[sdiy] Minimum parts count switch de-bouncer.

[Harry Bissell Jr]

Its that easy... but the two 10K resistors also form a
voltage divider that makes the input to the schmitt
gate 1/2 the supply voltage.

I'd suggest a 10:1 ratio... so maybe a 1K and a 10K
resistor.  1K would give 15mA

(when the cap is discharged... it can be thought of as
a short circuit. As Rene pointed out... it actually
has
some internal resistance, ESR)

So your Ohms law should essentially not include the
lower resistor (10K in series with .01 ohms ?)

(you slipped a decimal point,,, your last answer was
.75mA)

H^) harry

--- Dave Kendall <davekendall at ntlworld.com> wrote:

> Hi Harry.
> 
> >There should be a small resistor in series with the
> >switch, to limit that peak current. Just assume the
> >cap
> >is a short, and Ohms' Law will tell you the peak
> >current.
> 
> Ah, thanks for that...
> 
> So, the circuit below would give a positive going
> pulse drawing 7.5mA for
> about 10ms on a single button push.?
> 
> I = V(15) divided by (R)10,000 + 10,000
>  = 0.00075A
> 
> That seems too easy - does the Cap affect that
> equation. (I bet it does...)
> 
> Maybe 1uF is too high. Would it be better to use
> 100nF and 100K to get 10ms?
>                    
>                    
>                    
> I----------------I>o------------o  -I
>                     I           schmitt trigger
>  GND                I
>  I     sw           I   I------I 10KI-------I
>  I    ____          I   I                   I
>  I----o  o---I 10K I ---I                   I
>                         I             +     I
>                         I------II 1uF
> II----I----+15V
> 
> 
> Sorry if all this is a bit low level, but I've
> always been crap at theory
> and maths.
> It's all helping to bring a MIDI synced
> clock/envelope a bit closer to
> reality.
> 
> Hopefully other newbies are learning as much as me
> too...
>  
> cheers,
> 
> Dave (standing on the shoulders of giants, and
> asking WAY too many
> questions... :-)
> 
>