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Powering LED's

Discussion in 'Electronic Basics' started by [email protected], May 15, 2005.

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  1. Guest

    I'm working on making a very bright set of "headlights" for my
    remote-controlled car. I decided to go with white LED's because my RC
    car is power'd by a nitro engine that produces some good vibration
    which would probably make incandesent lights not last very long.

    I bought 20 white LED's which are spec'd for a forward voltage of 3.0V
    min, and 3.8V max. I read that I should wire them in series, and not
    wire them in parallel, but I'm planning to power them with 3 NiMH AA's
    (3x1.2v = 3.8V). Since that would be the same as the max forward
    current, can I get away with parallel wiring, and not using resistors?

    I am new to electronics, and from what I have read online, it seems
    that resistors are used when your power source produces more voltage
    than your electronics can handle. So if I am correct, 3 NiMH AA cells
    should produce a tolerable voltage at all times (perhaps too little
    when they get drained). So am I correct in assuming that I can wire all
    the LED's in parallel, without resistors, to the 3 NiMH AA's?

    For more information about the LED's I bought, the eBay page for them
    has tables of info on them:

    Lastly, I'm curious about how long my 3 NiMH AA's will provide good
    power to the LED's. It looks like my LED's will be drawing 20mA each,
    and I plan to use all 20 LED's, so that will be 400mA total. My NiHM
    AA's are 2400mAh, so my first assumption is that they would last about
    6 hours. But then I realized that would be about 6 hours to fully drain
    my batteries. So this leads me to the question of when my batteries
    will start to get to the minimum forward volage for my LED's: 3.0V. I
    have no idea how to calculate that... any tips?

    Thank you,
    --Farrell F.
  2. Tom Biasi

    Tom Biasi Guest

    Your LEDs are speced 3.0 to 3.8 Fwd. Wiring them in parallel means they
    would all have to drop the same voltage. There is no guarantee that this
    will be the case, in fact they most likely wouldn't be.
    So, no, do not wire them in parallel, and for the same reason as stated you
    would still be better off with a resistor otherwise you are relying solely
    on internal resistances.
    Since you do not have a lot of 'extra' voltage a small value (39 ohms)
    could be used on each LED then you could wire the combo's in parallel.

    To know more accurately than you have calculated the expected useful time of
    the lighting system, you would need to know two things.
    1. The current at which the LEDs produce a light that is not acceptable to
    2. The discharge curve of your power cells at the current that you are
    BTW: You seem to imply that you believe that the 3.0 to 3.8 volts are
    operating ranges. This is not the case. This is the range of the expected
    voltage drop across the LED when it is fully conducting.
    Good Luck,
  3. You may get away with it, but fair chance you cook the LEDs. For one
    thing, LEDs get more conductive as they get hotter, and you may get
    "thermal runaway".

    I would put a resistor in series with each LED and put the LED-resistor
    combos in parallel with each other. As for resistor value - I would guess
    22 ohms. You can try different ones - measure the voltage across the
    resistor and divide that by resistance to determine the LED current. That
    will probably indicate current better than a milliammeter, which on a
    suitable current range will probably add enough resistance to change the
    current significantly.

    - Don Klipstein ()
  4. Guest

    Thanks guys, so now I have a few more questions:

    To clarify, is this what Don is reccomending:


    and I'm still a bit confused about what resistor(s) to choose. I was
    reading this page on how to power LEDs:

    and it leads me to believe that resistors are specified for both
    resistance (ohms) and wattage (watts). Since now I will be using
    resistors, I was thinking of going with 4 AA NiMH batteries to increase
    runtime. So this leads me to the question of which resistors to choose.
    Is the 39ohms, or the 22ohms resistor better suited to this task? and
    what wattage? and just so I can better understand this, is it better to
    pick a resistor rated to handle more ohms and more wattage? or less?

    I noticed a section on that website I linked to above about combining
    both the series and parallel methods,
    and was thinking of maybe going that route. Is that a good idea, and if
    so, how would that affect my resitor selections.

    If I should go with the hybrid serial-parallel route, won't I need to
    have a battery that supplies more volage? 4 NiMH AA's would be 4.8V...
    not even enough for 2 LED's. (I plan to use all 20 LED's) I was
    thinking of perhaps buying a pair of rechargable 9V batteries, but 18V
    is only enough for about 5 LED's. I do not want to use too many
    batteries as it would weight-down my RC car... and since I have a large
    supply of rechargable AA's I would prefer to use them. Would I need to
    get another device to increase the voltage... what is it called, a

    Sorry, I'll bet I'm really confusing stuff up right about now. So back
    to basics... Is there a nice webpage or book that describes the basic
    electronics vocabulary, and basic devices (diodes, resistors,
    transformers, capacitors, transistors, etc...) I think this is my main
    problem, because the webpage I linked to above did a great job of
    defining how the equations worked, it's just that I don't quite know
    what voltage, wattage, current, power, etc. are, and I'm going off of
    guesses and past expierences.

    Thanks again,
    --Farrell F.
  5. Tom Biasi

    Tom Biasi Guest

    Try 39 Ohms and if the LED is not bright enough try a lower value.
    1/4 watt resistors are readily available and are more than sufficient for
    your case.

    Let's educate you a little so that you may better understand the suggestions
    that are given to you.

    The LEDs are diodes not light bulbs.
    The junction of the diode will emit light when current passes through it.
    The voltage across the junction will rise as current increases up to a
    point; at that point the voltage will remain fairly constant. This is the
    forward voltage the spec talks about.
    If you supply more voltage than that the junction will overheat and die.
    The resister in series will allow the extra voltage to "have somewhere to
    So the resister that you need is determined by the extra voltage that you
    have and the current that you need, R=V/I.
    You mentioned a 3.8 volts supply and LEDs that could max at 3.0 volts.Your
    resister would be .8V/.020A= 40 Ohms.
    39 Ohms is the nearest standard value. The power dissipated would be P=VI.
    ..8V*.020 A. = .016 Watts.
    For physical convenience I would use a 1/4 watt resister. If you use more
    voltage in your supply you would need to replace the .8v with the excess
    that you have.
    Forget about a transformer right now. The theory doesn't apply to what you
    are doing.
    Good Luck,
  6. Guest

    Thanks Tom!

    So since you reccomend that I should not go with a transformer, I guess
    my assumption of how to wire my LED's in my previous post is correct?


    No? and does it matter which "side" of the LED the resistor will be
    attached to?

    And now I think I will go with a 4 pack of 1.2V NiMH AA's as my power
    source, so will you please check my math and tell me if I am messing
    things up:

    * 3.0V is the min my LED's are spec'd for, so would it be bad if I aim
    to provide them with 3.6V, just a little below their 3.8V max?
    * 3.6V per LED, 4.8V power source leaves me with 1.2V to resist.

    1.2V/.02A=60 Ohms
    1.2V*.02A=.024 Watts

    And since I am not familiar with standard resistor specifications, I'm
    guess I should still go with a 1/4th Watt resistor, for approx. 60
    Ohms... and should I round up or down?

    I looked online at and found 100 Ohm 1/4 Watt
    resistors... do they sound about right?

    --Farrell F.
  7. Tom Biasi

    Tom Biasi Guest

    Your math is correct.
    You can get a standard resistor in 10% of 56 Ohms and in 5% of 62 Ohms. 1/4
    watt is fine.
    It doesn't matter which side the resister goes on but it does matter which
    side gets the plus and minus.
    Just some additional information: You indicate that you may still think the
    3.0 to 3.8 is a min/max supply rating for your devices.
    It is a range of expected forward drop voltage. The forward drop can be
    anywhere in that range for any given device. Measure them in operation. You
    may need to calculate a different resistor for each LED if you want uniform
  8. Rich Grise

    Rich Grise Guest

    No, Yes. ;-) I've always put the resistor in the + lead, just 'cause
    it feels better to have the LED cathode closer to ground.
    No, they're in series, and every point in a series circuit passes the
    same current.

    LEDs are rated by _current_ - the voltage is just what happens to
    appear across a given LED when the rated current is passed through it.
    In this case, you'd calculate the resistor value at the low end of
    that range, for a current of, say 20 mA. So 4.8V - 3.0V is 1.8V, divided
    by .02A is 90 ohms. 91 is the closest standard value. If your LED
    happens to be one of the ones that drops 3.8V at the rated 20 mA current,
    then there will be 1.0V across your 90 ohm resistor, which will give
    you 11 mA, but (A) The only LEDs I've done the experiment were so
    close to as bright at 11 mA as they were at 20 mA that it was practically
    the same, and (B) you're on the safe side for any LED within that range.

    As Tom Biasi has suggested, it would probably be a good idea to
    set up a circuit that provides a constant 20 mA, and measure the
    voltage drop across a batch of LEDs. It could be very educational.

    This is all assuming that your LEDs are rated for 20 mA, of course.
    If yours have a different spec, just use that value in the formulas.

    If you use a 12V supply, then for 20 mA with an LED that drops 3V,
    you'd use a (12 - 3)/.02 = 450 ohm resistor, which if your LED
    drops 3.8V, would still give you 18.2 mA, which is much closer
    to a constant current supply for your tests.

    Hope This Helps!
  9. Guest

    Thanks yet again :)

    A few more questions, hope you don't mind...

    By 5% and 10% I assume you mean the carbon film percentage? What does
    this mean, and how does it effect the results?

    And what exactly is "drop" voltage? The amount used by the LED?

    --Farrell F.
  10. Tom Biasi

    Tom Biasi Guest

    "Drop" is a term meaning the voltage across a device caused by the current
    going through it times the resistance.
    If you have 1 volt across the resister in this series circuit the LED will
    "see" 1 volt less than the supply, hence the term "drop".
    The percentage is how close to the specified value the resister really is.
    Ex.: A 100 ohm resister at 5% may really be 95 to 105 ohms. A 100 ohm
    resister at 10% may be between 90 and 110 ohms.
    (5% of 100 is 5 ohms 100 plus or minus 5 ohms.) Don't concern yourself with
    this for this application.
  11. The 5% or 10% is the tolerance on the resistance - a 5% resistor may
    be up to 5% above or below its marked value (but I think most metal or
    crabon film resistors will be within 1% of the marked value,
    regardless of the indicated tolerance)
  12. Rich Grise

    Rich Grise Guest

    Well, the voltage across the LED is the voltage "dropped" by the LED;
    the resistor will have its own voltage drop. It's not like volts get
    dropped off a cliff, or down a well, or something - it's exactly
    analogous to "pressure drop", which is used in pluming and HVAC
    all of the time. It's the differernce in pressure between one end
    of a component and the other. Volts don't get "used" other than
    to provide the pressure that pushes the current through the circuit.

    Think of a tall water pipe, with ports with pressure gauges arranged
    along its height. At the top, there's zero water pressure, and at
    the bottom, there's some amount of pressure just caused by the weight
    of the column of water. This is how altimeters work - at altitude,
    there's less air on top pressing down on itself. (and on you, and
    your altimeter).

    Now, if you have a 33ft. column of water, the water pressure at the
    bottom will be 15 PSI. The water pressure half-way up the column (16 1/2
    feet) will be half that, or 7.5 PSI. So there's a pressure drop of
    7.5 PSI from the top of the column to the middle, and another 7.5 PSI
    pressure drop from the middle to the bottom.

    That's all voltage drop is - a difference in pressure, which is called
    "electromotive force" - and as you remmeber from high school, force
    is just a push - it's there whether the object is moving or not.

    Hope This Helps!
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