ADCIN question

[tenaja]

If you are reading an 8-bit a/d converter, then your input will be from 0 to 255. The scaling depends on your reference. typically, for cheap setups, this is Vin. If your Vin is 3.0 volts, you divide that by 256 to see how many volts each bit is worth. 3/256=.0117. So, if you get a reading of 200, your voltage is just over 2.3. Just plug the math in to convert it in code, or for more compact code, simply use precalculated constants.

[Darrel Taylor]

IOFSCS_8MHZ     Hmmm, dyslexia day at Microchip

<img src="http://www.picbasic.co.uk/forum/attachment.php?attachmentid=2858&stc=1&d=122145428 7" /><!-- -->

Your device line is OK, just pointing out an oddity.
<hr>

> how do I figure out what value will get placed in the Vin variable?

1 decimal place ...

Volts = (Vin+1) */ 50

2 decimals ...

Volts = (Vin+1) */ 500

[Darrel Taylor]

I read that question again, and maybe you want the other way around.

> how do I figure out what value will get placed in the Vin variable?

With 8-bit resolution, the values will range from 0 to 255.

If the input is 2.5 volts, then the value that gets placed in Vin will be ...

255 / (5 / 2.5) = 127

For 0.7v that's ...

255 / (5 / 0.7) = 35

hth,

[Greg McFadden]

thanks! all!. This is all quite new to me. the last thing I programed was some cfd in fortran 95, so the hardware side of things is different.

I think I may end up going to a PIC12F683 just for flexibility, as I can still (barely) fit it on the circuit board I am making.