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how to connect up bulk leds ?

Discussion in 'Electronic Design' started by mark krawczuk, Feb 15, 2008.

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  1. hi, i want to make my own lead light for workin on cars etc.... i have
    ordered 100 , 10 mm white leds.
    what is the best way to connect them up ? in series or in paralell ? i will
    be using around 70-90 leds.

    mark k
  2. As LED's are driven with current rather than voltage, it's generally best to
    connect multiple LED's in series, together with a current-limiting
    The number of LED's you can connect in series depends on the available
    supply voltage and the type op LED's used.

    A white LED has a forward voltage drop of approximately 3.6 volts, so if you
    connect two in series, you need a supply voltage of at least 7.2 volts;
    with three in series the required supply voltage is 3 x 3.6 = 10.8 volts
    minimum and so on. Also, you have to add a few extra volts for the
    current-limiting resistor.

    So if you have a 12V supply, and you want to operate each LED at 20mA of
    current, these are the calculations:
    - You can connect three white LED's in series, resulting in a 10.8 volt
    forward voltage drop, with 1.2 volts remaining (12 - 10.8).
    - To get 20mA of current with a 1.2 volts voltage requires a resistor of
    1.2 / 0.02 = 60 ohms. So you cold use a 68 ohm series resistor.

    With 90 LED's in total, this means that you have to make 30 strings, each
    composed of three white LED's and a 68 ohm resistor connected in series.
    These strings are connected in parallel to the 12V supply, resulting in a
    total current of 30 x 0.02 = 0.6A.

    For 24 volts, you can double the amount of LED's in each string:
    6 x 3.6 = 21.6 volts. To get 20mA you divide once again the remaining
    voltage by the desired current: (24 - 21.6) / 0.02 = 120 ohms series
    resistor. Now you need only 15 strings of 6 LED's and one resistor each.
    The total current drawn is 15 x 0.02 = 0.3A.

    I would strongly recommend not raising the voltage any higher. And note that
    this is all based on a DC supply. If you have an AC supply, you could
    double the amount of LED's per string, with two LED's instead of one per
    3.6V voltage step, connected anti-parallel, and approximately half the
    resistor value to double the current (as each LED is switched for on only
    half the time). Also note that LED's don't take reverse voltages very well.

    Richard Rasker
  3. Guest

    The catch with this approach is that the "3.6V" drop per white LED is
    strictly nominal. The manufacturer's data sheet will also specify
    upper and lower limits to the LED voltage drop at 20mA (or whatever is
    approprate for the specific LED which may not always be 20mA) at room
    temperature - usually (but - not always - 20C).

    If you get a batch of LEDs with a lower forward drop, 68R may give you
    rather more current than you expect.

    There is also the problem that the forward voltage across an LED is
    temperature dependent and will fall by 2mV per degree Celcius increase
    in junction temperature. This means that you can get more current
    flowing through the LED as it warms up, and - potentially, higher
    power dissipation in the LED (if the voltage drop across the resistor
    is less than the voltage drop across the LEDs).

    For sufficiently long strings of LED's and sufficiently small
    resistors, the process can run away, melting your LEDs. A properly
    designed constant current driver can be immune to this problem.
  4. Indeed, with a very small voltage overhead (say less than 5%) and the
    resulting small resistors, these things must be taken into account.
    But as soon as you have some 10% or more of the total supply voltage across
    the current-limiting resistor, things will be fine in the vast majority of
    cases. But if you get a batch of unknown LED's, it can't hurt all the same
    to take a handful of LED's and measure the actual forward voltage drop at
    the desired current, and recalculate the series resistor accordingly.

    About temperature: with 25mA through a white LED, the internal power
    dissipation (~0.1 watts) is hardly capable of raising the temperature more
    than a few degrees. The increase in forward voltage drop versus increase in
    current (iow: the dynamic resistance) is usually much larger than the
    influence of temperature, and prevents runaway effects -- e.g. for a
    typical white LED, I found a forward voltage drop of 3.6 volts at 13mA; at
    20mA, the voltage drop had increased to 4.2 volts. Only when heating up the
    LED to approximately 70 degrees Celsius was I able to nibble off 0.1 volt.

    In over 25 years of using LED's, I have never blown even one using series
    resistors (I *did* destroy some by accidentally hooking 'em up in reverse,
    though). And several people I know even say they hooked up white and blue
    LED's directly to 5V power supplies without failure -- although I
    definitely wouldn't recommend this.

    Richard Rasker
  5. D from BC

    D from BC Guest

    Perhaps like this:

    --line AC---Rectifier---cap---Pot---LED1...LED?---Com

    An Idea for an Experimental Method
    0) Set Pot to 0ohm
    1) Make an LED string that adds up to the peak voltage for a safe peak
    2) Power it up and adjust the pot to compensate for self heating.
    2) Raise temperature of all the LEDs in an oven for the greatest
    ambient temp. 40C ??
    3) Adjust the pot resistance to maintain a safe peak current.

    Even trying to dodge a crapload of math it's still tricky :p

    D from BC
    British Columbia
  6. He might use a smaller series resistance and then add a variable pot and
    "extra bright" switch inline with all the other members(or even a 3-way
    switch). This will give him the option to increase the brightness if he
    needs(But he would need to know it won't last as long). The pot will allow
    for brightness controll. he could then, if he wanted, add a fan and/or temp
    sensor with slow shut off or reduced brightness.

    It would really suck to hook all those LED's up and turn on the power and 2
    mins later there all dead.
  7. Guest

    The Agilent HLMP-CW18 5mm white LED has a thermal resistance from
    junction to ambient of 240C/watt, which means that the junction is
    running 24C above ambient when dissipating 100mW.

    The numbers are usually available if you can be bothered to look. If
    they aren't, you'd better look for a more reputable manufacturer.
    For that particular LED. The HLMP-CW18 has a typical forward voltage
    of 3.4V at 20mA and a maximum of 4.0V, so your parts would seem to
    have been some kind of cheap rubbish.
    Very wise of you.
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