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ULN2803A (darlington array) anyone know what I'm doing wrong?

Discussion in 'Electronic Basics' started by Ben, Oct 4, 2006.

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  1. Ben

    Ben Guest

    Applying 5V to the inputs (power supply is also 5V), I'm only getting
    0.6V at the outputs. I thought it was a faulty chip, or perhaps damaged
    while soldering, but I've tried another one in a holder and got the same
    result. What could be causing this?
     
  2. Eeyore

    Eeyore Guest

    How is it connected ?

    What voltage did you expect to see ?

    Graham
     
  3. Phil Allison

    Phil Allison Guest

    "Ben"

    ** That is correct operation for a transistor in common emitter mode.

    Input high causes output to go low.



    ** What the hell did you expect ??

    The ULN 2803A darlington array is intended to drive relays, lamps, LEDs etc.

    YOU connect one side of the load to + supply and the array will connect the
    other to 0 volts when driven on.




    ........ Phil
     
  4. A ULN2803 is just a bunch of Darlington transistors, emitter to
    ground, and collector to the output pin. They are inverters, so a
    high (+5V) input will produce a low (near zero) output. Also, the
    things don't have anything to pull the outputs high, so, unless you
    supply an external pull-up (such as a resistor to +5V), a low input
    will also produce a low output.



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    peterbb4 (at) interchange.ubc.ca
    new newsgroup users info : http://vancouver-webpages.com/nnq
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  5. Phil Allison

    Phil Allison Guest

    "Peter Bennett"

    ** It won't be near zero volts.

    The voltage drive for the base of second device in the darlington pair comes
    from its own collector - so it will be at least 0.6 volts even at very low
    current.

    The data sheet gives saturation values for Vce up to 1.6 volts at 350 mA.

    http://www.datasheetcatalog.com/datasheets_pdf/U/L/N/2/ULN2803.shtml




    ........ Phil
     
  6. Ben

    Ben Guest

    Thanks for the explanation, I think I'd misunderstood what a Darlington
    array actually is. I was expecting a high input to produce a high
    output, with enough current to drive relays. Does such an IC exist?

    What value of pull-up resistor would you recommend for the Darlington array?
     
  7. Guest

    This is a perfect chip to drive your relays, the coil goes between the
    output and posative supply. You may find a circuit somewhere on the
    data sheet or do a search.
     
  8. Eeyore

    Eeyore Guest

    Do you know what a transistor does ?

    You're fixated on 'ICs' aren't you ?

    What do you need a pull-up for ? The load *is* the 'pull-up'.

    Graham
     
  9. BobG

    BobG Guest

    =========================================
    Be nice. Or just dont say anything. Some patient person will explain
    politely.
     
  10. Ben

    Ben Guest

    I'd need 8 transistors, so a single chip that does the job seemed like a
    good idea.
     
  11. Ben

    Ben Guest

    Yeah, this worked great - thanks :)
     
  12. Jamie

    Jamie Guest

    you don't need a pull up resistor..
    use the Array outputs as the Common/ground source when on.
    they are simply what is called open collector outputs.
    you can supply the relay coil with what ever voltage is required by
    the relay up to the max stand off voltage of the array of course.
    the + side of the relay coil does not need to be the same voltage
    as the Vcc (+ rail) of the chip..
     
  13. Jamie

    Jamie Guest

    Graham:
    the last time i looked at the name on this NG, it said "BASICS"!

    where i come from that means something. maybe you should show a little

    consideration for others that are at least trying, and not criticize

    at every turn.


    (Just an opinion from the peanut gallery)
     
  14. Don McKenzie

    Don McKenzie Guest

  15. jasen

    jasen Guest

    that's about right.
    probably you're using it incorrectly.
    the load should be connected between the output and a positive supply.
     
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