Blue lights (LEDs) consume more power than red (and green)?

[Metals]

Referring to LEDs as I'm not aware of any other "colour-producing" light, does blue light use more energy than red light?

Planck's formula (E=hf) leads me to believe that blue light would drain a battery quicker than red, due to its higher frequency. Surely a device emitting blue light would require more energy (more current), so wouldn't red and green lights which have a lower frequency consume less power than blue, resulting in a slightly longer battery cycle?

 

[Orodruin]

Your argument would only hold if all LEDs emitted the same number of photons per time unit. They do not.

 

[Nosebgr]

Alternate way of thinking (engineer's approach): Power = Voltage x Current. If you have a simple circuit that consists of a battery, an LED and a resistor, you can adjust the power output/consumption by changing the value of the resistor. Since LED's have negligible resistances, the power consumption would only depend on the resistor you connect the LED up with. You wouldn't connect an LED straight to battery anyways.
But assuming you connect up the LED so that the consume the same amount of power, and assume an imaginary world where the scale of the frequency difference really matters, then the blue LED would light up at a lower intensity than the red LED. We don't live in that imaginary world though, so no need to avoid blue light to save power.

 

[Metals]

Your argument would only hold if all LEDs emitted the same number of photons per time unit. They do not.

I see. Why is it that LEDs don't do that by the way? Say, with the same power source. And if they did, then blue light would have more power consumption?

But assuming you connect up the LED so that the consume the same amount of power, and assume an imaginary world where the scale of the frequency difference really matters, then the blue LED would light up at a lower intensity than the red LED. We don't live in that imaginary world though, so no need to avoid blue light to save power.

Why would blue be at lower intensity? Surely it would be higher considering it has more energy. So wave frequency is completely irrelevant for colour?

 

[Orodruin]

Why is it that LEDs don't do that by the way?

Why would they?

 

[Vanadium 50]

Why would they?

Exactly. What's wrong with a 5 mW LED needing 5 mW of power?

 

[Nosebgr]

Why would blue be at lower intensity? Surely it would be higher considering it has more energy. So wave frequency is completely irrelevant for colour?

In hindsight, perhaps I should have phrased it better. When I said "intensity", I meant it purely in terms of the photon density. I shouldn't have said intensity, because by definition it is the power per unit area. If we feed any two LED's the same power, regardless of their colour, the output intensities will be equal. (You get the same as what you put in). So, in terms of the actual definition of intensity, both LED's are the same as long as their input powers are the same.

However, in the same case, the red LED with the longer wavelength will light up brighter, because it has greater photon density than the blue one.

You can think of it as follows: There are two pools of same sizes, and they are both full. You empty one pool using 0.5L bottles, and the other one with 1L bottles. At the end, you will end up with more 0.5L bottles, (therefore a brighter LED) than the 1L bottles. But the total amount of water inside the bottles (energy/intensity) are equal. Hopefully this clarifies the ambiguity.

 

[Orodruin]

because it has greater photon density than the blue one.

Photon density is not directly proportional to how bright you experience something.

 

[Nosebgr]

Photon density is not directly proportional to how bright you experience something.

Why not? Intuitively that's how I always pictured it - assuming the angle at which the photons meet the eye are equal.

 

[Nosebgr]

Why not? Intuitively that's how I always pictured it - assuming the angle at which the photons meet the eye are equal.

Ok, I see now. The receptors in the eye - or a receiver, either way - doesn't necessarily sense / detect purely according to the number of photons. They would respond to the intensity - the rate of power transferred from a particular source. Then, both LED's would theoretically light up at the same brightness. Does this sound correct?

 

[calinvass]

If you have 2 lasers pd the same power consumption one red and one green, the green one is much more visible because the human eye responds better to green light. I don't know about blue light though.
I could have been that green lasers convert electricity to light more efficiently, but, apparently it is not the case.
However if your question has a different aspect, whether blue light led power output over power input is less that for a red led, then as long as they don't emit other light frequencies, like for example infrared, they should have the same ratio.

"Photometry deals with the measurement of visible light as perceived by human eye. " Wikipedia.
We can then compare luminous intensity of those two lights, by the same number of photons or by the same output power calculated using energy of a photon of hf.

 

[houlahound]

I believe we have the least respond to blue light. My neighbour is an insurance assessor who has never done a physics course. He claims blue cars are the hardest to see and hence are in more collisions that seem should never have happened.

 

[Metals]

In hindsight, perhaps I should have phrased it better. When I said "intensity", I meant it purely in terms of the photon density. I shouldn't have said intensity, because by definition it is the power per unit area. If we feed any two LED's the same power, regardless of their colour, the output intensities will be equal. (You get the same as what you put in). So, in terms of the actual definition of intensity, both LED's are the same as long as their input powers are the same.

However, in the same case, the red LED with the longer wavelength will light up brighter, because it has greater photon density than the blue one.

You can think of it as follows: There are two pools of same sizes, and they are both full. You empty one pool using 0.5L bottles, and the other one with 1L bottles. At the end, you will end up with more 0.5L bottles, (therefore a brighter LED) than the 1L bottles. But the total amount of water inside the bottles (energy/intensity) are equal. Hopefully this clarifies the ambiguity.

Ah so the energy output is whatever the battery supplies regardless of light, but because red light naturally has less energy, more red photons must be emitted in comparison to blue so that output energy is met?

But in the case of stars, more energy released will show blue flames because there isn't a battery limiting the output energy?

 

[Melissa Prostrollo]

Interestingly (what led me to this thread), I have a string of battery-operated LED snowflake lights, which change colors from red to green to blue to yellow, pink, purple, etc. I can tell when the batteries are nearly depleted because they start to appear only red and green, with the other colors failing. As the batteries get lower still, the green fails and they show only red.
Then, I replace the batteries, and the other colors function again, too.

 

[nikkkom]

IIRC higher-freq diode lights are using frequency-doubling, tripling, etc. Which is not 100% efficient (some power is lost as heat).

 

[jim hardy]

Planck's formula (E=hf) leads me to believe that blue light would drain a battery quicker than red, due to its higher frequency. Surely a device emitting blue light would require more energy (more current), so wouldn't red and green lights which have a lower frequency consume less power than blue, resulting in a slightly longer battery cycle?

you're right about the Planck's constant relation, but it shows up in the voltage not the current
http://arduino-info.wikispaces.com/Brick-Resistors-LEDs

How many electron volts in a blue photon ?

see http://electron6.phys.utk.edu/phys250/modules/module 1/photons.htm

• What is the energy of a photon of blue light (λ = 450 nm) and of a photon of red light (λ = 700 nm) in units of eV = 1.6*10-19 J?
  Solution:
  E = hc/λ.
  Blue light: E = (6.626*10-34 Js)(3*108 m/s)/(450*10-9 m) = 4.4*10-19 J = 2.76 eV
  Red light: E = (6.626*10-34 Js)(3*108 m/s)/(700*10-9 m) = 2.8*10-19 J = 1.8 eV

 

[Michael Skelton]

I tested a set of colour changing strip lights a while back using a multimeter in series and the blue used the least power, the green used a bit more and the red used the most. I assume it's because you need to use a resistor to drop the voltage more for red and green LEDs then you do for blue LEDs. The LEDs going out as the battery drains is because the forward voltage is higher for blue and green LEDs, not because the battery can't power them.

 

[jim hardy]

I tested a set of colour changing strip lights a while back using a multimeter in series and the blue used the least power, the green used a bit more and the red used the most.

Seems to be some confusion as to what is power.Thought experiment

Connect a red a green and a blue LED in series with a controllable current source. They'll have to be raw LEDs not the kind with an internal resistor that run on 5 or 12 volts.
Set the current source to 10 milliamps, which is typical current for small LEDs . That's 1/100 of an amp.
Since they're in series they all get 10 milliamps..

Measure the voltage across each LED .
Red one will have lowest voltage, see chart in post 16 above, maybe 1¼ volts
Blue will have highest maybe 3 volts

Power in Watts is Volts X Amps
so red consumes 1.25 X 1/100 = 12.5 milliwatts and blue consumes 3 X 1/100 = 30_{(edit - first time i typed in 3 milliwatts)} milliwatts.
What will be volts across the green one , if that chart is accurate ?

How much luminous power each puts out depends on its efficiency.

 

[dlgoff]

I've got a project going on that uses "Ultra Bright/Super Bright" blue LEDs with a luminous intensity (mcd) spec: average is 5000 ~ 6000. What this means from http://www.theledlight.com/technical1.html

LED light output varies with the type of chip, encapsulation, efficiency of individual wafer lots and other variables. Several LED manufacturers use terms such as "super-bright," and "ultra-bright" to describe LED intensity. Such terminology is entirely subjective, as there is no industry standard for LED brightness. The amount of light emitted from an LED is quantified by a single point, on-axis luminous intensity value (Iv). LED intensity is specified in terms of millicandela (mcd). This on-axis measurement is not comparable to mean spherical candlepower (MSCP) values used to quantify the light produced by incandescent lamps.

 

[Ryeisenman]

Thanks for explanations. I came here because my outdoor hiking LED headlamp has a separate red light. When battery gets weak, the white light gets dies. But I can still use the red LED on weak battery.

 

[valenumr]

Why not? Intuitively that's how I always pictured it - assuming the angle at which the photons meet the eye are equal.

I was thinking about this as well, but iirc, blues will seem brighter than reds and greens and yellows (perception). It actually annoys me that most modern things have crazy bright blue LEDs. Give me my '80s red led alarm clock.

But I *think*, and it's been a while, that blue LEDs tend to have a lower resistance than, say red LEDs, so one would put a larger resistor in series with the power source.

 

[valenumr]

Thanks for explanations. I came here because my outdoor hiking LED headlamp has a separate red light. When battery gets weak, the white light gets dies. But I can still use the red LED on weak battery.

The red light actually exists to not mess up your night vision. But I suppose that doesn't really answer your question. Perception of brightness is a separate topic I think.

 

[valenumr]

I was thinking about this as well, but iirc, blues will seem brighter than reds and greens and yellows (perception). It actually annoys me that most modern things have crazy bright blue LEDs. Give me my '80s red led alarm clock.

But I *think*, and it's been a while, that blue LEDs tend to have a lower resistance than, say red LEDs, so one would put a larger resistor in series with the power source.

More correctly after looking into it: blue (and white) LEDs tend to have a larger forward voltage than red ones. I shouldn't have called it "resistance", because it's not really the same (the voltage is nearly constant with current).

In any case, this might explain why your red LEDs worked with a low battery at least (smaller forward voltage).

 

[hutchphd]

More correctly after looking into it: blue (and white) LEDs tend to have a larger forward voltage than red ones. I shouldn't have called it "resistance", because it's not really the same (the voltage is nearly constant with current).

In any case, this might explain why your red LEDs worked with a low battery at least (smaller forward voltage).

This is exactly correct! The band gap in the materials for particular color LED has the appropriate energy for that photon. Assuming a similar quantum efficiency, the photon number output will scale with current for all of the colors but the energy will scale with the bandgap voltage. More energy from the blue at a given current.
Of course @jim hardy had it correct. RIP.

 

[Krail]

I just found this post while googling about my Christmas lights, and I think I've got some interesting anecdotal evidence for you all.

The little guys in question are color-shifting LEDs. There's a red, green, and blue, and it slowly color cycles each one so they blend through the color wheel.

When left on long enough, as the battery drains, the blue LEDs will just stop lighting up at all. When it gets really low, the greens die out, too.

Which leaves us with spooky little red lights that dim in and out.

 

[berkeman]

Welcome to PF.

Which leaves us with spooky little red lights that dim in and out.

Yep, those are the ones with the lowest Vf.