LED Driver with Darlington

Discussion in 'The Projects Forum' started by aconz2, Jan 5, 2011.

  1. aconz2

    Thread Starter New Member

    Feb 8, 2010
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    I am in the design stage of a multi-LED array to be driven by a TLC5947 (datasheet). Since my planned LEDs have a forward current of 100 mA, exceeding the 30mA that the chip can output, I was planning on connecting the output of each channel to the input/base of a Darlington uln2803 transistor array (datasheet). To my understanding, the TLC5947 has constant current output as low as 2 mA and the uln2803 has an input current max of 1.35 mA. Will this damage the uln2803 and if so, would a fine solution be to just add a resistor? All help is much appreciated
     
  2. #12

    Expert

    Nov 30, 2010
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    At first glance, the 5947 only dumps current to ground. The 2803 requires a positive voltage to enable it to dump current to ground. Directly connecting these would not work in the obvious manner. You must have a different scheme in mind. Please provide a bit of a drawing.
     
    Last edited: Jan 5, 2011
  3. CDRIVE

    Senior Member

    Jul 1, 2008
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    No, you're misinterpreting the Iin rating. You don't need a base resistor. For a 300mA load you only need to supply a base voltage => +3V. The data sheet simply states that you should not need to supply more than 1.35mA (Iin) to sink 500mA with an input voltage of +3.85V.
     
    Last edited: Jan 5, 2011
  4. aconz2

    Thread Starter New Member

    Feb 8, 2010
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    Yes it seems I have been mistaken because what I thought was a constant current output is actually a constant current sink. Thank you for pointing out that out. Any way to be able to use these two chips together?
     
  5. #12

    Expert

    Nov 30, 2010
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    You basically have to do an invert, one way or another. If you can reverse logic on the 5947, you can bias the darlingtons always on with pull up resistors to V+, then let the 5947 turn the LEDs off. Another way is to get (buy) PNP Darlingtons, keep them biased off with the pull-up resistors and run the 5947 as you expected to.

    I also wanted to mention that because the 5947 can dump 30 ma, you don't need Darlington quality transistors to run a 100 ma load. You could use single PNP's at a gain of only 3.3! or run the 5947 at 10 ma and expect a gain of 10 from the PNP stage. Maybe that will make buying the PNP stage cheaper.
     
  6. CDRIVE

    Senior Member

    Jul 1, 2008
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    The output pins (Out1 - Out23) appear to be 'Open Collector' so you will need to invert this. Please see attached circuits. The top one shows how to interface a PNP to the ULN2803 and the bottom one shows how to eliminate the ULN2303 and drive the LED directly from the PNP.
     
  7. aconz2

    Thread Starter New Member

    Feb 8, 2010
    16
    0
    Those two circuit suggestions were great, thank you. I am trying to avoid using too many excess components and it seems like an inefficient way to do things. I have also heard that PNP transistors are slower than NPN and this may have an effect because of the fast PWM signal (I could be wrong). I looked at the uln2981 (Datasheet) and I think this would be a solution to use.
    I think my options are down to two things: using the tlc5947 to control the uln2981 or other PNP transistors, or use a chip similar to the tlc5947 (onboard PWM and shift register type data input) to control the uln2803.
     
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