I have a circuit, that uses a voltage divider on a control pin, to turn it
on and off.  Basically if the voltage is present, and above a 1.32V
threshold, then the controller is activated.  I want to remotly control this
device, so I tied a N channel FET (BSS138) where the source is tied to the
control pin, and the drain is tied to ground, and the gate has a 1K pulldown
and then driven from a PIC.  If I drive the FET on, then the control pin
does goto ground, as expected.  However, if I tristate the PIC, the gate
floats to ground, but I don't see the control pin "isolated" for a better
word.  In other words, the voltage divider normally puts about 1.6V on the
control pin, but with the FET in circuit, and turned off, I get around 0.5V,
below its 1.32V threshold thus keeping it turned off.  So, figuring the FET
is in parallel with the 'lower' resistor, I went thru and figured (based on
48V power, a 294K and 10.2K resistor) that the FET was actually providing
around 5.5K and when in parallel with the 10.2K would give around 3.5K, and
feeding that back into the equation, would provide the 0.5V that I am seeing
on the pin.  So, recalculating that I need to make the upper resistor
smaller, figured a 90K would do the trick.  When I did this, the voltage on
the pin only went to 0.6V.  So is the off resistance of the FET related to
the current in the circuit?  Or am  I just going in the wrong direction in
assumptions here.

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