Keith_McClary
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Oh, I thought @chemisttree was making a cheesy joke.DennisN said:I have heard about green filters for the Moon
Oh, I thought @chemisttree was making a cheesy joke.DennisN said:I have heard about green filters for the Moon
I will test it and check.chemisttree said:Does the lens exhibit any chromatic abberation in daylight photos?
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.Devin-M said: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.
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.Devin-M said: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.
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.Devin-M said: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)?
Devin-M said: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.
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…Drakkith said: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.
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.Devin-M said: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.Devin-M said: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…
Drakkith said: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.
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.Devin-M said: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.
Devin-M said: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?
A 300k (80F) black body emits some infrared between 3 & 4 micrometers, which is in the detection range of the photodetector.collinsmark said: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.
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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.
Should some of this side-discussion be split off into the Thermodynamics forum?Devin-M said: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.
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.Devin-M said:A 300k (80F) black body emits some infrared between 3 & 4 micrometers, which is in the detection range of the photodetector.
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That sounds like a good idea to me.berkeman said:Should some of this side-discussion be split off into the Thermodynamics forum?
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.collinsmark said:That sounds like a good idea to me.
berkeman said: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 said:What if we just post further comments in a new discussion?
berkeman said: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.
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 said:@Devin-M what were you shooting with?
I got this one of the core of Andromeda a couple days before with identical exposure settings...Drakkith said: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!
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.Devin-M said:I got this one of the core of Andromeda a couple days before with identical exposure settings...
The bottom of a bottle, my optics prof used to say.Drakkith said:You can get a decent picture with two tin cans and a string.
Drakkith said: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!
Was this taken from a bortle 2 site?Devin-M said:Restacked... now you can see the nebulas... I think something very bad happened during the 1st stacking attempt...
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 said:Was this taken from a bortle 2 site?
I don't use DSLR's, so I'm not familiar with ISO settings. Why shoot at 6400 instead of 400ish?Devin-M said:It’s not a cooled camera & shot at 6400iso instead of a more ideal 100-400iso.
Drakkith said:I don't use DSLR's, so I'm not familiar with ISO settings. Why shoot at 6400 instead of 400ish?
That is not my experience (I shoot with a D810). I always shoot with as low an ISO as possible (ISO 64) to maximize the dynamic range, especially color information. Shooting with higher ISO values only increases the amount of noise in my stacked image.Devin-M said:As I understand it, it narrows the dynamic range of the sensor but puts the minimum detected values within the range of detection of the 14bit RAW files (on the Nikon D800 body). (otherwise certain various analog-to-digital detectable values will be normalized to 0 on the raw files). Above 6400iso (as I understand it, on the D800), the improvements are entirely digital as opposed to improvements in the analog to digital conversion).
Andy Resnick said:That is not my experience (I shoot with a D810). I always shoot with as low an ISO as possible (ISO 64) to maximize the dynamic range, especially color information. Shooting with higher ISO values only increases the amount of noise in my stacked image.
But there are only so many brightness values you can store in the 14 bit raw file.Devin-M said:narrows the dynamic range of the sensor
A good rule of thumb for deep sky (not planetary) astrophotography is once you decide on an exposure time (usually determined by how long your equipment can maintain proper tracking, or how long, on average, you can go without a stray cloud messing things up), you should generally increase the gain (ISO) as high as you can go without saturating stars. This generally minimizes the read noise, measured in electrons.Andy Resnick said:That is not my experience (I shoot with a D810). I always shoot with as low an ISO as possible (ISO 64) to maximize the dynamic range, especially color information. Shooting with higher ISO values only increases the amount of noise in my stacked image.
collinsmark said:A good rule of thumb for deep sky (not planetary) astrophotography is once you decide on an exposure time (usually determined by how long your equipment can maintain proper tracking, or how long, on average, you can go without a stray cloud messing things up), you should generally increase the gain (ISO) as high as you can go without saturating stars. [snip]
Andy Resnick said:That's true- and with my setup, I saturate pretty quickly. Typically, the brightest stars within a field of view saturate between 6s and 20s exposure times (ISO 64).
Edit- I forgot to mention your comment "And there's almost no sane reason why would ever reduce your aperture.", because for me, there are at least 2 good reasons. First, (slightly) stopping down the aperture makes the images less susceptible to poor seeing conditions. Second, (slightly) stopping down the lens improves the images by decreasing aberrations.
Andy Resnick said:Edit- I forgot to mention your comment "And there's almost no sane reason why would ever reduce your aperture.", because for me, there are at least 2 good reasons. First, (slightly) stopping down the aperture makes the images less susceptible to poor seeing conditions. Second, (slightly) stopping down the lens improves the images by decreasing aberrations.