slow fading LED using a 555 timer.

Discussion in 'The Projects Forum' started by ale2121, Feb 24, 2010.

  1. ale2121

    Thread Starter Active Member

    Mar 20, 2009
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    Hi,

    I'm using this schematic
    [​IMG]

    It's supposed to fade in and out on a timed interval, but in my case, it's simply fading on and then staying on permanently. Is this schematic incorrect?

    the only difference between my components and the components in this schematic are, i'm using a pn 3565 transistor and the 555 timer I've got is a TLC555.
     
  2. Bernard

    AAC Fanatic!

    Aug 7, 2008
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    The output capabilities of the 555 & C555 are quite different, but a 33k load should be no problem for either, so maybe you have a leaky cap not allowing cap v to rise to 2/3 of 9V, so reset level is never reached.
     
  3. Wendy

    Moderator

    Mar 24, 2008
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    Part of the problem may be the gain of the transistor. We've done something extremely similar here.

    [​IMG]

    You must use two separate transistors, and not one darlington. This is because the darlingtons have a built in resistor that we can't use between base emitter.

    A TLC should work, but the 1/3 and 2/3 points aren't quite as precision.

    The gain of the transistor controls the loading in a key point of the oscillator. The difference between one transistor and two is a gain of 150 (one transistor) is about 18KΩ loading and a gain of 22500 (two transistors) is about 2.7MΩ loading.
     
    Last edited: Feb 24, 2010
  4. SgtWookie

    Expert

    Jul 17, 2007
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    The TLC555 just can't source enough current from pin 3. It can only source 10mA when Vcc=15v. However, it can sink quite a bit more current than it can source.

    If you flipped the circuit around, using a PNP voltage follower to control the ground side of the LED, that would work.
     
  5. Wendy

    Moderator

    Mar 24, 2008
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    Actually, the two transistors will address that problem nicely, I think. I was still adding to my original post while you were typing.
     
  6. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    I agree that a discrete (2 transistor) Darlington is needed. Two PNPs or 2 NPNs should work equally well (with the LED installed in the right direction). You could also use a Sziklai pair, but the low-to-high illumination ratio might be less, due to having only one Vbe drop instead of two with the Darlington.
     
  7. Wendy

    Moderator

    Mar 24, 2008
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    Actually the numbers work surprisingly well for a 9V battery. 2.5V for the LED, 1.2 V for the BE drop, and 3V swing for the LED.

    At max voltage (which will turn off the LED on my schematic) the transistor is getting 6V, plus 1.2V (BE), which will leave 1.8V for the LED (which is off). One the other end the transistor base is 3V, add 1.2V BE, which leaves the resistor/LED dropping 5.8V on the LED circuit.

    This creates a smooth transition going all the way dark and back again. I've built this one.

    I think it is interesting the circuit is symmetrical, NPN and PNP work equally well either way.
     
  8. ale2121

    Thread Starter Active Member

    Mar 20, 2009
    71
    0
    Hey all,
    thanks for all the info.

    Here's how I worked it out:
    [​IMG]
    I avoided the 555 all together and went with this similar but simpler circuit.

    PS, someone on here showed me the gEDA site, where I could get all sorts of programs for drawing, simulating and PCB designing circuits. AWESOME.
     
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