B Infrared Detectors & The 2nd Law of Thermodynamics

[Devin-M]

A solar panel can’t be thermally radiating or reflecting or converting to internal energy the same power of incident electromagnetic radiation it receives since it’s also producing electric current.

 

[collinsmark]

A solar panel can’t be thermally radiating the same power of incident electromagnetic radiation it receives since it’s also producing electric current.

But if the solar panel is producing electricity, it's not in thermal equilibrium. It's kept cool by the Earth's atmosphere and by its backside, which isn't in the light.

Even if you put a solar panel in space, it's cooled by its own backside, which is facing the cold depths of space. One side is facing the hot sun, the other side facing the cold space. From that, yes, you can extract energy (because the Sun and space are at different temperatures).

Again, if you put the solar panel (the whole kit and kaboodle) squarely inside the Sun's photosphere (and allowed it to naturally heat up to the Sun's temperature), it won't work.

 

[Devin-M]

Are you saying if I point a solar panel at the sun then use a mirror to reflect light onto the back of the wafer it will stop working?

 

[collinsmark]

Are you saying if I point a solar panel at the sun then use a mirror to reflect light onto the back of the wafer it will stop working?

If you illuminate the solar panel by a constant amount, equally from all directions from blackbody radiation, and you allow the panel's temperature to naturally rise uninhibited (free from any heat conduction and heat convection), which pretty much means you'll need to move it outside Earth's atmosphere, and allow its temperature to stabilize at equilibrium, then I suppose yes, that is what I am saying.

 

[Devin-M]

I would radiate the amount of incident power onto the solar panel that doesn’t generate external current, but that radiation is less than the incident power.

Why can’t an HgCdTe detector do the same? Radiate the non-current generating incident 3.5 micrometer power, less than the amount received, same as with solar.

 

[collinsmark]

I would radiate the amount of incident power onto the solar panel that doesn’t generate external current, but that radiation is less than the incident power.

Why can’t an HgCdTe detector do the same? Radiate the non-current generating incident 3.5 micrometer power, less than the amount received, same as with solar.

I don't understand.

By sticking a 300 K an HgCdTe detector into a glass of water at 300 K, the detector is receiving and emitting photons at equal rates in all directions. How do you limit the photon flux from some directions but not others?

Before you say you could put an object in-between the detector and the water, realize that as the object reaches thermal equilibrium, it too will be emitting thermal photons at the same rate as everything else.

 

[Devin-M]

It isn’t emitting as many photons as it’s receiving since some photons are producing current in the wires that exit the water.

 

[collinsmark]

It isn’t emitting as many photons as it’s receiving since some photons are producing current in the wires that exit the water.

That's only half the picture.

The photodetector is also emitting photons that effectively sap the current through the photodetector. And these current "producing" photons that arrive at the photodetector, and those current "sapping" photons that are emitted by the photodetector are happening in equal amounts when everything is in thermal equilibrium.

Consider a few examples:

• Red LEDs are more sensitive at detecting red light than other wavelengths of light. It goes without saying that red LEDs emit red light more than other wavelengths (that's why it's called a "red" LED). So if you want to use a device to detect red light, you should use a red one rather than a blue LED.
• When light passes through cold hydrogen gas (or any particular gas for that matter), the gas absorbs spectral lines. But when the gas is hot, it emits light at those same spectral lines.

Similarly, even though HgCdTe diode is more sensitive to detect photons at certain wavelengths, I'm willing to bet experiment will show it's also more sensitive to emit photons at those same wavelengths.

Any of these things, when in thermal equilibrium with their respective surroundings, cannot extract useful energy.

It's true that some materials can take a single high frequency photon and convert that into several low-frequency photons more than the reverse. But this inequality can only be exploited outside of thermal equilibrium, which isn't what we're discussing here. Once everything reaches thermal equilibrium, the HgCdTe diode will be emitting as much energy as it receives, and no net current is produced.

 

[Devin-M]

How would the fact that the HgCdTe diode emits some infrared black body light of its own (at 300k) prevent it from converting some incoming infrared 3.5 micrometer photons from an external source into current through an external electrical circuit?

 

[hutchphd]

The arguments here are essentially similar to why a diode can not spontaneously convert thermal noise into useable power. Or why a "Brownian Ratchet" fails. These are perhaps easier to understand in detail. Feynman adresses them well as I recall. Basicly they all fail because whatever mechanism is used to select the "good" processes it uses up more power than it provides on average in steady state. I would need to see the experimental apparatus in detail to discuss the graphs.
Good question!

 

[Devin-M]

Just as a solar panel diode uses the photovoltaic effect to create electrical current from incoming visible light, the HgCdTe diode uses the photovoltaic effect to create electrical current from incoming 3.5 micrometer infrared photons.

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/

 

[collinsmark]

How would the fact that the HgCdTe diode emits some infrared black body light of its own (at 300k) prevent it from converting some incoming infrared 3.5 micrometer photons from an external source into current through an external electrical circuit?

It's not just some infrared blackbody light that it emits. It's the same amount of blackbody light as it absorbs.

When it absorbs photons it generates emf that can be used to generate current in the external circuit. When it emits a photons it produces an emf with opposite polarity, which acts in the opposite way. Both are always happening to some degree. But when the detector is in thermal equilibrium with its surroundings, they happen at equal amounts and the effective emf, when all is said and done, is zero. No net current is produced for the external circuit.

 

[hutchphd]

Just as a solar panel diode uses the photovoltaic effect to create electrical current

The photodiode inside the sun (were it sturdy enough) would be just a diode and then the fluctuations dominate.

 

[Devin-M]

It's not just some infrared blackbody light that it emits. It's the same amount of blackbody light as it absorbs.

Yes some of the absorbed IR light can be re-emitted from the detector itself and the rest of the energy it absorbs can be emitted as heat generated by the current through the resistance in the external wires coming out of the detector diode.

 

[collinsmark]

Yes some of the absorbed IR light can be re-emitted from the detector itself and the rest of the energy it absorbs can be emitted as heat generated by the current through the resistance in the external wires coming out of the detector diode.

That would require the photodetector to emit less energy into its surroundings than it absorbs; the difference going into the electric circuit.

And if the energy incident on the detector has a thermal spectrum, and the detector naturally emits radiation according to its thermal spectrum, it means that the detector must be colder than its surroundings. But we've established everything is at the same temperature.

Do you see the contradiction here?

 

[Devin-M]

I’m saying shine an IR 3.5 micrometer monochrome laser of whatever power you wish on the 300k HgCdTe detector and you get current responsivity in amps per watt through an external circuit in photovoltaic mode with 0v applied bias volts.

Now shine some 3.5 micrometer photons from 300k water onto it through the vacuum and aside from the number of photons, it can’t tell the difference between the individual 3.5 micrometer photons whether laser generated or 300k black body generated.

 

[collinsmark]

I’m saying shine an IR 3.5 micrometer monochrome laser of whatever power you wish on the 300k HgCdTe detector and you get current responsivity in amps per watt through an external circuit in photovoltaic mode with 0v applied bias volts.

The details of what happens depends upon the intensity of the laser (or whatever source). But if the intensity is above minuscule, the HgCdTe detector will start to produce current, but it will also heat up. And it will keep increasing in temperature and its efficiency will decrease.

If the detector is bathed in 3.5 micrometer light in all directions, and its temperature is not inhibited (no conduction, no convection, etc.), it will eventually stop producing current. Assuming it doesn't burn up, it will eventually become hot enough such that it is thermally radiating as much energy as it is receiving from its surroundings (including the 3.5 micrometer laser source) and no more current will be generated.

Now shine some 3.5 micrometer photons from 300k water onto it through the vacuum

In that case the photodetector will get no hotter than 300 K. There's simply not enough energy flux to increase its temperature more than that (i.e., not enough photons).

it can’t tell the difference between the individual 3.5 micrometer photons whether laser generated or 300k black body generated.

Good god. The number of photons makes a huge difference.

Consider a microwave oven. Microwave ovens utilize photons of around 12.2 cm. For reference, blackbody who's peak wavelength is 12.2 cm really cold compared to the 3.5 micrometer we were previously discussing.

Your popcorn can certainly tell the difference regarding the number of photons.

 

[Devin-M]

This is how many 3.5 micrometer photons the initially 300k water in the dark is emitting into the vacuum…
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/

 

[collinsmark]

This is how many 3.5 micrometer photons the initially 300k water in the dark is emitting into the vacuum…
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/

I'm not sure I understand what I should take from that.

Getting back to my previous example, deep space (far from any stars or galaxies) has a thermal signature from the cosmic microwave background corresponding to a black body of around a handful of Kelvin. It's really cold. And yet microwaves from a microwave oven of roughly the same ballpark wavelength regularly heats my hot dogs. They are not the same. The microwave oven has a much, much larger photon flux. And that makes a difference.

So, how do you increase 3.5 micrometer photon flux above what you would get by surrounding the area with 300 K water? There are several ways, but one way is to use a hotter source, and then filter off the other wavelengths. Sure, the hot source doesn't have a blackbody peak at 3.5 micrometers, it's much hotter. None-the-less, its flux around the 3.5 micrometer region is still plenty more than surrounding the target with 300 K water.

 

[Devin-M]

If all I care about is whether a single individual incoming IR photon pushes one particular electron and electron hole out towards the external circuit from the PN junction diode, why would it be of consequence whether that photon came from the IR laser or the surface of some 300k water in the dark?

 

[collinsmark]

If all I care about is whether a single individual incoming IR photon pushes one particular electron and electron hole out towards the external circuit from the PN junction diode, why would it be of consequence whether that photon came from the IR laser or the surface of some 300k water in the dark?

Because the 300 K water is radiating with a thermal spectrum. Its thermal flux is limited. It simply won't radiate more energy flux than some peak value.

The same can be true for the photodetector at 300 K (when in zero bias mode). It won't radiate more energy flux than some peak value.

And it turns out, with both the water and the photodetector at 300 K, their limited energy fluxes are the same value (when in thermal equilibrium with each other).

The IR laser on the other hand can have a practically unlimited energy flux.

It's a matter of comparing the 1 photon that came from the water with 10,000,000,000 photons that came from the IR laser.

 

[Devin-M]

It's a matter of comparing the 1 photon that came from the water with 10,000,000,000 photons that came from the IR laser.

I only want to compare a 3.5 micrometer photon from the water with 1 of the same wavelength of the 10,000,000,000 from the laser.

 

[collinsmark]

I only want to compare a 3.5 micrometer photon from the water with 1 of the same wavelength of the 10,000,000,000 from the laser.

Allow me to clarify: the laser also produces photons at 3.5 micrometer wavelength. All photons in question have a wavelength of 3.5 micrometers. The laser just produces more photons per unit area per unit time than either the water or the photodetector when the water and photodetector are at 300 K.

 

[Devin-M]

But if there’s water and a laser, both emitting 3.5 micrometer photons in different amounts, when the detector is producing current we don’t necessarily know whether a particular photon came from the water or the laser.

 

[collinsmark]

But if there’s water and a laser, both emitting 3.5 micrometer photons when the detector is producing current we don’t necessarily know whether a particular photon came from the water or the laser.

I agree with you that the detector doesn't care where the photons came from. What matters is their sheer numbers, i.e., the quantity of photons.

Situation 1: Detector and water, both at 300 K

• Water transfers a single 3.5 micrometer photon to the detector.
• Detector transfers a single 3.5 micrometer photon the water.

Result: Temperature remains the same, and no current is produced.

Situation 2: Detector and laser, where the detector is at 300 K along with its surroundings.

• Surroundings transfers a single 3.5 micrometer photon to the detector.
• Laser transfers 10,000,000,000 individual photons, each with a wavelength of 3.5 micrometers to the detector.
• Detector transfers a single 3.5 micrometer photon to its surroundings.

Result: Current is produced and the temperature of the detector tends increase, and will increase without some sort of intervention.

 

[Devin-M]

I agree with you except the current through the resistance in the external circuit releases heat of its own, and the detector can radiate some heat in different directions and/or increase in temperature.

 

[collinsmark]

I agree with you except the current in the external circuit releases heat of its own, and the detector can radiate some heat and/or increase in temperature.

I though we've already gone through this. If the water and the detector are in thermal equilibrium there's no energy left to be transferred to the electric circuit. There is no current. (This describes the case with no laser, no external light sources, and the detector is completely immersed in the water. All are at the same temperature.)

If you cool the detector to be at a lower temperature than the water, then yes, in that case you can transfer some energy to the electric circuit. But that won't happen spontaneously.

 

[Devin-M]

What happens when the detector is 301k, being shined on by the 3.5 micrometer laser (so is generating current through an external load with a particular resistance), and a 3.5 micrometer photon from the 300k water strikes the detector, but we don’t know whether the individual photon that generated some of the current came from the 3.5 micrometer laser or the 300k water?

 

[Drakkith]

How would the fact that the HgCdTe diode emits some infrared black body light of its own (at 300k) prevent it from converting some incoming infrared 3.5 micrometer photons from an external source into current through an external electrical circuit?

Let's assume the diode/sensor is connected to a capacitor, all at 300k. The sensor, wires, and capacitor are then placed in a closed box whose walls are 300k. If we look at what can happen then there are a few possibilities:

1. A photon is absorbed by the sensor and creates an electron-hole pair and generates current.
2. A photon is absorbed by the sensor and only creates heat.
3. A photon is emitted by the sensor via thermal motion.
4. A photon is emitted by the sensor via recombination of electrons with holes.

Numbers 2 and 3 don't really concern us except that they help keep everything in equilibrium.

Let's say option 1 happens. An electron-hole pair is created and a small amount of charge is stored on each plate of the capacitor.

But shortly thereafter option 4 happens. An electron combines with a hole and emits a photon. This creates a voltage in the opposite direction as option 1 and the charges are pulled off of the plates of the capacitor, leaving it neutral.

But there's no reason option 1 had to happen before option 4. Both are actually going on at the same time. Sometimes option 4 will happen 10 times before option 1 happens, and other times the opposite will be true.

If the entire sensor assembly is in thermal equilibrium with the environment then all of these possibilities are ongoing simultaneously. Due to the randomness inherent in these processes, sometimes the capacitor will have a small amount of charge stored on it and sometimes it will not, and sometimes the charges will be stored in the opposite polarity.

Overall, there will be no net current and no energy storage on the capacitor.

 

[Drakkith]

What happens when the detector is 301k, being shined on by the 3.5 micrometer laser (so is generating current through an external load with a particular resistance), and a 3.5 micrometer photon from the 300k water strikes the detector, but we don’t know whether the individual photon that generated some of the current came from the 3.5 micrometer laser or the 300k water?

You answered your own question. The photon is absorbed and we have no way of telling where it came from. What does this have to do with anything?

 

[Devin-M]

If there is a laser shining on the detector is the detector in thermal equilibrium with the environment?

 

[Drakkith]

If there is a laser shining on the detector is the detector in thermal equilibrium with the environment?

Certainly not. The laser is part of the environment.

 

[Devin-M]

In that case I’m not trying to look at a situation where the detector is in thermal equilibrium with the environment. Suppose the 300k IR detector is on the beach next to the ocean “aimed at” the water with a 3.5 micrometer laser aimed at it and the ocean has locally 300k surface water. The detector is out of thermal equilibrium & illuminated with 3.5 micrometer photons by both the laser and the 300k ocean. If the 300k detector generates current from both the laser and the 300k infrared spectrum from the environment, how do you prove the environment didn’t generate some of the current in the external circuit?

 

[collinsmark]

If there is a laser shining on the detector is the detector in thermal equilibrium with the environment?

No. Not in the sense that we're talking about here. The laser almost certainly has its own, external power source. It's not part of the closed system. And the detector, kept at a cool 300 K, is certainly not in thermal equilibrium with the laser.

Certainly not. The laser is part of the environment.

I would only clarify that if we expanded our system to include the laser in the closed environment, like letting it run off of a battery (so there are no external power source), this isn't something that can be maintained indefinitely in a cyclical nature (see @Chestermiller's post #5). Eventually the laser won't lase anymore.

[Edit: If we're OK with leaving the system open (rather than closed), such that the laser has its own external power source, and the detector is allowed to transfer some of that extra power out of the system to its electrical circuit, and there's another external circuit keeping the detector at a cool 300 K, then everything is still fine: the energy leaving the system through the detector to the circuit ultimately comes from the laser. That's OK. You could call that a form of equilibrium (although not the same sort of equilibrium we're talking about here). It's just that if you turn off the laser, and turn off the detector's cooling device, there's no other way for the detector to generate power.]

 

[Drakkith]

Suppose the 300k IR detector is on the beach next to the ocean “aimed at” the water with a 3.5 micrometer laser aimed at it and the ocean has locally 300k surface water. The detector is out of thermal equilibrium & illuminated with 3.5 micrometer photons by both the laser and the 300k ocean. If the 300k detector generates current from both the laser and the 300k infrared spectrum from the environment, how do you prove the environment didn’t generate some of the current in the external circuit?

It does. The sensor doesn't discriminate between laser and environment photons (for clarity I'm separating the laser from the environment in this example). The sensor is receiving X number of 3.5 micron photons per second, where X is made up of Y photons from the environment and Z photons from the laser, so X=Y+Z.

And that's okay. There's no problem with the environment generating no current in a scenario with no laser and then generating some current in a scenario with a laser. This is because all of Y was used to keep the sensor at 300k prior to the introduction of the laser. But now some portion of Y and some portion of Z are heating the sensor instead of generating current. There's no change in the number of 3.5 micron photons used in heating, the change is in where they come from.

 

[collinsmark]

@Devin-M, allow me to pose a question. If you understand the question and its answer I think it might clear up some confusion. So I won't give you the answer in this particular post. I want you to think about it.

Does the size of a constant-temperature, empty box, or a box with only solid objects (at same temperature), change the average photon power flux within the box? What, if anything, can you do to change the photon power flux within the box?

You have an HgCdTe IR photodetector of a fixed makeup (fixed size, color, material, etc). You also have a closed box, with black (opaque) inside surfaces, in which the detector can fit. You put the detector in the box. The box is always at 300 K. When you put the detector in the box, the detector is also at 300 K. For simplicity, let's assume that at this point, besides the detector, the rest of box contains vacuum (this vacuum part is only important because I don't want to get into more complicated gas laws and phase transitions and whatnot: the box can contain solid objects and that's it. The rest of the volume is vacuum).

Your goal: increase the photon power flux (photon energy per unit time, per unit area of the detector's surface) hitting the surface of the detector. Essentially, you're just trying to increase the rate that photons strike the detector.

Things you are not allowed to change:

• You cannot insert any form of energy into or out of the box. There are no holes in the box (besides a special-purpose airlock discussed below). That means no electrical wires coming into or out of the box. You cannot use electromagnetic fields to transfer energy into our out of the box. No radioactive energy transfer is allowed either. Gravitational waves are also not allowed.
• You cannot change the temperature of the box. The box, must at all times, remain exactly at 300 K. It's assumed that you can easily keep the box at a constant 300 K.
• You cannot change the size and shape of the detector. It must stay in its original form.
• You cannot manipulate the temperature of anything in the box after it is put it in the box.
• You cannot change the outside surface material/color/etc. of the box. The box must always remain opaque.
• You cannot put two separate items into the box that chemically react with each other. That would violate the "stored energy" rule (see blow).
• Same thing with radioactive material. Stuffing the box with ²³⁵U until it reaches critical mass is right out.
• You cannot shake up the box. (You can't jostle it back and forth, or strike it with hammers, or anything that would cause its contents to heat up through friction.) That includes spinning the box like a clothes dryer tumbler. That's not allowed.

Things you can change:

• You are allowed to slowly change the size of the box (slow enough not to excessively jostle the contents). The box must be big enough not to compress the detector or anything else in the box. But you can change its size. You can increase it to the size of a large city if you wish, or just small enough to hold its contents without squishing them. Or anything in between. The box is ideal in that if you change the size of the box, you don't need to worry about its surface temperature changing. This special box will always keep its properties (temperature, color, etc), even if you change its size.
• You can add (through an airlock) other solid objects into the box (e.g., marbles, dice, other photodetectors, kitchenware, statues, bridges, castles, lenses, mirrors, etc.) so long as (a) they are also at 300 K, and (b) they do not contain any stored energy and (c) they do not sublimate into gaseous form. (For example, you cannot add a battery or cheese, or a match, or a moose, or antimatter. You cannot add highly compressed, solidified nitrogen or any highly compressed gas [for two reasons, one because there is stored energy, and another because it could become gaseous].) The solid items must remain in their original, solid form. Also, the items must fit inside the box; it's OK to increase the size of the box before you put them in though, so they'll fit.
• You can slowly move items around within the box (slow enough not to cause friction/frictional heating though). In other words, you are allowed to move items closer to or farther away from the detector.

So what, if anything, can you do to increase the average photon power flux hitting the detector? Explain your answer.

Bonus question: how does this compare to the photon power flux leaving (emitted by) the surface of the detector?

 

[Devin-M]

@Devin-M, allow me to pose a question. If you understand the question and its answer I think it might clear up some confusion. So I won't give you the answer in this particular post. I want you to think about it.

Does the size of a constant-temperature, empty box, or a box with only solid objects (at same temperature), change the average photon power flux within the box? What, if anything, can you do to change the photon power flux within the box?

You have an HgCdTe IR photodetector of a fixed makeup (fixed size, color, material, etc). You also have a closed box, with black (opaque) inside surfaces, in which the detector can fit. You put the detector in the box. The box is always at 300 K. When you put the detector in the box, the detector is also at 300 K. For simplicity, let's assume that at this point, besides the detector, the rest of box contains vacuum (this vacuum part is only important because I don't want to get into more complicated gas laws and phase transitions and whatnot: the box can contain solid objects and that's it. The rest of the volume is vacuum).

Your goal: increase the photon power flux (photon energy per unit time, per unit area of the detector's surface) hitting the surface of the detector. Essentially, you're just trying to increase the rate that photons strike the detector.

Things you are not allowed to change:

• You cannot insert any form of energy into or out of the box. There are no holes in the box (besides a special-purpose airlock discussed below). That means no electrical wires coming into or out of the box. You cannot use electromagnetic fields to transfer energy into our out of the box. No radioactive energy transfer is allowed either. Gravitational waves are also not allowed.
• You cannot change the temperature of the box. The box, must at all times, remain exactly at 300 K. It's assumed that you can easily keep the box at a constant 300 K.
• You cannot change the size and shape of the detector. It must stay in its original form.
• You cannot manipulate the temperature of anything in the box after it is put it in the box.
• You cannot change the outside surface material/color/etc. of the box. The box must always remain opaque.
• You cannot put two separate items into the box that chemically react with each other. That would violate the "stored energy" rule (see blow).
• Same thing with radioactive material. Stuffing the box with ²³⁵U until it reaches critical mass is right out.
• You cannot shake up the box. (You can't jostle it back and forth, or strike it with hammers, or anything that would cause its contents to heat up through friction.) That includes spinning the box like a clothes dryer tumbler. That's not allowed.

Things you can change:

• You are allowed to slowly change the size of the box (slow enough not to excessively jostle the contents). The box must be big enough not to compress the detector or anything else in the box. But you can change its size. You can increase it to the size of a large city if you wish, or just small enough to hold its contents without squishing them. Or anything in between. The box is ideal in that if you change the size of the box, you don't need to worry about its surface temperature changing. This special box will always keep its properties (temperature, color, etc), even if you change its size.
• You can add (through an airlock) other solid objects into the box (e.g., marbles, dice, other photodetectors, kitchenware, statues, bridges, castles, lenses, mirrors, etc.) so long as (a) they are also at 300 K, and (b) they do not contain any stored energy and (c) they do not sublimate into gaseous form. (For example, you cannot add a battery or cheese, or a match, or a moose, or antimatter. You cannot add highly compressed, solidified nitrogen or any highly compressed gas [for two reasons, one because there is stored energy, and another because it could become gaseous].) The solid items must remain in their original, solid form. Also, the items must fit inside the box; it's OK to increase the size of the box before you put them in though, so they'll fit.
• You can slowly move items around within the box (slow enough not to cause friction/frictional heating though). In other words, you are allowed to move items closer to or farther away from the detector.

So what, if anything, can you do to increase the average photon power flux hitting the detector? Explain your answer.

Bonus question: how does this compare to the photon power flux leaving (emitted by) the surface of the detector?

As I understand your scenario, some percentage of the 3.5 micrometer photons emitted by the walls of the 300k box which reach the active area of the detector can produce electrons and electron holes in the detector which can then in turn create current in the detector’s external circuit.

When the detector itself emits 3.5 micrometer photons into the box, these emissions don’t result in current in the detector’s external circuit.

I reason that just by being in the box that has 300k maintained walls, some electrical current is being generated in the detector’s external circuit with no other items in the box, just from the 3.5 micrometer emissions of the 300k walls of the box.

 

[collinsmark]

As I understand your scenario, some percentage of the 3.5 micrometer photons emitted by the walls of the 300k box which reach the active area of the detector can produce electrons and electron holes in the detector which can then in turn create current in the detector’s external circuit.

When the detector itself emits 3.5 micrometer photons into the box, these emissions don’t result in current in the detector’s external circuit.

I reason that just by being in the box that has 300k maintained walls, some electrical current is being generated in the detector’s external circuit with no other items in the box, just from the 3.5 micrometer emissions of the 300k walls of the box.

Well, let's not jump to conclusions just yet about the current in the circuit, just yet.

If you wanted increase the photon power flux hitting the detector, how would you go about doing so? Would changing the size of the box make a difference? Would adding any new items (within the restrictions) make a difference? Would rearranging the items within the box make a difference? If so, how?

In other words, suppose you wanted to make the photon power flux on the detector in the box comparable to that with the previously discussed IR laser. How would you go about doing that, given the restrictions? Is it even possible?

(Edit: hint: What if you wanted to reduce the photon power flux on the detector in the box. How would you go about doing that, given the restrictions? Is that even possible? Given the restrictions, is it possible to change the photon power flux at all?)

(Another edit: another hint: What if you were to plaster all the inside walls of the box, from wall to wall, with HgCdTe IR photodetectors. How would that affect the photon power flux on the original detector? Would the original detector generate current? Would all the detectors generate current? Or would it make no difference whatsoever regarding the photon power flux?)

 

[Devin-M]

Well, let's not jump to conclusions just yet about the current in the circuit, just yet.

If you wanted increase the photon power flux hitting the detector, how would you go about doing so? Would changing the size of the box make a difference? Would adding any new items (within the restrictions) make a difference? Would rearranging the items within the box make a difference? If so, how?

In other words, suppose you wanted to make the photon power flux on the detector in the box comparable to that with the previously discussed IR laser. How would you go about doing that, given the restrictions? Is it even possible?

Lets suppose the detector is pointed towards a distant wall. To increase the photon flux on the detector I put a telescope in front of the detector and move the detector + scope closer to the distant wall.

 

[Drakkith]

Lets suppose the detector is pointed towards a distant wall. To increase the photon flux on the detector I put a telescope in front of the detector and move the detector + scope closer to the distant wall.

There's no need to move it closer. If the wall already fills the FoV of the detector then moving it closer does nothing. You're getting closer, sure, but you're viewing a smaller portion of the wall and the two simply cancel each other out.

 

[Devin-M]

Well if I put a curved mirror that reflects 3.5 micrometer light near the detector I can take photons that wouldn’t otherwise intersect with the active area of the detector and redirect them into intersection with the active area of the detector (before they hit the opposite wall).

 

[collinsmark]

Well if I put a curved mirror that reflects 3.5 micrometer light near the detector I can take photons that wouldn’t otherwise intersect with the detector before hitting the opposite wall and redirect them into intersection with the active area of the detector.

For basically the same reason that @Drakkith mentioned in the last post, that wouldn't make a difference either.

You might have missed an bonus hint in my last post, so I'll repeat it here:

What if you lined the inside sides of the box, from wall to wall, with HgCdTe IR photodetectors. In other words, the inside walls of the box are now made of photodetectors, all at 300 K. How would that affect the photon power flux on the original detector? Would the original detector generate current? Would all the detectors generate current? None of them? Would it make no difference whatsoever regarding the photon power flux, compared to the empty box?

 

[Drakkith]

One thing this thread has taught me is that thermodynamics is hard.

 

[Devin-M]

For basically the same reason that @Drakkith mentioned in the last post, that wouldn't make a difference either.

Just think of the James Webb Telescope mirror… you get more IR photons onto the sensor with the mirror than if you take the primary mirror away or make it smaller.

 

[collinsmark]

Just think of the James Webb Telescope mirror… you get more IR photons onto the sensor with the mirror than if you take the primary mirror away or make it smaller.

But the JWST is surrounded by both hot things (patches of angular area with stars, galaxies) and cold things (patches of angular area with no stars and no galaxies).

If JWST was in a box, and everything inside the box (including JWST's mirror itself) was all at the same temperature, the size or presence of the mirror wouldn't make a difference regarding the photon power flux reaching the instrumentation.

 

[Devin-M]

I thought the 3.5 micrometer photon count per second on the sensor’s active area would be directly proportional to the sensor’s surface area in this scenario. Adding the mirror effectively increases the effective surface area of the sensor.

Like with this RASA style telescope. If I want to collect as many photons as possible coming from a certain direction. I can either point the sensor towards the region emitting the photons, or I can point a large curved mirror towards the photon source with the sensor facing away from it as shown below. The latter option increases the photon count per second on the sensor and increasing the mirror size increases it further.

A solar concentrator works similarly.

 

[Drakkith]

Wherever you put the mirror it will block out part of the radiation that would have otherwise hit the sensor, so the net effect is that it does nothing. Same with a lens. For each bit of radiation that is bent towards the sensor, another bit is bent away from the sensor that otherwise would have hit it.

 

[Devin-M]

The active part of the PN junction is only on one side if the wafer so yes it may be receiving IR photons from all directions in the scenario but it’s more likely to generate current in the external circuit from received photons coming from preferred directions, ie the direction the sensor faces or the direction the mirror is pointing to reflect onto the sensor.

 

[Drakkith]

The active part of the PN junction is only on one side so yes it may be receiving IR photons from all directions in the scenario but it’s more likely to generate current in the external circuit from received photons coming from preferred directions, ie the direction the sensor faces or the direction the mirror is pointing to reflect onto the sensor.

This has nothing to do with what I said.

 

[Devin-M]

Yes the mirror will block IR photons coming from non-preferred (for current production) directions and enhance the IR photons coming from the directions which can generate current in the sensor’s external circuit. Only the IR photons coming in from one side of the wafer produce the useful current. So I propose to block the photons on the non useful side in order to enhance the photon count on the useful side.