Debouncing....at About 1Mhz

It should be a roughly 10 ns positive pulse at every transition of the input. The first such glitch clocks the r-c filtered data into the dflop. Any additional glitches do the same thing. There's not even a metastability hazard. This circuit is simple enough that a humanly-possible analysis can demonstrate, with high confidence, that it's safe.

You don't like it for personal reasons. That's a bad basis for engineering.

John

Reply to
John Larkin
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--
On the contrary,  I\'ve used a similar circuit for years as a clean
transition detector:


.IN>---+----------A
.      |        EXOR Y-->OUT
.      +--[R]-+---B
.             |
.            [C]
.             |
.            GND

The only thing I think that\'s wrong with your circuit is that at the
speed it\'s supposed to be used I think the EXOR chain is going to
generate a lot of hash as it follows the bouncing input.
Reply to
John Fields

I think I tried that. The R in the dampening cct gets fkn hot! It's not a low energy situation. The ring is from a few unavoidable parasitics..

However, this is uncharted waters for me.. I'll check for goofs..

I really don't believe there's anything I can do in the way of preventive measures. It'll bounce.

D from BC

Reply to
D from BC

JL, I'm tempted to replace that RC section with a D ff...

The D ff will hold the previous settled state. It can be clocked by a delay triggered by each edge.

That way.. the circuit works from 1Mhz all the way down to 1Hz and less.

Ok..if I hairballize your circuit a bit :) D from BC

Reply to
D from BC

[snip]

I've been thinking all those lines also. I had a "perfect" ratty-edge remover around here somewhere... now I can't find it :-(

Yep.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
         America: Land of the Free, Because of the Brave
Reply to
Jim Thompson

Le Mon, 05 Nov 2007 08:29:55 -0800, D from BC a écrit:

Why is it getting hot? What are the voltage and parasitics estimated values? What is the source?

If the average DC is high enough and the L/C impedance low enough so as to pose pb, then use an AC termination (series RC, where C is a feew times your C parasitics). Depending on the configuration you can also use source series termination.

Google about TL termination.

--
Thanks,
Fred.
Reply to
Fred Bartoli
[snip]

It's possible.. but I don't think so..

It's part of an smps controller.

I'll post the source of this problem as another topic.. It involves parasitics (inductance and capacitance), hysteresis limitations, current sensing and control loops.

I posted what I thought was an interesting and challenging problem. It also looked like a classic problem..so I thought there might be an already known solution.

The debounce problem was simple to describe... but surprisingly tricky for me to solve .

D from BC

Reply to
D from BC

Last time I studied key-debounce was for my old TRS-80 back in 1980, and they had a nice little solution.

Maybe this might work, I'll use 1 OR gate.

A ------------ OR ---------------- O/P B--- | |-- -------C-----| R gnd

Start with everything at 0.

Input A get's a "1", the O/P goes Hi to "1" and feeds to "B", doesn't matter now that "A" is bouncing about, the O/P stays Hi for the RC time, by the O/P feeding "B" via "cap C". (R can be the gate input resistance).

The gate resets when the RC times out ready for the next "1" delivered to "A". Regards Ken S. Tucker

Reply to
Ken S. Tucker

I don't know what you mean by "hash." If you mean emi, well, any gate that follows any signal generates "hash." Your circuits have lots of logic transitions, too. That's what logic is for, to process signals.

So you're saying that this signal is so bad it shouldn't be allowed to go through gates.

John

Reply to
John Larkin

Hey, Jim, how's the bod?

John

Reply to
John Larkin

That will break it of course, but that's your option.

It does already.

Sure. It's not my job on the line.

John

Reply to
John Larkin

Walking, mild pain, cleaning up the bed rash from 72+ hours lying on my back :-(

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
         America: Land of the Free, Because of the Brave
Reply to
Jim Thompson

--
Not a lot, just what\'s required to set in motion the chain of events
which accepts the first edge and suppresses the rest.

Your circuit sits there generating very high-speed, probably
high-current edges every time the input signal changes state.
Reply to
John Fields

Now, I haven't done a full analysis on this circuit, but the biggest drawback on being this edge triggered is the vulnerability of a noise spike, esp. on that positive edge, giving a false trigger. That was why I always appreciated those Moto debounce chips (for slower stuff) as it debounced both edges equally. You just had to live with the delays...

Charlie

Reply to
Charlie Edmondson

That's what this circuit was supposed to do.

"Probably high current?" Logic does tend to use power when it switches, I suppose.

Yup, sounds like.

Oh please. You can say that about any digital logic.

Joh

Reply to
John Larkin

Yeah, thinking about this, the problem definition becomes, instead of a four state machine, a two state machine:

Detect tranistion, change output state WAIT

Simple, and I am sure, full of gotchas, especially in that WAIT state... ;-)

Charlie

Reply to
Charlie Edmondson

It's problematic to say that any asynchronous machine has states. My circuit has only one flipflop, so I think it has two states: flipflop set, and flipflop clear.

John

Reply to
John Larkin

The wait state just has to exceed the worst-case undefined input time.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
         America: Land of the Free, Because of the Brave
Reply to
Jim Thompson

--
Then, unfortunately, since it didn\'t suppress the multiple edges
propagating through your EXOR chain, it\'s not working like it
should.
Reply to
John Fields

--
Yes, but for best noise immunity (if the period of the input signal
is known) the wait state will end just prior to the next expected
edge.
Reply to
John Fields

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