Dimmer Circuit

Discussion in 'The Projects Forum' started by Wehrdo, Oct 16, 2010.

  1. Wehrdo

    Thread Starter New Member

    Oct 16, 2010
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    I would like to create a circuit that I can dim an LED with a potentiometer.

    Here's the circuit:
    [​IMG]

    The problem is I don't know what values to use for the components. I'm not sure how much voltage and current I need to supply to the transistor.
    I am familiar with Ohm's law, but I don't' understand the current gain(hfe) or the saturation voltage on transistors.

    The reason I'm using a transistor in the circuit is because the one LED shown will actually be some arrays of LEDs with a power draw of around 2W.

    If somebody could explain to me how to figure out the correct component values, I would really appreciate it!
     
  2. wayneh

    Expert

    Sep 9, 2010
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    I think you'd do better to put the current limiting resistor at the "bottom", with the transistor atop that, and the LED array at the "top", connected to Vcc. Then the transistor will keep increasing the current until the voltage at its emitter - atop the resistor - is ~0.6 below the base voltage. The resistor values for setting the base voltage are not terribly important, except that the base current should not dominate the voltage. In other words the current thru the voltage divider should be more than your base current. Current gain will likely be ~100, give or take 10X. You need to read the datasheet for it. You didn't say what LED array current you need. Make sure you choose a transistor that can more than handle it, and don't forget about heat.

    The current limiting resistor should be chosen to give max brightness if the transistor resistance is zero, or a short.
     
  3. Wehrdo

    Thread Starter New Member

    Oct 16, 2010
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    I'm not sure what you mean with moving stuff to "top" and "bottom", but here's a more specific circuit to what I will be doing.

    [​IMG]

    D1-D6: 3V voltage drop and 70mA current draw
    T1: TIP120; DC Current gain minimum 1000. 65W power dissipation
    POT1: Potentiometer
    R1: Resistor

    My two questions: Does the value of the potentiometer matter, and what should the value of the resistor be?

    I thought about using PWM, but it seemed like overkill for something this simple.
     
  4. marshallf3

    Well-Known Member

    Jul 26, 2010
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    Won't work as expected, if each LED drops 9V you've got nothing left for the transistor junction Vce and even with the pot up full you won't be able to bias the transistor on.
     
  5. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
    102
    The range of the potentiometer that you can use is not critical because a darlington transistor is being used, any value of the POT from 1K to 5K will work. The relatively small transistor base current can usually be ignored in this case as it is around 0.1mA.

    Just pick the POT you already have and re-do the calculation for the new POT if necessary.

    The voltage drop across the transistor B-E is 1.2 to 1.3V and you also knows the maximum voltage across the LED string too.

    [​IMG]
     
  6. Bernard

    AAC Fanatic!

    Aug 7, 2008
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    A slight variation: R 1 is optional & is used to reduce large dead zone of Pot bottom end, 1k to 5 k. Avoid paralleling LED's unless Vf s are = or ballanced.
     
  7. Wendy

    Moderator

    Mar 24, 2008
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    Guys, this configuration will burn up the LEDs and the transistor!

    The OP is making the classic mistake, he is trying to use voltage, not current, to control the LEDs. Once the LEDs go over their Vf, they and the transistor will start to smoke. He does have a limiting resistor on the collector. This means the circuit will go to maximum current when the LEDs dropping voltage is reached (hopefully maximum is low enough not to allow anything to smoke). There will be no linear region.

    LEDs, 555s, Flashers, and Light Chasers

    [​IMG]

    This design is a variable constant current source.
     
  8. Wendy

    Moderator

    Mar 24, 2008
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  9. wayneh

    Expert

    Sep 9, 2010
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    See Bill's post for the pictures. By "top" I meant at the highest voltage, and by "bottom", closer to ground. Bill's drawing is exactly what I meant.

    You use the current limiting resistor to feed back a voltage proportional to the current across it. This in turn reduces Vbe and limits current, with control by the pot.
     
  10. Bernard

    AAC Fanatic!

    Aug 7, 2008
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    Note: Bill was not referring to this post as not workable.
     
  11. Wehrdo

    Thread Starter New Member

    Oct 16, 2010
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    0
    Sorry for the late reply. marshallf3: Thanks, I didn't think that a transistor had a voltage drop, even though that makes sense now, because it must have a minimum voltage to work correctly. Bernard: This looks like a good solution. The only thing I'm confused is about is what R1 does. What does putting a resistor between the pot and ground change about the pot? Bill Marsden: Thanks for letting me know I would burn stuff out with this setup. Those links that you posted helped clear a lot of things up about LEDs. You've got lots of good information posted around here. I spent about half an hour reading some of it.
     
  12. wayneh

    Expert

    Sep 9, 2010
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    R1 sets a minimum resistance even if/when the pot is turned down to zero ohms. A sort of insurance. In general you put a fixed resistor in series with a pot when you want to shift the useful range of the pot.
     
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