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555 or one shot

LESpark

Jun 30, 2018
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New here, I've just posted in the welcome forum,

I need to have a low powered/dim led come on for 1-2 seconds when powered then go off,

12v DC in, switched on, led on for 1-2 seconds then off

What's the best/most simple way to achieve this?

Thanks in advance
 

Bluejets

Oct 5, 2014
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Alternatively buy a ready made module complete with relay for a couple of dollars from Ebay.
 

WHONOES

May 20, 2017
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See attached circuit for a very simple timer.
How it works:
R1 and R2 serve to charge C1. Because C1 cannot charge instantaneously, R3 is pulled up to the +12V rail. turning on Q1 and Q2. Q1 and hence Q2 will turn off when as C1 charges, the voltage at the top of R3 reduces to about 4V.
As well as the time constant of the C1 R3 and R1 combination, the ratio of R1 and R3 determines the point at which Q1 turns off therefore the on period may adjusted by making R1 bigger or smaller.
The circuit was simulated using Simetrix and produced an on period of about 1.5 seconds. The simulation does not take account of the leakage current in C1 thus some adjustment may be need to its value to achieve the required period.
 

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WHONOES

May 20, 2017
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Forgot to mention, LED goes in collector of Q3.
 

LESpark

Jun 30, 2018
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That all seems really complicated, so this 555 one shot thing, I can't just put 12v to it, a switch and a LED on the output?
 

WHONOES

May 20, 2017
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What's difficult about a couple of transistors a few resistors and a capacitor.
 

LESpark

Jun 30, 2018
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Like I said above....Ebay.
Obviously you didn't read that.
I did, and I've ordered a 555 monostable one shot thing, but not enitrent sure what I do with that.

I want the 12v DC when turned on to make it light and then it go out, but the 12v won't be switched off
 

Bluejets

Oct 5, 2014
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That's correct.
Normally power is applied to the 555 unit, then when required, a trigger pulse is given which times out the 555 for the set time.
Which unit did you purchase?
 

WHONOES

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The thing is, you don't learn anything just buying bits off ebay. I thought that was the point of this forum or did you miss that?
 

Bluejets

Oct 5, 2014
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The thing is, you don't learn anything just buying bits off ebay. I thought that was the point of this forum or did you miss that?

Well actually the Op said they wanted the "best/most simple way to achieve this" and so the suggestion.


https://rover.ebay.com/rover/0/0/0?mpre=https://www.ebay.co.uk/ulk/itm/152752199872

This one, hope it's ok for what I need. Do I have to use a pulse to trigger it or can I simply trigger it when I apply 12v to it?

Doesn't appear to need a separate trigger as there is no input for such, however the one you are looking at appears to be a "delay-on" arrangement.
To me that means it will delay the turn on for the set time period and then switch the output on.
 

Bluejets

Oct 5, 2014
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Try this one...a lot cheaper also although you might have to wait a bit for delivery.

Notice the instructions are there also, hope this helps you out...cheers

Only drama I can see would be that the unit is for 0 to 60 seconds and it might be a bit fiddley getting 1 or 2 seconds. you could in that instance replace the on-board potentiometer with a 15 turn unit of the same resistance value. These also come on Ebay for a few cents each.

https://www.ebay.com.au/itm/DC12V-0...801139?hash=item41cbb96cf3:g:6M0AAOSwI0RZe~hl
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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To me that means it will delay the turn on for the set time period and then switch the output on

It appears that the output has the common, NC and NO terminals.

Using the NC terminals, these will stay closed for a short period after the power is connected. And that is essentially what is required.

Sure, the relay stays energised all the time, but as long as this power drain is not a problem, it's not a problem :)
 

LESpark

Jun 30, 2018
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It appears that the output has the common, NC and NO terminals.

Using the NC terminals, these will stay closed for a short period after the power is connected. And that is essentially what is required.

Sure, the relay stays energised all the time, but as long as this power drain is not a problem, it's not a problem :)
Right so, sorry to be a complete novis, but where do I put the 12v to? And where should the + and - of the led be? Do I need any resistors between the +ve output and led?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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One end of the device has a block with 2 connections. Connect 12V to that. Ensure you connect it the right way around.

The other end has a terminal block with 3 connections. Find the two which are connected when the power is off. They may be labelled "common" and "NC". Connect one of these to the +ve of your 12v, and the other to the positive terminal of your load. Connect the negative terminal of your load to the --ve from your 12V.

I just looked more closely at the module you linked to and noticed the terminals were labelled COM, CK, AND CB. COM will be the common connection. I have no idea whether CK or CB is the appropriate other one to use. There is no harm in using the wrong connection, it will just cause the LED to be on when it is supposed to be off and vice versa.

If your load is a simple LED then use a 1k resistor in series with it. Ensure the LED is placed in the circuit the right way around or you may damage it.

When you apply power the LED should remain on for a while before the relay clicks and the LED goes off. If the LED remains on too long adjust the module to set the time you require.
 
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