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Timer

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avz

Member
Hi all
I would like your help. I want to make a timer that when I wiil supply voltage to it, it will change the output status (go high and remain high until the supply voltage removed and/or giving it a pulse to change its status) after a delay of about 7-8 seconds. I believe that it can be done with a 555 but since I dont know it well, I would like your help and a short explanation about the circuit.
Thanks.
 

MrDEB

Well-Known Member
I looked at your site

really nice but the videos for beginners?? described in the text.
will be visiting often for sure.
 

MrDEB

Well-Known Member
On avz subject

need more info to know what, why, wanted outcome
simply turning on a switch and leaving on for 7 seconds then turning off is not a problem but your description is very vague.
NEED MORE INFO
 

confounded

New Member
If you want a timer that gives low output until it recieves a pulse then it goes high for 7-8 seconds then you can use a 555.

You need to use it as a monostable. Pin 2 of the 555 is the trigger, when it goes low, the output (pin 3) goes high for a time determined by 1.1RtCt (time in seconds) then stays output low until another trigger pulse.

As you want a positive pulse to trigger the 555 i've used a transistor to convert a high pulse in, to a low pulse for pin 2.

I'm quite new, i'm sure theres a better way to do this but i think this wil work!

if you want to interupt the output high before the 7-8 seconds you would need to take pin 4 low
 

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avz

Member
Thanks for all the answers.
I'm affraid I was misunderstood.
I want a dvice that will change its output position from let's say, low to high, about 7-8 seconds after supply voltage is applied to it, thats it. without any pulse/signal applied. then it remains in "high" position as long as the supply voltage is connected to it. I hope it is clearer now.
 

confounded

New Member
do you need a clean switch from 0v to your supply voltage or just the logic levels high from low?

If all you need is to reach a certain voltage after 7-8 seconds (enough to trigger 'high' input you can use a simple resistor and capacitor. The voltage out will continue to rise from 0V when power is first supplied until it reaches Vsupply. (63% of Vsupply after RC)

What are you using this for?
 

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MrDEB

Well-Known Member
see if I got this right?

you want to supply power to a circuit, THEN 7-8 seconds AFTER applying power the output goes high?
I think you just want a delay circuit that turns on a relay 7-8 seconds AFTER the power is applied.
How accurate do you want the timer?
My first idea is a 555 timer running in astable mode at 1 hz(power on all the time). when power is applied triggering the 555 to apply pulses to a 4017 which acts as a delay of 7-8 seconds. the 555 applies pulses to a 4017 that counts to 7 or 8 at a 1hz rate.
the 4017 in combo with a transistor etc to turn on the relay.
 

ccurtis

Well-Known Member
Thanks for all the answers.
I'm affraid I was misunderstood.
I want a dvice that will change its output position from let's say, low to high, about 7-8 seconds after supply voltage is applied to it, thats it. without any pulse/signal applied. then it remains in "high" position as long as the supply voltage is connected to it. I hope it is clearer now.
This is otherwise known as a power-up reset circuit, often used to reset a micoprocessor some time after the power supply voltage is stable after turn-on. The MAX16053 does it in one device with the addition of only one timing capacitor, assuming you don't care at what voltage level the delay starts at (Connect IN to the VCC). A 2 uF capacitor gives an 8 second delay. Output changes from low to high after time delay. Works with up to 16V power supply. Tie EN to VCC. Datasheet at:http://datasheets.maxim-ic.com/en/ds/MAX16052-MAX16053.pdf

A 555 can certainly be used as well. I offer one alternative.

Something you might want to consider is that the filter capacitors may maintain a voltage for some time after you turn power off. The result may be that if you turn the supply back on before the filter caps have discharged enough, the time delay circuit may not reset and its output will remain high immediately after you turn the supply back on, without any delay.
 
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avz

Member
Good Morning
Thanks for your responses. Yes ,Mr. Deb it is exactly what I need. sorry if I was not so clear at first. I hope that it can be carried out less compicated with only a 555. Curtis, you have mentioned that it can be done wite a 555, which is more popular and may be more simple then with the MAX. could you please be more specific about it (with a short explanation, please, because I want to understand it, not just build it).
Thanks.
btw. as I mentioned before - a second or two, more or less - doesn't matter.
 
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chingyg

New Member
Dude, what about this, it is a two stage 555 timer. basically you push a trigger the out put will delay for a set amount of time the it will go high for a set amount of time?
On my diagram it is driving the gate of an SCR, just ignore it.
It use an 556 chip which is just 2 555 together.
 

chingyg

New Member
lastscan.jpg

Here is the circuit diagram Sorry about my messy handwriting if you can read it, just tell me then i'll explain.
 
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avz

Member
Chingyg, thanks, but I'm affraid it not what I need. I need it to delay about 7 sec. from the time voltage is supplied to it without any intervention (no pushbuttom, switch or anything). then stay "high" or "on" not for a given period of time, but as long as the supply voltage is on.
 

confounded

New Member
hi chingyg, how does your circuit stop the rising edge of the 1st 555 timing pulse out from triggering the 2nd 555s trigger input?
I've added a sketch to show what i mean.
Could anyone tell me why it doesnt trigger on that rising edge?
 

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ccurtis

Well-Known Member
Good Morning
Thanks for your responses. Yes ,Mr. Deb it is exactly what I need. sorry if I was not so clear at first. I hope that it can be carried out less compicated with only a 555. Curtis, you have mentioned that it can be done wite a 555, which is more popular and may be more simple then with the MAX. could you please be more specific about it (with a short explanation, please, because I want to understand it, not just build it).
Thanks.
btw. as I mentioned before - a second or two, more or less - doesn't matter.
See attached circuit using a 555.

Simply speaking, the time delay is determined by multiplying R and C, which is called one RC time constant. In one RC time constant, the voltage across the capacitor (C) is 2/3rds the supply voltage, as it charges up through R. There is a flip-flop inside the 555 designed to be set when a pair of comparators, also inside, detect that the voltage at the joined inputs (Thresh and Trigger) is above 2/3rds the supply voltage, and reset when below 1/3rd the supply voltage. The output of the 555 is low when the voltage across C is higher than 2/3rds the supply voltage (flip-flop is set), and high when the voltage across C is lower than 1/3rd the supply voltage (flip-flop is reset).

The diode is there to discharge C quickly as the supply voltage drops (when you turn off the power supply), by bypassing R. The 0.01 uF capacitor acts to supress any external noise on the otherwise floating, unused CV input (though, internally, CV is not really floating). The RESET input is held high to prevent false resetting of the internal flip-flop from an otherwise floating input.
 

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chingyg

New Member
hi chingyg, how does your circuit stop the rising edge of the 1st 555 timing pulse out from triggering the 2nd 555s trigger input?
I've added a sketch to show what i mean.
Could anyone tell me why it doesnt trigger on that rising edge?
I know what you mean, I just used a switch of the second 555 output, triger, wait then switch on.
 
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