S, Erik Christiansen piše: > 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 Maybe you have just too little current in output. So just route unused gates paralel to last one. (pins 1,3,5,9 together and pins 2,4,6,8 together)
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