Controlling lamps/LEDs with transistors

Discussion in 'The Projects Forum' started by herg, Aug 4, 2012.

  1. herg

    Thread Starter New Member

    Aug 4, 2012
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    I'm working on a system that was designed to light incandescent lamps in a matrix. When the lamps are replaced with LED bulbs, they light dimly when they're not supposed to. These bulbs include the current limiting resistor and a rectifier to allow them to be used with AC.

    I've narrowed the matrix to a single bulb, 18VDC switched by a TIP107, GND switched by a TIP102, and the controlling circuit. I can turn on the 18V, for 300 usec, off for 1 ms, GND on for 300 usec, off for 1 ms. At no time are both transistors on, but the LED bulb still lights dimly. I have verified with a scope that the controlling logic timing is correct.

    Should I be looking at current leaking through the transistors when they're off? A capacitance source somewhere? Other ideas?
     
  2. JohnInTX

    Moderator

    Jun 26, 2012
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    Frequently, circuits driving incandescent lamps have a resistor across the switch (transistor etc) to pass a little current through the bulb even when its 'off'. This keeps the filament warm (but below incandescence) to eliminate some of the inrush current that happens when cold bulbs are first turned on. This current would be more than enough to light an LED. If you are staying with LEDs, the keep-alive resistors can be removed.

    If no keep-alives are found, you might look at the transistor drive to see if its biased partially on to do the same function.

    Finally, it could be a sloppy design, crud on the PCB conducting some bias into the transistor switch etc.

    Can you sketch a schematic of one of the drivers and post it?

    EDIT:
    If you can't eliminate the leakage current, you can add a resistor across the LED to shunt it and create a current threshold for the LED. The value of the resistor is selected so that with the leakage current passing through it, the resulting voltage is less than Vforward of the LED. For example for Vf = 1.5v and leakage current of 3ma, the resistor value would be 1.5/.003 = 500ohms. Make it a bit less than that and you are in business. It works because the R bypasses the leakage current around the LED. Since the voltage across the R (and LED) is less than Vf, the LED doesn't light. When the driver turns on, it provides more current. When the current through R is enough to cause the voltage drop across the R to exceed Vf, the LED turns on and any increase in drive will flow mostly through the LED. Nice.
     
    Last edited: Aug 4, 2012
  3. herg

    Thread Starter New Member

    Aug 4, 2012
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    No keep alive resistor in this case. Here's a schematic of a single lamp and the drivers.

    [​IMG]

    My breadboarded version doesn't include the feedback circuit, and I'm using a PIC where 374's are shown here. The base resistor on the TIP102 is 1k. Since this is a single lamp out of a 8x8 matrix, shunt resistors would be impractical. They would have to be added at each lamp rather than the PCB.

    I had thought about trying a similar approach of a resistor from the TIP107 collector to ground, but I'm not sure what to do about TIP102 if it's leaky. Maybe a resistor from 18V to the TIP102 collector to feed the transistor it's leakage current rather than it coming through the lamp. The weird thing is that the LED is fully off if I leave the transistor in the off position.

    The matrix throws another wrinkle into it. I can't wrap my head around all the interactions. Finally, I can't enforce LEDs only. It will likely be a mix.
     
  4. Wendy

    Moderator

    Mar 24, 2008
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  5. JohnInTX

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    Jun 26, 2012
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    That pretty much eliminates static leakage. A look at the datasheets for the darlingtons shows all kinds of fun stuff but no switching speed ratings. Since more gain usually means less speed (and you have a lot of gain), I think my next move would be to lengthen the both-off times and see if you notice any difference.

    Another thing to consider is an all-off time between switching rows/cols i.e. turn the current row AND col off, wait awhile then turn the next ones on. How long will depend on the switching speed of the transistors. This is commonly done when multiplexing LED displays to avoid ghosting.. Its something you wouldn't notice or likely need on an incandescent.

    Good luck!
     
  6. herg

    Thread Starter New Member

    Aug 4, 2012
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    Thanks, John.

    I tried increasing the off time, and it does seem to decrease the brightness, but it is still very noticeable with multiple ms of off time. I also tried the parallel resistor approach, and with that, I was able to limit the ghosting enough to make it workable.

    1k from point B to GND and 1k from 18V to point E is where I've settled. Even with only 30 usec between column off to row on, I can only see light with the room lights off. An extra 18mA drawn from the active column and 18mA on each active row is negligible in this system. A single lamp replaced with an LED will offset this. Now, I just need to build up the full matrix with the shunt/bleeder/whatever resistors to see how that works.
     
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