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Given two LED chips consuming same or very similar amounts of DC power,
One emitting light peaking at 630nm,
The other at 660nm,

In terms of output of (but light irrelevant to the sensitivity of human eye);

  • Which one would output more energy in form of light / photons?
  • Which one is more "efficient"?

Taking into consideration:

660nm means less energy per photon,
630nm means more energy per photon;

  • Which one outputs more number of photos?
  • Does larger wavelength of 660nm mean it takes longer between photon outputs?

Also:

If it takes longer to create a single photon of 660nm, which has less energy than 630nm;

  • Does the energy difference between 630nm and 660nm photon is turned into heat by LED?
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    \$\begingroup\$ Are you measuring the output power, or do you care about visibility? If the latter, remember the sensitivity of the eye will differ between 630 and 660 nm, and that will probably dwarf any differences between the LEDs. \$\endgroup\$ – Brian Drummond Dec 2 '14 at 15:04
  • \$\begingroup\$ I'm more concerned about understanding the output power. \$\endgroup\$ – xxfhzvar Dec 2 '14 at 15:25
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    \$\begingroup\$ If you are just concerned about output power, why not use a resistor? They are much cheaper than LEDs. \$\endgroup\$ – Phil Frost Dec 2 '14 at 15:26
  • \$\begingroup\$ I'm sorry I think I might have mislead you. I am concerned with output light in the visible spectrum, from around 400 to 700nm range. However, I am not concerned with the sensitivity of the human eye. Your answer is already adequate but could you update it with this additional information please? I'm very confused with the following Higher wavelength == less energy, right? Higher wavelength also == less number of photons coming out, right? So a 660nm LED would create less light / photons / light energy than 630 one I suppose since 630 means more frequent photons with more energy? \$\endgroup\$ – xxfhzvar Dec 2 '14 at 15:29
  • \$\begingroup\$ I mean, human eye is less sensitive to red than another color, I think. So an LED emitting a lot of RED (more photons, more energy thanks to more efficiency) might seem less bright to human eye than an LED emitting another color which might seem brighter but actually its output of photons and energy in terms of light will be less. \$\endgroup\$ – xxfhzvar Dec 2 '14 at 15:35
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Pedantically, assuming we are driving each of these 1 watt LEDs to their full power handling capability, they put out exactly the same radiant flux: 1 watt, or 1 joule per second. Then again, so does a 1W resistor.

The reason: with 1W of electrical power going in, there must be at equilibrium 1W of power going out. Some of it will be red light, and the rest of that electrical power will go into heating the component until blackbody radiation is enough to achieve thermal equilibrium. Within the normal operating temperatures of these components, that blackbody radiation will be in the far infrared spectrum, but the definition of radiant flux includes all electromagnetic radiation, even that which is not visible.

The LEDs will emit less photons than the resistor, because at least some of their radiation will be at a higher frequency. Higher frequency photons have more energy, so fewer are required.

But probably, you care just about the visible radiation from the LEDs. To do this, you have to define just what you mean by "visible". A standard definition is luminous flux:

In photometry, luminous flux or luminous power is the measure of the perceived power of light. It differs from radiant flux, the measure of the total power of electromagnetic radiation (including infrared, ultraviolet, and visible light), in that luminous flux is adjusted to reflect the varying sensitivity of the human eye to different wavelengths of light.

For sure, either of your 1W LEDs will have a much higher luminous flux than a resistor. The 630nm LED has a bit of an edge over the 660 nm LED as the eye is more sensitive to 630nm (peak sensitivity being somewhere around 550 nm). However, this is but one of many variables needed to answer your question. The particulars come down to the specifics of the device. Read the datasheets.


To address some of your additional concerns:

I'm very confused with the following Higher wavelength == less energy, right?

Less energy per photon, but this is really irrelevant for everyday purposes.

$$\text{energy per photon} = h\nu = \frac{hc}{\lambda} $$

Where:

  • \$c\$ is the speed of light in meters per second
  • \$h\$ is the Planck constant
  • \$\nu\$ is the frequency in hertz
  • \$\lambda\$ is the wavelength in meters

See Energy vs. power in transmitters from Amateur Radio Stack Exchange. Although that's about radio transmitters, it's still electromagnetic radiation just like visible light (but at a much lower frequency)

Higher wavelength also == less number of photons coming out, right?

Assuming for this comparison an equal radiant flux. You've got it backwards: higher wavelength means a lower frequency, thus less energy per photon. To attain a given radiant flux (power), you can have:

  • a higher number of lower energy (higher wavelength, lower frequency) photons, or
  • a smaller number of higher energy (lower wavelength, higher frequency) photons

being emitted each second. The number of photons emitted per second, multiplied by the energy per photon equals the radiant flux:

$$ \text{photons per second} \cdot \text{energy per photon} = \text{radiant flux} $$

I am concerned with output light in the visible spectrum, from around 400 to 700nm range. However, I am not concerned with the sensitivity of the human eye.

If this is the case, and you want to weight all radiation in this visible range equally, you can do that. In this case, there's really very little difference besides color between a 660nm and 630nm LED (or an LED of any other color). Any differences in efficiency by this definition will be a consequence of the device's construction. For example, the resistance of the leads, the shadowing of the LED die by the bond wires, the opacity of the lens, flaws in the manufacture, etc.

If it takes longer to create a single photon of 660nm, which has less energy than 630nm does the energy difference between 630nm and 660nm photon is turned into heat by LED?

No. Although a 630nm photon and a 660nm photon have different energies, the rate they are emitted will be different such that the radiant flux will be identical.

Think of it is this way: if you have a factory that consumes raw water at 100 liters per second and has to ship that water out at the same rate, you can:

  • ship 100, 1-litre bottles per second
  • ship 10, 10-litle bottles per second
  • or any number of other combinations, as long as your total water output rate is equal to 100 liters per second.

Each photon is a package of energy. The LED consumes electrical energy at some rate (1W), and it must output some other kind of energy at the same rate (probably emitting photons, either as visible light or blackbody radiation). It doesn't matter what size the packages are.

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  • \$\begingroup\$ Dear Phil Frost, I added a comment to the question to clarifying what I want to understand. Please tell me if I shall update the question accordingly. Your input and answer has great value as I am very confused with all I have been reading and it would help to fill a large gap, allowing me to understand things better. Thank you so much. \$\endgroup\$ – xxfhzvar Dec 2 '14 at 15:31
  • \$\begingroup\$ Interesting - I would have expected the eye to be less sensitive to 660nm, as it is further from the 550nm peak. \$\endgroup\$ – Brian Drummond Dec 2 '14 at 15:41
  • \$\begingroup\$ @BrianDrummond whoops...I got that backwards. Closer to 550 nm does mean more sensitive. \$\endgroup\$ – Phil Frost Dec 2 '14 at 15:45
  • \$\begingroup\$ @PhilFrost, updated my question. Thank you again for pointing out the flaws. I hope it clarifyings what I seek to understand. \$\endgroup\$ – xxfhzvar Dec 2 '14 at 15:55
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    \$\begingroup\$ @PhilFrost - Your answer explains it all and addresses all the points. I am also greateful for the further reading you've provided. I can not thank you enough for the time and effort you've put in. Thank you, so very much. \$\endgroup\$ – xxfhzvar Dec 2 '14 at 16:13
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Since you want to ignore the human eye response, (Lumens/ Lux and all that falderal.) and just count photons or Watt's, it's pretty easy. If you assume the same efficiency from both LEDs, then the output watts would be the same. But the longer wavelength would give you more photons. (Since each photon carriers h*frequency of energy.)

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  • \$\begingroup\$ If wavelength is longer, how can there be more photons? It sounds as if it should be opposite (shorter wavelength == more photon coming out). Also, I want to understand which LEd would be (or could be) more efficient due to limited known ways of LEDs creating light. \$\endgroup\$ – xxfhzvar Dec 2 '14 at 15:54
  • \$\begingroup\$ @xxfhzvar efficient for what purpose? Emitting a large number of photons of any kind? Looking bright to a human? Looking bright to a rattlesnake? Emitting electromagnetic radiation of any kind? Creating unicorns? \$\endgroup\$ – Phil Frost Dec 2 '14 at 16:10
  • \$\begingroup\$ Well we are assuming some ideal case. So the same amount of power from each LED. Then a longer wavelength means a smaller frequency. (since freq * wavelength = constant = speed of light.) And a smaller frequency means lower energy per photon, (the Planck relation I mentioned.) And you'll need more of them for the same power. In practice the efficiency of the LED's will depend on lots of solid state and process things and will be different. It would be very nice if the LED manufacturers would give light intensity in terms of Watts as well as lumens. \$\endgroup\$ – George Herold Dec 2 '14 at 16:11
  • \$\begingroup\$ @GeorgeHerold they do give light intensity in terms of watts. It's the maximum current multiplied by the forward voltage at that current. \$\endgroup\$ – Phil Frost Dec 2 '14 at 16:13
  • \$\begingroup\$ @PhilFrost those watts provided are input watts, not output (I think). If there were two LEDs, one emitting A nm (same eg. 630), the other B nm (eg. 660), both consuming W watts (eg. 75w), which one would be more efficient in terms of visible light output irrelevant to the sensitivity of the human eye? Which one would emit less heat due to inefficiency? (concerning only Red LEDs). \$\endgroup\$ – xxfhzvar Dec 2 '14 at 16:17
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The radiant flux is specified for each in their respective datasheets for a given forward current (I_fwd). In terms of efficiency (again referring to the datasheets), consider the thermal resistance of each package. Typically the lower Rth-ja is more efficient.

If the goal is ultimately to have the same "brightness" (radiated power of each), depending on how important this is to your application, it is very possible that you could calibrate this with a light detector (phototransistor amplifier). For example you could PWM the LEDs one at a time and have the duty cycle increase/decrease to zero in on the desired current you sink in the phototransistor.

Does this answer your question?

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Does a 1 watt 660nm LED put out more radiant flux than a 1 watt 630nm LED?

Not Likely... but for sure they will be different. Just like Fluorescent lamps, most LED lamps usually use phosphors to convert electromagnetic energy from the light source to output. So my guess is the LEDs that you have likely output a natural blue-white light that is converted by different phosphors in phosphor blends to output "dominant" wavelengths of either 630nm (red) or 660nm (deeper red). The amount of radiant flux/power will depend upon the Quantum Efficiency (QE) of the phosphor(s) used.

Some phosphors with a peak wavelength near 655nm can have a QE of 0.69 and other (more expensive) phosphors can have a QE of 0.86. The dominant wavelength of either is usually higher than 660nm. So to get dominant wavelengths of 630nm and 660nm, the LED manufacturer likely used a peak 655nm phosphor blended with the common/cheap Y2O3Eu3+ phosphor (peak 611nm) which fortunately has a high QE of 0.99.

So the reason I answered "Not Likely" is that the 630nm LED likely has a phosphor blend with more of the Y2O3Eu3+ phosphor which has a higher QE (note: other phosphors could be used in the blend). For "better than my guess" you need to get the light source report from that manufacturer... this will usually list radiant power (flux).

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