need mental check on BJT (darlington)

Discussion in 'General Electronics Chat' started by Gibson486, Jan 15, 2013.

  1. Gibson486

    Thread Starter Member

    Jul 20, 2012
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    I am trying to get max current through an LED by using a darlington.

    Since IE = (B +1)IB and IC = (B) IB, does that mean I would get a a little more (very little) by putting the LED on the emitter side as opposed to the collector?

    Also, I am a little confused....in saturation, does that mean I am getting the max amount of current? Or do I still get the max from the linear region since that is the region it is a current amplifier?
     
    Last edited: Jan 15, 2013
  2. Audioguru

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    Dec 20, 2007
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    A saturated transistor is a switch that is turned on. Its base current is fairly high then the resistance in the circuit determines the collector to emitter current.

    Beta is NEVER used when you want a transistor to saturate.
    Instead beta is used when the transistor is a linear amplifier with plenty of collector to emitter voltage.

    An example is the 2N5308 darlington. Its minimum beta is 20,000 but it saturates fairly well when its base current is 1/1000th its collector current.

    If the load is connected to the emitter then the base voltage must be higher than its power supply voltage.
     
  3. Gibson486

    Thread Starter Member

    Jul 20, 2012
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    When you say the load, do you mean just the LED? I have the resistor at the emitter currently, but the LED at the collector.
     
  4. MrChips

    Moderator

    Oct 2, 2009
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    The load is the resistor and the LED together in series. Put the resistor on the same side as the LED, on the collector side.
     
  5. Audioguru

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    Dec 20, 2007
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    You added a pic so now I also added a pic.
    Your darlington is a linear amplifier that is not saturated. It is getting VERY HOT!

    EDIT: The 10 ohm resistor should be in series with the collector, not the emitter. Then when the darlington saturates its collector switches to a voltage close to ground. The value of the base resistor can be much higher.
     
  6. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    When the Collector to Emitter Voltage is very low, < 1V, the transistor is in saturation and acting as a switch.

    With Darlingtons, you need about 1/100th the current you want to flow from C to E, with standard BJT, you need about 1/10th the current as a rule of thumb. If the transistor gets hot while switching an LED, it isn't being driven completely into saturation and acting like a switch.

    Placing the LED on the emitter doesn't make much of a difference, as you need to limit the current through the LED anyway, so the extra bit of current required to turn on the transistor is limited via a resistor in series with the LED anyway. If you are switching 1A, an extra mA or two from a Darlington B-E loop isn't going to change anything when dealing with a simple switching application.
     
  7. Gibson486

    Thread Starter Member

    Jul 20, 2012
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    Just an FYI, I am trying to get as much current a humanly possible. I now have a 1 ohm resistor in series at the collector. It sounds like alot, but this pulse is only 15uS on a 2% duty cycle at around 850Hz. I am glad to hear that I am at least in the linear region and not in saturation. If I decrease the base resistor, will I be able to get more current? What if I used a divider to bring it to the VBE on state?
     
  8. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    An LED will die from as much current as humanly possible. If you want max current, just use a switch. You do NOT want the Darlington in linear mode for this purpose, as it will dissipate a lot of power as heat. Wanting maximum current and wanting to be in the linear range of the transistor are mutually exclusive.

    Please explain your application and perhaps a better alternative is available from the one you are trying.
     
  9. Audioguru

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    Post the dsatasheet for the LED so we can see its maximum allowed current and its range of forward voltage.

    Your tiny little darlington is a high voltage, low current one that cannot dissipate much heat. Maybe you should use a logic-level Mosfet instead.
     
  10. crutschow

    Expert

    Mar 14, 2008
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    The transistor is a current amplifier in the linear region but that just means it amplifies a small base current into a larger collector current. You get the maximum collector current in saturation where the transistor is acting like a switch with minimum collector-emitter voltage drop.
     
  11. Gibson486

    Thread Starter Member

    Jul 20, 2012
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    LED is a osram LE RTDUW S2w

    It is for a camera application. We have a camera, but we need a bright light. We have a very high speed application, so the light pulse has to be short as possible (6 to 15 uS). If it is any longer, it will result is smearing.

    So yes, alot current will kill it, but with the pulse so short, I was hoping that i could pulse it at an extremely high current (like 2 or 3 amps).

    I tried mosfets. It worked, but dealing with the fast transients proved to be a huge pain in the butt.
     
  12. Audioguru

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    The maximum allowed surge current for the Osram LED is 2A for a duration of 10us when the temperature of its "S" is 25 degrees C. What is its "S"?

    Its forward voltage depends on the color. I don't know which color you have.
    It has 4 chips so in series then maybe you need a supply higher than 12V.
     
  13. Gibson486

    Thread Starter Member

    Jul 20, 2012
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    Turns out this worked...however, I got a question....

    Isn't there supposed to be two collector connections on this chip for each BJT? This is the first time I have ever used a darlington chip.
     
  14. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    A true Darlington has the collectors tied to the same node. If you return them to different nodes, you have cascaded emitter followers.:p
     
  15. Gibson486

    Thread Starter Member

    Jul 20, 2012
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    Thanks All....
     
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