Thanks to all who replied.  The biggest thing that I forgot from my
basic electronics is that the capacitor acts as a short initially.  So
then that clears up the question.  I did dig out my first year
electronics book last night.  This type of circuit is defined as a RC
differentiator.  As I can see, this type of circuit has many good uses:
Including a DC charge pump, for generating both a positive and negative
supply rail from a single positive rail.  So I fail to see how this can
be a bad design circuit.

-----Original Message-----
From: pic microcontroller discussion list
[mailto:PICLIST@MITVMA.MIT.EDU] On Behalf Of Olin Lathrop
Sent: Sunday, April 20, 2003 7:47 AM
To: PICLIST@MITVMA.MIT.EDU
Subject: Re: [EE]: Pulse circuit, can you explain how this works.

> IN ---|>o-----)|------|>o------ OUT
>                    |
>                    |_---/\/\/\/\--- GND
>
> To me, the capacitor is acting as a coupling capacitor, so I fail to
see
> how this is making the output a small pulse given the input state.
Can
> someone explain how this is producing a digital pulse on the output?

The C and R form a high pass filter into the second inverter.  In a
rough
sense, this means edges get thru but steady levels don't.

Start with the circuit with IN high and steady state.  IN goes low,
causing
a rising edge into the capacitor.  This step is coupled immediately to
the
output inverter, making it go low.  So far, the signal has just been
passed
thru.  Now consider what happens as IN stays low.  The input to the
capacitor stays high, but the output will exponentially decay towards 0.
This decaying voltage will eventually be interpreted as a low input by
the
second inverter, making it go high.  We have so far sent a falling edge
in
and gotten a low going pulse out.  In other words, and edge to glitch
converter.

Now IN goes high.  This causes the input to the capacitor to go low.
The
output of the capacitor is already at 0, so it will attempt to go to
-5V.
However, the input clamping diode in the second inverter will clip that
to -700mV or so.  The output is not changed, but now we're ready for IN
to
go low again and cause another pulse.

This type of circuit is somewhat of a kludge.  Wanting an edge to glitch
converter is usually a symptom of bad design elsewhere.  Note also that
the
second inverter should be a schmitt trigger unless the RC time constant
is
very low.

As an additional exercise, consider what the circuit does if R and C are
exchanged.


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