Delayed Pulse

Thread Starter

dieSTOOF

Joined Aug 23, 2013
10
Hi all

How can I create a delayed one-shot negative pulse that will auto reset after a few milliseconds.The inputs and outputs must look as follows.

Input:___________|""""""""""""""""""""""""""""""""""""""""|_____________
Output:""""""""""""""""""""""""""""""|________|"""""""""""""""""""""""""""""""""""""""
..........................|--50ms--||--20ms--|

Thanks
 

#12

Joined Nov 30, 2010
18,224
It is easy enough to use an LM556 chip to delay the start, and then do a low output, and return to high at the output.

The question is: How long do you want the device to ignore a second activation on the input? Until the power is removed and later restored? Forever? For 10 seconds?

It seems clear that you can not get your finger off a button in 70 milliseconds...or is this answering to something other than a button? What is the input device?

While you are answering, you can include things like the power supply voltage, how much current the load needs, and such as that.
 

Thread Starter

dieSTOOF

Joined Aug 23, 2013
10
Thanks for your quick reply

I have a PLC with relay outputs. I am using a CD4514 IC to decode 4-bit outputs to 16-bits(to increase my PLCs outputs).

The problem is that mechanical relays/switches bounce which will cause the IC to trigger multiple times as it recieves multiple pulses.

The relays is also a few milliseconds out of sync with each other which will cause incorrect triggering.

Lets say I want 0101 as input to the IC. Bit 2 triggers 3us before bit 4 which will cause a reading of 0100, then 3us later 0101 will be seen as input. This combined with bouncing is a disaster.

The IC has a Enable input which allows it to only read the 4-bit inputs when pin-23(enable) is low.

I want to connect the relay outputs with diodes to trigger this delayed pulse to give all the relays time to Turn on or aff as necessary, then the pulse must FALL to enable the IC to read the inputs.Then the pulse must go high again before the relays turn off so that no false readings can occur because of the out of sync relays.

My supply is 12v dc, the current is micro amperes becuase the IC is cmos.The Loads is Connected later through transistors
Thanks
 

#12

Joined Nov 30, 2010
18,224
Something triggers. Is it the PLC, and which direction is the trigger, high or low?
How much current is available to trigger the delay timer? A hundred microamps, a whole milliamp, 10 or 20 milliamps?

I am trying to find the sensitivity needed for the delay circuit input.

ps, there are ways to debounce in hardware, and software. Your question seems to indicate that you know your circuits limitations and asked for what you need, but I must ask if you are aware of debouncing with a debounce circuit or a programming method.
 

Thread Starter

dieSTOOF

Joined Aug 23, 2013
10
The PLC has internal relays as outputs(you can google Siemens logo! 0ba6).Software debouncing is thus not possible through programming(as far as I know).

The internal relays can be used as regular switches.It has +12V in and switches the 12 V to the output pin of the relay.

Because the relays is just basic switches the current can easily go over a 100mA but I would prefer a circuit that use less than 20mA if possible.

The delay before the falling pulse is triggered is also used as a debouncing circuit.
Thanks
 

#12

Joined Nov 30, 2010
18,224
OK. If I mess up, strantor is good with PLC's and he will straighten me out.;)
It takes more than 10 minutes to design these. Take your lunch now.:D
 

#12

Joined Nov 30, 2010
18,224
It would be good to go to "User CP" and type in your location.
After I post this, the others will examine it for misteaks.
 

#12

Joined Nov 30, 2010
18,224
Better wait. We have some very sharp people here and it's rare to do this many connections and get it perfect the first time.
 

Thread Starter

dieSTOOF

Joined Aug 23, 2013
10
Hi
The circuit works!!!!!.

I removed the .02uF cap at the input,otherwise the IC will only trigger once and not again if another input pulse is applied.

I managed to adjust the LOW time of the output.Is there a delay before th H-L state change occur?(I dont have an occiloscope to look at the output in detail,my eye is too slow to see with an LED)

Thanks agin for all your time and effort
 
Last edited:

#12

Joined Nov 30, 2010
18,224
When the input goes high, the first timer instantly starts clocking out 50 milliseconds. That is the delay before "low" happens. When that gets done, the second timer starts clocking out 20 milliseconds. When that gets done, the output returns to "high". At least, that is what I intended...

The input capacitor was intended to keep you from doing an immediate re-trigger. The 100k to ground should discharge the .02 capacitor in a couple of milliseconds so it is then ready for the next start signal. I guess there is something about the PLC that I don't understand. Because you figured out how to correct that, Good enough.
 

Thread Starter

dieSTOOF

Joined Aug 23, 2013
10
I guess I can Manipulate the delays with the R and C at the 1,2 pins and 13,14 pins.I havent connected the PLC yet.What I did is I built the circuit on a breadboard and used a jumper wire as a switch.

I think the reason that the bouncing of the wire didnt retrigger the Ic is because of the delay before the fall occurs.The delay time is also a debouncing mechanism(I think)
 

#12

Joined Nov 30, 2010
18,224
Yes. That's pins 1,2 and pins 12,13. The R and C connected there set the time. Time = 1.1 RC. The data sheet says any monostable circuit will not accept a re-trigger until it times out. That means you have about 45 milliseconds to get the (high) trigger signal off the input and let it fall back to zero volts.

I was assuming the PLC would just pulse the input high for less than 45 milliseconds. I don't know the minimum trigger time, but the propagation delay is at least 100ns, so you should probably keep the trigger high for at least 200ns...and the maximum time limit for high input is about 45 milliseconds.

Now that you say it won't re-trigger with the capacitor on the input, I come to doubt you have the 100k to ground properly. That resistor is responsible for discharging the input capacitor and it should be done in about 5 milliseconds. The 100k to the base of the input transistor is also a load on the capacitor and shortens the delay to about 3 milliseconds. Better re-check that.
 

Thread Starter

dieSTOOF

Joined Aug 23, 2013
10
I checked the resistors and cap.Everyting looks right(even changed the holes they were connected to),but it still was only working once.

However I found that if i put a 100k between ground and the input,it worked as you said.I guess that is because the PLCs outputs relays work only as normal switches.They are not pulled to ground when they are open.As far as i know th cap cant discharge if one end is open/not connected in any way to the other pin.
 

Thread Starter

dieSTOOF

Joined Aug 23, 2013
10
Hi

I said that the circuit is working.Actually it doesnt work completely.The delay before the faling pulse is triggered is super short.By changing the cap and resistor has almost no effect.when the cap and resistor has very high values the delay can still not be seen by the human eye.

I found a similar circuit online that works perfect.

Just add an inverter to the output like #12 did to create a falling pulse.
 

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