Willbe
New Member
That would be great
You're on.
-100 M makes me think current source.
-To get decent current through 100 M needs ~ 100 V.
-100 V is way higher than the base-emitter drop of a transistor so again I think Ideal Current Source.
-My old summary sheet from Teledyne Philbrick shows a I to V convertor where Eout = -Iin*R. For Eout ~15 V and negative Iin of 1 uA, R needs to be about 15 M, but since my doorbell xformer voltage doubler only puts out about 80 V this R has to be slightly bigger.
-The lightly loaded wall transformer puts out maybe 20% more than the rated 12 V, so size the LED driver resistor at (15 -2)/20 mA = 650, so use 680. 13^2 = 169 over 680 = ~1/4 W so use 1/2 W.
-You may need a decoupling cap, C1. It's good practice.
-You may also want to ground, either through a resistor or with a wire, all six unused opamp inputs.
-If you add a 1 M resistor in series with the WATER terminals this circuit will be much hardier.
-For the opamp output to swing positive the current has to be leaving terminal 2 so the doubler puts out a negative voltage.
-Re: the upstream post, the info I got from my 1988 summary sheet was the input bias current of the 324 was << 1 uA, but you may want to look into this. It should be << 1 uA, but accuracy in this case is not really important since the resistivity of even stinky air is probably >>> 100 M.
-C2 needs the + terminal to the right, if capacitors this small even need a polarity.
-Both AC sources go, of course, to the same wall plug.
-If the LED is dim because of high ambient lighting put it into a short section of heat shrink tubing.
-If you take any of this seriously you may want to build a test resistor from 5 ea. 20 M resistors. If you want to switch this in and out with a button, at 1 uA this might be "dry circuit switching", but you do have 80 V to wipe the switch contacts clean, so maybe a low current switch will actually work in this application.
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Last edited by Willbe; Today at 09:27 PM.
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