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ULN 2803

ULN 2803

This question is not answered
Nathan Hancock
Posted by Nathan Hancock
on Mar 03 2012 16:05 PM
Prodigy40 points

  I'm using a ULN 2803  to source power to a non-inductive motor.  It has 3 leads for voltage source, ground, and control.  The motor would be normally off.  It would only be on for about 1 sec. when a switch is pressed and then would be off again.  I'm using a 5.2V source.  The motor itself has about 3.3 ohms of resistance.  I find that if I put a 10 ohm, 2 watt resistor between any one of the uln's outputs and the voltage source I measure about 450mA between the output pin and the ground.  

  I'm using microcontroller with a switch on it's input that sources 20mA from its output.  It seems to me that a ULN’s output is always “on” as long as there is a load present.  Can the output be held normally low? If that is so and if the output from the microcontroller is connected to the input of the ULN 2803 can the transistor in the ULN 2803 be used turn the ULN 2803 high momentarily?

  Can six of these outputs be hooked together in parallel to produce 2.7A?  That’s over 12W of power.   Is there a wattage limit that the 2803 can handle or is just limited to 500mA of current per output?

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  • Ron Michallick
    Posted by Ron Michallick
    on Mar 03 2012 16:47 PM
    Mastermind30290 points

    Nathan,

    The ULN2803 is normally off. The collector output pin will not pass current and the output voltage will be high if a load is placed between the collector output and a positive voltage. A positive voltage placed on the base input will turn on the collector which will pass current to ground.

    Outputs can be paralleled for more output current. At no time should the total current exceed 2.5A (ABS MAX table). The average power dissipation should be limited so that the junction (Internal) temperature does not exceed 150C (105C preferred) (ABS MAX table).

    With a 5.2V supply and a collector voltage near 1V gives a motor current of 1.27A. Using all eight channels is 160mA per channel.
    1.27A * 1V = 1.27W dissipation. For DW package Tj = Ta + 74C/W * 1.27W = Ta + 94C. That is 119C at room temperature (25C) for continuous operation. The DIP (N) package runs cooler.

    Because the motor only runs for 1 second and is off much longer. This should not be a problem.

    Regards,
    Ron Michallick

    Regards,
    Ronald Michallick
    Linear Applications

    TI assumes no liability for applications assistance or customer product design. Customer is fully responsible for all design decisions and engineering with regard to its products, including decisions relating to application of TI products. By providing technical information, TI does not intend to offer or provide engineering services or advice concerning Customer's design. If Customer desires engineering services, the Customer should rely on its retained employees and consultants and/or procure engineering services from a licensed professional engineer (LPE).

     

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