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# Read Noise and Dynamic Range for a Newbie

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### #1 mccomiskey

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Posted 21 May 2019 - 09:51 AM

If this question has been covered elsewhere, I apologize, I was unable to find it.

I am trying to develop an understanding of the significance of the read noise of a CCD camera for the quality of an image.

I have read everything I can find, but there is am implicit logical leap in much of that material, that I need help making explicit.

Question1:

So far I understand that, simplistically, Full Well Capacity (Max signal) / Read Noise (Min signal) gives the effective dynamic range of a CCD.  More dynamic range is better, because it allows a finer recording of subtle differences in light intensity.  So, for example, with a FWC of 10,000 and a Read Noise of 10, there are 1,000 steps of dynamic range.  With a FWC of 10,000 and a read noise of 6e, there are 1,666 steps of dynamic range.

This is where I am having my conceptual difficulty.  I understand that the read noise establishes a floor for a signal (e.g. if read noise  = 10e, then one needs 11e to create a signal, and the dynamic range expressed in electrons would be 10,000-10 = 9,990.

I don't understand how that translates into a dividing the full well capacity by the read noise to determine steps of dynamic range.  There seems to be an assumption in this calculation that the size of each step of dynamic range of the sensor is equal to the read noise.

If read noise is 10e, is it not simply an adder to whatever number of electrons a photosite ends up capturing?  A photosite that captures 7,000e will read off with 7010e (on average) for example, while one that captures 600e will read off with 610e (on average).  In these terms, it seems an increase from 6e to 10e of read noise would be fairly de minimis.  However, expressed in terms of steps of dynamic range, the increase in dynamic range is 2/3rds by going from 10e to 6e.

Question 2:

What is the relationship between the read noise and the necessary exposure time?  I understand that generally, the advice is to expose for sub lengths sufficient to overwhelm read noise with sky noise, but how does that relate to the dynamic range of the camera?

Question 3:

If read noise is as important as it seems to be, then as between two similarly priced cameras with the same sensor and cooling capabilities (say the SBIG 16200 with 10e read noise and the FLI 16200 with 6e read noise) why wouldn't one always choose the camera wit h the lower read noise?  Note - not trying to start a conversation on the relative merits of these two brands (which both appear to be amazing), just trying to understand the theory here based on what I have been reading.

I know I must have a basic conceptual break here, so apologize if any of the questions above are wrongheaded, and appreciate any enlightenment this amazing community can bring!

### #2 ngc7319_20

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Posted 21 May 2019 - 11:23 AM

Some answers in BOLD.  I think most of the difficulties are in understanding the difference between adding a "signal of 10" to an image, and adding a random noise with a Gaussian sigma parameter (or standard deviation) of 10 to an image.  These are very different processes.

Question1:

So far I understand that, simplistically, Full Well Capacity (Max signal) / Read Noise (Min signal) gives the effective dynamic range of a CCD.  More dynamic range is better, because it allows a finer recording of subtle differences in light intensity.  So, for example, with a FWC of 10,000 and a Read Noise of 10, there are 1,000 steps of dynamic range.  With a FWC of 10,000 and a read noise of 6e, there are 1,666 steps of dynamic range.

Yes, this is sort of true.  But there are no discrete "steps" in the read noise.  It is a random noise, and can be characterized by a Gaussian distribution with a "sigma" of 10 or 6 whatever.

This is where I am having my conceptual difficulty.  I understand that the read noise establishes a floor for a signal (e.g. if read noise  = 10e, then one needs 11e to create a signal, and the dynamic range expressed in electrons would be 10,000-10 = 9,990.

No, it doesn't work that way.  It is not a hard "floor of signal."   The read noise is a random number that gets added to the signal.  It is not like "10e" gets added to the image everywhere.   It is more like a you rolled a pair of dice for every pixel, and added the numbers to the signal.  So some pixels get "2" added.  Some get "12" added.  Most get a number somewhere in between added.  It is random.

The description of "10e read noise" merely indicates  the noise is roughly like a Gaussian distribution with the "sigma" parameter of 10e.   But in an actual image one pixel might get a read noise of 2e added, another pixel might get 14e added, another might get 20e.    A rare few pixels might even get a noise of 58e or even 98e added.

A signal of your 1e would be completely lost in the 10e read noise.  Unless you took a very large number of images to average out the read noise...  The effective read noise of a stacked image is roughly (read noise) / square root (N) where N is the number of images stacked.  So to detect a 1e signal with 5 sigma certainty would need to have the average noise reduced below 0.2e, so you are looking at averaging around 2500 images.    10e  /square root(2500) = 0.2e.

I don't understand how that translates into a dividing the full well capacity by the read noise to determine steps of dynamic range.  There seems to be an assumption in this calculation that the size of each step of dynamic range of the sensor is equal to the read noise.

Again the read noise is a different random number that gets added to every pixel of the image.  Only is a statistical sense is the dynamic range equal to the (full well) / ( read noise).

There are no descrete steps in the read noise (except for electrons being quanta).

If read noise is 10e, is it not simply an adder to whatever number of electrons a photosite ends up capturing?  A photosite that captures 7,000e will read off with 7010e (on average) for example, while one that captures 600e will read off with 610e (on average).  In these terms, it seems an increase from 6e to 10e of read noise would be fairly de minimis.  However, expressed in terms of steps of dynamic range, the increase in dynamic range is 2/3rds by going from 10e to 6e.

No, the read noise is not adding "10e" to every pixel.  Again, some pixels get 20, some get 15, some get 13, some get 7.  It is all random.

Question 2:

What is the relationship between the read noise and the necessary exposure time?  I understand that generally, the advice is to expose for sub lengths sufficient to overwhelm read noise with sky noise, but how does that relate to the dynamic range of the camera?

This is more complicated.  And will depend on whether you need bright objects in the image to be accurately recorded.  If you want to properly record bright objects, you need to stay below the full well on the bright objects.  But this will mean short exposures -- so faint objects in the same image might be below the read noise.  This is exactly why you would want a camera with high dynamic range.

Question 3:

If read noise is as important as it seems to be, then as between two similarly priced cameras with the same sensor and cooling capabilities (say the SBIG 16200 with 10e read noise and the FLI 16200 with 6e read noise) why wouldn't one always choose the camera with the lower read noise?

There are other potential differences in the cameras, like full well capacity, quantum efficiency, spectral characteristics.  Also how long it takes to read out the image.  But yes, if all other things were equal, you would want the camera with lower read noise.  Example: say a camera had 100,000e full well and 10e read noise -- it would have more dynamic range than a camera with 6,000e full well and 6e read noise.  Which is preferable will depend on your application.

Sometimes a high full well capacity is more important than read noise.  If you were imaging a scene with a wide range in brightness, and needed to accurately record all the brightnesses, then full well is more critical than read noise.  On the other hand if you only cared about some faint nebula, and did not care that the bright stars were saturated, then read noise is more critical.

Note - not trying to start a conversation on the relative merits of these two brands (which both appear to be amazing), just trying to understand the theory here based on what I have been reading.

I know I must have a basic conceptual break here, so apologize if any of the questions above are wrongheaded, and appreciate any enlightenment this amazing community can bring!

Edited by ngc7319_20, 21 May 2019 - 11:50 AM.

### #3 BenKolt

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Posted 21 May 2019 - 11:37 AM

mccomiskey:

You are asking great questions.

I was in the middle of a lengthy response, but I see that somebody else just beat me to it.  I'll erase most of it now and just provide a couple of links for your perusal.

Nice link I've come across about the basics of CCD's including a nice discussion about dynamic range and bit depth: Starry Wonders Astrophotography

Here's an old thread that details many of the ways people determine optimal exposure time.  There are lots of factors involved.  Some methods simply use read noise to determine an optimum exposure, others also take into account sky fog, where others extend the consideration to include how long before stars are clipped.  The latter more or less includes dynamic range in the determination, at least implicitly.

I agree with the last poster's answer to Question 3.  There are a myriad of quantitative and measurable factors that ought to go into a camera purchase decision, where read noise is just one of them.  However, I would put lowest read noise high on that list, and all other things being equal, it would be my determining factor.

Best Regards,

Ben

### #4 bobzeq25

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Posted 21 May 2019 - 11:44 AM

If this question has been covered elsewhere, I apologize, I was unable to find it.

I am trying to develop an understanding of the significance of the read noise of a CCD camera for the quality of an image.

I have read everything I can find, but there is am implicit logical leap in much of that material, that I need help making explicit.

Question1:

So far I understand that, simplistically, Full Well Capacity (Max signal) / Read Noise (Min signal) gives the effective dynamic range of a CCD.  More dynamic range is better, because it allows a finer recording of subtle differences in light intensity.  So, for example, with a FWC of 10,000 and a Read Noise of 10, there are 1,000 steps of dynamic range.  With a FWC of 10,000 and a read noise of 6e, there are 1,666 steps of dynamic range.

This is where I am having my conceptual difficulty.  I understand that the read noise establishes a floor for a signal (e.g. if read noise  = 10e, then one needs 11e to create a signal, and the dynamic range expressed in electrons would be 10,000-10 = 9,990.

I don't understand how that translates into a dividing the full well capacity by the read noise to determine steps of dynamic range.  There seems to be an assumption in this calculation that the size of each step of dynamic range of the sensor is equal to the read noise.

If read noise is 10e, is it not simply an adder to whatever number of electrons a photosite ends up capturing?  A photosite that captures 7,000e will read off with 7010e (on average) for example, while one that captures 600e will read off with 610e (on average).  In these terms, it seems an increase from 6e to 10e of read noise would be fairly de minimis.  However, expressed in terms of steps of dynamic range, the increase in dynamic range is 2/3rds by going from 10e to 6e.

Question 2:

What is the relationship between the read noise and the necessary exposure time?  I understand that generally, the advice is to expose for sub lengths sufficient to overwhelm read noise with sky noise, but how does that relate to the dynamic range of the camera?

Question 3:

If read noise is as important as it seems to be, then as between two similarly priced cameras with the same sensor and cooling capabilities (say the SBIG 16200 with 10e read noise and the FLI 16200 with 6e read noise) why wouldn't one always choose the camera wit h the lower read noise?  Note - not trying to start a conversation on the relative merits of these two brands (which both appear to be amazing), just trying to understand the theory here based on what I have been reading.

I know I must have a basic conceptual break here, so apologize if any of the questions above are wrongheaded, and appreciate any enlightenment this amazing community can bring!

Question 2.

One would like read noise to be significantly lower than sky noise, to "bury the read noise in the sky noise".  Too many short exposures, and read noise becomes prominent.  Too few long ones and dynamic range suffers.

The obvious peak in the histogram is used as a surrogate for sky noise.  You convert it to electrons and compare it to the read noise.  People differ about the extent of burial.  Common goals are to have the peak equal to or greater than either 20 X RN, 5 X RN^2, 10 X RN^2.  Those goals may be selected or altered based on the dynamic range issues for the particular target.  It's an art.

There's a much more extensive, and hence better, explanation in this superb book.

https://www.amazon.c...h/dp/1138055360

Question 3.

All else being equal, lower read noise is better.  All else is never equal, read noise is just one of several considerations.

This is a complicated business, and each of your questions is deserving of some study on your part.  You'll never get everything you need from short posts here, it just can't be done.

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Posted 21 May 2019 - 01:06 PM

So far I understand that, simplistically, Full Well Capacity (Max signal) / Read Noise (Min signal) gives the effective dynamic range of a CCD.

As you probably know, dynamic range is usually expressed in dB or in f-stops, not as full well / read noise ratio - even though the ratio seems more intuitive.  Here's how to calculate dB and f-stop:

dB = 20 * log10(full well / read noise)

Steps on a calculator:  full well / read noise --> press log --> multiple by 20 to get dB

f-stop =  log2(full well / read noise)   ....    alternatively:  full well / read noise = 2^f-stop

Use a calc like this that uses log base 2  --> type in the result of full well / read noise and hit enter to get f-stop

### #6 jerahian

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Posted 21 May 2019 - 01:08 PM

First off, post #1!!  Welcome to CN

I totally understand your confusion.  It took me a while to wrap my mind around it, and I hope I can explain it appropriately.

1) Dynamic range (DR) is a measurement of signal to noise.  Basically, it represents the maximum signal to noise ratio.  I had read somewhere that the noise in the DR calculation is not just the read noise from the sensor, but the read noise from the system as a whole, which I believe can be captured by a dark frame (of 0 gain for the highest DR).  Obviously, the exposure duration of the dark frame will impact the read noise, as would be expected from heat.  To calculate this as an example, I have performed the following for my setup (ASI1600MM-COOL):

Given the ZWO specs:

At 0 gain, the FWC is 20,000 e-

At 0 gain, the DR range is 12.44 in stops (2^12.44 = 5557)

At 0 gain, the read noise (RN) is ~3.6 e-

FWC/RN = 5556 SNR

Opening up my 0 gain, 240s master dark in PixInsight, I calculate the SNR (using one of Herbert Walter's PI scripts Script > Image Analysis > SNR) to be 1505, which solves to a 13.29 e- RN (20,000/1505), effectively 10.56 stops.  I'd love for someone to correct me if I'm on the wrong path here, but it shows the "system" read noise is much higher than the isolated sensor read noise.  Sadly, I don't really know how to translate this to exposure times or to do anything useful with it for that matter!  I just tried to understand the maths behind it (hopefully correctly), to get a better clarity of what is what.

2) That said, check out this recent video by Dr. Robin Glover titled "Deep Sky CMOS Astrophotograhy".  It goes into some detail regarding SNR and exposure times.  I found this presentation to be enlightening.

3) I will leave this one for someone else, but my quick thought as to why one would choose a camera with higher read noise might be the cost of the systems and the comfort level with a less than "perfect" solution.

-Ara

### #7 Peregrinatum

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Posted 21 May 2019 - 01:24 PM

The only thing that gave me clarity on this issue was just getting data myself using my setup... here is a plot of Ha exposure times versus the mean adu of the image, and the peak star adu as a way to measure clipping.  The image was M101 in 19.2 moonless skies (not a bright or faint target) So from here I can select a "target adu" and get an idea of how much clipping would occur.  Target adu could be arrived at in several ways:  20*RN, 10*RN^2 etc. as was posted earlier.  Adjustments to the target adu and exposure times can be made depending on the sky brightness and/or the target brightness.

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### #8 Jon Rista

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Posted 21 May 2019 - 01:57 PM

If this question has been covered elsewhere, I apologize, I was unable to find it.

I am trying to develop an understanding of the significance of the read noise of a CCD camera for the quality of an image.

I have read everything I can find, but there is am implicit logical leap in much of that material, that I need help making explicit.

Question1:

So far I understand that, simplistically, Full Well Capacity (Max signal) / Read Noise (Min signal) gives the effective dynamic range of a CCD.  More dynamic range is better, because it allows a finer recording of subtle differences in light intensity.  So, for example, with a FWC of 10,000 and a Read Noise of 10, there are 1,000 steps of dynamic range.  With a FWC of 10,000 and a read noise of 6e, there are 1,666 steps of dynamic range.

This is where I am having my conceptual difficulty.  I understand that the read noise establishes a floor for a signal (e.g. if read noise  = 10e, then one needs 11e to create a signal, and the dynamic range expressed in electrons would be 10,000-10 = 9,990.

I don't understand how that translates into a dividing the full well capacity by the read noise to determine steps of dynamic range.  There seems to be an assumption in this calculation that the size of each step of dynamic range of the sensor is equal to the read noise.

If read noise is 10e, is it not simply an adder to whatever number of electrons a photosite ends up capturing?  A photosite that captures 7,000e will read off with 7010e (on average) for example, while one that captures 600e will read off with 610e (on average).  In these terms, it seems an increase from 6e to 10e of read noise would be fairly de minimis.  However, expressed in terms of steps of dynamic range, the increase in dynamic range is 2/3rds by going from 10e to 6e.

So, first thing, you are correct that dynamic range (in discrete steps of usable information) is FWC/RN. This is saturation point over noise floor...read noise isn't exactly min signal, it is the noise floor. So, for a camera that has 20,000e- FWC and 3.5e- read noise, you have 20000/3.5 = 5714 discrete steps of information.

You can convert this into decibels, or stops:

dB = 20 * log(FWC/RN)

stops = log2(FWC/RN)

stops = dB/6

So you have 75.14dB or 12.5 stops of dynamic range with 5714 steps.

NOW, onto your difficulty. You need more than 11e- signal to overcome read noise of 10e-. This is actually key to why dynamic range is useful. It tells us that the noise floor represents a DISTRIBUTION or BAND that determines the size of each DISCRETE STEP of useful information. You need to double the band, at a bare minimum, to be able to discern a new discrete level of information. So, to really be able to separate another value from your noise floor of 10e-, you need the signal to be 20e-. The read noise around this signal will distribute around its mean about the same as the read noise will distribute around the bias offset. So the two...the bias offset with read noise distributed around its mean and this new signal with read noise distributed around its mean...have become "separated" and are now discretely identifiable as different levels.

Now, you can repeat this process...to discern the next level of signal, it would at a minimum need to be another 10e- stronger than the previous, so you have 30e- vs. 20e- vs. 10e-, and three discretely identifiable levels of signal. Etc. etc. up through the saturation point.

Some visual demonstration is probably necessary here, so I'll try to come back later and add some images to demonstrate how this works in practice.

Question 2:

What is the relationship between the read noise and the necessary exposure time?  I understand that generally, the advice is to expose for sub lengths sufficient to overwhelm read noise with sky noise, but how does that relate to the dynamic range of the camera?

So, while read noise defines the width of the bands of discretely discernible information, there is really more to it than that in an ACTUAL signal. An actual signal has other noises...dark current noise (minior in cooled cameras, could be significant otherwise) and shot noise (significant, eventually dominant). The relationship between read noise and signal is what helps you determine necessary exposure time. A signal must be strong enough that its intrinsic noise, its shot noise, is the primary driver of noise distributing around the mean signal level.

So if you have read noise of 10e-, the weakest signal in the image should be 3xRN^2, and at worst the mean background sky level should never be smaller than that, which means you would need your background sky signal to be 300e-! Now, more optimally, if you want to get better SNR on faint signals...then you would likely want your background sky mean to be somewhere around 500-1000e-! Actually getting a 1000e- signal could actually be quite difficult, especially with higher resolution imaging...so you can see the potential challenge here. In some cases, exposing long enough to get a 1000e- signal may simply be impossible, or at least impractical given other considerations (i.e. sub loss, airplane/sat/meteor trails, etc.)

This should in turn demonstrate some of the value of low read noise cameras. With only 2e- you would need only 12e- at a minimum and 40e- optimally. So it takes less exposure time per sub exposure to reach the optimal criteria. In fact, with very low read noise, it actually becomes much easier to reach the more optimal 10xRN^2 criteria.

Now, how does this all play into dynamic range? While swamping read noise is one part of the equation, that is the "left side of the histogram" or "dark signal" part of the equation. We also have the "right side of the histogram" or "bright signal" part of the equation...notably, stars. If you expose long enough to swamp the read noise optimally...what happens to your stars? It may be that you have enough dynamic range such that you can achieve 10xRN^2 without even getting close to clipping stars (very high DR)...or you may clip a few of them (moderate to high DR)...or you may clip a lot of them (low to moderate DR).

Dynamic range is the relationship between the noise floor, the saturation (clipping) point, and the degree of fineness that you separate that total signal range into. A camera with higher read noise, say 9e-, but a huge FWC (100,000e-) and very high dynamic range (13.2 stops) could be an extremely capable camera when used right. Used right, as in with sufficiently long exposures to swamp the read noise, which even though read noise is high with so much dynamic range (huge FWC) you don't necessarily have to worry too much about clipping stars...which might require more expensive, specialized equipment. A camera with lower read noise, say 2e-, but a more limited FWC (8200e-) and moderate dynamic range (12 stops) would still be a capable camera, but would also be easier to achieve more optimal results with, and would not necessarily require extremely expensive equipment. In other words...a lower read noise camera is much more accessible to a much wider range of astrophotographers. The cost of this accessibility is the increased possibility of some additional clipped stars (and, in practice there is also usually a reduction in bit depth as well, which will produce a less-fine signal...something that can be managed with technique.)

Question 3:

If read noise is as important as it seems to be, then as between two similarly priced cameras with the same sensor and cooling capabilities (say the SBIG 16200 with 10e read noise and the FLI 16200 with 6e read noise) why wouldn't one always choose the camera wit h the lower read noise?  Note - not trying to start a conversation on the relative merits of these two brands (which both appear to be amazing), just trying to understand the theory here based on what I have been reading.

I know I must have a basic conceptual break here, so apologize if any of the questions above are wrongheaded, and appreciate any enlightenment this amazing community can bring!

In the case of the two examples you shared, they are the same sensor, but different qualities of readout electronics. Ignoring cost for the moment...there is ZERO reason to choose a 10e- RN camera over a 6e- RN camera if it is the same sensor. You lose dynamic range, and suffer for it. So cost aside, that is easy.

The differences here come into play when you factor in cost. Now, I guess the SBIG vs. FLI is a poor example here...looks like they cost the same (maybe the SBIG integrates a filter wheel?) So, yeah...really, there is no reason to get the camera with 10e- read noise. Remember...we really need to square the read noise to understand its full impact. So you are really comparing 100e- to 36e-...

A better example may be comparing the QHY or Moravian 16200 cameras with the FLI 16200. These other two cost about a third less or more than the FLI, but have read noise more similar to the SBIG. In these cases...COST would be the biggest factor. I mean, \$8000 is a lot of money, and with the FLI you still need a separate filter wheel which is a couple thousand more, and if you can get the same sensor at 9-10e- PLUS the filter wheel for \$3999, then that is a significant factor. The increase in read noise may well be a totally acceptable tradeoff, especially if you have the ability to use exposures long enough to sufficiently swamp the read noise. I think the SBIG probably integrates a filter wheel as well, I think many of their cameras do...and if that is the case, then there would still be a cost savings with the SBIG over the FLI.

Anyway...if cost is of no concern, or if there is no difference in cost and overall quality, then I would say read noise is indeed a primary driver of which camera to buy. Lower read noise has distinct benefits, especially if the FWC is maintained (since lower read noise with the same FWC means higher dynamic range!)

Edited by Jon Rista, 21 May 2019 - 02:02 PM.

### #9 kathyastro

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Posted 21 May 2019 - 02:17 PM

So far I understand that, simplistically, Full Well Capacity (Max signal) / Read Noise (Min signal) gives the effective dynamic range of a CCD.  More dynamic range is better, because it allows a finer recording of subtle differences in light intensity.  So, for example, with a FWC of 10,000 and a Read Noise of 10, there are 1,000 steps of dynamic range.

The part of the quote highlighted in green is correct.  Dynamic range can be expressed as a ratio, or as decibels or as an f-stop range.  Expressed as a ratio, it is the simple division you stated: the dynamic range is 1000:1.

The part of the quote in red is incorrect.  The only STEPS involved are ADU units.  The number of steps in the dynamic range is 10000 - 10 = 9990.

### #10 Jon Rista

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Posted 21 May 2019 - 02:25 PM

The part of the quote highlighted in green is correct.  Dynamic range can be expressed as a ratio, or as decibels or as an f-stop range.  Expressed as a ratio, it is the simple division you stated: the dynamic range is 1000:1.

The part of the quote in red is incorrect.  The only STEPS involved are ADU units.  The number of steps in the dynamic range is 10000 - 10 = 9990.

He actually had it right. Dynamic range in terms of discrete steps is FWC DIVIDED BY READ NOISE. You subtracted the read noise from the FWC, you need to divide it. In which case, the dynamic range is 1000 steps.

If you take 10000 ADU minus 10 ADU (which wouldn't be a read noise count...it would have to be something like an offset), you are not calculating dynamic range. You are calculating digital levels, which is not the same as dynamic range. You may have 9990 digital levels, digital numbers, digital units...but this does not account for the effect of read noise on the discrete bands of information that can be discerned within that range of levels. Levels is noise-agnostic, dynamic range is noise-aware.

Edited by Jon Rista, 21 May 2019 - 02:25 PM.

### #11 ngc7319_20

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Posted 21 May 2019 - 02:25 PM

The only thing that gave me clarity on this issue was just getting data myself using my setup... here is a plot of Ha exposure times versus the mean adu of the image, and the peak star adu as a way to measure clipping.  The image was M101 in 19.2 moonless skies (not a bright or faint target) So from here I can select a "target adu" and get an idea of how much clipping would occur.  Target adu could be arrived at in several ways:  20*RN, 10*RN^2 etc. as was posted earlier.  Adjustments to the target adu and exposure times can be made depending on the sky brightness and/or the target brightness.

What is the horizontal axis in the plot?

### #12 Peregrinatum

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Posted 21 May 2019 - 02:28 PM

What is the horizontal axis in the plot?

exposure time in seconds

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### #13 kathyastro

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Posted 21 May 2019 - 04:50 PM

He actually had it right. Dynamic range in terms of discrete steps is FWC DIVIDED BY READ NOISE. You subtracted the read noise from the FWC, you need to divide it. In which case, the dynamic range is 1000 steps.

If you take 10000 ADU minus 10 ADU (which wouldn't be a read noise count...it would have to be something like an offset), you are not calculating dynamic range. You are calculating digital levels, which is not the same as dynamic range. You may have 9990 digital levels, digital numbers, digital units...but this does not account for the effect of read noise on the discrete bands of information that can be discerned within that range of levels. Levels is noise-agnostic, dynamic range is noise-aware.

Then what exactly, do these "steps" represent in the real world?  The number 1000 in the example is just a ratio.  If you are saying that there are 1000 "somethings" (i.e. "steps"), what are they?  I see only a continuum, not 1000 discrete anythings.

The only thing I can see that comes in discrete steps is the ADUs, hence subtraction, rather than division, to count them.

Edited by kathyastro, 21 May 2019 - 04:50 PM.

### #14 freestar8n

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Posted 21 May 2019 - 05:16 PM

If every measurement is fundamentally limited by read noise (at least) then measurements close to each other are effectively providing the same value.  So there aren't 1000 discrete steps - but there are roughly 1000 distinguishable values you can measure.  Anything finer would be ambiguous and over-specified.

So I agree there aren't literally 1000 discrete steps - but that is sort of the idea.  I would just say the dynamic range is 1000 - by definition.

If you have a 16-bit ccd and the read noise amounts to the lower 4 bits - those bits don't really carry any information - and the steps they represent don't really say anything about relative signal strength since noise is overwhelming the precision.

Frank

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### #15 Jon Rista

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Posted 21 May 2019 - 05:25 PM

Then what exactly, do these "steps" represent in the real world?  The number 1000 in the example is just a ratio.  If you are saying that there are 1000 "somethings" (i.e. "steps"), what are they?  I see only a continuum, not 1000 discrete anythings.

The only thing I can see that comes in discrete steps is the ADUs, hence subtraction, rather than division, to count them.

I am hoping to actually create some visuals to explain, as it is tougher to explain with words. This is one of those cases where I think a picture is worth a thousand words, basically.

Note that it is a conceptual thing, these "discrete steps of information", and has to do with how noise limits the certainty of an estimate or a measurement as the differences between values encroaches on the distribution of the noise around them. Note that I use the term "information" here very explicitly, because I am not talking specifically about digital units or levels, I'm talking about information in a noisy data set...the concept of said information.

Edited by Jon Rista, 21 May 2019 - 06:48 PM.

### #16 kathyastro

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Posted 21 May 2019 - 05:27 PM

If you have a 16-bit ccd and the read noise amounts to the lower 4 bits - those bits don't really carry any information - and the steps they represent don't really say anything about relative signal strength since noise is overwhelming the precision.

Ah, that makes it clear.  Thanks.

### #17 mccomiskey

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Posted 22 May 2019 - 01:00 AM

My sincere thank you to everyone who has responded to my original post.  I have been reading the forums here for a couple of years as preparation for (re)discovering astronomy and starting astrophotography, and have always been impressed at the expertise and generosity of the forum members here.

There is a lot in the above, and it is helping me to better understand some of the questions I have been grappling with.

To try to synthesize (and ignoring shot noise, dark noise, and sky noise for the moment): read noise is, as its name suggests, random noise that creates uncertainty around the signal being measured.  This uncertainty manifests as additional electrons that add to the signal.  The number of additional electrons is itself uncertain, but can be statistically characterized by a distribution.

For a camera with a reported read noise of 10e, the central point of the distribution is 10e.  An implication is that any readings that are separated by less than 10e cannot reliably and consistently be distinguished from each other, and the minimum separation required  to be able to say with confidence that one reading is different from another is the 10e read noise figure.

So, for a sensor with a FWC of 10,000, if the read noise is 10e, and this establishes the minimum separation between electron counts needed to distinguish one photosite level from another, there are a maximum of 1,000 distinct different electron levels that the sensor is capable of resolving?  And a sensor with 6e would be cable of distinguishing 1,666 different levels, so would offer a 66% increase in the different levels of brightness it could record?

Is this correct?  Is there an analogy here to the Raleigh Criterion for resolution?

I still need to work through the implications of read noise on optimal exposure time, though from the above, it would seem that a lower read noise offers a similarly large decrease in required exposure time?

### #18 ngc7319_20

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Posted 22 May 2019 - 02:25 AM

More comments in BOLD:

For a camera with a reported read noise of 10e, the central point of the distribution is 10e.  An implication is that any readings that are separated by less than 10e cannot reliably and consistently be distinguished from each other, and the minimum separation required  to be able to say with confidence that one reading is different from another is the 10e read noise figure.

Still a few issues here. The read noise is not the average value or central point.  Typcially the signal from the detector is on a so-called bias level or voltage, and the read noise is +/- this bias level.  Here is an illustration of a so-called bell curve which describes a Gaussian statistical distribution.

This plot represents the histogram of pixel values you would get for a bias frame.  That is a calibration frame with the shutter closed and zero duration.  So all it contains is the bias voltage and read noise.  In this plot the read noise is the represented by "sigma", and is the WIDTH of the distribution or standard deviation of the noise distribution.  The "central point" as you call it is represented by the greek "mu" symbol or u, and it is the bias voltage of the detector.

So the "read noise" is the WIDTH of the noise distribution.  Not the central point.

An implication is that any readings that are separated by less than 10e cannot reliably and consistently be distinguished from each other, and the minimum separation required  to be able to say with confidence that one reading is different from another is the 10e read noise figure.

For any one image, yes, readings less that 10e cannot be reliably distinguished.  But you could average many images get hence reduce the effective noise.

Two signals that differ by the read noise generally would not be regarded as a significant difference.  In the noise distribution there is a 68% chance (34% x 2) the signals are really different when they differ by 1 sigma.  Usually that is not enough to be regarded as reliable.  For a "reliable" detection scientist would like to see a 3 sigma difference or more between the signals -- where the probability is more than 99% that it exceeds the noise.

So, for a sensor with a FWC of 10,000, if the read noise is 10e, and this establishes the minimum separation between electron counts needed to distinguish one photosite level from another, there are a maximum of 1,000 distinct different electron levels that the sensor is capable of resolving?  And a sensor with 6e would be cable of distinguishing 1,666 different levels, so would offer a 66% increase in the different levels of brightness it could record?

The terminology "distinct different electron levels that the sensor is capable of resolving" is not really appropriate.  There is nothing I would call "distinct levels" here.  It is all just statistical fuzz.  The full well is 10,000e.  And the noise distribution has a width sigma of 10e.  And this is all there is.  The ratio of full well / (sigma) is sometimes called the "dynamic range" and is 1000 in this example.  There are not 1000 distinct levels of anything.  It is just a ratio of two numbers.

Is this correct?  Is there an analogy here to the Raleigh Criterion for resolution?

I suppose there is a rough analogy.  Read noise is like the resolution of the telescope.  If two objects are closer than the resolution, you would not say they are distinct.  Similarly, if two signals differ by less than the read noise, you would not say they are different.  It is using the width of the telescope resolution as an analogy for the width of the read noise distribution.

I still need to work through the implications of read noise on optimal exposure time, though from the above, it would seem that a lower read noise offers a similarly large decrease in required exposure time?

Yes.  Reducing the read noise by say half reduces the exposure time by half -- at least for a single exposure of a very faint target (brightness comparable to the read noise).  That is the exposure time to reach some signal-to-noise ratio.  (For a bright target the shot noise must also be considered. And we are ignoring the noise from the sky background.)

If you are taking multiple exposures, then the improvement is even greater.  That is because each exposure or "read" of the detector comes with a dose of "read noise."  So if you need four 60 sec. exposures of a very faint target with a read noise of 10e, you would only need one 60 sec exposure if the read noise was magically reduced to 5e.  So read noise is pretty important.

Edited by ngc7319_20, 22 May 2019 - 02:29 AM.

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### #19 bobzeq25

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Posted 22 May 2019 - 09:29 AM

I still need to work through the implications of read noise on optimal exposure time, though from the above, it would seem that a lower read noise offers a similarly large decrease in required exposure time?

It's a bit more complicated than that.  Assume all other parameters are held constant.  If your goal is 10 X RN^2, then a decrease in RN from, say 10 to 5 will give an exposure reduction of 1/4.  If your goal is 20 X RN, that decrease in RN will give a change of 1/2.

But all other things are never constant.  The quantum efficiency of the two cameras is likely different.  Those goals are subject to modification to handle the dynamic range of a specific target.  Even sky brightness changes night to night, and, in light polluted skies, the variation due to altitude can be very large.

Two bottom lines.  I always check my exposure on real data from the target when I'm shooting, in real time.  While I may have a ballpark number from previous experience, I don't try to handle this theoretically.  But, rather, operationally.

Again, I recommend The Astrophotography Manual as _far_ superior to any number of short posts here.  This is complicated and any short posts leave a lot out.  Mine are intended just to get you started.

Doing AP well involves doing research/study.  Cookbooks are just not a good alternative.

Edited by bobzeq25, 22 May 2019 - 09:36 AM.

### #20 mccomiskey

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Posted 22 May 2019 - 01:51 PM

Hubble,

thanks for your several replies. Just to be clear, I am not looking for a cookbook nor attempting to avoid study or research. On the contrary I enjoy the process of reading and learning about Astrophotgraphy.   I Have read the Woodhouse book, which I agree is excellent.  I have read the sensors and exposure chapter several times. I have also read the relevant sections of the equally good book by Thierry Legault, and the Deep Sky imaging primer.  I have tracked down Craig Stark’s papers on signal and noise.  I can run the equations and see that higher read noise increase the minimal exposure time needed on a square function basis.

What I have not been able to get from reading these texts, and the reason I have turned to the Cloudy Nights community for help, is a conceptual understanding of  the relationship between read noise and dynamic range. I will continue to try to find explanations,  but was hoping that someone who had trodden this path before me might be able to provide guidance, as while I can manipulate the equations, I have hit a (temporary) wall in trying to develop an intuitive understanding of the concepts.

As to all other things being equal, I chose the SBIG and FLI cameras on purpose.  They use the same sensor, have the same well depth, roughly the same cooling capabilities and are roughly the same price.   Given these similarities, and the impact that a read noise difference of 10e to 6e seems to have when I run through the various equations, it struck me as curious that the FLI would not always be the default choice.  That caused me to question my understanding of this topic, hence the post.

### #21 Preston F

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Posted 23 May 2019 - 04:50 PM

I wouldn't define Dynamic Range as "steps of read noise".

If anything is a "step" its the Analog to Digital converter (see quantization noise).

The camera's gain determines the conversion from e to volts, which is then converted to digital number by an analog to digital converter.

Typically, mention of "volts" is never seen, and the gain is expressed in e/ADU (Electrons per Analog-to-Digital Unit).

Dynamic range is the ratio of the brightest non-saturated object in your filed of view to the dimmest object with an SNR of 1.

This is defined at a gain level where the FWC does not exceed the range of the ADC.

If you increase the gain, you will reduce the dynamic range because now the FWC doesn't matter, it hits the ADC limit first.

Very simple example to answer question #2:

Assume that your cooling is so good you have no dark current noise

FWC : 10000e

bright star = 1000e/sec

dim star = 1e / sec

Take a picture for 1 second exposure time:

bright star = 1000e + 10e read noise

dim star = 1e + 10e read noise

Take a picture for 10 seconds exposure time:

bright star = 10000e + 10e read noise

dim star = 10e + 10e read noise

Take a picture for 100 seconds exposure time:

bright star = 100000e (Saturated)

dim star = 100e + 10e read noise

To answer question 3:

You may not care about read noise as much for long exposures.  The camera you used as an example (SBIG 16200)  says it has a dark current of 0.1e/sec at 0C.

So for a 300sec exposure (5min) you would have an average of 30e of dark current, with sqrt(30) ~5.5e of dark current noise.

That is considered very low, uncooled cameras are much higher.

And then there is sky glow, which will easily exceed the read noise for long exposures when you have some light pollution.

Edited by Preston F, 23 May 2019 - 06:13 PM.

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### #22 ks__observer

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Posted 24 May 2019 - 05:19 AM

Reducing the read noise by say half reduces the exposure time by half -- at least for a single exposure of a very faint target (brightness comparable to the read noise).  That is the exposure time to reach some signal-to-noise ratio.  (For a bright target the shot noise must also be considered. And we are ignoring the noise from the sky background.)

If you are taking multiple exposures, then the improvement is even greater.  That is because each exposure or "read" of the detector comes with a dose of "read noise."  So if you need four 60 sec. exposures of a very faint target with a read noise of 10e, you would only need one 60 sec exposure if the read noise was magically reduced to 5e.  So read noise is pretty

Swamp formula is based on RN^2, so reducing RN by half obviously reduces exposure by 1/4.

You suggest integration time is reduced by reducing read noise.  It is not as direct as you say.

It just a matter of plugging numbers into the SNR formula:

SNR = Target/sqrt (target + sky + RN^2)

I like your bias guass visual.

Edited by ks__observer, 24 May 2019 - 05:19 AM.

### #23 gregj888

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Posted 27 May 2019 - 01:35 AM

mccomiskey,

I'll throw in another wrinkle... subs.  With higher read noise, subs would really effect the final SNR.  With CMOS and RN < 3e and <1e not uncommon, adding frames to get a greater dynamic range can be pretty effective.

I'll also admit that I cheat.  I use a CCD calculator,target and desired SNR to get my exposure and number of subs.  The one I use also figures in Sky noise (scintillation, shot, sky glow)  so is pretty complete.

### #24 spokeshave

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Posted 27 May 2019 - 08:44 AM

One of my pet peeves about how we discuss read noise is that IMO read noise should be more properly stated as electrons per square micron rather than electrons per pixel. This makes for a simpler and a more correct comparison of read noise. For example, at first blush, a sensor with 3.8u pixels with read noise of 2e- sounds much less noisy than a sensor with 10e- of read noise with 9u pixels. But that can be misleading. What's really important is how read noise affects SNR on the pixel level. Assuming that both sensors have the same QE, the signal acquired will vary with the square of the pixel area. The 3.8u pixel has an area of 14.44 square microns. The 9u pixel has an area of 81 square microns. So, for any given exposure, the 9u pixel with collect 5.6 times as many photons as the 3.8u pixel. Interestingly, the read noise for the 9u pixel is only 5 times that of the 3.8u pixel. So the 9u pixel will actually have a better SNR than the 3.8u pixel despite having massively more read noise. That's because it has slightly more massively more collection area than the 3.8u pixel. So the noisy 10e- RN sensor will actually produce a slightly "quieter" image than the 2e- RN sensor for the same exposure.*

One can look at this relationship by calculating the read noise per square micron of the pixel. For the 3.8u pixel the read noise per square micron is 0.139e-. For the 9u pixel, it is 0.123e-. As you can see, the 9u pixel actually has less read noise per pixel area than the 3.8u pixel, assuming equal QE. I'll call this "intrinsic read noise".

So how does "intrinsic read noise" compare to dynamic range?  Well, if you think about it, you'll recognize that FWC scales as the square of the pixel dimension and since:

DR = FWC/RN, then:

DR ~ (pixel_dimension_squared)/RN

Which is nothing more than the inverse of the intrinsic read noise:

RNi = RN/(pixel_dimension_squared).

So, when you are comparing intrinsic read noise, you are, for all intents and purposes, comparing dynamic range. However, they differ conceptually. People often state that the DR is a representation of the maximum SNR. But when you look at intrinsic read noise, I think it starts to become obvious that it defines the advantage one sensor has over another at any signal level.

*Obviously, image scale is also a factor.

Tim

### #25 ks__observer

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Posted 27 May 2019 - 09:23 AM

Re RNi:

This makes me wonder about exposure time and swamp read noise.

Without running the numbers i took it at face value what people say that low read noise = short exposure.

But the time it takes sky photons to hit say 10RN^2 is not just function of RN^2 but also 1/(pix size)^2.

So why has it been said exposure time for higher RN CCD's are so long compared to CMOS?

People don't seem mention the pix size factor.

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