LED driver to run off 4.8V

Discussion in 'General Electronics Chat' started by ian field, Oct 30, 2015.

  1. ian field

    Thread Starter Distinguished Member

    Oct 27, 2012
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    Does anyone have a favourite circuit for a LED driver from 4.8V they're willing to share?

    Preferably a very simple buck type. There isn't much room in the bicycle lamp housing - I was thinking of a "dead-bug" build that can squeeze in the dead space around the reflector.
     
  2. dl324

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    Mar 30, 2015
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    What will the LED be used for? Illumination, warning, ??
     
  3. ian field

    Thread Starter Distinguished Member

    Oct 27, 2012
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    In a bicycle lamp.
     
  4. dl324

    Distinguished Member

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    LED voltage and current? Is it battery operated or from a generator?
     
  5. ian field

    Thread Starter Distinguished Member

    Oct 27, 2012
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    I was sort of thinking I might plump for a white LED just to be boringly conformist, last time I looked the Vf was around the 3.4V mark.

    But since LEDs need current regulation, I didn't think of voltage as the foremost specification.

    4.8V is pretty much the nominal voltage of 4x nickel chemistry cells.
     
  6. dl324

    Distinguished Member

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    Current regulation isn't critical for many applications. Personally, I'd just go with a series resistor. If you want a current source (sink), you can do it with 2 transistors and 2 resistors. There was a thread last month about help understanding a current source. Don't pick the one started this month. That thread didn't have anything to do with a current source...
     
  7. ian field

    Thread Starter Distinguished Member

    Oct 27, 2012
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    A series resistor isn't quite a buck type regulator.

    The original bicycle lamp already has one of those - I'd like to improve on that.
     
  8. RichardO

    Well-Known Member

    May 4, 2013
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    How much current does the LED draw? The bike headlight I made is barely bright enough at 3 watts. I need more like 12 watts to ride at full speed. You will probably get better results because you are starting with better optics than I have.
     
  9. dl324

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    What would you gain by having a regulator that consumes power without providing illumination?
     
  10. ian field

    Thread Starter Distinguished Member

    Oct 27, 2012
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    Probably the same as most people can gain by replacing a power dissipating dropper resistor with an efficient step down SMPSU type design.
     
  11. bertus

    Administrator

    Apr 5, 2008
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  12. ian field

    Thread Starter Distinguished Member

    Oct 27, 2012
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    ATM: I don't know all the details, the LED package type could be 1W or 3W, but the specification for the bicycle lamp only claims to be 0.6W.

    The LED package has a heatsinking solder base as well as the solder lugs. In the lamp, that base isn't soldered, but was installed with thermal paste to an array of through plated holes for dissipation - I assume that's why it was de-rated to less than the probable 1W of the LED.

    A basic circuit idea would be a step in the right direction, I can tweak component values to get it just right.

    There's an SOIC-8 chip on the board that I can't identify any number even with a jeweller's loupe - that could possibly be the chip for the push button ON/OFF. If it wasn't for that - I'd just replace the 4x AA holder with a single D cell and use a blocking oscillator Joule thief circuit.
     
  13. ian field

    Thread Starter Distinguished Member

    Oct 27, 2012
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    A chip that I don't have, means placing an order and most likely making up a minimum order amount for a load of parts I'll never use.

    A circuit can be built up with parts that will do from the junk box.
     
  14. dl324

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    Mar 30, 2015
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    Here you go:
    upload_2015-10-31_12-16-47.png
    This is the current source that was discussed in the thread I referenced. Instead of dissipating the power in a resistor, you can dissipate it in 2 transistors and 2 resistors. The only advantage is that you can set a constant current which a resistor or buck regulator wouldn't have done.

    I guess another advantage would be that this is more efficient than any buck regulator.
     
  15. ian field

    Thread Starter Distinguished Member

    Oct 27, 2012
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    You need to go back to the drawing board!

    A linear regulator dissipates (wastes) energy just the same as a resistor.

    If you can't do it, stay out of the way and make room for someone who can.
     
  16. dl324

    Distinguished Member

    Mar 30, 2015
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    Believe it or not, I was actually trying to help you.

    Here's another chip that you'll probably have to order...
    http://www.linear.com/product/LTC1704

    If I respond to another of your posts, please remind me again not to waste my time.
     
  17. ian field

    Thread Starter Distinguished Member

    Oct 27, 2012
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    It is actually hard to believe, since most of what you suggested was contrary to what I asked - and some of it just plain wrong.

    Here's hoping you remember not to waste my time in future....................
     
  18. dl324

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    Again, sorry for trying to help and for wasting your time.
     
  19. NCSailor

    New Member

    Jun 15, 2013
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    I think Dennis was way too polite. You came here and asked if anyone had favorite circuit to solve your problem. You failed to provide many of the requested details. It appears to me that you already had a solution and were just here hoping that people would see how "smart" you are.

    Thanks for wasting all our time.
     
  20. ronv

    AAC Fanatic!

    Nov 12, 2008
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    You may be wasting your time with a buck converter. If the LED is 3.4 volts and the diode in the converter is .6, the current sense voltage .3 and maybe .5 for a FET and inductor it's a wash.
     
    RichardO likes this.
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