Darlington drivers

Discussion in 'Homework Help' started by wongkenji, Apr 6, 2007.

  1. wongkenji

    Thread Starter Member

    Mar 7, 2007
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    Hi,

    Could anyone explain to me how a darlington driver works? I've read the datasheet of a DS2003 but I'm not sure how it operates. Thanks in advance!

    Ken-ji
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    Here is a short explanation of a Darlington pair configuration that might be helpful. You will need to scroll down to a point near the bottom for the article to get to the description of the darlington pair configuration.

    For the most part, a darlington pair is just a way cascading two transistors to achieve a higher current gain than can be had with a single device. One thing affected by this configuration is that the collector-to-emitter voltage of the output transistor, when it is fully saturated, cannot go as low as a single transistor stage.

    hgmjr
     
  3. wongkenji

    Thread Starter Member

    Mar 7, 2007
    21
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    That seems rather simple, I suddenly feel so dumb. I want to use the ULN2003 block for the darlington driver, do I have to connect pin 9 (common) to anywhere?
     
  4. hgmjr

    Moderator

    Jan 28, 2005
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    To answer that question I need to know what are you planning to uses this part to drive?

    hgmjr
     
  5. wongkenji

    Thread Starter Member

    Mar 7, 2007
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    Oh right.. I'm planning to use it to driver a 15V inverter, don't have the diagram with me now but I'll get it scanned and will post it up later.
     
  6. hgmjr

    Moderator

    Jan 28, 2005
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    Just to be on the safe side, I would recommend you tie pin 9 to your positive voltage supply.

    The diodes are there for those application where the part is used to drive an inductive load such as relays. They prevent the flyback voltage associated with such loads from destroying the transistor.

    hgmjr
     
  7. wongkenji

    Thread Starter Member

    Mar 7, 2007
    21
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    Hmm ok, here's the circuit diagram for the inverter btw: http://kenji.britney.googlepages.com/IMG_2548.jpg

    I need to ask another question, since the chip is has an open collector output, what value of the pull-up resistor should I use if I'm connecting it to a 15V supply?
     
  8. hgmjr

    Moderator

    Jan 28, 2005
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    Are you connecting this device to the input at J2 on the schematic you have referenced?

    hgmjr
     
  9. wongkenji

    Thread Starter Member

    Mar 7, 2007
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    Yes I'm connecting it to J2.
     
  10. hgmjr

    Moderator

    Jan 28, 2005
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    That being the case, the design already has 3K pull-ups (the schematic is a little fuzzy so I am guessing here) to 15V located on the board. With 3K pullups to 15V that would mean that around 4.3 to 4.5 mA will be set up in the opto_isolators located on the board.

    The original designer seems to have felt comfortable with that value.

    Does that answer your question?

    hgmjr
     
  11. wongkenji

    Thread Starter Member

    Mar 7, 2007
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    Erm not really the question I was asking, let me rephrase. I designed this circuit: http://kenji.britney.googlepages.com/delay-schematic01.jpg to be connected to J2 on the previous schematic and was wondering whether the pull-up resistor values I used was correct. Could I reduce them into 1 resistor also? Thanks very much!
     
  12. hgmjr

    Moderator

    Jan 28, 2005
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    I think I have it clear now.

    You want to take the output from the circuit in the schematic referenced above to be used to drive the circuit in the earlier schematic. You would be connecting the signals at J3 of your design to the signals on J2 of the older schematic.

    In that case, you should be able to get rid of all of the pull-up resistors in your design. The pullups on the older schematic are all the pull-ups you need. Make sure that you connect the grounds between the new design and the older design.

    hgmjr
     
  13. wongkenji

    Thread Starter Member

    Mar 7, 2007
    21
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    I'll be connecting J1 to J2. I still won't need the pull-up resistors right?

    Ken-ji
     
  14. hgmjr

    Moderator

    Jan 28, 2005
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    Just so that you are aware of it, the output on J3 of your design will be constantly low when it is not connected to the second board. That is because the second board supplies the pull-up to your board.

    You can use a high valued pullup (around 15K or greater) on your board if you want to be able to make sure that the signals are active without the necessity of connecting it up to the second board.

    hgmjr
     
  15. wongkenji

    Thread Starter Member

    Mar 7, 2007
    21
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    Oh I see, I'll have higher pullups on my board then. I plan to input pulse width modulated signals into J3, that's why I'm using a low voltage supply for it.
     
  16. hgmjr

    Moderator

    Jan 28, 2005
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    If you encounter any problems, please don't hesitate to return to AAC for further assistance.

    hgmjr
     
  17. wongkenji

    Thread Starter Member

    Mar 7, 2007
    21
    0
    Yup I sure will, thanks very much for your help!

    Ken-ji
     
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