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![]()
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![]()
Steve
It's not a bug, it's a random feature.
There's no problem, only learning opportunities.
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
Last edited by malc-c; - 12th April 2008 at 10:12. Reason: Afterthought :)
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.
Steve
It's not a bug, it's a random feature.
There's no problem, only learning opportunities.
I don't see why any PIC should be used here anyway![]()
Steve
It's not a bug, it's a random feature.
There's no problem, only learning opportunities.
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