Advice on Input circuitry for 0-10v or Pot

[mister_e]

Nope... range of 0-2.5 volts only... because this extra 10K in parallel with your POT.

2x1% 10K in serie, your ADC in the middle... a 0.1uF cap and that's it.

If the source handle this load, and is not that far of the PIC (i doubt because of this ol' 0-10v naming), this shouldn't cause any problem... as usual, software filtering/averaging are your friend.

[Acetronics2]

Looks Darrel had written something (LOL ) on that subject ... nooooooo ????

http://www.pbpgroup.com/modules/wfse...p?articleid=25

....

Boooooooooooooooooooooooooooo

Alain

[malc-c]

Nope... range of 0-2.5 volts only... because this extra 10K in parallel with your POT.

2x1% 10K in serie, your ADC in the middle... a 0.1uF cap and that's it.

So how would he adjust the voltage, would placing the 10K pot across the 10v supply, and then placing the 2 x 10K resistors between the wiper and gnd, and then tapping off in the middle work. My thoughts would be that the tap in the middle of the resistors would therefore be half of what ever voltage is coming out of the wiper ?

[mister_e]

Oh well, there's some voltage divider and ohm law that i'm not going to explain on a friday

But to me, when at all possible, i try to avoid any kind of trim pot, software calibration is interesting... but there's alway extra work to be done... and the PIC Vdd, and voltage source have their own error margin as well. So i stick to the simple 2 resistor voltage divider

If you really want to stick with a Trim Pot, i would suggest something like...
<img SRC="http://www.picbasic.co.uk/forum/attachment.php?attachmentid=2493&stc=1&d=120795014 3">

[malc-c]

If you really want to stick with a Trim Pot, i would suggest something like...

I thought that was what the original poster wanted...

What I can't work out is how to get a 10K pot and a 0-10v input (obviously only one is used at a time) I was thinking a pot between Gnd and the input
bill.

Steve, I'm still confused over this, using your circuit effectivly gives 10K and 11.1K (2k + 9.1K) when the pot is at min (ie wiper diectly touching the 10K resistor) which would give 5.26 v, but at the opposite throw, it would be 12K and 9.1K which would give 4.31v to the PIC. I assume that you would therefore use the 0.95v difference as the range to measure for the ADC ?

[mister_e]

The pot is there for calibration purpose only. inject 10V at the input, then adjust the pot to reach 5v... a bit useless in most case

[malc-c]

So in essance its a straight divider, 5V (+ or - 0.25 ish of a volt).

So how would you then take the 5v output and use this to give a voltage range the ADC can use to intern generate the PWM to dim the LED. I assum that any pot between 5v and gnd will thus be in parallel with the 9.1K, hence your comments regardig Ohms Law

EDIT:
Or have I missed something, in that the 0-10v stated in the original post is already a varible voltage between zero and 10v, rather than a fixed 10v supply ? If so then I can see your logic

[mister_e]

i'm not sue if i follow you malc ... not sure you follow us too no pun intended

You take the output and plug it directly to the PIC... straight, not much. Sugarless, fat-free, not much, zip, zero,nada. Want to use it to dim a LED? easy...

ADCIN 0, ByteVar
HPWM 1,ByteVar,Freq

If i haven't missed anything in the OP request, he have an external source which provide voltage between 0-10V... and want to monitor it (for xyz reason) with a PIC, hence why the voltage divider.