# Noise levels as a function of pixel size

Discussion in 'Digital Photography' started by Alfred Molon, Dec 13, 2005.

1. ### Alfred MolonGuest

Is there some formula or scientific basis to estimate noise levels as a
function of pixel size? At the moment I'm assuming that noise levels
decrease with the square root of the size increase, i.e. doubling the
pixel area should lower noise levels by the square root of 2. Is this
correct? I'm only referring to "pixel" noise, i.e. not read noise, A/D
conversion noise and other external noise sources.

Alfred Molon, Dec 13, 2005

2. ### David J TaylorGuest

The photon-limited noise is proportional to the square root of the number
of photons collected. So, if everything else is equal, a pixel which is
double the linear size will connect four times the number of photons, and
the noise will be twice as much. Therefore the signal to noise ratio will
be doubled.

Doubling the area will increase the signal-to-noise ratio by a factor of
square root (2), with the signal being doubled and the noise being root 2
higher.

David

David J Taylor, Dec 13, 2005

3. ### Daniel SilevitchGuest

Photon counting is a Poisson statistics problem, so the noise goes like
the square root of the signal. The signal/noise ratio thus goes like
1/sqrt(signal). Doubling the linear size of a pixel increases the area
(and hence the signal) by a factor of four. The noise only doubles, so
s/n doubles.

More generally, the s/n ratio scales with the linear size of the pixel.

-dms

Daniel Silevitch, Dec 13, 2005
4. ### Lourens SmakGuest

No. Try a Kodak 14n and a Canon 5D both at iso 1600, and you'll see.

Lourens

Lourens Smak, Dec 15, 2005
5. ### Scott WGuest

There are limits to noise performance base on pixel size. Not every
camera comes close to these limits, but no camera passes them.

Scott

Scott W, Dec 15, 2005
6. ### misoGuest

I don't see why this would be any different than low noise amplifier
design, so double the area and the SNR improves by 3db (linear square
root of two).

I don't know if cameras do this, but you could sample the sensor with
the shutter closed, then sample with the shutter opened. A certain
amount of noise will be cancelled. [Double correlated sampling.]

miso, Dec 15, 2005
7. ### WannabeSomeoneGuest

Yes, there is scientific basis to estimate noise levels as a function of
pixel size.

The intrinsic noise from the CCD or CMOS pixel is mainly a function of
thermal energy, and independent of pixel size. The signal is a result of
light hitting the pixel surface, which is a function of pixel size because a
larger pixel collects more light. Therefore a larger pixel will have a
higher signal to noise level. So if you get the same signal output, a larger
pixel will have less noise than a smaller pixel.

Wannabe
=======

WannabeSomeone, Dec 15, 2005
8. ### Tony PolsonGuest

And the 'full frame' 6 MP Contax N Digital has noise levels that are
even worse than the Kodak!

Far worse, in fact, even at low ISO.

;-)

Tony Polson, Dec 15, 2005
9. ### SMSGuest

See "http://clarkvision.com/imagedetail/does.pixel.size.matter/"

SMS, Dec 16, 2005
10. ### prepGuest

You have an upper limit on the well capacity, and that sets the best
possible SN. Increasing size increases well capacity if the doping
profiles are not changed, so that is a plus in high light conditions.
But the larger area pixel will collect more photons in low light as
well increasing SN there.
Nope. You now have 1/sqrt2 noise and 1/2 signal. You have removed any
dark current bias if you do it right though.

--
Paul Repacholi 1 Crescent Rd.,
+61 (08) 9257-1001 Kalamunda.
West Australia 6076
comp.os.vms,- The Older, Grumpier Slashdot
Raw, Cooked or Well-done, it's all half baked.
EPIC, The Architecture of the future, always has been, always will be.

prep, Dec 16, 2005
11. ### Kennedy McEwenGuest

Not quite. When you subtract the dark frame you have virtually no
effect on the signal, since all that is being removed is the dark
current, not half of the signal. However, you have an *additional*
noise source of equal value (2 uncorrelated frames) to the original
frame and thus increase the total noise to sqrt(2) of the original.
An estimate of SNT from your figures would give the correct effect, but

Kennedy McEwen, Dec 16, 2005
12. ### Kennedy McEwenGuest

Not quite. When you subtract the dark frame you have virtually no
effect on the signal, since all that is being removed is the dark
current, not half of the signal. However, you have an *additional*
noise source of equal value (2 uncorrelated frames) to the original
frame and thus increase the total noise to sqrt(2) of the original.
An estimate of SNR from your figures would give the correct effect, but

Kennedy McEwen, Dec 17, 2005
13. ### Dave MartindaleGuest

An aside: one way to reduce this extra noise is to take several dark
frames and average them. That reduces the noise *in the average dark
frame* by a factor of sqrt(N). With even a modest value of N, the
additional noise due to dark frame subtraction is reduced well below the
noise in the image itself.

I think this is common in astronomical imaging, but digital cameras only
use a single dark frame.

Dave

Dave Martindale, Dec 17, 2005
14. ### John FrancisGuest

Still not useful.

That technique works in astronomical imaging because what's being
averaged is a complete image, with both (deterministic) signal
and (random) noise.

If you take several (random) noisy dark frames, and average them
together, you'll end up with an average noise close to zero. But
that doesn't help you if you've only got a single frame with signal;
although you know the average noise is zero, that doesn't give you
a better estimate of the noise in your single image frame.

John Francis, Dec 17, 2005
15. ### Kennedy McEwenGuest

Exactly, and this is one reason why I was apparently a little pedantic
about Prep's reasoning - this would not be apparent from his reasoning,
however it is clear from the correct reasoning that if you reduce the
noise on the dark frame *before* subtracting it you can avoid degrading
the overall noise in the final result.

Kennedy McEwen, Dec 17, 2005
16. ### Kennedy McEwenGuest

True, it doesn't improve the noise in the single frame - nothing you do
subsequently can reduce that. However, using the technique Dave
suggested, you can correct for the dark current sourced noise without

We need to distinguish between different noise types to discuss this
sensibly. Misho used the term "noise" generically, however the approach
he suggested would improve the spatial noise (fixed pattern dark level
noise) at the expense of increased temporal noise (random noise changing
from frame to frame). The method Dave suggested avoids increasing the
random noise, thus permitting the spatial noise to be eliminated,
leaving only the temporal noise of the single image frame. This usually
results in a significant improvement in CCD/CMOS sensor sourced images.

Kennedy McEwen, Dec 17, 2005
17. ### JPSGuest

In message <dnvskj\$eop\$>,

JPS, Dec 17, 2005
18. ### Dave MartindaleGuest

Yeah, "noise" is too general a term. Fixed pattern noise is noise
because it interferes with the desired signal, but it's not noise
because it isn't really random (either spatially or temporally). Photon
and electron noise is random, and is noise without any argument.

Dave

Dave Martindale, Dec 18, 2005
19. ### JPSGuest

In message <do2do4\$nve\$>,
.... and will always be there, even if all other forms of noise can be
--

JPS, Dec 18, 2005
20. ### Roger N. Clark (change username to rnclark)Guest

And photon noise is the dominant noise source in any good
modern digital camera. At the lowest end, read noise
contributes.

But what we can hope to see is improvements in quantum efficiency.
A 2 to 3x improvement in quantum efficiency would help
the small size P&S sensors perform more like current
large pixel DSLR sensors. And the DSLR sensors would
be proportionally better too!

Roger

Roger N. Clark (change username to rnclark), Dec 18, 2005