On Sat, 11 Dec 2010 09:12 -0800, "Kirk Wallace"
<kwall...@wallacecompany.com> wrote:

> 
> The circuit works properly to detect non-AC input. It's just when the +5
> input is connected, there is a one pump charge that gets through which
> is large enough to bring the output cap up a couple of volts in a spike.
> I am concerned that when the machine power is turned on and there might
> be a glitch on the output. A slightly larger output cap absorbs the
> glitch, but cap also charges up well enough with real AC input (or more
> than one pump in a row).
> -- 

This is exactly why the "pump cap" is ideally a lot smaller than the
"reservoir" cap.  That way it takes quite a few cycles to build up 
enough voltage in the reservoir cap and turn on the output.

But you can only get away with a small pump cap when the pumping
frequency is much faster than the output time constant.

I like the water pump analogy that someone posted earlier.  Continuing
that analogy:  the pump cap is the cylinder of your pump.  The output
cap is the bucket the pump empties into.  The load resistor is a leak
in the bucket.

If you have a big pump cylinder, one stroke of the pump can put a
lot of water in the bucket - you don't want that.  A small cylinder
needs a lot more strokes to fill up the bucket, which is good.

But if the cylinder is too small, pumped too slowly, or the leak
is too big, then the pumping might not be able to keep up with
the leak.  In that case, make the leak smaller or pump faster if
you can.  Making the pump cylinder bigger should be the last resort.

Note that if you make the leak too small, then it will take a long
time for the bucket to empty when you stop pumping.  It's all about
the tradeoffs...

John Kasunich
 
-- 
  John Kasunich
  jmkasun...@fastmail.fm


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