Led Mountain Bike Headlights
Help me understand the wiring for this LED light?
Im wiring this up for a mountain biking headlight:
http://search.digikey.com/scripts/dksearch/dksus.dll?Detail?name=475-1295-ND;enterprise=36
Voltage Rating— 20.8V
Current – Test— 700mA
How do I know what Input i can put into these? Does it mean I use a 20.8 volt battery pack? What about the mA?
Anything else I should know about wiring this up?
This module appears to be just six LED’s wired in series.
The forward voltage of 20.8V will be the typical voltage that these units will operate at. LED’s operate like a zener diode in that they will draw more and more current in an effort to keep the applied voltage at the operating voltage. Any excess voltage applied over its operating voltage will cause the LED to draw more current in an effort to hold the voltage at its operating voltage. More and more current will then be drawn by the LED until either the power supply cannot supply any more current or the LED burns up.
To prevent excess current from destroying the LED a series limiting resistor is used to limit current flow to a safe operating level. (700ma in this case)
So for your LED selection 20.8 VDC you will apply 24VDC (easily obtained from batteries).
24 – 20.8 = 3.2V You will need to limit current flow to 700 ma AND drop this excess 3.2 volts. Using Ohms law you can compute the series limiting resistor
3.2 / .7 = 4.57 Ohms
You now need to compute how much power in watts this resistor will need to handle.
P = I² * R = .7 *.7 * 4.57 = 2.2 watts double this for a safety factor
and you will need a 4.6Ohm 5 Watt series resistor to operate this module with a 24VDC power supply or battery.
Digikey sells one 4.7W-5-ND that will be close enough.
As a final check the LED module can handle 27W max so compute its expected wattage as 20.8V * .7 = 14.56 W. Well below its 27W limit which will give a decent safety factor.
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