Electronic – Calculating resistor value and power rating for LED driving


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?

Best Answer

This is a basic electronics calculation, do it a hundred times before you move on.

It's Ohm's Law:

\$ V = I \times R \$

or, put differently:

\$ 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. There are switchers which output a current instead of a voltage however, and they're made for the job. The LT3474 needs only a couple of external components, can drive 1A and can handle input voltages up to 36V. Efficiency for 1 LED at 800mA is slightly above 80% (for two LEDs it achieves near 90%).