How to trigger a 5 sec delay without using 555 ic

Discussion in 'General Electronics Chat' started by tpny, Oct 15, 2012.

  1. tpny

    Thread Starter Member

    May 6, 2012
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    I want a pushbutton when pressed and released to trigger a 5 second (or x seconds whatever) voltage level change (high to low or low to high doesn't matter) without using a timer ic. Any guidelines? Thanks!
     
  2. RRITESH KAKKAR

    Senior Member

    Jun 29, 2010
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    You can use a relay timed by a charged capacitor or in place of relay you can use transistor..
     
  3. tpny

    Thread Starter Member

    May 6, 2012
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    Yes, I knew I had to have a cap in there, just didn't know how to put it in a circuit to serve this purpose. Do you have a drawing? Thanks!!
     
  4. crutschow

    Expert

    Mar 14, 2008
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    3,227
    What voltage(s) do you have for power?
     
  5. tpny

    Thread Starter Member

    May 6, 2012
    216
    0
    between 14~16Vdc
     
  6. KJ6EAD

    Senior Member

    Apr 30, 2011
    1,425
    363
    You might be able to use a three inverter oscillator and a counter.
     
  7. ScottWang

    Moderator

    Aug 23, 2012
    4,853
    767
    You can use two NPN transistors connect them similar as darlington, but the each C of two NPN connecting a 10K to +V.

    The circuit as this:
    +V → Start Key → 1K → Cap([-] toGND) [+] → 1M Ω → two BJT → C output of second NPN.

    The similar as darlington driver on the top of right side of page 2.
    Take off the Rbe,Re resistors from the first NPN Tr1.
    http://panasonic-denko.co.jp/ac/e_d.../common/catalog/mech_eng_cau_appli.pdf?via=ok
     
  8. crutschow

    Expert

    Mar 14, 2008
    12,991
    3,227
    You could use a single MOSFET which has the advantage of a very high input impedance, allowing a larger value of resistance and a smaller value of capacitance then a BJT.

    The circuit would be +V\rightarrowStart Key\rightarrowcapacitor\rightarrowMOSFET gate & resistor with resistor\rightarrowground. Connect the MOSFET source to ground and the drain through a resistor to +V (resistor value determined by load). This will cause the MOSFET drain output to go momentarily low and then back high when the Start Key is pressed. Try R = 10MΩ and C = 0.5μF to start.

    To reset the capacitor you will need to add a resistor, 1MΩ or less, from the switch output to ground.
     
  9. Dodgydave

    Distinguished Member

    Jun 22, 2012
    4,969
    744
    The relay will operate immediately the push switch is pressed, and stay operated until the capacitor discharges,time depends on coil resistance and capacitor value, bigger values = longer time delay.
     
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