I have seen many resistances with difference Wattage i.e 1/4W, 1/2 W, 1 W, 2 W ,3W etc. If I have 3 W Load which is an (large 3W) LED in my case, which Resistance should I use? Also The LED I am using needs 3.5V DC and 0.8 A current. I have a Battery which outputs 8V. How can I calculate the Value of the resistance which can drop 8V to 6V? This is a basic electronics calculation, do it a hundred times before you move on. \$ V = I \times R \$ \$ R = \dfrac{V}{I} \$ The voltage is the remainder after the 3.5V drop caused by the LED, so that's 8V - 3.5V = 4.5V. The current seems to be 800mA (though I see also 350mA here and there). \$ R = \dfrac{4.5V}{0.8A} = 5.6\Omega \$ Don't just pick a common 1/4W resistor. You should always, but especially with high currents like this, check what power it will consume. \$ P = V \times I = 4.5V \times 0.8A = 3.6W \$ So the answer is a 5.6\$\Omega\$/5W resistor. That's much of a waste however. Both LED and resistor see the same current, then their power ratio is the same as their voltage ratio.

And the efficiency is 3.5V/8V = 44%, excluding the LED's own efficiency. A linear voltage regulator to bring down the 8V is no solution; it will dissipate the 3.6W just the same as the resistor. A switching regulator would help, but you'll have to keep its output pretty close to the LED's 3.5V to be maximum efficient. Windows 7 Oem AlienwareThere are switchers which output a current instead of a voltage however, and they're made for the job. Used Hayley Paige Wedding DressesThe LT3474 needs only a couple of external components, can drive 1A and can handle input voltages up to 36V. Ford Focus St Fog Light BulbEfficiency for 1 LED at 800mA is slightly above 80% (for two LEDs it achieves near 90%). The Watts dissipated in a resistor is the current flowing through it times the voltage dropped across it.

The variable missing in your question is the current (which you have added now) the LED needs to operate at. It will depend on your LED and there will be a range which is ok, more current for brighter and less for dimmer. Resistance = (Power supply voltage - LED voltage drop)/led current. Once you know those values you can see how many watts your resistor will dissipate. Check out wikipedia on leds It is possible I am misunderstanding your specs, or that they are wrong. But if they are correct, you need an LED driver instead. First thing: 350 mA at 3,5 V gives about 1,2 W, so are you sure about your specs? EDIT: OK, now we have the right current. Second: in my opinion driving a 3,5 V LED with a 8V source and only a resistance is a waste of power, because you will dissipate more power on the resistance than on the LED, so you will need at least a 4.5*0.8 > 3,6W resistance. One way could be using two LEDs in series, or use a voltage regulator; but if you are sure that you want to use this configuration, i whink that you need at least a 4W resistor with value of 4,5/0.8 = 5,6 Ohm circa (To remain in the E12 series standard values).

Maybe a better solution would be a PWM regulation with a capacitor, but you would need a wave generator... Typical LEDs need about 20mA of current. Ohm's law says 1k Ohm resistor is needed for that current draw on 8 Volts to run your LED... you can adjust that resistance slightly for brightness. Your comment about a 3W load confuses me. A single LED doesn't pull 3 Watts. More info would help. Anyway, if it's just 8 Volts with one LED pulling 20mA then you need a resister rated for 1/8W or better. Current through the LED is dependent also on its temperature. When the LED lights up, its temperature rises. Unfortunately as the temperature rises, LED will draw more current at the same voltage - which means even higher temp, higher current, malfunction. This is why high-power LEDs should be always used only with constant-current source and not just series resistors. You can find many LED driver ICs, to which you need only few components. Another reason against series resistor in your design is that the heat dissipated in the resistor would deplete the battery much quicker.

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