LED Surprise

[Newz2000]

I was surprised to find that LEDs with a number of different intensities all have IF of 20ma. I'm used to incadencent bulbs where more current means more bright. But with this, LEDs ranging from 80mcd to 9,000 mcd all have the same IF.<p>Also, I'm not confident my calculations are correct, I think it is because there's an element of the formula I don't know...<p>If I grab a blue LED and seek to get IF of 20ma and the Vf is 3.6v I'd use this formula to get R, the value of the resistor to put in series:
R = (Vs - Vf)/IF<p>So, for a 5v circuit I'd do:
(5 - 3.6)/.02 = 70. To be safe I take a 100ohm resistor, or if I'm going to have a duty cycle of < 100% I might choose a 68ohm. But can I use a 1/8th w resistor or must I use a 1/4th w, or, will that even be enough? I can't remember how to figure this out. <p>Tell me if this is right...
Since 3.6v are dropped across the LED that means 1.4v are dropping across the resistor... So using ohm's law, 1.4v = i * 100ohm then i = 1.4/100 = .014
To find the power consumed, I'd use P = i * e, which yields P = .014 * 5 = .07w<p>70mw is much less than 125mw (1/8th watt) if my calculations above are right. Am I going about this the right way?<p>Thanks. Sorry if that mess of math is incomprehensible. If I need to rewrite it, let me know.

[Chris Smith]

Not all LEDS draw the same current, but it is possible to be brighter and still draw a lesser current. <p>The doping techniques, the die materials, and cavity designs have come a long way since the days of RED LEDs only. <p>Phosphorescence, mixed dies, and other techniques have added to the increased luminosity and color of the modern Led while maintaining a low current draw.

[Robert Reed]

Assuming Vr remains constant regardless of If and using a5V source:<p> 100 ohms--14 ma--0.02 watts (20 mw)
68 ohms---20 ma--0.029 watts (29 mw)<p> 0.125 watt resistors are more than good enough.

[philba]

You can get 75 and 82 ohm 5% resistors - through hole and surface mount. At least Mouser has them. In 1%, you can get 69.8 and 71.5. But you wont damage anything by using 68, either.<p>By the way, you calculate the wattage on Vs-Vf, not on the total voltage.

[rshayes]

One reason for the wide intensity is the shape of the package. The power which can be dissipated by the LED is mostly determined by the package and the material of the LED. Wider band gap materials can operate at slightly higher temperatures, which offsets their higher forward voltage. With the common plastic packages, 20 milliamps is a fairly conservative current that gives reasonable light output. With a given material and type of LED, the efficiencies and total optical output will be similar. The difference is whether this output is concentrated in a narrow angle or spread over an entire hemisphere. This is usually done by using the package to either form a lens or to diffuse the light.

[Enzo]

Incandescent bulbs make light by heating something so hot it glows or incandesces. The more power run through it the hotter it gets, so the brighter it gets.<p>LEDs do not work by heating something up until it glows. It works by having electrons cross energy levels. LEDs can vary as to the efficiency of turning this into visible light. They don't generate heat. Well, not much anyway.<p>So it is not a legitimate way to look at it to determine voltage drop and current. The drop is not due to resistive effects - Ohm's Law does not apply. The fact that the blue one has a higher voltage across it is just an artifact of the chemistry inmvolved with making blue color light. It took them a long time to get blue LEDs to work at all.

[Dean Huster]

"They don't generate heat. Well, not much anyway."<p>My students made them generate a lot of heat on a frequent basis. Half of the LEDs in their parts kits had a blacked look about the emitting area and also had a nasty shift in color in the orange direction. Needless to say, you could never convince some of those clowns that series resistors are necessary -- after all, the LEDs lit up just fine for them without one, didn't they?<p>Dean

[cato]

Your calculations are correct (at least the thinking behind them is, I didn't check your math)..up to a point. <p>You are assuming that IF is actually going to be the IF on the data sheet (at least I think your are assuming that). This might not be the case. Dig deeper into the spec sheet to worst case your current needs (to get the lighting you need) and for your heat dispation calculation. Don't forget that the resistor wattage is rated for room temp assuming a particular mounting arrangement, so if you might be in a warmer environment, you might need more wattage rating head room. Lumin output will also vary with temperature. Be sure to account for that when you pick a drive current.

[John Brown]

This is something that always confuses me. If different LEDs produce different amount of light with the same power consumption, and they "don't generate much heat", then where does the rest of the power go in the lower-light-output LED?

[MrAl]

Hello there,<p>Calculating the wattage of a resistor isnt too
hard if the voltage across it isnt changing and
you know what it is.<p>For your circuit, the voltage ACROSS the resistor
is (Vs-Vf) so you know this voltage.
To calculate the power (wattage) there are a
few formulas you can use:<p>P=(Vs-Vf)*(Vs-Vf)/R
where
R is the resistance of the resistor in ohms.<p>There's also:<p>P=(Vs-Vf)*I
where
I is the current in amps.<p>There's also:<p>P=I*I*R
where
I is the current in amps and
R is the resistance of the resistor in ohms.<p>You can use whichever formula you wish.
Once you calculate P (the wattage) you should
double it to select the correct resistor size.
For example, if you calculate P to be 0.25 then
you should double this (which gives you 0.50) so
the correct resistor size should be 1/2 watt.<p>The final consideration is whether or not the
resistor will be in an enclosed area such as
a box of some kind (as in a flashlight).
The simplest way to determine if the resistor
size is large enough and has enough cooling is
to stick a temperature probe inside the box
in order to measure the temperature of the resistor.
Run the light for a few hours and watch the
temperature rise of the resistor. If it gets
above 70 degrees C you might want to drill a
few holes in the box to allow cool air to pass
around the resistor in order to cool it better.
Going to a physically larger resistor might
help but you'll have to repeat the test to make
sure. Keep in mind that some resistors that get
too hot will smoke and burn
Flameproof resistors are about the best.<p>
Take care,
Al

[EPA III]

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by Dean Huster:
"They don't generate heat. Well, not much anyway."<p>My students made them generate a lot of heat on a frequent basis. Half of the LEDs in their parts kits had a blacked look about the emitting area and also had a nasty shift in color in the orange direction. Needless to say, you could never convince some of those clowns that series resistors are necessary -- after all, the LEDs lit up just fine for them without one, didn't they?<p>Dean<hr></blockquote><p>Dean, <p>You are obviously using a power supply with too low of an output voltage. If you want to demonstrate the necessity of a series resistor, I would suggest a 24 Volt or even a 48 Volt supply. <p>And make then pay for the blown LEDs. <p>Paul A.

[Dean Huster]

We used 5V supplies, Paul. Low enough voltage that the LEDs worked for them without current limiting but enough current available (1 amp) that they could wreck them after time.<p>What? Make students pay for what they foolishly or deliberately break? The school where I taught had an administrator who considered an idea like that to be an affront to a student's personal rights. They could destroy a $5000 oscilloscope and not even have their hand slapped. If a student was lazy and refused to read the instructions for an experiment and screwed things up, it was always the instructor's problem, not the student's. The worst part of teaching was usually the school system. The best part of teaching was finding the motivated student and helping them develop.<p>Dean