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

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1. Sep 8, 2016

### 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?

2. Sep 8, 2016

### Orodruin

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

3. Sep 11, 2016

### 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.

4. Sep 13, 2016

### Metals

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?

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?

5. Sep 13, 2016

### Orodruin

Staff Emeritus
Why would they?

6. Sep 13, 2016

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

7. Sep 14, 2016

### Nosebgr

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.

8. Sep 14, 2016

### Orodruin

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

9. Sep 14, 2016

### Nosebgr

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

10. Sep 14, 2016

### Nosebgr

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?

11. Sep 18, 2016

### 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.

Last edited: Sep 18, 2016
12. Sep 18, 2016

### 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.

13. Sep 30, 2016

### Metals

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?

14. Dec 12, 2016

### 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.

15. Dec 17, 2016

### nikkkom

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

16. Dec 17, 2016

### jim hardy

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 ?

17. Nov 1, 2017

### 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.

18. Nov 1, 2017

### jim hardy

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.

Last edited: Nov 1, 2017
19. Nov 1, 2017

### 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