Connect with us

LED Parrellel Current limiting Cap. Question........

Discussion in 'Electronic Basics' started by Baronvonrex420, Aug 1, 2003.

Scroll to continue with content
  1. Let me start by "Yes, I know I'm supposed to have a current limiting resister
    to each LED", but I don't. This project requires over 100 LED's and I need to
    reduce drilling where I can.

    I need to drive 5 and 6 3VDC, 20 ma LEDS in par. but I can't find a formula
    for calculating the resister value. The life expectancy of this product is
    less than 100 hours and current will be supplied with a 7812 regulated 12 VDC
    current from a car battery.

    I tried 33 ohm 1/2 watt, 100 ohm 1/4 watt, and tonight I tried a 10 ohm 1/2
    watt resister. The LEDS work great but the resisters are hotter than heck!

    What do I need to do to bring the resister temps down or calculate the actual
    value I need? I need to encase this in resin and I can't afford the heat.

    Any help would be appreciated! TIA!

  2. Okay, let's take the case of 6 LEDs. I don't know why you MUST put them in
    parallel, which is particularly wasteful of power (meaning lots of heat made for
    nothing), but let me suggest a better strategy.
    Since your supply is 12 volts, and you have LEDs that run at 3 volts, why
    not put three LEDs in series with a single resistor, and do this twice? Now,
    you greatly simplify the picture and you don't drop 9 volts in the resistors.
    Here is how you calculate your resistance and the power dissipation.
    Three LEDs in series will drop 9 volts, leaving 3 volts to drop in the
    resistor. You know that in series, the current consumption remains unchanged,
    so 20 mA is the figure we use. Since R = E/I, we divide 3 volts by .02 amps and
    get 150 ohms. Now, since P = EI, the resistor will have to dissipate (3 volts x
    ..02 amps) watts as heat. This is 0.06 watts. Almost any small resistor can
    handle this with no trouble, even the little 1/8 watt ones.
    You will put two of these assemblies in parallel, meaning that you will have
    6 LEDs and 2 resistors for one of these assemblies. Total power turned to heat
    is a mere 0.12 watts per pair of LED strings.
    With your 100 LEDs on at once, if you were running all of them in parallel
    and dropping 9 volts through your resistor, you would have been dissipating 18
    watts as heat if and only if your resistor was (9 volts / 2 amps) or 4.5 ohms.
    But it is very likely that you would also have had a few dim ones, and not only
    that, 3/4 of the power from your battery would be heating up a resistor instead
    of supplying illumination.
    If you wanted to string them in threes (as I mentioned above), then fully
    3/4 of the power would go to lighting and only 1/4 would be wasted.


    Chip Shults
    My robotics, space and CGI web page -
  3. Probably your best first bet is to not use a 7812. You are just
    asking for dissipation trouble. 100*20mA is 2A and you have to
    get rid of (12V-3V)*2A = 18 Watts one way or another. Pulse
    width modulation probably wouldn't be a lot of use here, either.

    Instead, use a switcher and set it to output your 3V rail. 2A
    is quite reasonable. You won't be able to avoid the LED
    dissipations of 6 watts or more, but your switcher will
    definitely help out on the hot resistors. Can you live with
    brightness variations?

  4. Thanks Sir Shults!
    That makes a lot of sense and hopefully keeps the ambient temp down. I'll look
    at a re-design and see how I can incorporate this into my design...

    My company web site might be of interest to you, Robotic Technology Systems:

    Thanks Again!
  5. LEDs really don't have a simple linear resistance. They have a voltage drop
    instead. Given the v drop of each LED, you can approximate the resistance
    needed by (12-N*Vled) /.020, where N is the number of LEDs in each parallel
    branch. This is assuming a 12V voltage across the branch.

    You may also want to consider that LEDs can be flashed very quickly without
    appearing to flicker, with a resulting large savings in power. I just hacked
    up a circuit using a 555 and a 4017, both of which will run with Vcc=12V.
    The 555 drives the clock of the 4017. You can use this to drive 10 branches,
    with a correspondingly large savings of power. I'm not sure of your
    application, so I don't really know if it will work for you, but I am
    driving 4 red LEDs from a single branch, and they all light up nicely. If
    the LEDs aren't bright enough with 1/10 of the time, you can use one of the
    outputs of the 4017 to drive the reset, which will limit the number of
    branches, and so give each branch a bigger 'duty cycle'.

    One final note, most voltage regulators won't work well unless you have an
    input voltage that is 2 or 3 volts above the output voltage. You may want to
    run your system at 9 volts using a 7809, rather than 12 volts. If 12 volts
    is absolutely required, there are low dropout voltage regulators that can
    run with a smaller input/output ratio.

    Bob Monsen
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day