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LED lamp

Discussion in 'Electronic Basics' started by [email protected], Mar 13, 2009.

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  1. Guest

    is it possible to make an LED lamp with a 35v / 25mA supply? if so can
    anyone teach me?
  2. Yes, easy (assuming your supply is DC). Use a 1.5K 1W resistor in series
    with the LED connected to your 35V supply.
    If you don't like the resistor getting too hot then try a 5W resistor or a
    couple of 1W resistors in series or parallel to give you the same value.
    Find the brightest LED you can that runs at 25mA.
    You can also puts extra LEDs in series, dropping your resistor value
    Your supply is capable of running better LEDs like say a 1W luxeon, but that
    would need a suitable DC-DC converter, not nearly as easy unless you buy a
    pre-built module.

  3. Eeyore

    Eeyore Guest

    Do you mean a single LED or multiple LEDs ?

  4. Guest

    Depends on the LED you want to use and what % of rated output you can
    accept, supply ALL details then we have something to go on.
  5. Nobody

    Nobody Guest

    1. Decide which LEDs you want to use, and look up their forward voltage.

    2. Divide 26V (75% of 35V) by their forward voltage to determine how many
    LEDs to use. E.g. if the forward voltage is 3V, 26/3 = 8.66, so use 9 LEDs
    in series.

    3. That will drop 9*3 = 27V, so you need a resistor to drop the remaining
    35-27 = 8V. At 20mA, you would need 8V/0.02A = 400 Ohms, for which the
    closest E24 (or E12) value is 390 Ohms.

    4. The power dissipation in the resistor is I*V = 0.02A * 8V = 0.16W, so a
    1/4 Watt resistor will suffice.

    5. Connect all of the LEDs and the resistor in series, i.e.:

    [+] o---/\/\/\---|>|---|>|---|>|---|>|-- ... --|>|---|>|---o [-]

    where /\/\/\ is the resistor and |>| is an LED.

    Where does the 75% figure in #2 come from? So that typical variations in
    the forward voltage of the actual LEDs and in the supply voltage don't
    substantially alter the current drawn. A lower figure will be more robust
    (i.e. less variation in current for a given variation in forward voltage)
    but less efficient.

    If you can find actual min/max forward voltage values for the LEDs,
    and the actual min/max output voltage of the power supply, then choose the
    number of LEDs such that the difference voltage to be dropped by the
    resistor doesn't vary too much, as the current through the LEDs will vary
    in direct proportion.
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