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Constant signal triggering a pulse

itsonlyaname

Feb 2, 2015
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Feb 2, 2015
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Greetings,

I'm looking for something that when receiving a (constant) signal produces a single pulse, does nothing when the signal falls off, and produces another pulse when it comes back.
Is there a chip or (combination of) logic gates that does this?


Some more background info:
Package receiver sends a pulse when it's ready to take a package. (1)
Laser beam interrupt thingy sends a signal when a package crosses it's beam on the conveyor belt. (2)

When (1) the conveyor belt always receives a "start" signal.
When it's going and (2) produces a pulse, it stops.
Then when (1) gives a signal again, the belt must continue even though the laser beam is currently interrupted (that's why i need the constant -> pulse), UNTIL it reaches the next package, giving a pulse and stopping again.

(packages always have some space inbetween)

Thanks in advance
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
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8,393
Hi there and welcome to Electronics Point :)

So you want a circuit that will generate a pulse when an input signal changes from OFF to ON, and another pulse when it changes from ON to OFF?

This can be done with two monostables, one rising-edge-triggered and one falling-edge-triggered.

Here's a simple solution for your case.

272543.001.GIF

This circuit requires a DC power source between 5V and 15V. It is based around a CD4093B quad NAND gate IC with Schmitt trigger inputs, which is represented by the four D-shaped symbols.

The input signal comes in on the left and is inverted by U1D, then again by U1C, producing true and inverted versions. These are fed into two C-R differentiators (pulse generators) using C1 and R1, and C2 and R2. The output of these differentiators are combined by U1B which produces a high-going pulse each time the input signal changes state. U1A is not used.

CD is a decoupling capacitor and should be a 100 nF (0.1 µF) ceramic, connected as closely and directly as possible between the VDD and VSS pins (14 and 7) of U1.

The pulse durations are determined by C1 and R1 (when the input goes low) and C2 and R2 (when the input goes high). C1 and C2 should be the same, and R1 and R2 should be the same. The pulse duration t (in seconds) is very roughly equal to C1 (in farads) × R1 (in ohms). R1 should be in the range 10 kΩ to 1 MΩ (for example, 100 kΩ) and C1 can be chosen according to the desired pulse duration.

Say you want a 0.01 second pulse. The formula t = R1 × C1 can be rearranged to C1 = t / R1. So:
C1 = t / R1
= 0.01 / 100,000
= 1×10-7
= 0.1 µF. (C2 is the same.)

If you need an exact pulse duration, use preset potentiometers ("trimpots") for R1 and R2.
 

itsonlyaname

Feb 2, 2015
2
Joined
Feb 2, 2015
Messages
2
Thanks for your response.

Close, but it only needs to pulse on the rising edge, nothing on falling edge.
Though, isn't there some ready-made circuits around that already do this? It would be impractical to create one myself.
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
Then you just need a single edge-triggered monostable. You can use that circuit; just delete C1, R1 and the associated 10k resistor and link U1 pin 5 to the positive rail. You may prefer to buy a pulse relay that generates a pulse of defined duration when power is applied to it.
 
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