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.

active| newest | oldestCharles,

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.

Note that no matter what, the author will still need to drop that voltage difference @ I

_{f}, 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.Now, if your source voltage is above the forward voltage of the diode (LED), then the required resistor is

R

_{l}= (V_{source}- V_{f}) / I_{f},where R

_{l}is the limit resistor, V_{f}is the forward drop of the LED, and I_{f}is the forward current of the LED at V_{f}Then to calculate the required power rating for the resistor

P

_{R}= R_{l}x I_{f}so, if you had a 5V source and a V

_{f}3V, i_{f}700mA LED, thenR

_{l}= (5-3)/0.7 = ~2.8 ohm, andP

_{R}= 2.8 *0.7 = 2WIn all likelihood, you could also power them using a pulse-width modulator to drive the 3.6 V

_{f}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.

Here's a very useful Instructable detailing driver circuits for various types of LED.