Adam, thank you for the reply.  However, the circuit I have given is a
"black box", I am only exposed to the pins A and B.  I think I can still put
your idea to work by adding another resistor (something smaller than 10K
perhaps?) at point A and measure the voltage at that point.  Does this sound
right?  Also, I am not familiar with op-amps, so I wouldn't know where to
start; therefore I am trying to avoid using an op-amp (unless it becomes
necessary).

Regards,

Omer


-----Original Message-----
From: pic microcontroller discussion list [mailto:PICLIST@MITVMA.MIT.EDU] On
Behalf Of M. Adam Davis
Sent: Saturday, April 17, 2004 12:12 AM
To: PICLIST@MITVMA.MIT.EDU
Subject: Re: [PICLIST] [EE:] Reading different resistor values

This circuit uses large value resisters, which makes it difficult to read
when directly connected to an analog input pin, but by no means impossible.

You have a simple resister divider.  Hook the analog input pin to the
junction of R1 and R2 (where B1 connects).  As the switches open and close
the voltage level at that junction varies.

If the voltage is one volt, then these are the voltages at that junction you
can expect:
B1, B2 open -->  .745
B1 closed --> .109
B2 closed --> .647
B1, B2 closed --> .093
B3 close --> 0

If the voltage going in is higher, then just multiply the above values by
that voltage, so for 5 volts it would be 5 * .745 = 3.72 and so on.
You can measure this directly with a multimeter at that point.

The PIC may measure slightly different voltages for both open and B2 closed
since there is an impedance limit in the PIC - it doesn't like to measure
voltages through resistors more than 10k (It's a little more complex than
that, but this will get you started).

However, it will still be able to detect the change and correctly identify
the various states of the switches.  If it doesn't work very well adding an
op-amp to buffer the voltage will set it straight.

I hope this helps!

-Adam

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