There is a 10k maximum input impedance requirement for the 16F PIC's, whi=
ch=20
you will exceed with this setup.   IIRC, it's about 1/4 of that for the 1=
8F=20
series.

You should be able to do this successfully with a lower-resistance LDR.  =
I=20
picked one up from digikey ages ago with a range of 4k - 11k ohms, which =
when=20
combined with the fixed resistor in the voltage divider, will let the PIC=
=20
input see an impedance in it's required range.  (The PIC input sees an=20
impedance of the parallel combination of both resistors in the divider).

Alternatively, if you have an LDR with a slightly higher resistance than =
will=20
meet the PIC's input impedance specs, you could try paralleling the LDR w=
ith=20
another resistor, but you'll lose some linearity (which should not really=
=20
matter if all you need is a threshold).

Cheers,
-Neil.



On Sunday 18 April 2004 01:05 pm, Ake Hedman scribbled:
> I am in need for a low cost light detector that has a use settable
> trigger point for the light/dark threshold. My first thought has been t=
o
> just set up a voltage divider in the following way.
>
>         +5V
>
>          | | 500K
>           |
>           |--------- PIC ADC
>           |
>          | | LDR
>
>          ---
>
> The LDR has >500K resistance in darkness and 3K-20K at 10 Lux.
>
> One could use a PIC output pin to feed the divider instead of the +5V
> and just activate this when a measurement is needed.
>
> Is there any drawbacks with such a scenario? Is there a better way to d=
o
> this in a low cost way?
>
> /Ake
>
>  ---
> Ake Hedman (YAP - Yet Another Programmer)
> eurosource, Brattbergav=E4gen 17, 820 50 LOS, Sweden
> Phone: 46 657 413430 Cellular: 46 730 533146
> Company home: http://www.eurosource.se      Kryddor/Te/Kaffe:
> http://www.brattberg.com
> Personal homepage: http://www.eurosource.se/akhe

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