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Thread: Could someone look at this circuit?

  1. #1
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    Could someone look at this circuit?

    Hi I am workign on a circuit that will case a delay of a few seconds before causing a relay to set off the power switch on my laptop I am unexpierienced whit making circuits and I was hoping somone could take a look at it and tell me weather or not it will work/explode. thanks, explination below the image.




    I am afraid i dont really have any tools for drawing this stuff so thats about the best i could do this is designed after another similar circuit a found online. the acc wire runs into a 1 amp fuse(F1) then a 1 amp rectifier diode(D1). then the wire splits and a 18v capacitor(C1) causes the delay depending on the uf and the other wire runs into the relay normally. I hope this makes sense. as I said i am pretty new to wiring things like this. So anyone who knows what there doing and is willing to comment on weather or not this will work I would be very appreciative to you. thanks.

  2. #2
    Constant Bitrate
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    look for some timer, like this one: http://magnecraft.thomasnet.com/item...=prod&filter=0

    You set it up as OFF DELAY timer, set delay for let's say 30 sec, and it will do the job for you.
    It's cheap and it will work without any additional parts.
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  3. #3
    FLAC
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    Quote Originally Posted by dupa2 View Post
    look for some timer, like this one: http://magnecraft.thomasnet.com/item...=prod&filter=0

    You set it up as OFF DELAY timer, set delay for let's say 30 sec, and it will do the job for you.
    It's cheap and it will work without any additional parts.
    cheap....

    http://www.newark.com/jsp/displayPro...K=821TD10H-UNI

    ok, if you say so. I mean compared to your usual industrial time-delay relays, sure it's cheap. but compared to a cap, a resistor and some elbow grease, it's a lot.

    ___________

    In order to make a delay circuit, you will need to know your relay's dropout voltage, and coil side resistance.
    The relay-coil side of your circuit is much like the example circuit in this tutorial, with the coil of the relay replacing the resistor. The idea is that with supplied power connected, the cap will charge to supplied voltage, and will stay there until voltage supplied lowers below the voltage supplied by the cap. Once this happens, tthe capacitor will discharge, but it needs a path for current to flow in order to do so. that path in your circuit is through the coil of the relay. The time that it takes to discharge is inversely proportional to the current that can flow I=V/R. if it discharges at a lower current it will last longer. But keep in mind, as it discharges it's voltage drops. The result is that the voltage decays at a non-linear rate. The rate can be found by multiplying the series resistance (coil) times the capacitance of the cap (uF). The result is time in seconds and is called the time constant. Time constant=R*C. For our purposes it can be assumed that it takes 5 time constants for a cap to discharge completely, once power is removed.

    Each time constant lowers voltage by about 63 percent of the previous value.

    if coil resistance is 200 ohms and dropout voltage is 1.2V and is at 13.8Volts and capacitance is (2 4700uF capacitors in parallel) 9.4mF.

    RC= (.0094)(200) = 1.88 seconds
    this is the time that it takes for the capacitor to discharge to 37% of the previous voltage which was 13.8. V=13.8*.37=5.106

    If you know the voltage you want it to decay to and want to know how long it will take to decay to that value, use this eqn. V(t)=V0e^-t/RC.

    t= (RC(-ln(v(t)/V0))

    I'm attaching a excel file that you can use to find an appropriate capacitor or combination of capacitors in parallel to do the job.
    There are also many calculators online for this.

    If you find you want to change the total resistance to influence RC, than maybe
    add a small resistor in series with the relay, but keep in mind, this will lower the voltage across the coil. Be sure to not reduce voltage so low as to cause the coil not to pick up. Pickup voltage is usually much higher than dropout voltage.

    Just click the picture if you want it.

  4. #4
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    Thank you thats very helpfull especially the excell sheet. So it seams like i can just put a capacitor before the relay and maybe a fuse before that for saftey and it will give me a delay. Now if i can get the delay I want without using a resistor is there any advantage to using one or am i just as good without it? Right now i view the circuit as:

    12VDC----1 AMP fuse---4700UF CAP-------- Relay coil 400ohms--------Ground

    according to the spreadsheet this will give me about a 3.6 second delay which is what im looking for Thanks for all the help so far youve saved me quite a bit of

  5. #5
    FLAC W3bMa5t3r's Avatar
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    you can check out the circuit in the laptop auto turn on thread link in my sig. By changing the capacitor and resistor values you can change the delay.

    Cheers.

  6. #6
    FLAC
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    Quote Originally Posted by W3bMa5t3r View Post
    you can check out the circuit in the laptop auto turn on thread link in my sig. By changing the capacitor and resistor values you can change the delay.

    Cheers.
    Nice, I never noticed that before. I like it; simple, open source, and functional.

    OP, if you want a momentary on signal, I'd recommend doing it like that. The circuit you have will be such that the relay will turn on, delayed after power is applied, be on until power goes away, and then drop out after a delay. I don't think I paid enough attention to how you wanted to use it.

    Also I didn't have a section in the spreadsheet that shows the pick up time based on the capacitor, it requires a pickup voltage and a slightly different equation. I'll add it sometime this morning.

  7. #7
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    great thanks both of you guys. I think with a little more research it will soon be time to start playing around with this stuff.

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