On Thu, 30 Jan 2003, Tim Webb wrote:

*>If you had two electrodes equal in length stuck into a water tank, would
*>the resistance across the two electrodes change as the water level
*>changes?

Water has a high dielectric constant (80 times that of air). Thus two
electrodes (wires) dipped into it will form a capacitor that changes
linearly with the amount of immersion. If the wires are too close then
water will form drops between them, in air, and appear to be 'more' then
there is.  If the wires are insulated with thin insulation then no
corrosion will occur. Since it's a capacitor it will be read in ac anyway.

So a good recipe would be to take a tube and a wire (the wire running
inside the tube, without touching it), both painted with waterproofing
paint, and dip them into the liquid. Any circuit that depends on
capacitance will allow readout, including the 555 based ones (use a CMOS
555). There is no need to have the 555 free running, it can be used as a
monostable MMV triggered by the PIC. The tube wall to wire distance should
be about 10mm to be sure no water drops can bridge it, and several holes
should be drilled in the tube (and painted) to allow water and air to
circulate.

My (probably wrong) calculations show that a 2mm wire in a 10mm i.d. tube,
20 cm long, should give about 0.5 pF in air and 41 pF in water, with the
capacitance changing linearly with water level. Neglecting the influence
of the paint dielectric, which is supposed to be very thin. I used the
plate capacitor formula for this (not the one for cylindrical caps) so the
numbers are slightly off.

Peter

PS: It is 1:30AM here. If I wrote bs please bear with me

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