"John S"
** At last, someone with a tad of common sense.... Phil
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- posted
11 years ago
simple capacitor test circuit
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- posted
11 years ago
I think it's one of those fringey free-energy hydrogen generators.
-- John Larkin Highland Technology Inc www.highlandtechnology.com jlarkin at highlandtechnology dot com Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom timing and laser controllers Photonics and fiberoptic TTL data links VME analog, thermocouple, LVDT, synchro, tachometer Multichannel arbitrary waveform generators
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- posted
11 years ago
He's all wet. ;-)
-- Politicians should only get paid if the budget is balanced, and there is enough left over to pay them.
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11 years ago
My mistake. Thanks Jeroen. And, I believe you are correct that it will still not be easy to measure (if at all).
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- posted
11 years ago
G > What made you choose water?
MAT > He's all wet. ;-)
Even if it had mineral oil dialectric it's not much plate area.
John Larkin > I think it's one of those fringey free-energy hydrogen generators.
I don't think stainless fits that idea. And they usually get both hydrogen and oxygen though they could intend to throw away (vent) the oxygen.
I always thought it was neat how you can take water and break it into hydrogen and oxygen but then if you combine those two gases and ignite they can be quite energetic.
Is the energy from that process worth the cost of the electricity used to break up the hydrogen and the oxygen?
I took it for granted that it was a lossy energy conversion.
When they reconsidered this stuff for automobile use, what exactly killed that idea off?
The explosive hazard? Conversion costs? What else?
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- posted
11 years ago
Of course not. You can't win. You can't break even. I'm surprised by the very question. What are you?
Nothing beats the convenience of a tank full of liquid fuel.
Jeroen Belleman
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- posted
11 years ago
That's the most accurate way; known C and unknown form half the bridge, and two resistors (really, a potentiometer ) form the other half. The capacitor ratio sets the voltage division on the capacitance half, and resistor ratio sets it on the resistor half. An AC source drives both (I'd use a wall-tumor transformer to start with), and an AC meter determines a null if they're balanced. Adjust the potentiometer for best null. C1/C2 = R1/R2 when the output is minimum.
The AC meter is unnecessary, really; a speaker would do (you just listen to the signal).
Less accurate, but also effective, you could oscillate with a known resistor, in a NE555 (or otheer '555) astable circuit, and use charge-pump circuitry (two diodes and a capacitor or two) into the milliamp inputs on your multimeter. If the frequency is low enough, replace the charge pump with a flashing light or ticking speaker and use a stopwatch.
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- posted
11 years ago
--
Because of the resistive loss of the water and the phase shift it
causes on the DUT side of the bridge, that won't work well.
This should: (View using a fixed-pitch font.)
AC>---+-------------------+
| |
[C1] [C2]
| |
+--[NULL DETECTOR]--+---+
| | |
[DUT] [C3] [R1]
| | |
AC>---+-------------------+---+
where C1 and C2 are the same value and C3 and R1 are selected to get a
null.
Ideally, C3 would be a capacitor decade box and R1 could be a pot,
both adjusted to get a good null.