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[Devin-M]

My understanding is that it's the thermal motion of the electrons in the sensor that leads to the generation of dark current, not IR radiation. A single IR photon doesn't have enough energy to cause an electron to jump the energy gap, but a lucky collision/interaction between several electrons/ions can give an electron enough energy to jump the gap and into the area of the pixel well that stores the photoelectrons prior to readout.

No. Prior to exposure the pixel wells undergo a charge separation process that puts them in a high-energy state. Photons, or random interactions from thermal motion, cause electrons to jump an energy gap and get caught in this charged well. Given enough time or photons the well becomes completely saturated and can no longer collect charge. You won't extract more energy out of this than it took to separate the charges in the first place.

A solar panel operates somewhat differently and I don't really know enough to explain it well. However, I do believe that the solar panel needs to be at a lower temperature than the emitting object it is capturing radiation from. Besides, the solar panel itself and the surrounding environment is a temperature reservoir, so there's more than one.

Could you comment on the graph on this webpage… I might be misinterpreting it but I believe graph (a) shows a particular detector which is at 300k (~80F) operating temperature generating current from mid-infrared light up to 4000 nanometers with zero bias, which I take to mean the detector is operating in photovoltaic mode with no outside voltage applied…

https://www.researchgate.net/figure/a-g-The-spectral-responsivity-measured-at-zero-bias-ie-photovoltaic-mode-for-the_fig3_346511011

“(a)-(g) The spectral responsivity measured at zero bias (i.e. photovoltaic mode) for the Te-hyperdoped Si photodetector at different temperatures. The room-temperature spectral responsivity of a commercial Si-PIN photodiode (model: BPW34) is included as a reference (brown short dot). (h) Illustration of the below-bandgap photoresponse in the Te-hyperdoped Si photodetector. Te dopants introduce deep-level states (intermediate band) inside the Si band gap, which facilitate the absorption of photons with sub-bandgap energies. Process I: VB to CB (Eph ≥ Eg); Process II: VB to IB (Eph ≥ Eg-ETe); Process III: IB to CB (Eph ≥ ETe, only measurable at low temperatures where the thermal contribution is neglected).”

 

[Drakkith]

Could you comment on the graph on this webpage… I might be misinterpreting it but I believe graph (a) shows a particular detector which is at 300k (~80F) operating temperature generating current from mid-infrared light up to 4000 nanometers with zero bias, which I take to mean the detector is operating in photovoltaic mode with no outside voltage applied…

Not really. I'm not an expert in the area of photodetectors and solid state physics and such. I'll try to remember to give it a read tomorrow or the next day if I can, but I might not have time.

 

[Drakkith]

Could you comment on the graph on this webpage… I might be misinterpreting it but I believe graph (a) shows a particular detector which is at 300k (~80F) operating temperature generating current from mid-infrared light up to 4000 nanometers with zero bias, which I take to mean the detector is operating in photovoltaic mode with no outside voltage applied…

Quickly skimmed through the article just now. I come to the same conclusion as you.
Note that at 300K an object barely emits any radiation in the 1-5 micrometer range. You have to get warmer for that. You can use the calculator here to see the spectrum emitted by an object at a given temperature. Use 1 micrometer as the upper limit and 5, 10, or 20 as the lower limit to get a good looking graph of the region of interest.

 

[Devin-M]

Quickly skimmed through the article just now. I come to the same conclusion as you.
Note that at 300K an object barely emits any radiation in the 1-5 micrometer range. You have to get warmer for that. You can use the calculator here to see the spectrum emitted by an object at a given temperature. Use 1 micrometer as the upper limit and 5, 10, or 20 as the lower limit to get a good looking graph of the region of interest.

Thanks for the calculator! According to its output, with inputs for the emissivity of water (0.96) at 300k (~80F), ordinary room temperature water is emitting some blackbody infrared radiation from 3-4 micrometers— at wavelengths the “Te-hyperdoped Si photodetector” also @ 300k can generate current from in photovoltaic mode… I must be missing something because why couldn’t I just generate a small amount of electricity by submerging these room temperature photodetectors in room temperature water to harvest the 3-4 micrometer infrared black body radiation photons by photovoltaic means? Wouldn’t that conflict with the 2nd Law of Thermodynamics? I shouldn’t be able to generate any useful work from a single temperature reservoir, was my understanding.

 

[Drakkith]

I must be missing something because why couldn’t I just generate a small amount of electricity by submerging these room temperature photodetectors in room temperature water to harvest the 3-4 micrometer infrared black body radiation photons by photovoltaic means? Wouldn’t that conflict with the 2nd Law of Thermodynamics? I shouldn’t be able to generate any useful work from a single temperature reservoir, was my understanding.

That I can't answer. I'm certain the 2nd law isn't being violated, but I couldn't tell you how or why it isn't.

 

[collinsmark]

I must be missing something because why couldn’t I just generate a small amount of electricity by submerging these room temperature photodetectors in room temperature water to harvest the 3-4 micrometer infrared black body radiation photons by photovoltaic means?

Maybe I'm missing something myself. But It's my understanding that you couldn't generate any electricity by simply submerging the photodetector in water because there wouldn't be a light source in that situation.

It's my understanding of the test setup that the photodetector is placed and held at a given temperature, then it is exposed to a light source of a specific wavelength and specific intensity (with a proportional power reaching the detector, measured in Watts) and the current of the photodetector is measured (measured in milliamps). That is used to generate a single point on a single graph. For any given situation (wavelength of the light source and temperature of the photodetector), the current of the photodetector is proportional to the power of the light source. Which is why the measurements are in units of mA/W.

At least that's my understanding. The power is ultimately coming from the light source. The 2nd Law is not violated. The current vanishes as soon as you turn off the light.

 

[Devin-M]

Maybe I'm missing something myself. But It's my understanding that you couldn't generate any electricity by simply submerging the photodetector in water because there wouldn't be a light source in that situation.

View attachment 296237

It's my understanding of the test setup that the photodetector is placed and held at a given temperature, then it is exposed to a light source of a specific wavelength and specific intensity (with a proportional power reaching the detector, measured in Watts) and the current of the photodetector is measured (measured in milliamps). That is used to generate a single point on a single graph. For any given situation (wavelength of the light source and temperature of the photodetector), the current of the photodetector is proportional to the power of the light source. Which is why the measurements are in units of mA/W.

At least that's my understanding. The power is ultimately coming from the light source. The 2nd Law is not violated. The current vanishes as soon as you turn off the light.

A 300k (80F) black body emits some infrared between 3 & 4 micrometers, which is in the detection range of the photodetector.

 

[berkeman]

Wouldn’t that conflict with the 2nd Law of Thermodynamics? I shouldn’t be able to generate any useful work from a single temperature reservoir, was my understanding.

Should some of this side-discussion be split off into the Thermodynamics forum?

 

[collinsmark]

A 300k (80F) black body emits some infrared between 3 & 4 micrometers, which is in the detection range of the photodetector.

View attachment 296248

Yes, but the photodetector is also emitting infrared too -- the same amount that it receives when its own temperature is at 300 K, along with everything else in the surroundings being at 300 K, and when no external light source is present. Without the presence of the external light source the net current is zero. At least that's my understanding.

I don't know the test setup, but here's how I imagine it:

A broadband blackbody radiation source is involved; an incandescent bulb will do. The light from the source passes through a slit followed by a diffraction grating, thus splitting up the light (including infrared light) into a spectrum. The intensity along the spectrum is measured and calibrated (perhaps with a small, calorimeter device). With this information, the light intensity along specific wavelengths of the spectrum is known.

The photodetector can then be placed along the spectrum for measurements. Changing the wavelength is just a matter of moving the photodetector spacially to a different part of the spectrum produced by the light source + diffraction grating.

But again, if I'm imagining the setup correctly, the current in the photodetector will vanish when the light source is turned off.

**** Edit *****
Reading into the research paper a little more (https://www.researchgate.net/publication/346511011_Silicon-Based_Intermediate-Band_Infrared_Photodetector_Realized_by_Te_Hyperdoping), it states in the Device Measurement seciton: "A Globar (SiC) source coupled with a TMc300 Bentham monochromator equipped with gratings in Czerny-Turner reflection configuration was used as the infrared monochromatic source. Its intensity is spatially homogenized and was calibrated with a Bentham pyrometric detector."

The "TMc300 Bentham monochromator" utilizes a diffraction grating turret. So my imagined setup, albeit a bit simplistic, was conceptually accurate.
****************

Should some of this side-discussion be split off into the Thermodynamics forum?

That sounds like a good idea to me.

 

[berkeman]

That sounds like a good idea to me.

Can you folks suggest which posts I should split off into the Thermo forum? I don't want to mess up the astrophotography part of the discussion.

 

[Devin-M]

What if we just post further comments in a new discussion?

 

[berkeman]

That would be good too. I'm only able to move posts, not copy/paste posts. So maybe start a new discussion in the Thermo forum based on the discussion here. It's a pretty interesting discussion, IMO.

 

[collinsmark]

Can you folks suggest which posts I should split off into the Thermo forum? I don't want to mess up the astrophotography part of the discussion.

What if we just post further comments in a new discussion?

That would be good too. I'm only able to move posts, not copy/paste posts. So maybe start a new discussion in the Thermo forum based on the discussion here. It's a pretty interesting discussion, IMO.

I would suggest, if @Devin-M agrees, for @Devin-M to create a new thread in the appropriate forum (Thermodynamics?) with basically a copy-and-paste copy of Post #1558 as the original post.

Then copy over posts #1561 - Onward, to the new thread. The other posts are good for the astrophotography thread.

Post #1559 by @Drakkith is a toughy though; that post could go either way.

 

[Devin-M]

Moved here:
https://www.physicsforums.com/threads/infrared-detectors-the-2nd-law-of-thermodynamics.1011817/

 

[Devin-M]

My most recent astro-photo-session was a terrible failure. Long story short I spent hours in the cold & this was the only remotely salvageable shot from the whole session and it was done with my iphone. I call it “Tree with iPhone in Bortle 2 conditions.” I was attempting to shoot a dim target at 2130mm focal length at the limit of exposure time (90 sec per subframe) I can do without too much tracking issues and also at the limit of iso sensitivity I can reasonable achieve (6400iso). Unfortunately that wasn’t enough exposure time so while the stars came out ok there was no nebula visible in the final photos. I thought I’d be able to bring it out in post processing but it was just a bunch of noise. I’m going to have to choose a brighter target, switch to a much shorter focal length with bigger aperture, or shell out thousands of $… for now I’m ruling out the 3rd option.

Edit: I was too embarrassed to show the final image taken through the telescope but here it is, for posterity:

Here are a couple more photos of the same region through a much more capable scope of NGC 2170:

Source:
https://www.astrobin.com/2nzzzj/0/

Source:
https://en.wikipedia.org/wiki/NGC_2170

 

[Drakkith]

@Devin-M what were you shooting with?

 

[Devin-M]

@Devin-M what were you shooting with?

It was a 2130mm f/14.2 Maksutov Cassegrain w/ Nikon D800 dslr fitted on a Star Adventurer 2i mount (slightly modified to go at least 3x over the weight limit).

 

[Drakkith]

Dear god, f/14.2?!
It'll be the heat death of the universe before you get a deep sky photo finished!
Get a focal reducer for that scope!

 

[Devin-M]

Dear god, f/14.2?!
It'll be the heat death of the universe before you get a deep sky photo finished!
Get a focal reducer for that scope!

I got this one of the core of Andromeda a couple days before with identical exposure settings...

 

[Drakkith]

I got this one of the core of Andromeda a couple days before with identical exposure settings...

M31 (Andromeda Galaxy) is amongst the brightest of the deep sky objects. You can get a decent picture with two tin cans and a string.

Seriously, my F/8 scope is something close to 4x 'faster' than yours and I still think it's too slow!

 

[Ibix]

You can get a decent picture with two tin cans and a string.

The bottom of a bottle, my optics prof used to say.

 

[Devin-M]

M31 (Andromeda Galaxy) is amongst the brightest of the deep sky objects. You can get a decent picture with two tin cans and a string.

Seriously, my F/8 scope is something close to 4x 'faster' than yours and I still think it's too slow!

My counterweight is a 600mm f/9…

 

[Devin-M]

Does anyone else think I've captured a very strange transient signal?

view in WorldWideTelescope

 

[Devin-M]

Here's a cleaner re-process... (ps this is a stacked image... 84x 90 sec (2.1hrs), 6400 iso, 2130mm f/14.2)

view in WorldWideTelescope

 

[Devin-M]

Mystery solved... it's not in the individual frames so it must be nothing more than a stacking misalignment.

 

[Devin-M]

Restacked... now you can see the nebulas... I think something very bad happened during the 1st stacking attempt...

view in WorldWideTelescope

 

[Drakkith]

Restacked... now you can see the nebulas... I think something very bad happened during the 1st stacking attempt...

Was this taken from a bortle 2 site?

 

[Devin-M]

Was this taken from a bortle 2 site?

Yes it’s in bortle 2 conditions just outside of Shingletown, California, USA. It’s literally the same image I already posted but restacked… I think some very bad stacking misalignment happened the first time but it was quite hard to notice for some reason. I think with 90 second exposures a lot of the individual shots come out fine but over time the field drifts in the viewfinder so first I cropped all the source TIFs to the same field of view and then I restacked them without any dark or flat calibration frames and finally histogram stretched the stacked TIF in Adobe Lightroom.

 

[Drakkith]

Okay. It just looks really noisy for 2 hours of data in a bortle 2 spot. But I don't have your equipment, so this may be entirely normal for your setup.

 

[Devin-M]

It’s not a cooled camera & shot at 6400iso instead of a more ideal 100-400iso.