Oyvind Tjervaag wrote:
> I am looking into making an Ohm meter which will measure the high
> resistance of wood. This is to make a wood moisure meter. The
> resistence of wood goes down as the moisure goes down.

No, it's the other way around.

> Is it
> possible to do it using a simple potential divider as shown below?
>
>
> ------o 5V
>>
>>
> <  Wood of unknown resistance
>>
>>
>> -----> to PIC A/D
>>
>>
> <  Known resistance
>>
>>
>> -----o gnd

Not as shown.  The PIC A/D has too low an input impedence.  I would use a
higher voltage going into the wood with around a 2Mohm current sense
resistor to ground.  The voltage from the current sense resistor would be
buffered and amplified by an opamp with CMOS inputs, then go into a PIC A/D.
Since this is a high impedence circuit, special attention needs to be given
to leakages, guard traces, etc.

In other words, the circuit is as you diagrammed above, except 5V is
preferably higher, the known resistance is about 2Mohm, and a high impedence
buffer with gain is inserted before the PIC A/D input.  It might also be a
good idea to divide the actual voltage and feed it into another A/D input so
that its absolute value can be factored out.  Varying that supply from the
PIC will also allow for larger dynamic range of measurement.


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