Phosphor conversion of photons in LEDs & photon recycling efficiency

Genzod

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I know that phosphor in an LED converts some of the blue photons supplied to it into yellow photons to create white light.

I'm wondering what happens when other colors of photons are supplied to the phosphor. What would happen if green photons enter the phosphor layer? Yellow photons?

Thanks.

Edit: Also interested in understanding photon recycling efficiency that increases LED luminance. How are the losses in a waiven collar accounted for when photons are sent back to the die for recycling? In the set up below (credit: SMA@ Taschenlampen Forums), the gain is 2.2x over the fraction of light defined by the flux ratio of solid angles for a 30 degree half angle beam opening--a 55% efficiency of a potential of 4x. The reflector is 99% efficient but it seems to pass quite a bit of light despite the rating.

P2152023.JPG
 
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brickbat

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Re: Phosphor conversion of photons in LEDs

Generally a phosphor provides photons at a wavelength (color) related to its molecular and crystalline structure, and not related to the wavelength of the incoming light. If the wavelength of the incoming light becomes too long, the phosphor won't emit any light.
 

Genzod

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Re: Phosphor conversion of photons in LEDs

Generally a phosphor provides photons at a wavelength (color) related to its molecular and crystalline structure, and not related to the wavelength of the incoming light. If the wavelength of the incoming light becomes too long, the phosphor won't emit any light.

So if you have a typical blue emitter with yellow phosphor, the phosphor crystalline structure is designed to create yellow photons. But if the wavelength of the emitter increases, the generation of yellow photons will decrease until a point is reached where no more yellow photons will be generated, is that correct?

If so, is that because the energy difference between the emitter source photons and the phosphor photons is approaching zero? Hence a hypothetical red or orange emitter can't give rise to yellow photons in that yellow phosphor?
 

The_Driver

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Re: Phosphor conversion of photons in LEDs

So if you have a typical blue emitter with yellow phosphor, the phosphor crystalline structure is designed to create yellow photons. But if the wavelength of the emitter increases, the generation of yellow photons will decrease until a point is reached where no more yellow photons will be generated, is that correct?

If so, is that because the energy difference between the emitter source photons and the phosphor photons is approaching zero? Hence a hypothetical red or orange emitter can't give rise to yellow photons in that yellow phosphor?

Normal white LEDs use YAG:Ce phosphor. Here you can see the absorption spectrum of this type of phosphor. It should answer your question.
This absorption and emission phenomenon is called Stokes Shift.

Here you can find a nice article on the topic.
 
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ssanasisredna

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Re: Phosphor conversion of photons in LEDs

So if you have a typical blue emitter with yellow phosphor, the phosphor crystalline structure is designed to create yellow photons. But if the wavelength of the emitter increases, the generation of yellow photons will decrease until a point is reached where no more yellow photons will be generated, is that correct?

If so, is that because the energy difference between the emitter source photons and the phosphor photons is approaching zero? Hence a hypothetical red or orange emitter can't give rise to yellow photons in that yellow phosphor?

That is correct, and unfortunately at least at this point in time, only one photo out for each photon in, and as the emitted photons are less energetic, there is an efficiency loss equal to the ratios of the wavelengths.
 

Genzod

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Re: Phosphor conversion of photons in LEDs

That is correct, and unfortunately at least at this point in time, only one photo out for each photon in, and as the emitted photons are less energetic, there is an efficiency loss equal to the ratios of the wavelengths.

The Driver said:
Normal white LEDs use YAG:Ce phosphor. Here you can see the absorption spectrum of this type of phosphor. It should answer your question.
This absorption and emission phemnomenon is called Stokes Shift.

Here you can find a nice article on the topic.

Thanks everyone.

Up until now, I wasn't sure where the extra energy was going from the shift. I thought perhaps part of the increased surface brightness of an LED undergoing the photon recycling that arises from a wavien collar was coming from the ratio of wavelengths. But now it would appear redirection of the formerly wasted wall light undergoes about a 15% loss from reflection and a tint shift toward yellow of any higher energy photons that make it back into the phosphor, representing a further energy loss due to conversion. The gain in surface brightness in LED due only*to the super-positioning of redirected photons that eventually escape through the collar's hole toward the lens.

*Unless I'm overlooking something else, of course. ;) Any comments regarding that?
 
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brickbat

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Re: Phosphor conversion of photons in LEDs

...I wasn't sure where the extra energy was going from the shift....

High energy photon enters a phosphor, low energy photon exits. The energy difference shows up as heat in the phosphor, no?
 

Genzod

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Re: Phosphor conversion of photons in LEDs

High energy photon enters a phosphor, low energy photon exits. The energy difference shows up as heat in the phosphor, no?

Exactly. In my mind, I had imagined two boundaries describing the energy balance:

1) For every 100 blue photons phosphor generates 120 yellow photons (assuming wavelength ratio 1:1.2)

2) For every 100 blue photons phosphor generates 100 yellow photons and the extra energy from the wavelength ratio represents an efficiency loss, e.g. heat.

(or something in between)

Answer confirmed above by ssanasisredna as item 2.
 
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The_Driver

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Re: Phosphor conversion of photons in LEDs

The Wavien Collar redirects the light onto the phosphor. The blue part is "recycled". It can come back out at a different angle (converted light is scattered isotropically in phosphor) and is converted to yellow-green. 75% of the emitted light (of a domeless LED) is collected by the collar and the remaining 25% is amplified by a max of 120%. Thus the efficiency in terms of lumens is 55%.
See here for more details (in German).
 
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Genzod

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Re: Phosphor conversion of photons in LEDs

The Wavien Collar redirects the light onto the phosphor. The blue part is "recycled". It can come back out at a different angle (converted light is scattered isotropically in phosphor) and is converted to yellow-green. 75% of the emitted light (of a domeless LED) is collected by the collar and the remaining 25% is amplified by a max of 120%. Thus the efficiency in terms of lumens is 55%.
See here for more details (in German).

That's a very decent article. Vielen Dank!

I had likewise introduced the Lambert cosine through each dA area element in the hemisphere and integrated to get the same sin2​
θ result, so it was reassuring I was working with the correct formula.

With 99% reflectivity, I would think the gain to have been close to 4X, not 2.2x, so I can see where you get the 55% efficiency. Is that inefficiency due the low critical angle of the phosphor and air, resulting in reflections that repeatedly bounce themselves off the phosphor and back to the reflector ad infinitum to eventual absorption?

If so, would it help greatly to have a roughened phosphor to improve chances of reentering the phosphor so they can reemerge at a steeper angle and have a better chance of exiting the hole? I think that's the case with the newer XPG2 and XPG3 phosphors. I'm wondering if gain has been improved using a collar with those LEDs?
 

The_Driver

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Re: Phosphor conversion of photons in LEDs

With 99% reflectivity, I would think the gain to have been close to 4X, not 2.2x, so I can see where you get the 55% efficiency. Is that inefficiency due the low critical angle of the phosphor and air, resulting in reflections that repeatedly bounce themselves off the phosphor and back to the reflector ad infinitum to eventual absorption?

I don't know the answer to that.
Possible reasons:
- Phosphor saturated by the heat of the die underneath and the additional heat from the light reflected by the collar
- the die underneath the phosphor could be overheating from the reflected light causing it to emit less blue light which in turn reduced total amount of available light
- reflection as you mentioned (what I know: at 90° angle the phosphor reflects ~3% at the phosphor-air-junction)

If so, would it help greatly to have a roughened phosphor to improve chances of reentering the phosphor so they can reemerge at a steeper angle and have a better chance of exiting the hole? I think that's the case with the newer XPG2 and XPG3 phosphors. I'm wondering if gain has been improved using a collar with those LEDs?

sma tested the collars with a de-domed XP-G2 (this LED had the highest known luminance of all LEDs in the beginning of 2015). The phosphor of the XP-G2 is pretty rough. The newer XP-G3 on the other hand has a much lower luminance because the effective die size is much larger. Its die is basically 3D, it also emits light to the sides. Generally it can be said that the XP-G3 is a fundamentally different LED compared to the XP-G2.
 
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Genzod

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Re: Phosphor conversion of photons in LEDs

I don't know the answer to that.
Possible reasons:
- Phosphor saturated by the heat of the die underneath and the additional heat from the light reflected by the collar
- the die underneath the phosphor could be overheating from the reflected light causing it to emit less blue light which in turn reduced total amount of available light
- reflection as you mentioned (what I know: at 90° angle the phosphor reflects ~3% at the phosphor-air-junction)...

Okay, great to understand other avenues of potential losses. At least heat can be managed.

I'm a little confused about your last mention. The critical angle (from the vertical) for air and phosphor is about 34 degrees, right? So wouldn't very shallow angles of approach (coming from 90 degrees from vertical) mostly bounce right off the phosphor, like that reflected at the base of the reflector (90 degrees)? It seems like you're saying 97% of photons approaching LED from the bottom of the reflector is absorbed (3% reflected). I'm thinking 3% absorption and 97% reflection at that horizontal vector. Maybe our angle conventions are opposite, or I'm just not familiar enough with the physics here.
 

The_Driver

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Re: Phosphor conversion of photons in LEDs

Okay, great to understand other avenues of potential losses. At least heat can be managed.

I'm a little confused about your last mention. The critical angle (from the vertical) for air and phosphor is about 34 degrees, right? So wouldn't very shallow angles of approach (coming from 90 degrees from vertical) mostly bounce right off the phosphor, like that reflected at the base of the reflector (90 degrees)? It seems like you're saying 97% of photons approaching LED from the bottom of the reflector is absorbed (3% reflected). I'm thinking 3% absorption and 97% reflection at that horizontal vector. Maybe our angle conventions are opposite, or I'm just not familiar enough with the physics here.

I meant that when you shine blue light from straight-on (vertical) toward a flat piece of phosphor, 3% of the light is reflected back at the phosphor-air junction. Very shallow angles probably increase this percentage for a flat piece of phosphor.

Here's a picture of a nicely de-domed Cree LED. Around it you can see the reflective LED package. The exposed phosphor-silicon-mix reflects much less light than the package, it has a much rougher surface. All flashlights with Wavien collar that achieved very high luminance values used de-domed XP-G2 LEDs or now Osram Black Flat LEDs (no dome from the factory, but much smoother surface compared to de-domed Cree LEDs).

Here you can find some more detailed findings regarding the optical properties of the phosphor-silicone-mix used in LEDs.
 

Genzod

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Re: Phosphor conversion of photons in LEDs

I meant that when you shine blue light from straight-on (vertical) toward a flat piece of phosphor, 3% of the light is reflected back at the phosphor-air junction. Very shallow angles probably increase this percentage for a flat piece of phosphor.

Okay, haha! Our angle conventions were different and opposite. You were describing a downward perpendicular approach, hence 90 degrees from base, and I was thinking 90 degrees measured from the vertical axis--photons moving horizontally.

Here's a picture of a nicely de-domed Cree LED. Around it you can see the reflective LED package. The exposed phosphor-silicon-mix reflects much less light than the package, it has a much rougher surface. All flashlights with Wavien collar that achieved very high luminance values used de-domed XP-G2 LEDs or now Osram Black Flat LEDs (no dome from the factory, but much smoother surface compared to de-domed Cree LEDs).

Here you can find some more detailed findings regarding the optical properties of the phosphor-silicone-mix used in LEDs.

I'll definitely read that publication, thanks!
 

Genzod

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Re: Phosphor conversion of photons in LEDs

I was looking over this reference TheDriver supplied. In the post, SMA is describing his experiment where he determines the luminance gain arising from a large wavien collar centered on a dedomed XP-G2. I'll describe what I think is going on here with this configuration, and I hope someone of experience would jump in to guide and correct my thinking.

It's hard to say exactly whether the XP-G2 SMA@Taschenlampen Forums pulled out of his scrap box was bare phosphor, had a sealant over the bare phosphor or if the dedome was a slice job with some silicone remaining over the bare LED. If someone here knows SMA, that information would be helpful in my understanding of photon recycling. This fact is critical to understanding the behavior of reflected photons returning to the emitter surface.

Assuming the XP-G2 has a seal and the sealing material has a refractive index of 1.5 like silicone, the critical angle between air and silicone would be about 42 degrees. (measured from the vertical axis :)) If the seal is perfectly smooth (assume for the moment), I would expect all photons approaching the phosphor at 42 degrees or more would simply bounce off the barrier and head to the opposite side of the collar for another 99% reflection and 1% absorption. It would take a number of reflections back and forth like a PONG game to be completely absorbed by the mirror.

Photons approaching at less than 42 degrees from vertical axis (regardless of wavelength) would enter the seal, then the phosphor, and some with enough energy (shorter wavelength) would be converted to lower energy photons. Those surviving, converted and unconverted photons would take another stab at a favorable exit angle that allows escaping the dome or get stuck in a reflective loop until completely absorbed by the mirror.

In essence, only the region of the mirror from 30-42 degrees supports increasing the luminance of the LED. The region from 42-90 degrees would eventually be absorbed by the mirror.

Of course this is not a perfect world with perfect surfaces, so I would expect the line to be less tightly defined (blurred/transitioned). Roughness of the LED surface would change angles of incidence to favor more re-absorption and emission, so I would anticipate the useful area of the mirror might extend beyond 42 degrees a bit.

If I use the following formula:

GAIN = K* sin2​γ / sin2​θ

(Ratio of flux through solid angles times a correction factor: constant * [critical angle of the die's top layer / half angle of the collar port].)

Where γ=42 degrees (gamma) is the critical angle of the first layer of the LED and θ=30 degrees (theta) is the half-angle of the collar's port (both measured from the vertical axis), and GAIN is about 2.20 in the posted experiment from Taschenlampen Forums, K becomes 1.23. If I set K to 1 and predict the angle that divides the useful from non-useful regions of the mirror, I get 48 degrees, which is close to the critical angle, deviated only by 6 degrees. I would therefore suspect surface roughness is responsible for the 6 degree offset in prediction of that line.

If the phosphor were bare, I would expect the critical angle to be around 34 degrees from vertical. The phosphor surface would probably be rougher than a sealed surface so I would expect the line to be lower than 40 degrees but probably not beyond 48 degrees.

This is all just conjecture from a photon noob, and I'm not trying to assert that that is what is actually going on here. But at the moment, it is what I'm thinking is happening as I'm trying to understand this physics.

Anyone here with some expertise that can give some knowledgeable insight, guidance or correction? Thanks.


EDIT:
Conceptual error: I re-examined my calculations for photons approaching the die at shallow angles.

Snells law can't describe an angle of approach that prevents photons from entering a slower medium (air to phosphor) like it can in reverse (phosphor to air) to describe TIR.

If photons in air coming from 70 degrees (from vertical axis) impact the phosphor surface, refraction into the die takes place 31 degrees from the vertical axis. Even 90 degrees has a refractive angle of 34 degrees from normal, as weird as that seems to my intuition (due to particle wave duality). So refraction is always taking place from all angles of approach coming from the collar to the die.

 
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JoakimFlorence

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Re: Phosphor conversion of photons in LEDs

I'm wondering what happens when other colors of photons are supplied to the phosphor. What would happen if green photons enter the phosphor layer? Yellow photons?
Above about 500nm (green-cyan) YAG phosphor has a very low level of absorption.

Not directly relevant to your question but a while back I was doing some research into blue-green (cyan) phosphors, and with the types of phosphors used in LEDs (including the advanced new ones that have just been developed) it generally takes an excitation source at least 40 nanometers less than the peak emission you hope to get from the phosphor. Try to excite the phosphor with a wavelength that's just barely less than the normal peak wavelength emission of the phosphor and it will end up shifting the wavelength distribution towards longer wavelengths, so that there will still be that characteristic gap in the spectrum. Although the minimum of the gap might be as high as 45 percent of the peak phosphor value.
 

Agpp

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Re: Phosphor conversion of photons in LEDs

As to losses: Light converted by phosphor and emitted at wrong angle will not be reflected back and therefore will be lost.
 

The_Driver

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Re: Phosphor conversion of photons in LEDs

I was looking over this reference TheDriver supplied. It's hard to say exactly whether the XP-G2 SMA@Taschenlampen Forums pulled out of his scrap box was bare phosphor, had a sealant over the bare phosphor or if the dedome was a slice job with some silicone remaining over the bare LED. If someone here knows SMA, that information would be helpful in my understanding of photon recycling.

In the post, SMA is describing his experiment where he determines the luminance gain arising from a large wavien collar centered on a dedomed XP-G2. I'll describe what I think is going on here with this configuration, and I hope someone of experience would jump in to guide and correct my thinking.

I know him personally. He used a cool-white XP-G2 and de-domed it using the classic way (imersion in nitro paint thinner or similar). This removes the silicone lens on top of the LED, but does not (to my knowledge) alter the silicon-phosphor-mix underneath. But you can ask him yourself, he is also a member here on the forum.
 
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