Infrared with a solar panel

[Stefano]

Hi, in this topic I will talk about infrared, but not the normal kind of infrared we are used to. I tried pushing the sensor to its limit, at 1150 nm.

Not all people know this, but every photodiode (and a solar panel is a photodiode) can also emit light through electroluminescence. And every LED (Light Emitting Diode) can be used as an inefficient photodiode (and even as a solar panel if you have a lot of them). Of course, a solar panel was never meant to be used like I did, and it is very inefficient at giving off light.

Silicon has a bandgap of ~1.12 eV. This means that when an electron crosses the PN junction of a solar panel, it has the potential energy of a photon with a wavelength of about 1100 nm. Occasionally (not wery often in this case) when an electron makes his "energy jump" it releases it as a photon. Experimentally the peak wavelength emitted from a silicon solar panel is around 1150 nm. The sensor of a normal camera is made basically with the same material, and this makes it very insensitive at this wavelength. My camera has probably tens of times more sensitivity at 365 nm than at 1150 nm.

 

Having said this, I did some experiments with a solar panel. For all shots I used a Panasonic DMC-F3. I didn't have any IR longpass filter to use, and that would have helped. All shots were made without any filter and keeping the camera in the same position.

Almost all of them have the same settings (f-stop: f/2.8, ISO 80, 60 s exposure), but for simplicity I kept them in the descriptions. Luckily my camera has a "starry sky" mode that allows me to take photos with exposures up to 60 seconds, but at ISO 80 and without white balance (I don't know why but this is it). To give an idea of what 60 seconds can do, I took this images of the sky:

 

f-stop: f/2.8, ISO 80, 60 s exposure

 

f-stop: f/2.8, ISO 80, 60 s exposure

 

This is a solar panel, lit up with (I think) 0.5 A of current.

 

f-stop: f/2.8, ISO 80, 60 s exposure

 

You can see that it is damaged, and I can not see anything with the naked eye. This technique is used to inspect solar panels and find defects. The camera didn't focus well, and I can not do it manually.

 

And now the main experiment I did. We know that water has some absorption peaks in the infrared (970 nm, 1200 nm, 1450 nm, 1950 nm, around 2900 nm and around 6100 nm). The longer the wavelength, the stronger they are. The electromagnetic absorption spectrum by water is something worth a topic per se, as it is a very, very big topic that would require pages of discussion and research. Andy did a series about photographing water in the first main absorption peak (970-980 nm), showing that it starts appearing darker than normal even in a thickness of a few centimeters, and Cadmium tried a Schott RG1000 + S8612 stack, which resulted again in dark water. With a solar panel I should see the beginning of the 1200 nm peak, as water becomes quite transparent again in the 1000-1100 nm region.

 

 

This is my simple setup. I have the solar panel as the illumination source, a paper sheet that acts like a diffuser, the water container and the camera. The thickness of the water in the direction light rays had to travel was about 7.5 cm. The other things in the background are some of my LEDs on their heatsinks. I chose this (dusty, I know) closet because I needed complete stray light shielding, as even the display on the power supply could have spoiled the image. I took 23 photos, 60 s exposure each, and stacked them in one image. I ran the solar panel at 1 A (current-limited), and since it required between 8.3 and 8 V to do so (the voltage drop goes down as it heats up) it dissipated 8+ watts of power, almost completely as heat. Since the camera required 2 exposures to make one photo (the actual one and probably a dark frame), I could turn off the power supply to allow the solar panel (wich reached temperatures above 50 °C and probably even above 60°C) to cool down a bit. I ran it basically with a 50% duty cicle. The final image I got was equivalent to a 23 minutes exposure.

 

f-stop: f/2.8, ISO 80, 1380 s equivalent exposure

 

This is how low the sensitivity is on this region of the spectrum. I have probably the same sensitivity in UVB. The light from the camera LCD (that displayed basically a black image) was enough to contaminate the image.

 

this is a single exposure without the paper sheet. I like the warped look that came out

 

f-stop: f/2.8, ISO 80, 60 s exposure

 

And this one with the solar panel only

 

f-stop: f/2.8, ISO 80, 60 s exposure

 

This is a visible reference, taken with a white LED flashlight

 

f-stop: f/2.8, ISO 1600, 1/250 s exposure

 

And this is an edited version of the final image. I increased the brightness and the contrast by 150% (first by 100% and then by another 50%), and I made it B&W by lowering the saturation to -100.

 

 

I don't need to say that water here appears noticeably darker than usual. Not as much as in SWIR at 1450 nm, where it is pitch black, but darker than under 940 nm LED light.

[Andy Perrin]

It's hard to know if you are seeing what you think you are seeing. My experience with my silicon sensors is that I never see anything past 1100nm and definitely not 1150nm. It does not sound like you did any kind of filtration of the lens to block out of band light, so you really don't know what you are seeing here, although the dark water does suggest it's something toward the infrared end.

 

Regarding the solar cell: Objects will start to emit light by blackbody in the near infrared also, and you said the solar cell is quite warm. There is also a technical reason why you shouldn't be seeing much electroluminescence from the panel, and that is because silicon is an indirect bandgap semiconductor, and those don't make light by recombination very well (LEDs are always direct bandgap materials for this reason).

 

 

https://en.wikipedia.org/wiki/Direct_and_indirect_band_gaps

 

NIR blackbody radiation from the warm panel in the 900-1000nm range is a better bet. While objects have to be quite hot to glow to the naked eye, with such a long exposure I can imagine you'd get some NIR glow.

 

If you want to make this more persuasive, get a filter from Thorlabs for the 1100+ range (a good one, with OD5+ blocking):

https://www.thorlabs.com/thorproduct.cfm?partnumber=FELH1100

[Stefano]

It's hard to know if you are seeing what you think you are seeing. My experience with my silicon sensors is that I never see anything past 1100nm and definitely not 1150nm. It does not sound like you did any kind of filtration of the lens to block out of band light, so you really don't know what you are seeing here, although the dark water does suggest it's something toward the infrared end.

 

Regarding the solar cell: Objects will start to emit light by blackbody in the near infrared also, and you said the solar cell is quite warm. There is also a technical reason why you shouldn't be seeing much electroluminescence from the panel, and that is because silicon is an indirect bandgap semiconductor, and those don't make light by recombination very well (LEDs are always direct bandgap materials for this reason).

 

 

https://en.wikipedia...irect_band_gaps

 

NIR blackbody radiation from the warm panel in the 900-1000nm range is a better bet. While objects have to be quite hot to glow to the naked eye, with such a long exposure I can imagine you'd get some NIR glow.

 

If you want to make this more persuasive, get a filter from Thorlabs for the 1100+ range (a good one, with OD5+ blocking):

https://www.thorlabs...number=FELH1100

Yes, a solar panel produces light very inefficiently, but I can see (with a camera) that it turns on and off instantly. I don't have to wait for it to heat up. And I think that even at 1000+ nm an object must be very hot to see the thermal radiation. I probably have to heat up something to at least 300°C to start seeing something with a full spectrum camera, and surely 60°C aren't nearly enough, even for a SWIR camera probably. Keep in mind that the total radiated power from a blackbody is proportional to the fourth power of the (absolute) temperature, so an object at 200°C (detectable with a SWIR camera but probably not with a silicon one) emits 4 times the radiation a 60°C body emits (at the same surface area and emissivity). I have a soldering iron capable of reaching ~500°C (too much to solder correctly), and I can barely see it with my naked eye. With a full spectrum camera it is bright enough to illuminate objects at ~10 cm.

Also my chinese BG39 starts to be slightly transparent at this wavelengths.

 

I should test the minimum temperature an object has to have to be seen with my camera.

[Stefano]

I tried to test the threshold at which my camera starts peaking up thermal emission, but I didn't have a way to measure temperature correctly (my infrared thermometer showed a reading of ~250°C when my filament (see later) was red hot).

 

So I did a quick experiment: I heated up the filament of a homemade "lamp" (made with a 26 gauge kantal wire) to the point it was red hot, and I did a 60 s exposure to see how water looked like under that light.

 

Setup, as usual

 

Detail of the lamp

 

Visible reference (f-stop: f/2.8, ISO 80, 1/8 s exposure)

 

Infrared image (f-stop: f/2.8, ISO 80, 60 s exposure)

The camera was my usual Panasonic DMC-F3, and I didn't use any filter.

[Andy Perrin]

Yeah, 60C May be the surface temperature but not necessarily the temperature of the thin wires in the solar array? And they would not take long to heat up. But it’s possible you are seeing the luminescence I guess. What I don’t believe is that you are seeing 1150nm without much better evidence than I’ve seen so far. Possibly some of the luminescence is at shorter wavelengths if it is luminescence.

[Stefano]

If you search for the spectrum of a solar panel electroluminescence (I can maybe post an image if I can do it) you can see that it is basically a distorted gaussian curve (like LEDs) and it goes from ~950 nm to ~1300 nm, with a big portion below 1100 nm (it reaches 50-60% of the peak at 1100 nm). If my camera really sees past 1100 nm (that's the doubt we have), surely the sensitivity it has below 1100 nm is much greater, and the image is made almost completely from those wavelengths (kind of like what happens with UV). I can try (it's not easy) to see approximately where the peak is with a diffraction grating, using my 850 and 940 nm LEDs as a reference. Since you are one of the very few people here (if not the only one) that has an actual SWIR camera (I can only dream that for now), it would be interesting to see a comparison or something like that. If you don't have a small solar panel like I do I think that a normal silicon photodiode should do the same. Thorlabs also sells some SWIR LEDs, even if they are not very powerful. I don't have them (but I plan to buy them one day), but I should not be able, for example, to see anything pointing the camera directly to a 1450 nm LED.

Also, can double photon absorption occur in a camera sensor? It is a very rare event, but it can make the sensor sensitive to SWIR wavelengths, and there is a SWIR CMOS camera commercially avaiable for 1199 € that uses this phenomenon, though Its sensitivity decreases exponentially towards longer wavelengths.

[Andy Perrin]

I’m not at home for a few days, but when I return I could try doing a side by side test with the same camera settings on my Sony. It should be possible to see what is what.

 

Re double-photon absorption: does not happen naturally at any usable rate, but you can find sensors that are coated with a phosphor. Please read my SWIR sticky for more info on all that. I don’t recommend them. Instead you might consider a vidicon (FindRScope) on eBay. Some go out to 1550nm.

[dabateman]

I have a strong feeling you are seeing the 975nm water absorbing peak with this setup.

I have a slightly leaky long pass 1100 filter which cuts off at 1050 to 1080nm, which I stacked with a tight 950nm filter and my water was clear. So I missed the 975nm water absorbing peak and don't have enough signal to catch the 1200nm water absorbing peak.

I think you are seeing 975nm.

 

[Andy Perrin]

That is what I think also. But I will try to do a test if I have the right filter.

[Andy Perrin]

I found this link which has a graph of the luminescence spectrum. I would say it’s almost certain to be the 975nm peak. Incidentally they show the silicon CCD cutting out at 1200nm, but this is not the case for normal camera sensors (possibly the fancy backside illuminated ones could do that? Or if you have a cooled sensor.). Our normal cameras seem to go blind at 1100nm or so.

 

https://www.pveducation.org/pvcdrom/characterisation/electroluminescence

 

[dabateman]

This the spectrum of a solar panel that I could find:

 

I didn't see one with spectrum versus applied voltage or amps, can't remember which would have more effect.

 

There is a lot from 950nm to 1050nm that I think your camera will be detecting.

 

Ha Andy just beat me.

[Andy Perrin]

Amps, LEDs are current-controlled devices. But that would mostly affect the intensity not the spectrum. There is probably a temperature effect though, just like with Convoys.

[ulf]

I have a soldering iron capable of reaching ~500°C (too much to solder correctly), and I can barely see it with my naked eye. With a full spectrum camera it is bright enough to illuminate objects at ~10 cm.

 

Here are some images of the tip of my Weller soldering station at a few temperature settings:

https://www.ultravioletphotography.com/content/index.php/topic/2928-drones/page__view__findpost__p__25475

[Stefano]

Your image of the solar panel spectrum is the exact one I was referring to when I said it reaches 50-60% of the peak at 1100 nm. I have to say that the spectrum is quite broad

[Stefano]

Here are some images of the tip of my Weller soldering station at a few temperature settings:

https://www.ultravio...dpost__p__25475

This pretty much confirms that you need at least 300°C to see incandescence with a full spectrum camera. The Draper point for the human eye is ~525°C. I start to see incandescence in UV probably at least at 1000°C.

[Stefano]

Amps, LEDs are current-controlled devices. But that would mostly affect the intensity not the spectrum. There is probably a temperature effect though, just like with Convoys.

Usually the bandgap of an LED widens at low temperatures and narrows at high temperatures. There is a famous experiment where you put an orange LED in liquid nitrogen and it turns green, and if you overload a green LED (I tried that) it can turn orange. A red LED (also tried) becomes infrared. The same should happen with a solar panel.

[Andy Perrin]

Yeah, so the state of things at the moment seems to be:

1) panel really is luminescent (fast startup time and the 300C required for blackbody radiation would destroy the cells probably)

2) but the luminescence is a wide spectrum and definitely is in range of a camera with a 1100nm cut-off (give or take a few nm)

3) So the conclusion is that it’s the 975nm water peak, since that is where the camera gain spectrum is overlapping the luminescence spectrum.

 

By the way, did you see my fun this summer with the 975nm peak? I got a bandpass filter for the peak and took some pics:

https://www.ultravioletphotography.com/content/index.php/topic/3505-dark-water-and-the-nir-absorption-peak/

https://www.ultravioletphotography.com/content/index.php/topic/3524-dark-fountains-980nm-bandpass/

https://www.ultravioletphotography.com/content/index.php/topic/3514-newton-upper-falls-980nm-bp10-filter/

[Stefano]

By the way, did you see my fun this summer with the 975nm peak? I got a bandpass filter for the peak and took some pics:

Yes, I saw everything. It's been months since I started following this website daily, before becoming a member. That's why I know you have a TriWave camera

[Stefano]

About the spectrum with a diffraction grating, I tried and the solar panel is too weak. It is reasonably bright when looked at directly, but I can not see the spectrum. I don't want to make another set of 23 expoosures, it is very time-consuming and the last time I completely discharged the battery which was fully charged. With LEDs, instead, it's easy to see the spectrum.

[Stefano]

This is probably the last thing I will post in this decade.

 

I did an experiment with 3 infrared LEDs an a halogen lamp, to see how far my camera can see. I made 4 photos of the four spectrums, and stacked them in one image. I used a black pen ink filter to pass infrared only.

 

Setup

 

Final image

 

lamp spectrum image only (if you think I went too far, try to guess where the spectrum ends).

 

I calculated the wavelength "counting" the pixels (Paint did that for me).

 

If you want I can post the other 3 images and the calculations I did.

 

It is 23:00 in Italy.

[Andy Perrin]

Did you account for the fact that it’s not linear (wavelength proportional to sin(angle) rather than the actual angle for large angles)?

 

Also, why is there so much other light in your image? If you want to test the limits if the sensor you have to get rid of all the background light to maximize signal-to-noise.

 

(You definitely have to get a real IR filter one of these days too. They are not expensive even for Hoya. “Black pen ink” filters, while very creative, won’t give any kind of quantitative results and may have leaks from uneven ink thickness etc. This probably is irrelevant for the current test, but just generally speaking...)

[Stefano]

The wavelength is proportional to the linear "displacement" (as you said to sin(angle)). If you diffract some laser beams with a wall in front you will see multiple dots on it. If you just point your laser at the wall and look at the spot with the same diffraction grating you used before, with your eye at the same distance from the wall as the diffraction grating you used before was, you see the same image with the same dots (virtual this time) on the wall, in the same position you saw them before, as strange as it may seem. So there wasn't any angle to consider, everything is linear, because if I projected the same image on a flat surface it would have been the same.

 

About the filter... yeah, I have to get a real one. Strangely I have a proper glass UV bandpass filter (even if the quality is not the best possible, but I have it), and I never had a glass or interference IR longpass filter. Even from China they are not very expensive. Consider that this test was very quick and rough, I did it only to see approximately how far could my camera see into the NIR.

[Stefano]

Anyway I too think that I am probably seeing the 975 nm peak, as the sensitivity at 1100-1150 nm (if it gets there) is extremely low. This can be seen in the test I did. If I had a 980 nm LED I could have made a comparison, for now I only have the 940 nm one that I used in the test (measured at 946 nm) and water isn't so dark there.

[Andy Perrin]

The wavelength is proportional to the linear "displacement" (as you said to sin(angle)). If you diffract some laser beams with a wall in front you will see multiple dots on it. If you just point your laser at the wall and look at the spot with the same diffraction grating you used before, with your eye at the same distance from the wall as the diffraction grating you used before was, you see the same image with the same dots (virtual this time) on the wall, in the same position you saw them before, as strange as it may seem. So there wasn't any angle to consider, everything is linear, because if I projected the same image on a flat surface it would have been the same.

Yes, I get that it doesn't matter whether you project through the grating or take a photo through it, that's not what I meant. I was referring to the Bragg equation:

In this derivation below, d is the grating line spacing, m is the order of the spectrum (=1), and I labeled the rest on there.

Only when x is much smaller than L is it approximately linear.

[Stefano]

Yes, I get that it doesn't matter whether you project through the grating or take a photo through it, that's not what I meant. I was referring to the Bragg equation:

In this derivation below, d is the grating line spacing, m is the order of the spectrum (=1), and I labeled the rest on there.

Only when x is much smaller than L is it approximately linear.

I thought about this and I noticed I made a mistake: I confused the L line with the red one, and so x isn't linearly dependent on the wavelength. The function sin(x) can be approximated to the line function y=x, for x approaching 0. But this makes me question something:

 

1) does a spectrometer with a linear CCD or CMOS array have to compensate for this? I know that usually they use a prism to split light into its components, but I think that not all of them use this method.

 

2) I used "Theremino Spectrometer" together with a USB camera and a diffraction grating to make a crude spectrometer. I know that it can not be as precise as a $2000 one, but since (as far as I know) it simply counts the pixels on the image the camera sees, is it inaccurate? I measured my LEDs this way, and I got pretty reasonable values.

 

3) can the angle θ be greater than 90°? Since λ is directly proportional to sin(θ), what happens as it gets big? The codomain of sin(x) is limited, so something strange should happen there. I don't know.

 

About the background light in my image, I tried to reduce it, but I didn't want to seal everything, since it was a quick experiment meant to be indicative only. In a real spectrometer (I tried making one, it is the one I showed in "Strange spectrums with a diffraction grating") everything is light-tight and the inside is painted black.