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Sink or Source from Darlington Array?

Discussion in 'Electronic Basics' started by mark risher, Feb 24, 2004.

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  1. mark risher

    mark risher Guest

    Hello,

    I am relatively new to this; please forgive a series of beginner questions
    regarding darlington arrays.

    My project is to use a MSP430 microcontroller to control a high-current LED
    (~300 mA). I think a Darlington array (ULN2003) can help me out here, but
    I'm confused about whether to source or sink from it.

    I plan to connect as follows:

    -Connect an output pin from the micro to one of the darlington's inputs
    -Connect the corresponding darlington output to the cathode of the LED
    -Connect the GND pin of the ULN2003 to ground

    Questions:
    1) Is the above connection correct?
    2) If the darlington is sinking, as above, is there a voltage drop to
    consider? Will it "lift" the ground from 0V to 1.4V?
    3) Do I set the micro's output low or high?
    4) Do I need to limit the current going into the darlington to avoid melting
    the micro? The datasheet doesn't seem to list the amount of current it will
    draw, only the beta I'll get given a certain current.

    Thank you so much,
    /m
     
  2. Ian Bell

    Ian Bell Guest

    I think probably not. IIRC the ULN2003 sinks so you need to connect its
    output to the LEd and the LED to +5V via a current limiting resistor:

    -.- +5V
    |
    --
    \/ LED
    --
    |
    ---
    | |
    | | Resistor
    ---
    |
    |
    . ULN2003 OP

    HTH

    Ian
     
  3. mark risher

    mark risher Guest

    I think probably not. IIRC the ULN2003 sinks so you need to connect its
    That's what I meant by "connect the darlington output to the LED cathode,"
    but it's entirely possible I phrased it incorrectly. Thank you for the
    confirmation on that point.

    Will the Darlington's voltage drop matter when it's sinking in the
    configuration you drew below? If B-E drop is 1.4 volts, does that mean a
    load between +5 and the output of the darlington will see 3.6V?

    Thanks,
    /m
     
  4. James W

    James W Guest

    Vce will be ~0.3V
     
  5. Gareth

    Gareth Guest

    Yes roughly.

    Look at the datasheet, there should be a graph of Collector-Emitter
    Saturation Voltage against Collector Current. This is the voltage
    between the collector and emitter when the darlington is saturated
    (fully on). Looks more like 1.2V than 1.4V at 300mA. I assume you can
    just adjust the resistor value to fix this problem?

    --
     
  6. Ian Bell

    Ian Bell Guest

    I assume you mean C-E drop not b-e. In a normal transistor the saturated
    c-e voltage is less than 0.4V. However in a darlington there is usualy an
    extra b-e drop of 0.7V to add in giving a total of around 1V. And yes it
    does mean the load will see %V less this voltage drop. You need to take
    this into account in calculating the resistor value needed to set the LED
    current.

    Ian
     
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