555 timer delay switch!

Discussion in 'The Projects Forum' started by ben, Dec 6, 2013.

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  1. ben

    Thread Starter New Member

    Aug 20, 2006
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    Hi wondered if you could help

    I have purchased a NE555 Timer Delay Switch 1-300 second 5-12v DC input TIME OVER ACTION Circuit board off ebay. I wonder if you could help me wireing up? I know i need a permenate live a neutral and a live with A swich so i can activate the timer. once the rocker switch has been activated the timer should run for 1-300 seconds and then it switches off. But im UNSURE where the wires should go?
    This link shows a picture of the board

    http://www.ebay.co.uk/itm/331044972...X:IT&_trksid=p3984.m1439.l2648#ht_1239wt_1395

    Its for my nephews farm yard.
    hope you can help!
    thanks ben
     
  2. ben

    Thread Starter New Member

    Aug 20, 2006
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    or at least could someone tell me which terminal should have the permenate live?

    At one end there are terminals called GND & VCC and the other end of the board termilals CB, COM, CK
     
    Last edited: Dec 6, 2013
  3. Dodgydave

    AAC Fanatic!

    Jun 22, 2012
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    What you have there is a power on timer, when the DC power of 5 to 12V is applied on the Orange terminals, the relay will activate after the set time by the preset resistor, the relay terminals are on the Green connector marked Com, NC, NO.

    So if you want your load to switch off after the set time, you need to use the Com and NC terminals, if you want the load to switch on after the delay, use the Com and NO terminals.
     
  4. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    In case you missed it, here it is again:

    This board expects an input of 5 to 12VDC. It is capable of switching up to 250VAC at 10A at the output terminals.

    Where are you going to get the low voltage DC input?
     
  5. elec_mech

    Senior Member

    Nov 12, 2008
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    Before you connect anything, allow us to draw you a wiring diagram. Before we can do that, can you elaborate on your set-up?

    From what you've posted so far, you want to flip a switch to turn the timer on, correct?

    And this switch will control AC power, such as from an outlet? Or will it come from a set of batteries? If the former, you'll need to use a wall wart to drop the voltage down and convert it to DC in the 5-12VDC range.

    What is the load you want the timer to control, e.g., a light, a heater, a motor, etc?

    What type of power does the load require, e.g., 240VAC, 120VAC, 9VDC, etc.?

    What is it you want the timer to do: a) turn on load for x seconds then shut off, b) turn on load after x seconds and stay on?
     
  6. ben

    Thread Starter New Member

    Aug 20, 2006
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    basicly i want to press a switch on which then turns on some leds on for a set time and then they turn off. the voltage is around 7v (6x 1.2v batteries) very little load on batteries maybe 15LEDS x 20mA. So what your saying is positive and neutral go to the orange terminals. Then the leds leads go to the Com and either NC or No?

    To activate the process again i would have to EITHER switch the power supply off and on again or by pressing a momentary switch? hope that makes sense!
     
  7. Dodgydave

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    Jun 22, 2012
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    Yes, you will need to remove the power to the timer to repeat the process, and yes you connect the leds via the com and nc.
     
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  8. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    What is your 'power supply'?
    Are you planning to power the whole circuit from the batteries? If so, do you know how much current it draws?
    Are you aware that LEDs need to be current-driven rather than voltage-driven?
    Depending on the Vf of the LEDs, 7.2V may not give enough headroom for driving them in strings of 2-in-series-per-string. If so, a 4.8V supply would waste less power.
     
    Last edited: Dec 6, 2013
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  9. elec_mech

    Senior Member

    Nov 12, 2008
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    We're still at a bit of loss as to where power is coming from. You initially mentioned hot and neutral which are terms typically used for AC mains.

    I assume what you want to do is use a single battery pack to turn on some LEDs for a short time after you flip a switch. Is this correct?

    If so, I think the diagram below will accomplish this. Since you mentioned a farm yard, I assume these are white LEDs. If you are wiring these yourself, I've shown how they should get wired based on a 7.2VDC supply voltage.

    As Alec_t mentioned, depending on the forward voltage requirement of the LEDs, you may have to wire them all in parallel (as shown below). Unfortunately, this will consume the most amount of current and each LED will draw ~20mA, 15 x 20 = 300mA current draw from the batteries.

    Is this an LED light with a solar rechargeable battery pack? Did this come as a set together?

    As Dodgydave mentioned, the circuit you purchased is designed to be triggered at power up, meaning the timer will start as soon as power is applied. Once the timer is finished, the only way to restart the timer is to remove then re-apply power. You will have to use a latching switch. If you attempt to use a momentary switch, the timer will stop as soon as you release the switch since power will no longer be applied.

    If you're feeling adventurous, we can show you how to add a couple of passive components so you could use a momentary switch. We'll need good, clear pictures of the front and back of the circuit.
     
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  10. ben

    Thread Starter New Member

    Aug 20, 2006
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    the timer switch seems to be working fine, the only problem is once the timer has been activated and it runs my leds for the set time the led on the circuit board comes on and continues to stay on untill the circuit is re-activated. can i remove the led to stop it draining my battery pack?? hope you can help
     
  11. elec_mech

    Senior Member

    Nov 12, 2008
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    So the LED on the board comes on after the timing cycle has finished?

    In any event, yes, you can remove the LED - it should not affect operation.

    Glad to hear it is working as you wanted. :)
     
  12. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Even if you remove the LED, you will still likely drain the battery because of the standby current it takes to operate the 555. The circuit wasn't meant to be powered with a battery. Locate an old AC-powered DC output wall-wart power-supply from an old cell phone, etc...
     
  13. jason barsby

    New Member

    Nov 14, 2014
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    DRL circuit.jpg
    As a follow up to this, and in case it helps anyone, I have just bought and used one of these circuits for a car lighting application.
    I recently fitted DLR (Daytime running lights) to my car. They need to be fed from the ignition so that they are on all the time the ignition is on.
    That's fine, but they also (evidently) have a low voltage protection circuit, to protect the LEDs I suppose.
    So what happens is that you turn on the ignition, the DRL lights come on, you start the engine (putting a strain on the battery and lowering the voltage available to the lights) and they go off! ...and remain off until the end of your journey. not much use.
    So I used this circuit, to create a delay before the lights are switched on.
    The thing to bear in mind is that you use the Com and CB for switching the lights, but your power feed to the common terminal MUST COME FROM THE IGNITION TOO, because whilst the CB terminal comes live after the delay (which is what we want) it is ALSO live at all times the circuit is unpowered. so if you take your 12V feed from the battery, the lights would be on all the time, all day and all night.
    (When I say "comes live" I do of course mean the connection is made between CB and Com)
    I hope that helps someone.
     
  14. bertus

    Administrator

    Apr 5, 2008
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    Hello,

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