Howdy,
The debounce process involves a time delay in which multiple on/off
signals are ignored. this happens very fast. After the delay, the
circuit is left at the present switch position. It is possible to set
up a debounce of one second or more with a big enough capacitor and
resistor. We're talking 5 to 10 megohms and 3 to 5 microfarads. The
Cmos cookbook has charts to find the values. Right now, on a project
I'm doing , I have a 1 meg resistor with a 1 microfarad capacitor.
that gives me 0.8 seconds. I also have to run a 2 meg resistor
between the switch and capacitor to positive to make the signal stop.
Otherwise it will stay on as long as the switch is held down. In my
project, several switches may stay on at the same time, so I am
forced to make it a pulse to give the 4017 separate signals even
though 2 or 3 switches may be in the on position. Please note that I
am using ground or "0" as my switching signal. If using positive
or "1" then the resistors would go to ground. In your situation, 10
microfarads and 300 Kohms would give you 2 to 3 seconds. You need to
figure out what your time is going to be and adjust components from
there.
A more elegant approach would be to use a short debouncer into a
flip-flop. Then, when you push the button you will get either a
signal or no signal, depending on how you wire it. Upon release you
would get the opposite. Thus you could set up your flip-flop to
output on the press and "flop' to no signal for the release or vice-
versa, depending on the response you desire. Hope this helps and, GET
THAT BOOK!!!!
Rig
The debounce process involves a time delay in which multiple on/off
signals are ignored. this happens very fast. After the delay, the
circuit is left at the present switch position. It is possible to set
up a debounce of one second or more with a big enough capacitor and
resistor. We're talking 5 to 10 megohms and 3 to 5 microfarads. The
Cmos cookbook has charts to find the values. Right now, on a project
I'm doing , I have a 1 meg resistor with a 1 microfarad capacitor.
that gives me 0.8 seconds. I also have to run a 2 meg resistor
between the switch and capacitor to positive to make the signal stop.
Otherwise it will stay on as long as the switch is held down. In my
project, several switches may stay on at the same time, so I am
forced to make it a pulse to give the 4017 separate signals even
though 2 or 3 switches may be in the on position. Please note that I
am using ground or "0" as my switching signal. If using positive
or "1" then the resistors would go to ground. In your situation, 10
microfarads and 300 Kohms would give you 2 to 3 seconds. You need to
figure out what your time is going to be and adjust components from
there.
A more elegant approach would be to use a short debouncer into a
flip-flop. Then, when you push the button you will get either a
signal or no signal, depending on how you wire it. Upon release you
would get the opposite. Thus you could set up your flip-flop to
output on the press and "flop' to no signal for the release or vice-
versa, depending on the response you desire. Hope this helps and, GET
THAT BOOK!!!!
Rig
--- In cgs_synth@yahoogroups.com, Richard Brewster <pugix@...> wrote:
>
> Oops. Typo. That is a CD4010 non-inverting CMOS buffer. Not 4001.
>
> -Richard
>
> Richard Brewster wrote:
> > R1
> > Switch
> > ---/\/\/\/\/----
> > +V | |
> > | |\ 4001 |
> > <-----o------| /o----------o OUT
> > |/
> > Gnd
> >
> > The switch is between +V and ground. The 4001 is a non-inverting
> > buffer, powered by +V. Let the switch be centered. On power up
OUT may
> > be at ground or +V. Whichever it is, since it feeds back to the
input
> > through R1, it will stay put. If it's at ground, then a
momentary flick
> > of the switch to +V will cause the output to follow. The switch
can
> > remain at +V, bounce off the contact, or move back to center (no
> > connection). OUT remains at +V. Nothing changes until the
switch is
> > moved back to GND. At that instant OUT becomes zero volts and
will
> > remain so, even if the switch is lifted (bounces). It is a
simple
> > flip-flop.
> >
> > Richard Brewster
> >
> >
> > rafe127 wrote:
> >
> >> I guess it is about time I picked up the CMOS cookbook, I will
order
> >> it tomorrow.
> >>
> >> Explain to me, if you will, the basic premise of this kind of
> >> debouncing. Will it still work for my purposes if the open and
close
> >> of the switch are as much as one second apart? I'm just curious
what
> >> the principle is.
> >>
> >> Thanks again,
> >> Rafael
> >>
>