Hi Rob.
It isn't perfectly clear to me just how you are expecting your circuit to
work but there do seem to be a few gotchas present!

First off: the zeners. You drew 2 zeners back to back. Thus they will clamp
positive voltages at 3.3V + 0.6V = 3.9V; is this what you were thinking of?
 The top zener (assuming the input never goes negative) does nothing but
act like a forward biased diode.

Second: when your transistor is on, the PIC MUX/IN will see ground (not
really, see the third point).  When the transistor is off, the bottom of
the 5 ohm resistor will be at +12V and the 5k6 resistor will drop this down
to the zener clamp voltage of 3.9V.  You say you are trying to determine if
the 5 ohm part is in the circuit; if it is not, then when the transistor is
off, the PIC MUX/IN pin will be floating.  Hard to say where it will end
up, could go high or low.

Third: the transistor has a load line that would like to go from zero
current to 12V/5ohms = 2.4AMPS.  For a single transistor you should count
on beta of no more than 50 (depends highly on what part you use). A beta of
50 means that the base current would need to be 2.4AMPS/50 = 48mA. With a
1K base resistor, you will need a buffer capable of output swing up to
about 50 volts to supply the 48 mA.  If you are driving it from a five volt
supply, you will need to rethink this resistor.  If you are doing all this
at relatively low speed, a fet might be a better choice than a bipolar.  A
fet will require almost zero current to it's gate. You could probably
eliminate the buffer then.

Good luck!
Tom M.

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