Re: The mechanism behind bouncing...


"John Larkin" <jjlarkin@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:kcl9s21bv0ulr6m2sdjlnkffdftp8mh68c@xxxxxxxxxx
On Sat, 03 Feb 2007 08:13:46 GMT, "Jon Slaughter"
<Jon_Slaughter@xxxxxxxxxxx> wrote:


"KILOWATT" <kilowatt"nospam"@softhome.net> wrote in message
news:45c3aa6d$0$31564$c3e8da3@xxxxxxxxxxxxxxxxxxxx
Hi... thanks for your attention.

I just wish to know the precise reason why for example, a digital
counter
may count many pulses on it's clock input when the clock is feed via a
non
noise-free source like a mechanical switch. It is because when the
contacts
makes/breaks, arcing (i've read somewhere that there can be a
possibility
of
arcing even at low voltage) occurs, or if it's because of the very rough
surface (microscopically-speaking) of the switch contacts, were the
metal
molecules grinds (and possibly flexes) together, during switch
activation?
TIA for your reply.



The atoms of the two materials are not configured in such a way that there
is complete contact. If they were then the materials would be fused. Since
there are not fused and they slide there is friction involved and this
friction causes the contacts to move farther a part and then closer
together. So the average distance between the constants is changing
significantly campared to when is not moving and they are making good
contact. So now the electric field is changing because of the distances
changing between the contacts. As the contacts move farther away the field
becomes weaker but now we have a capacitive effect. This effect creates a
force between the contacts that attract them. One now has a kinematic
force
pulling the contacts away(so it can slide), one of friction that wants to
stop the slide, and one of capacitance that is attractive(I'm sure there
are
more too).


Sorry, but that's all nonsense. At low voltages and currents, switch
contacts bounce for purely mechanical reasons.


Um, and you seem to think that mechanical bouncing is some real thing. Its
an abstract concept. There is no real think as bouncing. When a ball bounces
you think that the surfaces are idealized. No, they are governed by quantum
mechanics. Believe it or not, doesn't matter ot me.

If your field theories were true, the applied voltage would radically
change the bounce waveform. It doesn't. Try it.


hmm. so the waveforms are exactly the same? They do not scale with voltage?
You seem to think that ohms law doesn't apply here? V = IR or did you not
learn that? What do you really think these waveforms will look like? Do you
think they will be perfect unit step functions? Get into the real physics
of it and stop trying using idealized descriptions of the behavor.

http://www.ece.uci.edu/rfmems/publications/papers/mems/C021-EUMTT99.pdf

http://www.scienceprog.com/dealing-with-switch-bounce-problem/

Do you really think that the effects I'm talking about are going to
radically change the macroscopic scale? The time scale is femto or less and
the forces are fN or less. (although the above analysis are still idealized)

Your logic is like saying a resistor behaves exactly the same no matter what
conditions. Your a bafoon in thinking that everything is some simple
mathematical equation that you learned in cal 101.

You think that a resistor doesn't change its "waveform" with voltage? R =
V/I. What happen when V is very low? what about when V is very high? What
about most of the time? Most of hte time R is the APPOXIMATELY constant.
THATS RIGHT!!! A resistor doesn't change its "waveform"(which is wrong way
to put it as switches and resistors are not waveforms) for a wide range of
voltages. If this what the case then they would be practically useless.

Did you ever take quantum mechanics 101? Hell, even basic physics supplies
an approximately correct answer. Coulomb's law for the contacts state
something like

Sum(k*Qi*Qj/r_ij^2,i=1..N)

(this is better described in a statistical quantum fashion but I don't want
to confuse your little brain)

Do you think that distances in this equation have some special meaning
between when a switch is open and not? Sure there is a point where the
strenght of the field drops off almost to 0 but it is not instantaneous.

The OP asked for the microscopic answer and not something you read out of an
electricians manual.

(BTW, show me some waveforms's from two different switches using the same
voltage and lets see if they are even close. Hell, show me two waveforms
from the same switch using the same voltage and lets see if they are even
close.)


Anyways, So there are all these forces that are interacting and the end
result is this oscillation of the contacts moving toward and away from
each
other. One always gets "arcing" but thats kinda relative turn. (In some
sense all electronic flow is "arcing".)

Metallic conduction is not "arcing." Arcing is gaseous conduction.
Vacuum tunneling happens too, but the range is just on the order of an
atomic diameter, not important for things like switch contacts.



So we cannot have an arc in a vacuum without any gas? Hmm, can you prove
this? I think this would go to explaining a lot about vacuum tubes(I guess
they don't "arc" or must contain a gas(a significan't amount to explain the
arcing)).

It may not be significant to you but you are not the genius you think you
are. People like you are satisfied with any explaination that doesn't
confused them people like the KILOWATT want to know the real reason why
things work. You take it on faith(sure, you might look at a few switch
characteristics using an oscilliscope but then you do not care to go
farther) while he wants to know the real reasons. The difference is one of
religon and one of science.

The same factors that cause friction are at work with a mechanical switch.
You can ignore this all you want and thats fine. But don't try to act like
the world is some idealized place that is perfectly described by few simple
mathematical equations. (all equations are wrong to some extent and some are
better than others. Usually the better equations are more complicated).

Your logic is like "A diode is a switch" while mine is "A diode is a device
that can behave like a switch but this is because of the properties of the
material. (then I'd have to talk about doping, holes, drift, junctions,
valence electrons, pauli exclusion principle, etc...). Rarely is any simple
explination the full explination.

If he asked what was switch bounce then your answer is good enough. What he
asked was what was the microscopic reason for switch bounce. Now I didn't
talk about quarks because obviously that level is to low and there is no
need(as far as we know). Even the theory of friction at the atomic level is
not know that well. But this is the best level to explain it because you
get at the heart of the reason. Now my explinations might not be perfect but
just because you think they are wrong doesn't mean they are. I also never
said the effects were significant, but there is a macroscopic effect.

Its fine if you want to act like the world is not made up of atoms and
idealize everything. I have no problem with that. But when someone else
wants to know more then don't try to make them believe what you believe.
Only thing I can think of is that your afraid that if they go and explore
that they might prove you wrong. So you care more about looking right than
being right.

Did you factor in resistance into your switch? Didn't think so. Do you know
that resistance depends on voltage? (doesn't matter how, just that there
exists two different voltages that produce two different values of
resistance). You know that voltage and heat are related? (Even directly.
Not that its significant. Even absolutely zero cannot stop an atom from
moving.)

Anyways...



.

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