.43
4.32
5
4.92
5
5

I have tried changing the board I'm using aswell as the input pin.

Has anyone experienced anything similar ?

Have a look at this thread

Please use code tags when posting code.

for (int thisReading = 0; thisReading < numReadings; thisReading ++)
    ac_input[thisReading] = 0;

they're already zero; this is pointless.

What are you expecting exactly? A 65 Hz signal would have a period of 15 mS, and you have a 2 mS delay between readings. What is your circuit?

gmbroh:
I am trying to measure the voltage of an 5volt AC rectified voltage at 65Hz

How have you connected the Arduino to the source?

Where does the rectified signal come from - is it possible that it includes some smoothing?

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.

michinyon:
The phrase "5 volt rectified AC voltage" is gibberish.

To me that seems like a reasonable way to describe a DC signal which has been obtained by rectifying an AC signal.

michinyon:
Your DC voltage has a peak of 5 volts and a minimum of about 4.5 volts.

Do you mean that this is what would be expected from a rectified sin wave? If so, how did you predict the minimum value?

gmbroh:
I am trying to measure the voltage of an 5volt AC rectified voltage at 65Hz. Using the UNO R3 Below is the current code.

Is that full-wave or half-wave rectification?

Read this before posting a programming question

Please edit your post, select the code, and put it between [code] ... [/code] tags.

You can do that by hitting the # button above the posting area.

And with or without a filter capacitor?

Lefty

See reply #2:

What is your circuit?

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

F0008TEK.JPG

retrolefty:

[quote author=Nick Gammon link=topic=148820.msg1118351#msg1118351 date=1360809591]

gmbroh:
I am trying to measure the voltage of an 5volt AC rectified voltage at 65Hz. Using the UNO R3 Below is the current code.

Is that full-wave or half-wave rectification?

And with or without a filter capacitor?

Lefty
[/quote]

It is full wave rectification without a filter capacitor.

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.

The input to the arduino is 0 to 5volts ac I dont understand why you think its dc.

Because your scope trace clearly shows everything above 0V.
Alternating current implies a current reversal.

analogRead works.

However you have built delays into your code.

delay (2);

That's 2 mS. Printing the previous result:

1/960 * 6 = 6.25 mS

Taking the reading:

0.104 mS

So you are basically sampling every 8.354 mS.

The period of a 65 Hz wave would be:

1/65 = 15.3 mS

Your sampling isn't fast enough.

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.

AWOL:

The input to the arduino is 0 to 5volts ac I dont understand why you think its dc.

Because your scope trace clearly shows everything above 0V.
Alternating current implies a current reversal.

At first the code was working really well. All I was doing was changing the delay to get different sample rates. Please see below

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?

gmbroh:
At first the code was working really well. All I was doing was changing the delay to get different sample rates. Please see below

The problem is that the output on the serial monitor only shows the peak values of the voltage. That is from 4-5v instead of zero.

The chart you posted (reproduced output) shows the graph reaching zero. What are you talking about?

The chart of the reproduced output is from when the code was working but not I am not getting those values.