low-cost LED-based "one sun" light source

No passive system can exceed the surface temperature of the source. My colleagues used liquid metal thermal interface to a water-cooled copper cold plate. Worked great--they installed hundreds of them in Saudi.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net

lly

The surface of a solar cell will reflect more incident light at a glancing angle, so collimation DOES affect its sensitivity. Nearly-parallel light from th e sun has 0.5 degrees of angular spread, while diffuse illumination comes from 180 degrees.

Solar cells with reflectors (concentrator cells and arrays) receive light f rom the full angle subtended by the reflector as seen from the cell (presumably at the focus).

Hi,

I think a passive system that was constructed with non linear optics, could potentially exceed the surface temperature of the source, but not sure if that is still considered a passive system.

cheers, Jamie

Hi,

Here's another one, probably pretty expensive but nice setup:

cheers, Jamie

Sure is. They're used for testing avionic and automotive displays. Expensive beasts (>$50K, IIRC).

There's no way you're going to get any harmonic yield from sunlight, because the photon occupation number is way too low.

However, you can run a diode laser from a solar panel and focus it down to a spot less than a wavelength across.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net

Wow - I never knew that. Is it really true that any EM wave can be focused to less than a wavelength across? That's a very cool thought - please say more!

Sure, though it sort of depends how you measure the size. A converging spherical scalar wave with uniform illumination in a cone of some given numerical aperture produces a focal spot given by

E(r) = 2 E_0 J1(k r NA)/(k r NA),

and an intensity

I(r) = |E**2|

(In air, NA is the sine of the half angle of the cone, so it goes from 0 to 1.)

The intensity falls by 50% at a radius xNA/lambda = 0.252, so the FWHM is about 0.504 lambda/NA.

The distance between the first nulls is 0.61 lambda/NA.

This is just the 2-D version of electrical pulse width vs bandwidth.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net

Yeah, I can see that, but we are discussing solar cells.

Hmmmm... I'm going to have to think about that one. :)

--

Rick C

Hi,

If you focus sunlight and bandpass filter it, to get high photon density at a specific narrow frequency range, I think that could potentially trigger some type of (maybe theoretical) non-linear crystal to give some small amount of second harmonic output, this would qualify potentially as a higher temperature than the source I think depending on the frequency, but not sure.

Also you mentioned that NO passive system can exceed the surface temperature of the source, but if the source is already a narrow frequency high photon occupancy source, couldn't the non-linear optics output exceed the surface temperature of the source? I don't know what source could qualify for this.

There are some interesting thermal to electric converter designs that do something similar, by converting thermal photons to visible photons and use photovoltaics to generate electricity, I think these suffer the same required high photon occupation requirement as you mentioned to do the second harmonic generation.

cheers, Jamie

Ok here is a way to do it I think:

Just get a small body of water with "phenol at the air-water interface":

Then focus the reflected sunlight off this body of water, and there should be *some* second harmonic generated (SHG) light from whatever frequencies are sensitive to SHG from the phenol (which I am assuming are occurring in the wideband sunlight). Then highpass filter out the frequency doubled light, and technically this system can exceed the surface temperature of the source.

I know the SHG output goes up with the square of the photon occupancy or something, but still there is some output even for wideband sunlight I think.

cheers, Jamie

Hi,

For a black body source with characteristic thermodynamic equilibrium:

One question I have is why is this graph the "unique stable distribution for radiation in thermodynamic equilibrium" for a black body?

From the page:

It gives the formula to derive the above black body graph, but still it seems a bit mysterious.

Also I wonder if the curve on the graph for say a 5000K temperature black body can be adjusted in any way?

One way to get a black body to emit with a different curve than on the graph is with meta-materials, ie for thermophotovoltaics.

It is interesting all black body emitters aren't the same I guess, and may react differently to external fields etc to give different curves than shown on that graph I guess, ie an "active" black body.

Also the black body graph shows that even with a lower temperature, ie an infrared black body, there is still a tiny bit of UV that will be in the spectral output, so if this is filtered out, you can always get a UV signal from an infrared black body I guess..

Really I think it wouldn't be that hard to develop a black body geometry that takes this into account and can preferentially output visible light while reabsorbing near 100% of energy with infrared wavelengths. I think the larger the geometry, the lower temperature is required to do this too maybe. It is neat to imagine a black body that has a different graph than what Planck's law shows.

Here's one idea for a black body geometry to preferentially emit visible light:

Take an infrared black body source, and screen it with layers of one of these:

If the layers are lined up then the source should be visible through the holes in the screen from outside, and if the holes in the screen are large enough for visible light, but too small for infrared light, then the output should be skewed to have excess visible light compared to the central black body source.

The main problem might be to reduce the thermal heating of the layers from black body heating layer by layer, but if the inner surface of each layer is reflective to infra red and the outer surface absorbs infra red, that should reduce the thermal loss a bit.

So if this "black body screen" surrounds a black body source it should ideally insulate it well to keep the black body source emitting infra red and at the same time not insulate the visible light that is given off by the black body. The hard part is that an infra red black body is giving off less visible light than infra red light, which probably explains why this idea isn't practical :D

While I am at it, one more question: there are some carbon nanotube absorbers, which can absorb almost all frequencies of sunlight, and increase their temperature, so could these exceed the temperature of a source?

cheers, Jamie

It's called a "laser". ;)

Lasers aren't thermal sources, and (since you have to apply pump power exte rnally) they aren't in thermodynamic equilibrium.

And it really isn't worthwhile going through all sorts of complicated scena rios. I've posted a detailed thought experiment a few times--search Google Groups for "sufficiently small heat engine" and it'll come tight up.

Cheers

Phil Hobbs

Thermal radiation sources have a limited intensity versus frequency, that accurately represents the temperature of the source. A 'high photon occupancy source' like a laser beam is NOT a thermal source, so you can make temperatures greater than the 'temperature' of the laser.

I put 'temperature' in quotes, because the innards of a laser do not enjoy the kind of thermal equilibrium that defines a temperature.

Hi,

If you filter out the longer wavelengths of a thermal radiation source, then it no longer represents the temperature of the source, and so just by taking away part of the thermal spectrum, that should allow a temperature greater than the temperature of the source I think.

Conversely if you have a black body at 5000K and illuminate it with a

4000K black body, that should reduce the surface temperature of the 5000K black body, effectively shifting the thermal equilibrium to a lower temperature.

cheers, Jamie

Physics Today has an article about the most extreme confinement of light so far by cryogenic gold nano particles.

--
Regards, 
Martin Brown

This blows my mind. Can you explain more, or point me to a resource to study this?

All it means is that if you have two objects at different temperatures irradiating each other, more heat flows from the hot one to the cold one than vice versa.

Nothing to see here, folks, move along. ;)

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net

It sounds like a complex scheme, which ignores the three-dimensional nature of radiation from a source. TIn thermodynamics jargon, a hot result from a cool source has the ability to do work (i.e. is equivalent to a useful perpetual motion machine).

The 'focusing' of light from (for instance) the sun will NEVER produce a point of light, but rather some variant on a disc (i.e. an image of the visible disc of the sun). The size of that disc, and the ratios set by collection surface area, set limits on possible intensification in optical systems. Even if you replicate by filtering the spectral distribution of a hotter source, you do NOT replicate the watts/square meter output of such a source. The target has to get more watts/square meter input than it radiates out, and that never is the case when the target is as hot as the source (or hotter).

That's because the real concern is watts/square meter/steradian. Lenses raise watts/square meter, but NOT watts/square meter/steradian.