Infrared Detectors & The 2nd Law of Thermodynamics

  • B
  • Thread starter Devin-M
  • Start date
  • #1
Devin-M
816
659
This started as an astrophotography question and morphed into a thermodynamics question & it was suggested to start a new thread in Thermodynamics.

Essentially the question is how do I reconcile the behavior of particular infrared photodetectors with the 2nd Law of Thermodynamics?

For example suppose I take a well-insulated tank of room temperature (300k) water and lower in a PN junction HgCdTe infrared photodetector (also at room temperature, 300k) operating in Photovoltaic mode with zero applied bias voltage. Water at 300k has a black body spectrum which includes the emission of 3.5 micrometer infrared light. IR detectors made from HgCdTe (operating at 300k) in the paper linked to below exhibit "current responsivity" when absorbing infrared radiation, also at 3.5 micrometers (with 0v applied bias voltage).

If I can generate any current from the black body radiation of room temperature water with an HgCdTe IR detector, also operating at 300k with 0v bias voltage, how do I reconcile it with the 2nd Law of Thermodynamics? I thought I shouldn't be able to produce any useful work from a heat reservoir which initially has no temperature gradient.

https://www.researchgate.net/publication/343856156_Higher_Operating_Temperature_IR_Detectors_of_the_MOCVD_Grown_HgCdTe_Heterostructures

https://www.speakev.com/cdn-cgi/image/format=auto,onerror=redirect,width=1920,height=1920,fit=scale-down/https://www.speakev.com/attachments/5000b9e5-011d-4ef0-af59-bedb15ff822e-jpeg.156575/

https://www.speakev.com/cdn-cgi/image/format=auto,onerror=redirect,width=1920,height=1920,fit=scale-down/https://www.speakev.com/attachments/94567243-aafc-4537-af56-e65cc0fd1755-jpeg.156576/

Black body spectrum of room temperature (300k) water:

https://www.speakev.com/cdn-cgi/image/format=auto,onerror=redirect,width=1920,height=1920,fit=scale-down/https://www.speakev.com/attachments/aea0d9fd-14b5-4308-a81c-f5fb6ef06d83-png.156577/

Quotes from the original discussion:

Core of Andromeda Galaxy, 2.5Mly
2130mm f/14.2 29x 90sec (43.5min) 6400iso, Bortle 2

View attachment 295965

It’s been a while since I’ve taken any astro-photos (its been cold, cloudy and rainy for quite a long time.) I managed to find a break in the clouds last night while the moon was down and captured this photo of the core of Andromeda Galaxy, which is around 2.5 million light years distant. I shot using a 150mm diameter Meade Maksutov Cassegrain on a Star Adventurer 2i mount with some modifications to go probably 3x over the weight limit with a Nikon D800 camera body. I aimed the telescope by taking test shots and plate solving through a website on my phone and then making adjustments. I used the 29 best images although I think I got around 60 total, only about half had round enough stars with 90 second exposures at this focal length on this mount. My bortle 2 dark sky location is near Shingletown, California.

View attachment 295970
View attachment 295971

Quick question... Sometimes I shoot dark calibration frames to remove noise from the final image on the way home (for example I might take 10x shots of 5min each at 6400iso with the lens cap on). Because I'm too impatient to wait around at the cold, dark sky location, the camera is heating up from for example 35F (outdoor temp) to 70F in the car while I drive. This results in the amount of noise in each dark frame increasing as the temperature of the camera increases, as seen in the animation below. My conceptual understanding is the photodiodes in the CMOS sensor are operating much like solar panels, and are somewhat sensitive to infrared light while taking exposures even when the lens cap is on. My question is can solar cells or CMOS sensors convert infrared light into electricity, and if so how does that reconcile with the 2nd Law of Thermodynamics? For example, I thought I shouldn't be able to extract useful work from a single temperature reservoir... but if a solar panel can convert infrared light into electricity, isn't it extracting work from a single temperature reservoir (suppose it's immersed in an insulated tank of water as a heat source)?

https://www.speakev.com/attachments/13c208ea-b300-4a24-a496-9e9f9fae4c25-gif.156383/

You will almost certainly get better results by shooting your darks in concert with your lights. Shoot a couple of light frames, then shoot a dark, then a few more lights, then a dark. Rinse and repeat. The better the temperature match, the better the dark frame subtraction.


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.

An alternative that you might want to try in your off time (i.e., a cloudy night), is to realize that you can reuse DARK frames (for the given camera). Take your camera and an accurate thermometer outside and let the camera the acclimate to the outside temperature. Once the camera is in thermal equilibrium, start taking DARKs. Lots of DARKs.

Vary the exposure times in a controlled, roughly exponential way -- exposure times that you might likely use for your LIGHT frames. For example, 40 sec, 60 sec, 90 sec, 120 sec, 180 sec, 240 sec, 300 sec, etc.

The whole while, keep meticulous records of the ambient temperature. Also keep a record of the camera's ISO setting you are using for each dark (only use ISO settings that you would use for LIGHT frames). Sit down, have a beer in the cold. This will take a while. Maybe have two beers. Repeat as much as you can. As the night continues, the temperature is likely to drop, so you should expect to have several different temperature points for your DARK frames.

Later, organize these DARK frames. Rename each DARK frame such that it has
  • Specific camera taking the photos
  • ISO setting,
  • Exposure time,
  • Temperature for that particular dark frame, and
  • Some unique identifier so you don't accidentally overwrite existing DARK frames
in the file name.

Repeat the next cloudy night. Put the camera in the refrigerator and repeat there too. If you live in a cold climate, and foresee yourself taking astrophotos in really cold weather, try the freezer as well.

Eventually, you'll have a DARK library with many frames for each particular ISO setting + Exposure time + temperature. Ideally, you'll have many DARK frames for each combination. Organize these on your comptuer. Once you have that, all you need to do when taking LIGHT frames is to record the temperature at the time (stash your thermometer in your camera bag), and your corresponding DARK frames will be waiting for you at home.

(Of course, the data is specific to the specific camera. You cannot share DARK frames between cameras.)

---------------

For personal reference, I have a few cooled cameras (dedicated, astrophotography cameras). Having a temperature controlled camera makes the process a lot easier. But even then it still takes a while. Whenever I get a new cooled camera, the first thing I do set the camera next to the computer and let it take DARKs for days. Literally days. But once it's done I don't have to take DARKs again for maybe a year or two. And I only use 2 temperature settings (0 deg C and -5 deg C).

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


View attachment 296153
(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).

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.

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.
View attachment 296212
View attachment 296216

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.

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.

View attachment 296248

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

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.
****************


That sounds like a good idea to me.

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.
 
  • Like
Likes berkeman

Answers and Replies

  • #2
Twigg
Science Advisor
Gold Member
873
469
Ok, so I definitely haven't read all of the OP, since it was quite long, but I have a few comments.

First off, the 2nd law of thermo says you can't extract power from a single temperature reservoir. However, notice that in your situation, the photodiode has to be connected to a load, or no current will flow. I suspect that the load takes the role of a cold reservoir in this case.

Since the diode is running in photovoltaic mode, it builds up a terminal voltage that is related to the incident infrared power. If you were to short the terminals out (i.e., connect a load of ##0 \Omega## which would not dissipate any power out of the system), that voltage would go to 0 and there would be no electrical power in the system. This is fully consistent with the second law. To drive a current, you need to dump power via the load's resistance.

I see one problem with this explanation, which is that there isn't an obvious steady state. The water dumps infrared power into the MCT, and so the water cools off. The load dissipates that power and heats up. So if I had two cups of water, and I put the MCT in one cup and the load in the other cup, it would seem that I could indefinitely heat the cup with the load in it and cool the cup with the detector in it without doing any work, but this is a different violation of the second law. Not clear to me how to resolve this one off the top of my head. My gut feeling is that there must be some reciprocity between the blackbody radiation of the water and the thermal noise of the load, but I could be BS-ing.
 
  • #3
DaveE
Science Advisor
Gold Member
2,841
2,488
I also think someone smarter than me needs to figure out the heating of the PD system, even when shorted. You can't make a bunch of electrons move around without making heat, or perhaps altering the PD junction and changing the responsivity. PV mode doesn't actually exist until someone invents a superconducting semiconductor junction.
 
  • #4
Devin-M
816
659
I also think someone smarter than me needs to figure out the heating of the PD system, even when shorted. You can't make a bunch of electrons move around without making heat, or perhaps altering the PD junction and changing the responsivity. PV mode doesn't actually exist until someone invents a superconducting semiconductor junction.

This graph shows the “spectral responsivity measured at zero bias (i.e. photovoltaic mode)” of a ”Te-hyperdoped Si photodetector.” (top left chart is the detector at room temp 300k)

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


View attachment 296153
(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).
ured-at-zero-bias-ie-photovoltaic-mode-for-the-png.png
 
  • #5
22,332
5,205
The 2nd law of thermodynamics does not say that you can't extract energy from a single temperature source and do work. It says you can't do it by a system operating in a cycle.
 
  • Informative
  • Like
Likes DaveE, Devin-M, berkeman and 1 other person
  • #6
Devin-M
816
659
So it won’t violate any laws of physics if I tried to generate electricity from the black body radiation of room temperature water with a room temperature HgCdTe PN Juction IR detector as described in the original post?
 
  • #7
22,332
5,205
So it won’t violate any laws of physics if I tried to generate electricity from the black body radiation of room temperature water with a room temperature HgCdTe PN Juction IR detector as described in the original post?
Right.
 
  • Like
Likes Devin-M
  • #8
DaveE
Science Advisor
Gold Member
2,841
2,488
So it won’t violate any laws of physics if I tried to generate electricity from the black body radiation of room temperature water with a room temperature HgCdTe PN Juction IR detector as described in the original post?
Right. But you can't expect that to go on forever. The system will be changed by the extraction/redistribution of energy. Systems out of equilibrium can absolutely do work... until they can't anymore.
 
  • #9
Devin-M
816
659
I had envisioned harvesting energy continuously day & night from ocean heat so that less energy storage is needed for intermittent renewables like wind & solar.
 
  • #10
Charles Link
Homework Helper
Insights Author
Gold Member
5,415
2,715
From what I can see, the researchers likely made an error somewhere. You simply do not get a response out of an IR photodiode when everything is in equilibrium, which it should be if everything, including the surroundings, is at the same temperature. (The I-V characteristic curve for a non-illuminated photodiode passes through the origin).

The kind of response they are getting is what you could expect if the photodiode were thermoelectrically cooled.=Edit: scratch that.

Edit: Looking it over more carefully, the researchers say the photodiode is "backside illuminated", but the details are missing with what they illuminated it. I do think it may be necessary to get the extra details, before jumping to any conclusions on whether there is actually an error in their work. In any case, it is necessary to illuminate the photodiode by something more than the ambient to get a response out of it.
 
Last edited:
  • #11
Devin-M
816
659
From what I can see, the researchers likely made an error somewhere. You simply do not get a response out of an IR photodiode when everything is in equilibrium, which it should be if everything, including the surroundings, is at the same temperature. (The I-V characteristic curve for a non-illuminated photodiode passes through the origin).

The kind of response they are getting is what you could expect if the photodiode were thermoelectrically cooled.=Edit: scratch that.

Edit: Looking it over more carefully, the researchers say the photodiode is "backside illuminated", but the details are missing with what they illuminated it. I do think it may be necessary to get the extra details, before jumping to any conclusions on whether there is actually an error in their work. In any case, it is necessary to illuminate the photodiode by something more than the ambient to get a response out of it.
Both of these papers (by different teams) are showing results indicating responsivity with zero applied bias voltage (photovoltaic mode) from 3.5 micrometer infrared light with the detector at 300k (80F)…

https://www.researchgate.net/publication/346511011_Silicon-Based_Intermediate-Band_Infrared_Photodetector_Realized_by_Te_Hyperdoping

(page 22)

https://www.researchgate.net/publication/343856156_Higher_Operating_Temperature_IR_Detectors_of_the_MOCVD_Grown_HgCdTe_Heterostructures

(page 11)
 
Last edited:
  • #12
Devin-M
816
659
My understanding is “backside illuminated” means the electronics for each pixel are on the opposite side of the wafer from the photosensitive material so you have more active area for a given sensor size.
 
  • #13
Charles Link
Homework Helper
Insights Author
Gold Member
5,415
2,715
My understanding is “backside illuminated” means the electronics for each pixel are on the opposite side of the wafer from the photosensitive material so you have more active area for a given sensor size.
This is kind of typical of my experience with a lot of the specialized journal articles=they seem to have omitted the details on what they used for a calibration source to characterize the photodiode. It is likely they used a calibration lamp (with known spectral characteristics) along with a spectrometer, but they certainly didn't describe their procedure in detail.

In any case, they needed to use something more than the ambient background, because the response in that case would be zero.
 
  • #14
Devin-M
816
659
This is kind of typical of my experience with a lot of the specialized journal articles=they seem to have omitted the details on what they used for a calibration source to characterize the photodiode. It is likely they used a calibration lamp (with known spectral characteristics) along with a spectrometer, but they certainly didn't describe their procedure in detail.

In any case, they needed to use something more than the ambient background, because the response in that case would be zero.
It says here on wikipedia:

”Cooled infrared detectors



Without cooling, these sensors (which detect and convert light in much the same way as common digital cameras, but are made of different materials) would be 'blinded' or flooded by their own radiation.

https://en.wikipedia.org/wiki/Thermographic_camera
 
  • #15
Charles Link
Homework Helper
Insights Author
Gold Member
5,415
2,715
It says here on wikipedia:

”Cooled infrared detectors



Without cooling, these sensors (which detect and convert light in much the same way as common digital cameras, but are made of different materials) would be 'blinded' or flooded by their own radiation.

https://en.wikipedia.org/wiki/Thermographic_camera
This article gives some good general information, but it is lacking in specifics. It is necessary to cool InSb photodiodes, but from what we can see with the papers in your original post, (and post 11), they have gotten HgCdTe to work well at room temperature.
 
Last edited:
  • #16
Devin-M
816
659
The way I interpreted the paper was that you wouldn’t use the detector uncooled because it would be “flooded” but still if you shine an infrared laser on it, it will make even more current than just what it produces from the room temperature black body spectrum.
 
  • #17
Drakkith
Staff Emeritus
Science Advisor
2022 Award
22,260
6,337
From pages 14-15 of the article linked in the first post:
Device Measurement: A vacuum pump is used to avoid moisture condensation at low temperatures. 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 infrared light emitted from the Globar (SiC) source is modulated by a mechanical chopper at 87 Hz before entering the monochromator. The shortcircuit photocurrent was extracted with the help of a SR830 digital signal processing (DSP) lock-in amplifier. For dynamic photoresponsivity experiments such as time-resolved photoresponsivity measurements, a 1.55 µm-light emitting diode (LED) (Thorlabs, 1550E) together with a long-pass filter at 1.3 µm was used. The LED was powered using a homemade current-drivers circuit coupled to the frequency of the output Transistor-Transistor Logic (TTL) signal of the lock-in amplifier. Therefore, the frequency of the TTL signal is adjustable to generate pulsed light and thus perform a frequency scan of the responsivity. Moreover, the 15 output power of the LED is also adjustable with a maximum incident power density of 1.83 mW/cm2 . The cut-off frequency of the LED is 0.1 GHz, which is sufficiently higher than that of the Te-hyperdoped Si p-n photodetector under investigation. In the time-resolved experiments, a low-noise current amplifier was employed to amplify the photocurrent signal from the device under 1.55 µm-LED excitation. A digital oscilloscope was used to record the time-resolved photocurrent.

I admit I'm not familiar with most of the terms and equipment, but it looks like they've illuminated the device via LED or Globar source and not used ambient radiation of the environment. Per a quick google search, a globar is: a solid light source made of silicon carbide; it is heated by resistor heating method by a current source. The silicon carbide rod is heated to temperatures of 1000 to 1650°C (1832-3002°F) and then combined with a downstream interference filter to produce radiation having wavelengths of 4 to 15 µm.

So they have an LED and an adjustable IR light source, which means that the 2nd law isn't violated since more energy is being radiated to the device than away. Any work done by the device is less than what was done by the radiation sources.
 
  • Like
Likes collinsmark and Charles Link
  • #19
collinsmark
Homework Helper
Gold Member
3,238
2,130
Yes, I've been saying that for awhile now. In the paper, the photodetector was illuminated by a monochromater (utilizes a diffraction grating). The source to the monochromater is a relatively broadband light source. The diffraction grating isolates very narrow wavelength bands coming from the source while specific measurements are made. (And it's bright enough such that at any isolated wavelength, it's is still brighter than any background IR.) The whole while, the photodetector is held at a constant temperature.

My point is, if you turn the light source off, the current in the photodetector will vanish.*

*(assuming the photodetector is otherwise in thermal equilibrium with its surroundings, neglecting the light source, of course, which may as well not even exist after it's turned off.)
 
Last edited:
  • Like
Likes Drakkith and Charles Link
  • #20
Devin-M
816
659
My point is, if you turn the light source off, the current in the photodetector will vanish.*

*(assuming the photodetector is otherwise in thermal equilibrium with its surroundings, neglecting the light source, of course, which may as well not even exist after it's turned off.)
If the detector is in a tank of 300k (80F) water, there is still a 3.5 micrometer light source… the black body radiation spectrum of the room temperature water…

https://www.speakev.com/cdn-cgi/image/format=auto,onerror=redirect,width=1920,height=1920,fit=scale-down/https://www.speakev.com/attachments/aea0d9fd-14b5-4308-a81c-f5fb6ef06d83-png.156577/
 
  • #21
Charles Link
Homework Helper
Insights Author
Gold Member
5,415
2,715
If everything is at the same temperature, the radiated power (off of the photodiode) is equal to the incident power absorbed from the surroundings, and this 300 K background has no effect on the photodiode.
 
  • #22
Devin-M
816
659
If everything is at the same temperature, the radiated power (off of the photodiode) is equal to the incident power absorbed from the surroundings, and this 300 K background has no effect on the photodiode.
I thought the 3.5 micrometer light reaching the detector is absorbed and then “radiated” from the PN junction as a conduction electron & an electron hole.
 
  • #23
Charles Link
Homework Helper
Insights Author
Gold Member
5,415
2,715
It would take a little effort with a solid state book like Streetman's that covers photodiodes in detail, but it would be fairly straightforward to show that there is no photocurrent from a photodiode in a thermal equilibrium state.

You are trying to make an argument for something that is physically and thermodynamically impossible, and you really need to put more effort into seeing that it doesn't work, than trying to claim that it is supported by literature that is making no such kind of claim.
 
  • Like
Likes russ_watters and Lord Jestocost
  • #24
Devin-M
816
659
An ordinary (visible spectrum, silicon) solar panel would produce electron + electron hole pairs in a well insulated tank next to a hot object emitting visible spectrum black body radiation.
 
  • #25
Charles Link
Homework Helper
Insights Author
Gold Member
5,415
2,715
If it has a hot object in its vicinity, that is not the case of thermal equilibrium that we have been discussing. In the case of the photodiode whose spectral response curve was measured, the hot object was a globar, as @Drakkith pointed out. There would be nothing to measure from the 300 K surroundings if the photodiode is also at 300 K, without any other sources present.
 
  • Like
Likes russ_watters and Lord Jestocost
  • #26
Devin-M
816
659
The room temperature water is the hot object illuminating the IR sensor with 3.5 micrometer wavelength photons.
 
  • #27
Charles Link
Homework Helper
Insights Author
Gold Member
5,415
2,715
The room temperature water is the hot object illuminating the IR sensor with 3.5 micrometer wavelength photons.
Please read the last sentence I added to my previous post. There is really no point in discussing this further. This discussion for me has run its course.
 
  • #28
Devin-M
816
659
I just don’t understand why it can generate current with the photovoltaic effect from a powered 3.5 micrometer infrared laser / light but not a room temperature 300k / 80F black body like water emitting the same 3.5 micrometer wavelengths.
 
  • #29
Charles Link
Homework Helper
Insights Author
Gold Member
5,415
2,715
If you raise the water temperature to 301 K while the photodiode is at 300 K, then you will get some small photocurrent. You get zero photocurrent when the water is at 300 K.
 
  • #30
Devin-M
816
659
The infrared light emitted from the Globar (SiC) source is modulated by a mechanical chopper at 87 Hz before entering the monochromator. The shortcircuit photocurrent was extracted with the help of a SR830 digital signal processing (DSP) lock-in amplifier.

So why did they extract the “shortcircuit photocurrent” with the help of a “SR830 digital signal processing (DSP) lock-in amplifier” when they did the 300k measurement with no bias voltage ?
 
  • #31
Charles Link
Homework Helper
Insights Author
Gold Member
5,415
2,715
Even with a strong source such as a globar, when a monochromator is used at moderately high resolution to only have a small spectral increment ## \Delta \lambda ## incident from the globar as each reading is taken, the incident power on the photodiode is still rather small. With the lock-in amplifier with an optical chopper you can then get a very accurate measurement.
 
  • #32
Devin-M
816
659
If I put the incident power source in a box separated from the 300k operating temperature detector by a vacuum, and I expose the detector to 3.5 micrometer light from a laser or from a 300k cup of water what is the difference?
 
  • #33
collinsmark
Homework Helper
Gold Member
3,238
2,130
If I put the incident power source in a box separated from the 300k operating temperature detector by a vacuum, and I expose the detector to 3.5 micrometer light from a laser or from a 300k cup of water what is the difference?

It's not just about the wavelengths of photons hitting or leaving the photodetector (although, yes, that's part of it), it's also about the quantity.

In the case of the 300 K cup of water with with the photodetector submerged in the water, also at 300 K, the photodetector receives some level of power flux incident on it, but it also radiates an equal amount, such that the net power flux reaching the photodetector is zero.

In the case of the 3.5 micrometer laser (which can also be applied to the case with the broadband light source + monochromater), the power flux is much greater. The amount of photons incident on the photodetector is much greater. But the photodetector, held at 300 K, doesn't naturally radiate any more than it did in the glass of water example. A small fraction of that extra power gets converted to photodetector current, a lot of it is reflected (or passes through it; we are talking deep IR here), and the rest of it goes into the mechanism keeping the photodetector cool (the mechanism that keeps the photodetector and its immediate surroundings from heating up).
 
  • #34
Devin-M
816
659
In the case of the 300 K cup of water with with the photodetector submerged in the water, also at 300 K, the photodetector receives some level of power flux incident on it, but it also radiates an equal amount, such that the net power flux reaching the photodetector is zero.
If I understand the PN junction correctly, the material absorbs the infrared photon but it does not re-emit an infrared photon. Instead that photon jumps an electron from the valence band to the conduction band in the material, and since it's a diode it only has one way it can go.
 
  • #35
collinsmark
Homework Helper
Gold Member
3,238
2,130
If I understand the PN junction correctly, the material absorbs the infrared photon but it does not re-emit an infrared photon. Instead that photon jumps an electron from the valence band to the conduction band in the material, and since it's a diode it only has one way it can go.
In terms of overall energy flux (regardless of particular wavelengths), it works both ways.

Photodetectors can emit light, contrary to their typical operation.
LEDs can detect light, contrary to their typical operation.



When everything is in thermal equilibrium (no external light sources, no external current through the device), the device operates right in-between. It is neither a detector nor source. It doesn't receive any more energy than it emits.
 

Suggested for: Infrared Detectors & The 2nd Law of Thermodynamics

Replies
20
Views
727
  • Last Post
Replies
33
Views
647
Replies
5
Views
1K
  • Last Post
Replies
2
Views
397
Replies
2
Views
408
Replies
2
Views
159
Replies
3
Views
421
Replies
100
Views
4K
Replies
2
Views
563
Top