Infrared Detectors & The 2nd Law of Thermodynamics

[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.

 

[Drakkith]

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.

The mirror is in front of the sensor. Obviously it will block radiation coming from the preferred direction.

 

[Devin-M]

This telescope only has 1 mirror in the back. The mirror area is larger than the red sensor which is on the front pointing into the telescope. The sensor faces away from the source of the photons but the mirror faces towards it and focuses the photons by changing the angle of flight of some of the incoming photons.

 

[Drakkith]

This telescope only has 1 mirror. The mirror area is larger than the senor which is on the front. The sensor faces away from the source of the photons but the mirror faces towards it and focuses the photons by changing the angle of flight of some of the incoming photons.

Irrelevant. This isn't the setup we were discussing. Please stop posting pictures that serve no purpose.

 

[Devin-M]

In the setup we are discussing then I will increase the useful IR photons per second onto the sensor by increasing the area of the sensor.

 

[Drakkith]

You can do that, but it will accomplish nothing. You still won't get current.

 

[Devin-M]

You can do that, but it will accomplish nothing. You still won't get current.

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.

So do you disagree with Post #7?

 

[Drakkith]

No, and I don't think anything we've said contradicts Chestermiller's post. If everything is in equilibrium, detector, environment, etc, then you won't be getting any current. If not, then you might. You can add mirrors or change the size of the detector all you want, it won't change anything.

 

[Charles Link]

I'm still following this discussion, and I think @Chestermiller (post 7) needs to look more carefully at what you @Devin-M were asking. You don't get any photocurrent from a photodiode inside an enclosure when everything is at thermodynamic equilibrium. The participants in the discussion have been very patient, but don't seem to be having much luck in getting the concept to sink in that there is no mechanism for any power generation from the photodiode or solar cell when it is inside an enclosure with everything at the same temperature.

 

[Drakkith]

You don't get any photocurrent from a photodiode inside an enclosure when everything is at thermodynamic equilibrium.

Indeed, and the key here is equilibrium. Part of the difficulty here is that no one is quite sure exactly where that equilibrium is. Is it when everything is at the same temp? Should the sensor be just ever so slightly lower in temp or higher in temp given its ability to absorb a particular range of wavelengths very well? Does adding a mirror change the equilibrium? (my argument is no)

Like I said already, thermodynamics is hard. Especially when you get beyond black bodies and perfect reflectors and such.

 

[Devin-M]

I’m saying when an IR laser shines on the detector you get current in an external circuit. This means the detector is not in thermal equilibrium. The detector is not able to discriminate whether one of the IR photons it used to generate power came from the laser or was emitted by the black body spectrum of some room temperature water that happens to be nearby.

 

[Drakkith]

I’m saying when an IR laser shines on the detector you get current in an external circuit. This means the detector is not in thermal equilibrium. The detector is not able to discriminate whether on of the IR photons it used to generate power came from the laser or was emitted by the black body spectrum of some room temperature water nearby.

I don't think anyone ever doubted this.

 

[Devin-M]

Well if it’s possible one of the photons from the IR radiation from the black body spectrum of the nearby water created some power in the external circuit of the detector when the laser was on, then it should also be a possibility when the laser is turned off.

 

[Drakkith]

Well if it’s possible one of the photons from the ir radiation from the black body spectrum of the nearby water created some power in the external circuit, then it should also be a possibility when the laser is turned off.

Why?

 

[Devin-M]

Because the laser being on or off has no bearing on whether a single 3.5 micrometer photon from whatever source can produce an electron + hole pair thereby generating dc current.

 

[Charles Link]

But you don't get any photocurrent from the photodiode when the laser is off, and everything including the water is at 300 K. This is a claim you tried to make in the OP, and when we looked into it in more detail, we saw that the researchers that you had referenced did not make this claim either. They had a high temperature globar for their additional IR source, above the 300 K ambient. The 300 K ambient by itself, with water or no water, or any other material at 300 K, would not generate any photocurrent.

 

[Drakkith]

Because the laser being on or off has no bearing on whether a single 3.5 micrometer photon from whatever source can produce an electron + hole pair thereby generating dc current.

We already know the detector converts photons into electron-hole pairs, even under an equilibrium condition.

So again, why should the detector generate current while the laser is off?

 

[Devin-M]

The dc current from the laser is a quantum effect…. one photon pushes one electron into the external circuit. Why would identical photons be assumed to have different effects within the PN junction? You’re saying a 3.5 micrometer photon from one source pushes an electron into the dc load and an identical 3.5 micrometer photon from a different source doesn’t.

 

[Charles Link]

@Devin-M You need to study a book like Streetman's in detail=the section on photodiodes. It's been a number of years since I studied it in detail, but as I recall there are several things going on including electron-hole recombination, along with diffusion currents. When all of the various contributions to the junction current are summed, the part that generates the extra current seen as photocurrent is the extra photons, above the ambient, that are absorbed, generating minority carriers in the p-material (electrons) that get swept across the junction. It may be worthwhile for you to try to get a copy of Streetman's book and work through his calculations.

 

[Devin-M]

But if I have a detector and a cup of water floating through space (separated by vacuum), if even 1 IR photon from the water’s black body spectrum strikes the detector, that puts the detector out of thermal equilibrium, does it not?

 

[Drakkith]

Why would identical photons be assumed to have different effects within the PN junction? You’re saying a 3.5 micrometer photon from one source pushes an electron into the dc load and an identical 3.5 micrometer photon from a different source doesn’t.

I'm saying no such thing. In fact, I've said exactly the opposite more than once. I've already explained, in detail, my understanding of the situation. Now I'm asking you why you think the detector should generate a current, and your answer, so far, completely ignores statistics, thermodynamics, and most of the operational details of the detector. This in spite of those topics being brought up time and time again in the 100+ posts in this thread. That's not a good way to learn.

But if I have a detector and a cup of water floating through space, if even 1 IR photon from the water’s black body spectrum strikes the detector, that puts the detector out of thermal equilibrium, does it not?

No, in fact the reverse is true. If the detector received zero IR photons from the water then something would be out of equilibrium in some way.