# I am trying to find out why I need a 2 Watt resistor with a 3 Volt 700 milliamp Current LED?

Hello: My name is Charles and I am 51 years old,all of this Computer Technology is all NEW to me as I am OLD SCHOOL if you know what I mean.I have had a Nervous Breakdown due to Stress,and have lost a lot of my Technical Memory.I am trying to pick back up on the Electronics for a pass-time,but things don't seem to make sense like they did before.I wanted to know how I can ask questions I have about the Electronics I am wanting to work on.As of right now,I need to see if someone can explain when using OHM's Law (Power is equal to Voltage of your Input times the Current needed in your Circuit).I have a 3 Watt Red L.E.D. which should run at 2.5-3.0 Volts with a 700 Milliamp Current.If using a 3 Volt supply times .700 would equal a 2.18 Watt Power for the Resistor.That does not sound right to use a 4 Ohm - 2 Watt Resistor for a 3 Volt Input.Is this really right or have I gotten confused somewhere?Thank you very much if you can help me out,Sincerely Charles.

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7 years ago

Charles,
Sorry to hear of your troubles.

Part of the problem here is that you're probably not that used to these MONSTER LEDs - These beasts take a lot of current to deliver a lot of light - that leads to a need for a big power resistor too, to drop the excess voltage. In fact, as Sean says, you might get away without a resistor at all, but you may find that the LED isn't quite as bright as you'd expect, and that it get a little dimmer as the temperature increases. You'd be better providing a few more volts of head room and using a simple circuit like this to drive the LED. Here R = 1.22/0.7 or roughly 1.8 Ohms, 1Watt - the LM317 will need to be supplied from 6V.
wccolvin (author)  steveastrouk7 years ago
Hello Steve: I want to thank you and the others here for your help,most people try to keep their Knowledge to theirselves.They think they had to go through the Complete Course to learn,that everyone else should have to do the same.People just don't want to help others like they did many years ago.Thanks again to you all it gives hope to this World we live in Today.
7 years ago
Actually, the reason you got a clear answer is that you had clearly done a lot of the leg work yourself. I hate replying to people who say "This doesn't work, fix it for me". Some people DO need the complete course !!
7 years ago
Agreed on the constant current source...was mainly trying to give a refresher on the basic thought/math process rather than alternate advise on method.

Note that no matter what, the author will still need to drop that voltage difference @ If, so the same basic power losses will still be present for a linear circuit, whether that's thru a single resistor or the resistor and LM317.
7 years ago
ABSOLUTELY !! Time to throw a switching solution into the standard circuits, but the a working one needs a good layout....and Instructables isn't really the place to do that.
seandogue7 years ago
If you're powering an LED (just a special diode after all) with a forward current of 0.7A @ 3VDC forward voltage, then you really don't need a limit resistor, (unless the mfg says it's required that is) since your voltage drop is direct across the LED (diode) and equal to the value required by the LED.

Now, if your source voltage is above the forward voltage of the diode (LED), then the required resistor is

Rl = (Vsource - Vf) / If ,

where Rl is the limit resistor, Vf is the forward drop of the LED, and If is the forward current of the LED at Vf

Then to calculate the required power rating for the resistor

PR = Rl x If

so, if you had a 5V source and a Vf 3V, if 700mA LED, then

Rl = (5-3)/0.7 = ~2.8 ohm, and

PR = 2.8 *0.7 = 2W
wccolvin (author)  seandogue7 years ago
So I still need the 2 Watt Resistor even with a Small Voltage!
7 years ago
Well, no, not with a 3V source. But for source voltages in excess of the LED forward voltage, you will have to drop that energy somewhere, either by using a passive solution like a resistor or a step down voltage regulator or constant-current regulator. Each has their advantages and disadvantages.

In all likelihood, you could also power them using a pulse-width modulator to drive the 3.6 Vf LED ~directly from a 5V source. I can't say with 100% assurance, since I don't have the electrical specifications for the LED in front of me to verify, but at a guess they have a pulsed mode that will allow such a configuration.

For instance, with that 5V source I showed in the last part of the discussion above, a 555 timer can be used to create a PWM that will power the LED, using a MOSFET driver to source the the 700mA to the LED (if P channel) or sink the current thru the LED (if N channel) with very minimal losses in the FET (since they have very low on-resistances), as long as the maximum pulse mode ratings aren't exceeded in either frequency or (on) pulse width.

In general, for high power LEDS, using a passive step-down technique is not advised, for the reason that so much power is lost to heating the passive element. Instead, either dedicated LED driver ICs are used or a regulation circuit like those noted above is used to minimize system losses.