LED current vs. amp current?

Discussion in 'The Projects Forum' started by tracecom, Oct 26, 2012.

  1. tracecom

    Thread Starter AAC Fanatic!

    Apr 16, 2010
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    I have a class D amplifier as shown in the attached pic, and I am mounting it in a box. As you can see, the amp has a red LED, but the LED is facing the wrong way, and I want to replace it.

    I don't have a schematic, but with a 12 VDC power source, the Vf across the LED measures 1.66 V, and the current drawn by the entire circuit is 22.2 mA. Because there is no input to the amp, I presume almost all of that current, say 20 mA, is for the LED.

    Is that logical?

    ETA: I see from the datasheet that the PAM8610 chip has a quiescent current of 20 mA. Does that mean that the LED is only drawing 2.2 mA?
     
    Last edited: Oct 26, 2012
  2. ErnieM

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    Apr 24, 2011
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    A LED getting just 2.2mA glows a bit, you can plainly see it, though it doesn't illuminate much about itself. It may well be getting more current as the quiescent current in the PAM is probably a max spec.

    But the LED is going to take what it's given, this is a place where you can over think things: just about any RED LED will drop in replace this, excepting those special devices with internal resistors.

    Just unsolder this LED, move it to where you need it, then solder it back. Or buy a new one and solder that to where this one was.

    If you get the new one backwards it will not light but nothing gets burnt, just reverse it and it should work. (That means you can just touch wires as a test before you solder).

    Good luck.
     
  3. tracecom

    Thread Starter AAC Fanatic!

    Apr 16, 2010
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    I do that a lot, and often confuse myself in the process. :)

    I replaced it with a blue LED, which is actually brighter than the red one was. The overall current draw dropped to 21.4 mA; it worked out fine.

    Thanks.
     
  4. ErnieM

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    Apr 24, 2011
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    That makes sense. There is probably just a resistor in series with the LED, and by changing color you increased the voltage the LED takes, so the resistor gives it a little less current.
     
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