LED lamp

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 accordingly. 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.

Dave.

Do you mean a single LED or multiple LEDs ?

Graham

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.

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

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

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

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.