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Old 13-07-2010, 02:08 AM
ericwbenson (Eric)
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Join Date: Sep 2009
Location: Adelaide, Australia
Posts: 209
Adding my two cents here:
If you examine the CCD equation and rearrange it a bit, you can make some observations about binning and other stuff. First the simplified equation for the SNR of square object (easier math!) spanning at least a few pixels in width (so we are above the critical sampling) recorded on a CCD:

SNR = S / sqrt(S + Nsky + Nd + npix˛*nr˛)

S: Signal [e-]
Nsky: sky signal [e-]
Nd: dark signal [e-]
nr: CCD read noise [e-/pixel]
npix: object digital width [pixel]

So we can expand the terms in the equation into other measurable/comparable factors using:

S = QE * t * A * w˛ * Cobj

QE: CCD average quantum efficiency [e-/photon]
t: time [sec]
A: telescope collection area [m˛]
w: object angular width [arc sec]
Cobj: average photon rate from object hitting the earth [photon/sec/m˛/arcsec˛], this is the related to object brightness quoted in [magnitudes/arcsec˛] with a conversion factor since flux is actually measured in Jansky's, we'll skip that part for now! Suffice to say that a 22 mag star = 14 photons/sec for every square meter of telescope area.

Nsky = QE * t * A * w˛ * Csky

Csky: average photon rate from sky background [photon/sec/m˛/arcsec˛], due to light pollution, moon, aurora and sky glow, again it is related to what a sky quality meter reads in [magnitudes/arcsec˛]

Nd = t * Cd * npix˛
Cd: average dark signal per pixel from heat [e-/sec/pixel]. This is a function of CCD temperature and usually halves for every ~6 deg. C. drop. For Kodak chips it is in the vicinity of 0.1 e-/pix/sec at -20C.

So the telescope and CCD parameters give us the image scale:
Image scale = 206 * p / f [arcsec /pixel]
p: CCD pixel size [µm]
f: telescope focal length [mm]

and now we can relate the object digital size to its angular size in the sky:
npix = w * f / (206 * p)

Introducing these factors into the original equation gives:

SNR = QE * t * A * Cobj * w˛ / sqrt[ (QE * t * A * w˛ * Cobj) + (QE * t * A * w˛ * Csky) + (t * Cd * npix˛) + (npix˛*nr˛)]

grouping terms and replacing npix

SNR = QE * t * A * Cobj * w˛ / sqrt[ (QE * t * A * w˛) * (Cobj + Csky) + (t * (w*f/(206*p))˛) * (Cd + nr˛/t)]

pulling out the t and w from the sqrt and dividing the nominator:

SNR = QE * sqrt(t) * A * Cobj * w / sqrt[ (QE * A * (Cobj + Csky) + (f/(206*p))˛ * (Cd + nr˛/t)]

So if you can ignore light pollution, dark current and read noise (perfect camera in outer space), telescope area, integration time, CCD QE, object brightness are all equally important as is the angular area of the object.
Pixel size only matters if QE*A*(Csky+Cobj) is small as compared to dark current and read noise. Of course pixel size matters very much for system resolution!
Therefore when the object is faint under a dark sky or when using a narrow band filter (where Cobj is not affected but Csky is greatly reduced) increasing the pixel size (i.e. binning) reduces the effect of read noise. N.B. the dark current noisee remains unchanged since binning 2x2 makes Cd 4x bigger!

Lets look at some real numbers:
Csky (naked eye limiting mag ~ 6.0) ~ 21 mag/arcsec˛ = 18 photons/sec/m˛ in green filter
Cobj (e.g. a 12th mag. galaxy 30"x30" in size) ~ 19 mag/arcsec˛ = 82 photons/sec/m˛ in green filter
A (for a 10" scope)= pi * 0.254˛ / 4 = 0.05 m˛
QE = 0.5
f = 2500 mm, p = 9 µm
Cd = 0.1 e-/pix
nr = 15 e-
exposure time t = 600 sec

Inside the sqrt we have:
0.5 *0.05*(88+14) + (2500/(206*9))˛ * (0.1 + 225 / 600)
0.025*100 + 1.8 * (0.1 + 0.375)
2.5 + 0.855
So the read noise is more important than the dark noise, but the shot noise from the galaxy itself dominates all other noise sources. In light polluted skies (18 mag/sq" = 300 photons/sec/m˛), the Csky term easily dominates above all else, making binning, pixel size, and to some extents cooling, sorta moot.

BTW Csky is easily measured if you know your CCD QE and gain [e-/ADU]. Take an exposure of a few minutes, record the pixel value in an empty area, subtract any software pedestal (MaxIm/CCDSoft add 100 ADU for technical reasons), multiply by the gain to get electrons, divide by QE, telescope area and exposure time and you get flux in photons/sec/m˛.

EB
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