Simple HIgh Current Led Driver

Discussion in 'The Projects Forum' started by bobank, Aug 2, 2007.

  1. bobank

    Thread Starter Member

    Nov 19, 2005
    21
    0
    Hi, guys,
    does anybody have schematics for simple high current led driver, something that can work from 9v batteries?
    I used NUD4001 but it doesn't have tollerance for span in voltage ( like you can not start using it from 9v-when battery is full up to 6-7 volts) and it is tight thermal design. Plus it is not in production anymore.
    I would like to have simple circuit, with not too many external components, to have current at least 700mA.
    Anything in mind? Like any IC that, with few external would make good driver.
    Thanks
    Boban K
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    9 volt batteries are characterized for maximum drains of 15 ma. I don't think you want to try placing 50 batteries in parallel. 700 ma is a substantial draw for a battery, especially if it is expected to supply that current for a long time. The fairly common 12 volt 6 AH lead battery is reasonably priced, and would be a better candidate.
     
  3. mrmeval

    Distinguished Member

    Jun 30, 2006
    833
    2
  4. bobank

    Thread Starter Member

    Nov 19, 2005
    21
    0
    I use LM 317 when have enough voltage to allow big drop voltage, like in automotive. LM317 is great regulator but not efficiant in batteries as require couple volts above working voltage.
    Answer for letter before this from beenthere: There are lot of drivers that pull 1 - 1.5A from small batteries. My NUD4001 design pulls easily 700ma from single 9V battery, problem is that is little voltage to work with, between operating and starting loosing brightness. If you increase voltage than it is heat problem.
     
  5. John Luciani

    Active Member

    Apr 3, 2007
    477
    0
    If your are powering a single white LED (Vf=4V) from a 9V battery source the only way to avoid the power dissipation problem is to use a buck regulator. Check Linear Technology and Maxim for buck regulators. Otherwise you will need to dissipate 1.4W to 3.5W
    across your current regulation circuit. When the battery is fully charged you will be
    dissipating more power across the regulation circuit than the LED.

    (* jcl *)
     
  6. bobank

    Thread Starter Member

    Nov 19, 2005
    21
    0
    For 9V battery, it is to high difference beteween input voltage and working voltage. I can add resistor or second led. With second LED dissipation is not so bad and I have more light.
    3 x 9V battery = 27V. Using 7 Luxeon Rebels at little bit less than 700mA ( using 2 parallel NUD4001). It works great. I am not sure what is power dissipation ( I did calculate before starting work but don't have it in mind now). So far i've been using it for 4-5 hrs on-off work. Still full or close to full power. It desn't overheat.
    My only problem, if this is a problem is that I want to have more working voltage span, what makes batteries last longer. There is in NUD4001 data sheet circuit with transistor and shotky diode, where transistor with heatsink will take 99% of current. Have to try that one but my time is very limited, will use next free time to make one. Already have Circuit drawing.
    I am looking from practical point. I am not designing driver. I want to have reliable, constant brightness light for night work on, what locksmiths call "impressioning". That is a way of making working key for car or house by inserting blank key, obtain marks and file on those marks. This is the highest knowlege in locksmithing trade because it is very hard and require training and inteligence. It is very hard to work on day light, impossible on night. With light I made, i am able to work on most cars and make a key. Very bright, uniform, pure white light.
    Boban
     
  7. John Luciani

    Active Member

    Apr 3, 2007
    477
    0
    You can't get more working voltage span unless you use either a regulator or sacrifice efficiency. Except for the -2mV/DegC Vf drop the Vf of the LEDs is constant at apx. 4V.

    For one LED you dissipate 5V across the current sink and 4V across the LED.
    You have a 4-9V voltage span but waste half your power at 9V.

    For two LEDs you dissipate 1V across the current sink and 8V across the LEDs.
    You have a 8-9V voltage span but waste only ~11% power at 9V.

    If you want to sacrifice efficiency use the 9V battery, a single LED and MOSFET
    current sink that can dissipate 3.5W (5V * 700mA). For a simple high power current sink schematic take a look at the load schematic in my document ---- http://www.luciani.org/geda/util/matrix.pdf

    (* jcl *)
     
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