555 LED driver. stand alone capability

Discussion in 'The Projects Forum' started by Fenris, Oct 7, 2009.

  1. Fenris

    Thread Starter Active Member

    Oct 21, 2007
    288
    2
    Hi all

    I have been drafted in to knock up a small LED driver circuit for a couple of props for a large event next month.

    So straight into Bill's e-books! A 555 circuit with 2 lots of super bright LED's in parallel either side of pin 3. (See diagram) I know that parallel isn't ideal but we are looking for a quick and dirty fix if needs be.

    I understand that the 555 can 'sink' or 'source' around 200mA so given the layout and component specs I should be OK shouldn't I?

    The timing is going to be set such that one set of LED's are flashing so fast they appear to be on solid. The other side will pulse at a lower rate and will in fact be controlled by a switch so will only join the primary LED's when selected. There will also be 2 additional UV LED's that will also only come on when switched on but these will be powered directly from the supply.

    Something along the lines of a 2 way DPDT type switch to select the operation wanted and a button to 'fire'.

    My colleague was intending to use 9V but I don't think they would be up to the job or at least for very long. I have suggested 12V minimum to ensure efficient and at least a days worth of 'zapping' the public.

    The 'prop is the blaster from an NSD Dalek. This will be used off the dalek. The arm ball is plastic and will be drilled with lots of small holes for sound to exit from the speaker that is to be installed inside. A small 386 amp will also be inside. The LED's will be mounted on a bespoke PCB at the hot end of the blaster. The driving circuit, sound and battery will be in a case that clips to the handle end with the switches and fire button. 3 connecting cables will hooup the controls to the speaker/amp and lights.



    This will all be coupled up with a sound module for the 'blaster' sound.

    Have I got it right? I thought I would have to use a PNP and NPN transistor at first to drive each set of the LED's.
     
  2. Bernard

    AAC Fanatic!

    Aug 7, 2008
    4,170
    395
    For parallel operation , Vf should be same within each group or use seperate resistors for ea. LED. 555 out to gnd. & +12V are not likely =. Thirsty little buggers.
     
  3. Wendy

    Moderator

    Mar 24, 2008
    20,765
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    You're much better off with resistor per leg. I've seen it done the way you show, but I don't recommend it.

    *****************

    You ever get that LED tester I showed you a while back (or your local version). You can mix and match Vf to increase the chance of success.
     
  4. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Bernard is correct; if you are planning on using just one current limiting resistor for multiple LEDs in parallel, you will have to match their Vf's. Otherwise, one of the LEDs will "hog" more of the current, and quickly burn out. Then the remaining LEDs will very rapidly burn out in succession.

    Here is a simple circuit that will enable you to test your LEDs for their Vf:
    [​IMG]

    If you want a different current through your LEDs, change R1 as:
    R1 = 1.25 / DesiredCurrent, where 10mA <= DesiredCurrent <= 1.5A

    National Semiconductors' datasheet for the LM555 is here: http://www.national.com/ds/LM/LM555.pdf
    Look on page 5 for charts regarding voltage drops at various source/sink currents, and tables on page 4.

    Note that when sourcing or sinking current at over 100mA, you will lose around 2v dropped in the 555's output.

    So if you are using 12v from batteries, you'll lose around 2v across the 555. Just a guess, but your average Vf @ 40mA will probably be somewhere around 3.9v. If you're running three LEDs in parallel, you will have 120mA current flowing through a single current limiting resistor.

    Rlimit = (12v - (2v+3.9v)) / 120mA = (12v - 5.9v) / .120A = 6.1 / .12 = 50.833 Ohms. The closest standard value of resistance is 51 Ohms.
    Power dissipation is 6.1v x 120mA = 0.732 Watts. It is accepted practice to double the wattage requirement, so 1.464 Watts; 2 Watts is a standard value.

    However, if you are going to be running the LEDs at roughly a 50% duty cycle, you can get away with using 1 Watt resistors.
     
  5. Fenris

    Thread Starter Active Member

    Oct 21, 2007
    288
    2
    Hi Chaps

    Thanks for the info.

    I was vaguely aware of the 1 resistor per leg in parallel so I will go with that to keep things in good health long term.

    I haven't got the gadget yet Bill but Sgt's Circuit looks similar to an LED tester that turned up in a google search on LED matching.

    Presumably if I calibrated a moving coil display and used it in conjunction with Sgt's circuit it give me the Vf of an unknown LED and also become my own permanent tester??

    regards

    Fenris
     
  6. Wendy

    Moderator

    Mar 24, 2008
    20,765
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    Yep, it is minor overkill, LEDs aren't that complex. I used a variable pot in series with a 100Ω resistor and a 9V long before I bought the tester. The 100Ω let me measure current, and I adjusted the pot to whatever.

    If you can get your hands on 1/8W resistors you can use them to distribute and balance the load, using larger ¼W resistor further back.
     
  7. Fenris

    Thread Starter Active Member

    Oct 21, 2007
    288
    2
    Hi Bill

    I remember you mentioning your 'device' for LED matching. So in this direction - A home made LED Vf finder and a matching device all in one.

    All I need do is ensure a couple of sockets for the LED's to match them for brilliance side by side. I presume that you need to know the mA that the LED's are rated for to prevent frying anything?

    If you didn't know the mA of an LED you can find it's Vf at a save current level with the tester. But how would you safely work out what mA the unknown LED could be driven at?

    regards

    Fenris (on a tangent)
     
  8. Wendy

    Moderator

    Mar 24, 2008
    20,765
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    That one is a good question. For the most part 20ma seems to be a magic number for both old and new LEDs, but like most generalizations I bet somewhere their is an LED waiting to be fried by it.
     
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