ULN2003 very confusing problem

Discussion in 'General Electronics Chat' started by ja7me, Jul 26, 2012.

  1. ja7me

    Thread Starter New Member

    May 24, 2012
    18
    1
    Hi
    My name is Jay, I am new to electronics and have stumbled across a problem regarding the ULN2003 Darlington driver IC. Usually this IC is used to sink current, therefore lighting a bulb or relay that is connected to the supply voltage.
    However my question is:
    Can the ULN2003 switch a PNP power transistor, so that it could provide a current source??
    I have provided a drawing of my setup, and gave you it in writing below:
    From what I have read this should be possible by connecting the transistor in the following manner:
    - PNP base to output of ULN2003.
    - PNP collector will be the current source, such as driving a bulb, a 1ohm resistor is also tied to ground from the collector.
    - PNP emitter to positive supply rail.
    - Resistor from PNP base to positive supply rail, to ensure transistor is switched off.
    PNP transistors are turned on by negative voltage to base, the ULN2003 should do this. However it doesn’t, and I am very confused by this!
    Here is some more info about my setup:
    - PNP transistor I have been experimenting with is the D45H8 power transistor (10A).
    - Output of PNP goes to a 20W, 12V halogen bulb.
    - Supply voltage to PNP = 12V.
    - ULN2003 input voltage = 12V.
    - Resistor from PNP base to supply rail = 100K.
    - Pin 8 of ULN2003 connected to 0V
    - Pin 9 of ULN2003 connected to supply rail = 12V

    Thankyou

    Jay
     
  2. Sensacell

    Well-Known Member

    Jun 19, 2012
    1,131
    267
    You are on the right track.

    You need a resistor in series with the base lead to limit the current flow.

    The 1 ohm resistor in parallel with the lamp? is that another load? it serves no purpose other than getting hot.
     
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  3. ja7me

    Thread Starter New Member

    May 24, 2012
    18
    1
    Thanks for your rapid response, I will try the base resistor and eliminate the 1 ohm resistor. I will get back to you if this works or not.
     
  4. ja7me

    Thread Starter New Member

    May 24, 2012
    18
    1
    Ok, I have just tested the same circuit but without the 1 ohm resistor, and put a 100 ohm base resistor in place between the output of ULN2003 and PNP transistor as recommended. However it still does not switch, i get a constant 12 volts at the PNP base.

    I then decided to try just a resistor at the ULN2003 output. The ULN2003 seems to switch correctly in this situation going from positive to ground.

    Is the load i am trying to drive to much for the ULN2003 to handle?

    Anyone have any idea what might be going on?

    Why would the ULN2003 not always ground something that is connected to its output?
     
  5. Sensacell

    Well-Known Member

    Jun 19, 2012
    1,131
    267
    What happens to your collector? if you have a load connected, it should turn on when the ULN2003 output goes low.

    Note that the base-emitter junction effectively 'pins' the base voltage about 0.7 volts below the power supply voltage, the base voltage will always stay high in this circuit. You calculate the base resistor value knowing that one end of the resistor is at 11.3 V, the other end is at about 1 volt, (output saturation voltage of the ULN2003) you have about 10.3 volts across the base resistor when the ULN is on..

    Your lamp is going to draw about 1.6A, (20W/12v) assume a worst-case gain of about 50 from your drive transistor (from the data sheet) 1.6/50= .032A, double that to get that transistor fully saturated, say 60 ma.

    The base resistor should then be 10.3V/.06= 170 ohms.

    The ULN2003 can sink 500 ma from one output, it should have no trouble with 60 ma.
     
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  6. praondevou

    AAC Fanatic!

    Jul 9, 2011
    2,936
    488
    You measured at the base. Measure again at the ULN output.

    Should give you a bit more than 100mA base current with a 100Ohm resistor. According to the datasheet of the transistor it has a minimum DC gain of 40 so it should be ok for the lamp.
     
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  7. Audioguru

    New Member

    Dec 20, 2007
    9,411
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    An incandescent light bulb draws 10 times the normal current when it turns on.
    The normal current is 1.6A so your circuit must provide 16A when the lamp turns on.
     
  8. gerty

    AAC Fanatic!

    Aug 30, 2007
    1,153
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  9. praondevou

    AAC Fanatic!

    Jul 9, 2011
    2,936
    488
    Does it? If the base current is limited wouldn't the current through a initially too low load resistance just be limited too? The transistor would not be saturated...

    If you can initially provide only 4A to the lamp it would slightly light up, heat up and consequently increase it's resistance.
     
  10. #12

    Expert

    Nov 30, 2010
    16,298
    6,811
    or put in the relay you mentioned. It won't have the start surge problem.
     
  11. ja7me

    Thread Starter New Member

    May 24, 2012
    18
    1
    Thanks for your input everyone, a 100ohm resistor at the base of the PNP transistor does appear to work, the 20W bulb lights. When i tried before i dont think i had a solid connection.
     
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