On Saturday 19 April 2003 08:56 am, James Williams wrote:

> I can seen to figure out how this is working.  The circuit is a
> simple inverter + a series capacitor to another inverter.  Also
> from the input of the second invertor, between it's pin on the
> output of the capacitor, a resistor is used to pull the pin to
> ground. The circuit looks something like this:
>
> 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?
>
> When I look at the circuit, it seems like no DC logic signal will
> get through to the output.  When the input is low, the capacitor
> will charge through the resistor.  Which leaves the output of the
> capacitor still at ground level, correct?  Then when the input is
> high again, this causes an instantaneous discharge of the
> capacitor, still leaving the output of the capacitor at ground
> level, so I fail to see how this is used as a way to generate a
> fixed length pulse when the input goes low.
>
> What am I missing?

The fact that you've got a voltage divider. When the IN goes low, the
cap is still discharged (i.e. there's 0 volts across it). At that
instant, the resistor end is pulled high. However, that causes
current to flow into the capacitor. As the current flows, the cap
charges up, which means that the voltage across it increases. Shortly
after IN goes low, the capacitor is charged up to V+/2 or so, and the
voltage at the top of the resistor is at the threshold of the OUT
inverter. The signal then switches back to the quiescent state.

An even more amusing circuit uses the same RC network, but uses an
exclusive-OR gate to produce a pulse at *both* the low-to-high and
high-to-low transitions.

--
Ned Konz
http://bike-nomad.com
GPG key ID: BEEA7EFE

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