Hi,

Lets look at the limiting case.  Set the bottom resistor open.  Label
the junction of the two 2K resistors and the cathode of the diode "N".
You imply 5V at the top of the string and we will assume 0.6V across the
forward biased diode.  The current law states that (steady state) the
current into a node sums to zero.  The equation is:

(N/2000)+((N-5)/2000)+(((N+0.6)-5)/200)=0

This simplifies to:

N = 49/12 = 4.0833V

Of course if the bottom resistor is shorted the remainder of the circuit
is irrelevant and the output voltage is zero.

Regards,
Dave

Benjamin Bromilow wrote:

> However, if it gets really cold the resistance will go >200 and the voltage
> goes >2.5v. The resistance could easily go much higher and the PIC ADC will
> see more than Vref+0.3v (from the 16F877 datasheet)....
> So I drew up this circuit. Theoretically if the middle of the second pot
> divider is 2.5v, any voltage on the first greater than this will cause the
> diode to conduct... But where will the "volts" go???
> My gut instinct on this circuit is that it will become an oscillator very
> quickly... Any ideas??

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