On Wed, Dec 08, 2010 at 07:00:27PM -0800, Kirk Wallace wrote:
>
> Thank you John. I really appreciate the time you take to help. Using
> your information above, I redrew the schematic, which I'll revise as I
> go. It's at the bottom of the page here:
> http://www.wallacecompany.com/machine_shop/Shizuoka/
Kirk, apropos the comment there that it's not quite clear how the charge
pump works, does it help at all, just for a more physical visualisation,
to imagine C1 as a liquid pump's cylinder, with the top plate being the
piston, driven up and down by IC1A, and D1 & D2 as fluid non-return
valves? C2 is the reservoir into which this pump is pumping "positive
charge" [1]. If C1 isn't driven up and down, charge isn't going to flow
out of ground to C2, and its charge decays through R2.
The only change I'd make to the circuit is to connect R1 to +5v, so that
the default state of IC1A output is low. That is because a capacitor is
more likely than not to go short circuit in the rare event that it
fails. That would give a false "OK", in the absence of input. (It'll
work fine as-is if there's no component failure.)
Dunno if my attempt to paint a picture really ended up less laboured,
and more descriptive that going through it in purely electronic terms,
but we can always do that if this attempt isn't as clear as it should
be.
Erik
[1] While it's electrons going the other way really, it's more
convenient to discuss "positive charge" when talking about
accumulating a positive voltage.
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