Thanks Marshall and Wayne for your replies.
Wayne, I am looking at a schematic of your circuit. Is the LED the current
limiter? If so according to Marshall I need no more than 1ma per square
inch, I am assuming of electrode surface area. Lets say the bridge will
allow about 50.6V (voltage drop from diodes) to the rest of the circuit.
LED's drop about 2V so the resistor will drop about 48.6V. At 8K that means
a current of about 6ma, that LED will be dim. So that means I need at least
6 square inches of electrodes in the water.
12 gauge wire is 1/12" in dia, or .08333", so the Area of the wire
electrodes equals:
I want 6 square inchesof surface area. Area of the ends of the wires plus
the area of the sides.
6 =(2*pi*r^2)+(2*pi*r*h) = 2*pi( r^2 + r*h)
6/(2*pi) = r^2 + r*h
.955 = r^2 + r*h
R equals .041665"
so,
.955 = .041665^2 + .041665*h = .00174 + .041665 * (unkown h or total length
of electrodes)
.955 -.00174 = 0.95326 this is inches squared.
square rooted .955 equals .976
.976/.041665 = 23.43 inches total lengths of electrodes.
Or 11.7 inches each electrode underwater. If I use 12" that should put me
under the 1ma per square inch requirement.
Does this seem right to the CS generator veterans on the list? Should I be
worried enough to boggle my own mind trying to do this?
Thanks,
Kel
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