Trigger pulse from a push button

That Schmitt-trigger ascii-schematic had tabs in it. Sorry. This should be a working one:

+V | \ / 2R | \ \ | \ / | \ / R | | ____\ +---/ ---/\/\--+---------| / / +---- | | | _/_/ / \ ----- C | / ----- | / | | / \

--
Rikard Bosnjakovic                         http://bos.hack.org/cv/

Anyone sending unwanted advertising e-mail to my address will be
charged $250 for network traffic and computing time. By extracting
address from this message or its header, you agree to these terms.

If you have any 4050 CMOS buffers, you can easily have 6 debounce circuits.

....

--- The reason the circuit didn't work was because you were charging the capacitor with the switch, but never discharging it, so the best you could do would be to get one pulse out of it; the first. The right way to wire it up if you want a positive pulse out of it is like this:

+V>----> | S1 | O--+--[C1]---+-------+------>OUT | | |K [R1] [R2] [DIODE] | | | GND>--------+---------+-------+------>GND

-- What happens is that when the switch is made, the fast rising edge of the waveform appearing across R1 will be differentiated by the capacitor and will appear across R2. However, if the arm of the switch continues to remain connected to +V the capacitor will charge up to +V and the voltage across R2 will drop to zero volts because the voltage on the input of the cap isn't changing, and DC can't get through a capacitor. When the switch is released, or if it bounces, the end of the capacitor connected to R1 will discharge from +V to 0V, and, since the end of the cap connected to R2 was sitting at zero volts, that end of the cap will go _negative_ when the cap discharges, because that's the direction the other end of the cap went. That negative pulse will be clamped to about -0.7V by the diode, and the 'K' denotes the diode's cathode.

When the switch is made, a positive spike with a voltage of +V will appear instantaneously across R2, and as long as S1 stays made, the spike will start to decay and will reach +V/3 in one "time constant", which is the product of R2 and C1. So, if you wanted a pulse that would rise to +V and then fall to one-third of +V in, say, a millisecond, you'd choose the values of R2 and C1 to make that happen.

Let's say you had a 1µF capacitor laying around and you wanted a 1ms pulse. Then, since:

T = RC,

you could rearrange it and write:

T 1 E-3 second R = --- = -------------- = 1000 ohms C 1 E-6 farad

and that would be the resistance you'd need.

But... there's a problem in that you will _always_ have the switch contacts bouncing around when they make and break, with the result that instead of a nice neat single pulse coming out of C1 you'll have a lot of garbage which will enventually settle down, but which will cause a lot of clocks to be counted up if you're trying to use the circuit to generate clocks for a counter. What to do? Debounce the switch, and since you've got some logic laying around, you can do this:

+V>----> | S1 | O | +--[1N4148>]--+----A | | Y------>OUT +---[1kR]-----+----B | | [100kR] [0.1µF] | | GND>-----+-------------+-------------->GND

ABY is a gate, an AND or an OR if you want a positive-going output when you press the switch and negative-going when you release it, a NAND or a NOR if you want the opposite.

what happens is that when you press the switch the cap charges up in a few microseconds (probably when the contacts first touch) through the diode, but it takes about a tenth of a second for the cap to discharge through the resistors when the switch is released, so as soon as you press the switch "Y" goes high, but it won't go low until

100 milliseconds after you release the switch. That way it debounces both the make _and_ the break.

-- John Fields

One problem I have is not knowing what you have on hand or can get in a shop just up the street. You can do this with a comparator or a couple of transistors. The gate circuit John Fields suggested will work well too, as long as it's a CMOS gate. TTL family inputs will almost float high, and don't take much of a pullup, but you have to pull them down pretty decisively; so you'd want the switch to ground, and maybe 4K7 to 10K to

+Vcc and to the input - in other words, kinda turn his circuit upside down and sink current rather than sourcing voltage. :-) But only for TTL-style inputs.

Good Luck! Rich

The 555 makes for a good general-purpose switch debouncer. The two inputs TRIG and THRESH switch at 1/3Vcc and 2/3Vcc respectively with TRIG making OUT high when pulled below its threshold and THRESH making output low when pulled above its threshold. So if you use the circuit below, switch closure pulls the node voltage below TRIG threshold causing the output to go high immediately and it stays that way until switch release. Then when switch opens, capacitor charges up through R towards Vcc and triggers THRESH when it reaches 2/3Vcc causing output to go low and stay there. The equation for the time it takes to charge C from 0 to 2/3Vcc is 1.1*R*C- therefore, in order to debounce the 555 output, this charge time must exceed worst case switch bounce. Making

1.1*R*C~50ms will do it most of the time. Using C=0.22uF and R=220K is close enough, making 1.1*R*C=53ms. The 555 always puts out a logic compatible pulse and this will work just fine with edge triggered TTL.

View in a fixed-width font such as Courier.

. .. .. VCC .. | .. R | .. +---[220K]-----+---+ .. | | | .. | +-------------LM555 .. * | | RST VCC | .. R | | | .. +---[100]------+----+-----|TRIG | .. | | | | OUT |----> PULSE .. | | +-----|THRESH | .. | | | | .. |o | | | .. -| === | | .. |o 0.22U | CTL GND | .. | | +-----------------+ .. | | N/C | .. | | | .. +--------------+---------------------+ .. | .. --- .. /// . .. . .. .. * added to limit discharge current thru switch . .. . .. . .. | .. OPEN +------- ------ .. | || | .. | || | .. | || | .. SWITCH | || | .. | || | .. | || | .. CLOSE + ---------- o o o - .. | .. -------------------------------------- .. -> | | | | . .. . ..

...

--
Nice one!  I didn\'t even think of TTL. You\'re right, of course!

--
This is better, since there's no need for the diode:

+V>----> |
      S1 |
         O
         |
         +----[1kR]----+----A  
         |             |       Y------>OUT
         |             +----B
         |             |         
       [1MR]        [0.1µF]      
         |             |        
GND>-----+-------------+-------------->GND