On Fri, Jun 23, 2006 at 11:33:41PM -0400, Spehro Pefhany wrote:
> At 11:15 PM 6/23/2006 -0400, you wrote:
> >I built a quick circuit to take rough light level readings. It's a
> >18f458 PIC with a CdS photocell/1k resistor voltage divider connected to
> >AN0. The 1k resistor goes to 5v, the CdS phtocell to ground.
> >
> >I noted the datasheet specifies that signal sources connected to the ADC
> >pins should have an impedance of less than 2.2k. My understanding is
> >that the voltage divider chain would not qualify for that in low light,
> >as the CdS photocell's max resistance in darkness is around 300k.
> 
> The Thevenin equivalent source resistance (assuming a regulated Vdd line)
> is the LDR in  parallel with 1K , so it will always be less than 1K.
> 
> Rs = Rx || 1K = (Rx * 1K)/(Rx + 1K)

Thanks, I think I understand it a lot better now.


With a fair bit of work I was able to kludge together an equation that I
think describes the final voltage vs. lux graph accurately with gnuplot.
Not exactly the most linear of things, but at least I can see that my
intuitive choice of 1k isn't that great a pick... If I got everything
right, something more like 220ohms would still give me sensitivity at
both ends, 1k ends up at a zero slope long before the CdS saturates.

-- 
pete@petertodd.ca http://www.petertodd.ca
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