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Led lighting in "2 steps" (555 blinker circuit)

Discussion in 'LEDs and Optoelectronics' started by ericd, Oct 26, 2012.

  1. ericd

    ericd

    3
    0
    Oct 26, 2012
    Hi, I just built an LED blinker circuit using a 555.

    I'm using a 1uF capacitor, and 1M resistors for both Ra and Rb (aiming for about 50% duty).
    It's blinking with about the desired frequency; however the LED is lighting up in "2 steps" (comes on dim at first, then goes full brightness).

    I've posted a video of it here:

    I'm assuming this is due to some sort of voltage ramping up on the pin3 (output) of the 555?
    What do I need to do so that the led lights up "cleanly" (i.e. no partial/dim lit up period)

    Thanks
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    I've seen damaged LEDs do this.

    Also note that Ra=Rb does not give you 50% duty cycle. It gives you about 33% (or 66% - depending how you have the output wired). Can you link to your circuit so I can be sure?

    Also 1M is getting a little large for resistors used with a 555. Try 100k and a 10uF capacitor.
     
  3. ericd

    ericd

    3
    0
    Oct 26, 2012
    Hi Steve, I tried several different LEDs and the same thing happened with them all.

    I'm using a slightly modified circuit that allows a duty cycle < 50% -- here's a link to it:
    http://www.williamson-labs.com/555-circuits.htm#timing
    ( the one the right).

    I actually fixed the issue by putting a 0.047uF capacitor between pin5 (control voltage) and ground (some other circuit diagrams just left pin5 floating).

    Can you briefly explain why 1M resistors are a bit large for the 555?

    Thanks
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Ah, OK. The 555 is pretty noisy when it switches. The Glitch on the supply can cause other logic to misbehave. In this case it's apparently causing the 555's comparators to mis-trigger. The capacitor on pin 5 maintains the input voltage on the comparators even if the power supply voltage goes glitchy.

    The specs say that "maximum value is R = RA + RB ≉ 3.4MΩ, and for VCC = 15 V, the maximum value is 10MΩ." see here.

    The reason is the threshold current. As the resistor values increase, the current flowing through this pin can exceed the current flowing through the resistor. It's best to keep well away from this as it will affect the timing before it stops it.

    The specs are assuming the worst case for this, so it's almost always safe. You just don't want to be bitten by it on the rare occasions it doesn't. If you can keep away from it you'll be safer.

    In your case Ra + Rb is not an issue because of your diode. In your case it's only Ra that's important, and 1M << 3.4M.

    I tend to keep the Ra + Rb under 1M if possible.
     
  5. ericd

    ericd

    3
    0
    Oct 26, 2012
    Cool, thanks Steve. I'll try with the 100k and 1uF like you suggest.
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    It likely won't make any difference, but remember that if you divide the resistors by 10 you need to multiply the capacitor by 10 (so 10uF)
     
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