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Powering an LED Array On and Off at 1kHz

  1. Dec 20, 2011 #1
    So I have a question about a circuit I'm making for class. I'm using this PowerSupply is a power supply I'm using in the lab. The 1kHz square wave is 700mV pk to pk with a 350mV offset. LEDarray is a red LED light array (http://www.ledtronics.com/Products/ProductsDetails.aspx?WP=2003).

    When I connect the light to the power supply directly, it draws 230mA at 12V. But when I connect the light as shown in the schematic with the intent to switch it on and off at 1kHz, not enough current is drawn to fully power the LED array (only a fourth of the light turns on, and not very bright). Only 10mA is drawn.

    I'm not sure how to make the light draw 230mA while switching on and offat 1kHz. Is the issue with the power supply not being able to switch that fast at that amperage, the switch not being able to put 230mA through at that frequency, or something else entirely? Help please?

    Thank you

    Capture.png
     
    Last edited: Dec 20, 2011
  2. jcsd
  3. Dec 20, 2011 #2

    MATLABdude

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    Welcome to PhysicsForums!

    The problem is that your frequency generator is connected across the base and emitter of the BJT, and isn't amplifying your signal. You also don't mention the duty cycle of the square wave (percentage of each period which is high), which will impact the amount of current drawn, assuming that you're able to supply the 12 V across the LED.

    I would look into using the BJT as a common emitter amplifier, and modifying your circuit such that the BJT sinks current, rather than trying to source it (at 12 V).
     
  4. Dec 20, 2011 #3
    Clarifying what MATLABdude said, connect the emitter directly to battery -. Connect your signal source between battery - and the base and connect the cathode of the LED to the collector and the anode to battery +.
     
  5. Dec 20, 2011 #4
    Thanks for the replies! Not sure why I never tried this. :P
     
  6. Dec 20, 2011 #5

    vk6kro

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    You will also need to add current limiting resistors like this:

    http://dl.dropbox.com/u/4222062/LED%20driver%202.PNG [Broken]

    The transistor shown as NPN will need to be a high gain, power transistor, but not a Darlington, as there is not enough voltage drive for a Darlington.
     
    Last edited by a moderator: May 5, 2017
  7. Jan 25, 2012 #6

    vk6kro

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    If you check the data sheet for the 2N2222, notice that the gain drops a lot for high current and also for a saturated transistor.

    If this transistor had about 1 volt across it and it was supplying 230 mA, the current gain could be about 30. So, this means the base current would have to be about 7.6 mA.

    In this case, the position of the LED doesn't matter since you have the signal source isolated from the power supply.

    Some things you could try:

    Increase the offset voltage. At present, only the very peak of the input signal is causing any LED current to flow.

    Increase the power supply voltage to about 15 volts so that the voltage across the transistors is greater.
    Gradually increase the square wave amplitude from zero while monitoring the LED current.

    Put 4 2N2222's in parallel. This will give more gain because each transistor would not have to draw so much current.
     
  8. Jan 26, 2012 #7

    sophiecentaur

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    To be fair to bluurple, his original diagram is fine - but only as long as either his power supply or his oscillator is 'floating'. The elements in the circuit that includes battery, LED and transistor need not actually be in the order we'd normally expect as long as the transistor can be turned on and off by the oscillator.
    In real life, of course, we would try to stick with just one 'ground' and to use the same power supply for everything so the suggested circuit would be better. But, in principle . . . . . ..
     
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