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

Omer YALHI wrote:

>Hi,
>
>I have the following resistors setup, as the magnet around them moves, B1
>and B2 reed switches close or open, and thus the total resistance changes
>(B3 is for sensing strong magnets and short circuits the system).  I need to
>find out different states of the 3 reed switches.  I am able to do this by
>connecting a capacitor and timing the charge time it takes, thus I can
>differentiate different states.  I do this every 500ms.  I had to use a
>special capacitor not to be effected by the temprature changes.
>
>I am wondering if I can differentiate the switch states by analog reading?
>The current changes but not the voltage; how would I do this?  Note that
>power consumption is extremely important.  With current setup I am down to
>10uA (averaged).
>
>Please let me know if this is a stupid question also.
>
>Thanks in advance.
>
>Omer
>
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