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LED sign help

Discussion in 'LEDs and Optoelectronics' started by mortyBox, Jun 1, 2015.

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  1. mortyBox


    Jun 1, 2015
    Hello all.

    I am trying to make an LED sign with the following schematic. I just cant figure out what value of resistor I need for the transistors. (R1, R2 and R3). I think everything else is solid.

  2. Harald Kapp

    Harald Kapp Moderator Moderator

    Nov 17, 2011
    Welcome to electronicspoint.

    In a way it is. You may see the LEDs blink on shortly, after that the LEDs will stay off. All of them. That is because the bases of the transistors are pulled low by the resistors so the transistors are all off.

    Google LED chaser for a suitable circuit, eg. this one. You may also need to include a "start" button as shown here.
  3. mortyBox


    Jun 1, 2015
    Thank you for your reply.

    I realize the resistors are pulling the transistors out of commission. That's what they are supposed to do. The caps are supposed to put it high again. It is my understanding that I just need R1, R2 and R3 to be the right value to bias the transistors.

    Basically, I'm asking what value of R1, R2 and R3 do I require to properly bias these transistors?

    PS... I have already found the circuit you referenced, but it is based on a 9V supply. I require 24 VDC
  4. Arouse1973

    Arouse1973 Adam

    Dec 18, 2013
    Hello mortyBox
    Can you explain how the circuit is supposed to work? Or how you want it to work.
  5. mortyBox


    Jun 1, 2015
    OK... there are 3 banks of LED's. Bank 1 should power on for a short period, then turn off as Bank 2 illuminates. Then Bank 2 should turn off as Bank 3 illuminates. And should repeat that sequence indefinitely.
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    One of the problems with this circuit is that it may reverse bias the transistor's base/emitter junctions.

    I would recommend either placing a diode in series with the base, or reverse biased across the base/emitter junction.

    The values of R1, R2, and R3 are probably not very critical. Why not start with 10k?
  7. mortyBox


    Jun 1, 2015
    cant find a start button here...
  8. hevans1944

    hevans1944 Hop - AC8NS

    Jun 21, 2012
    You are totally wrong in your understanding of this circuit, which can be found here:
    First, the authors of this circuit say that there will always be TWO LEDs on at any particular time. One will fade off as another fades on, then second one of those two will fade off and the first one will fade on again, like this: OOo, oOO, OoO ... OOo as cycle repeats. I think you said you want your LEDs to illuminate like this: Ooo, oOo, ooO ... Ooo as cycle repeats. Perhaps you want the transitions to fade in and out, but you didn't say you wanted two sets of LEDs to be on at the same time. The upper-case O is an LED that is on, and a lower-case o is an LED that is off.

    The circuit above apparently does not do what you want, but the only difference I see between it and your circuit is you have the "bias" resistors terminated at ground, providing NO bias to turn any of the transistors on, whereas the circuit above provides some forward bias through variable resistor VR1. Also, the above circuit can use ANY DC voltage for the power source, not just 9V. You would simply adjust the current-limiting resistors for the LED strings accordingly, and perhaps use transistors with a higher Vce maximum rating and/or a higher Ice maximum current rating.

    @(*steve*) raises a valid point: once the capacitors have charged, whichever transistors are conducting will reverse-bias the base-emitter junction of the remaining transistor to turn it off using the charge stored on the capacitor. Transistors are not designed to have their base-emitter junctions reverse-biased and can fail from avalanche breakdown when this occurs. Power transistors may be somewhat less susceptible to this failure mode, but it is poor engineering practice to intentionally reverse-bias the base-emitter junction of ANY bi-polar junction transistor by more than a few millivolts. The datasheet for the transistor you are using should provide the maximum base-emitter reverse bias voltage allowed.

    IMO, it would be better if you were to build a simple three-transistor ring oscillator and use the signals developed by its three transistors to drive the bases of the power transistors controlling your LED strings. This separates the problem of driving current through the LEDs from the problem of building a circuit to control that current. There are many examples of ring oscillators on the Internet. An even better solution would be to use constant-current LED drivers and eliminate the power-wasting series current-limiting resistors.
    Harald Kapp likes this.
  9. Martaine2005


    May 12, 2015
    I have just built (copied) this scenario (circuit) from a youtuber.
    Works as you want it, led matrixes.
    Only difference is it uses a 555 and a cd4017 counter. Very easy circuit and very cheap to build.
    It just utilizes transistors to drive the higher output led matrix.
    hevans1944 likes this.
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