[PIC] Concept of Sampling AC to use for alternating current calculations using PIC 16F877A

Hello,
I am a making a monitoring unit which will take in input from a user (current) and will sample it and will give me the output.

Now I searched the internet and I understand the part where I will have to decrease the voltage value so I did using a voltage divider circuit and I getting a 5V peak to peak and I've added an offset as well but what I don't get is that what will I detect or write in my code?

Can someone please help me just with the pseudocode so that I can get a direction. I know I will have to sample it but how or what the concept will be be behind it. Please help.

Hi,

for sampling AC there are some points to consider.

* What is it for?
(just to display the value? or to make realtime signal analysis... and a lot of inbetween)
* only one channel?
* How often do you need an output value?
(for a display you need it about three times a second. For an electricity meter mabe every 15 minutes is enough..)
* What precision do you need?
( lets say for a mains voltage meter with a range of 300V +/-1% is sufficient, this means +/- 3V)
* What resolution do you need?
(example: for a full range three digit 7 segment display you need a resolution of 1:1000)
* is the input signal a known waveform?
(sine, square, triangle or unknown?)
* is the AC input DC-free?
(for most loads connected to mains this is true)
* is the AC frequency and it´s stability known?

***************
For most measurement (you speak of monitoring) equipment you are on the safe side with (AC-coupled input and) RMS calculation.
On the other side here is a lot to care of....
* selecting the right sampling frequency (calculate an appropriate analog anti alias filter, oversampling, undersampling...)
* interrupt driven sampling (for fixed timing and low processor load)
* signal conditioning (overrange detection, offset and gain correction)
* squaring the value
* filtering the value
* dividing the value
* calculate the square root

And ... there are a lot of ways to do this.

Hope this helps so far.
Klaus

Thank you very much, very informative and the dividing the value and calculating it's square root worked.