Aggregate.org publishes slides on ISO-less camera by Hank Dietz and Paul Eberhart from University of Kentucky presented at SPIE Electronic Imaging conference:
What is called ISO in a camera is a sensor gain, therefore as gain also amplifies noise and reduces the saturation capacity, the SNR must decrease with ISO and therefore the dynamic range.
Dark noise will be roughly proportional to the gain and photon shot noise at the saturation capacity will be roughly proportional to the square root of the inverse of the gain and saturation capacity is inversely proportional to gain.
If i take the basic DR formula for a linear sensor with perfect DSNU and PRNU correction then i get DR=(SAT_CAP-DARK_CURRENT*EXPOSURE_TIME)/SQRT(q*DARK_CURRENT*EXPOSURE_TIME+READ_NOISE^2) [read my book about dynamic range].
If we neglect the dark current component then roughly DR(gain)=DR(no_gain)/gain^2 or in dB DR(gain)[DB]=DR(no_gain)[dB]-2*Gain[dB]; or -12dB of DR each time the gain is doubled
I don't know the ISO stuff well but i assume that ISO 200 is twice the gain of ISO 100 and that the DR values are reported in f-stops or bits.
SONY A100 10.9 f-stops = 109 bits = 2^10.9 values = 20*log(2^10.9) = 65.62dB 10.2 f-stops = 61.41 dB (-4.2dB) 8.9 f-stops = 53.58 dB (-7.8dB) 7.7 f-stops = 46.35 dB (-7.2dB) 6.9 f-stops = 41.5 dB (-4.8dB) Average loss for gain step of 2 = -6dB !!!
Digital cameras are not linear, their response is an s-curve and the results have a very wide variation so it is hard to take a conclusion. Maybe there is so much processing involved that it does not becomes as bad as expected. Or maybe my understanding of the ISO setting is wrong or maybe the DR is not reported in f-stops.
Arnaud, I think that the DR drop at gain step of 2 goes from 3dB to 6dB, depending on amplifier architecture. The simplified definition of DR is highest detectable signal over lowest detectable signal.
Voltage domain: the highest detectable signal doesn't change due to gain while the noise level doubles. This reduces the DR by 6dB.
Charge domain: the max detectable charge reduces by half while the noise in electrons remains constant (the noise in volts doubles but so does the conversion gain, so same noise in electrons). Also in charge domain you get (as expected) -6dB in DR at gain of 2.
Now, if you use a gain amplifier directly connected to the pixel SF you get noise bandwidth reduction. The noise in volt now increases as sqrt(2). So the DR drop at gain step of 2 is now only 3dB..
The degradation is mainly generated by the system noise. Its influence decreases with gain. This gives ISO effect. The different ISO effects in these cameras reflect the system noise level since the source follower inside the pixel. -yang ni
I think it is all about digital gain ("ISO-less") vs analog gain (not "ISO-less"). If the read noise in electrons vs gain is fairly constant and quantization is not an issue (enough sampling bits) then there is no need for analog gain, digital gain can do the job with the additional benefit of a higher dynamic range.
What is called ISO in a camera is a sensor gain, therefore as gain also amplifies noise and reduces the saturation capacity, the SNR must decrease with ISO and therefore the dynamic range.
ReplyDeleteDark noise will be roughly proportional to the gain and photon shot noise at the saturation capacity will be roughly proportional to the square root of the inverse of the gain and saturation capacity is inversely proportional to gain.
If i take the basic DR formula for a linear sensor with perfect DSNU and PRNU correction then i get DR=(SAT_CAP-DARK_CURRENT*EXPOSURE_TIME)/SQRT(q*DARK_CURRENT*EXPOSURE_TIME+READ_NOISE^2) [read my book about dynamic range].
If we neglect the dark current component then roughly DR(gain)=DR(no_gain)/gain^2 or in dB DR(gain)[DB]=DR(no_gain)[dB]-2*Gain[dB]; or -12dB of DR each time the gain is doubled
I don't know the ISO stuff well but i assume that ISO 200 is twice the gain of ISO 100 and that the DR values are reported in f-stops or bits.
SONY A100
10.9 f-stops = 109 bits = 2^10.9 values = 20*log(2^10.9) = 65.62dB
10.2 f-stops = 61.41 dB (-4.2dB)
8.9 f-stops = 53.58 dB (-7.8dB)
7.7 f-stops = 46.35 dB (-7.2dB)
6.9 f-stops = 41.5 dB (-4.8dB)
Average loss for gain step of 2 = -6dB !!!
Digital cameras are not linear, their response is an s-curve and the results have a very wide variation so it is hard to take a conclusion. Maybe there is so much processing involved that it does not becomes as bad as expected. Or maybe my understanding of the ISO setting is wrong or maybe the DR is not reported in f-stops.
Arnaud, I think that the DR drop at gain step of 2 goes from 3dB to 6dB, depending on amplifier architecture. The simplified definition of DR is highest detectable signal over lowest detectable signal.
ReplyDeleteVoltage domain: the highest detectable signal doesn't change due to gain while the noise level doubles. This reduces the DR by 6dB.
Charge domain: the max detectable charge reduces by half while the noise in electrons remains constant (the noise in volts doubles but so does the conversion gain, so same noise in electrons). Also in charge domain you get (as expected) -6dB in DR at gain of 2.
Now, if you use a gain amplifier directly connected to the pixel SF you get noise bandwidth reduction. The noise in volt now increases as sqrt(2). So the DR drop at gain step of 2 is now only 3dB..
You are right, SAT_CAP is not affected by gain. Too quick in my calculations.
ReplyDeleteBut in the end the data shows a nice -6dB per ISO level
The degradation is mainly generated by the system noise. Its influence decreases with gain. This gives ISO effect. The different ISO effects in these cameras reflect the system noise level since the source follower inside the pixel.
ReplyDelete-yang ni
I think it is all about digital gain ("ISO-less") vs analog gain (not "ISO-less"). If the read noise in electrons vs gain is fairly constant and quantization is not an issue (enough sampling bits) then there is no need for analog gain, digital gain can do the job with the additional benefit of a higher dynamic range.
ReplyDelete