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Creating a square wave with current strong enough to light a powerful LED

  1. Sep 1, 2011 #1
    Hello! I haven't touched base in electronics in about a year (only took one course too)

    I'm currently working on a project and I'm wondering what the best solution might be to creating a square wave which is 5 milliseconds high, 12 milliseconds low (low as in ground not negative voltage).

    This square wave most power an LED from on of those powerful LED flashlights. The LED itself has to be on 5 milliseconds off 12 milliseconds.

    What do you think is the best way of doing this? Again really rusty on this stuff... Would a 555 be able to do this? - with a transistor to up the current?

    The LED light itself was originally hooked up to 3 AAA batteries (4.5 V total)
    and the circuit was a simple Vin -> 0.75 Ohm Resistor (purple -green - silver -gold)-> LED >- ground.

    do transistors switch on and off as rapid as I would like? or would they lag behind and not fully drop the voltage or raise it for the LED?

    Any other efficient cost effective solutions are welcome. :D
    If someone could even provide a schematic I would be indebted to them (or because I know the layout of the 555 - just what values to use) - also.. I sort of forgot how to even use transistors correctly.. or I am using them and I'm just not sure o_O... Pitiful.

    thanks,

    Lith
     
  2. jcsd
  3. Sep 1, 2011 #2
    Anyone?

    from jost looking at these formulas:
    [URL]http://upload.wikimedia.org/math/a/6/2/a62b875aeb779e3c160ba5eef1de33f5.png[/URL]
    [URL]http://upload.wikimedia.org/math/5/b/9/5b928ccff98539e6b678d2dd3ca0b553.png[/URL]

    It appears you can't have the high value be less then the low one... if the 555 is rigged this way:
    [URL]http://upload.wikimedia.org/wikipedia/commons/thumb/3/3d/555_Astable_Diagram.svg/220px-555_Astable_Diagram.svg.png[/URL]

    Help?
     
    Last edited by a moderator: Apr 26, 2017
  4. Sep 2, 2011 #3

    jim hardy

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    Well set aside the formulas for a minute and think about how the 555 works.

    the cap charges through both resistors but discharges through only one.

    And the 555 inverts, so while cap is low(charging) output is high. That's why high cant be less than low, it charges slower than it discharges. Do formulae agree with that?

    I see two ways around your dilemma:

    What if you bypassed R2 with a diode/resistor combo to speed up charge time?

    What if yu used a second timer (556 is two timers in one package) as an inverter? Tie second one's TRIG and THR together and to first one's OUT, set first one for 5 low 12 high and let second one invert? That'd be more precise because no diode drop in timing circuit.

    old jim
     
  5. Sep 2, 2011 #4
    Thanks! I'll try the two 555s in the morning (its 2am now -_-)

    If it doesn't work out well do you think there are any other alternatives?
    I am really looking for cost effective method (if I'm making a bunch but I also want it to be precise).

    Would it be possible to use a 741? Just throwing ideas around... the more I play around the more I learn.

    Also what is the best way to hook up the LED once I have the square wave I want going?

    Thanks again,

    Lith

    EDIT: I forgot to mention that I am also trying to use the least amount of energy possible to run this circuit.
     
    Last edited: Sep 2, 2011
  6. Sep 2, 2011 #5

    jim hardy

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    how much current do you want to put through the LED?

    Look into LED Drivers madeby TI, National, Linear Technology
    they use switching technology for maximum efficiency - remember energy is stored in inductor vs dissipated in a resistor...
     
  7. Sep 2, 2011 #6
    Well in my first post i said i got the LED from a LED flashlight being powered by 3 AAA batteries (look back at it for more info). The light was pretty bright though.. 120 Lumens I think?

    If I set up the 555s what could I do to get the full power from those batteries to power the LED (again using the square wave) 5 milliseconds on 12 off.

    I'm looking up the LED Drivers.

    Thanks,

    Lith
     
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