Thursday, June 02, 2016
e2v Sapphire WVGA Sensor
e2v Sapphire EV76C541 WVGA CMOS sensor is said to be a viable replacement for traditional industrial CCD sensors due to an excellent performance in low-light conditions and an electronic global shutter.
Subscribe to:
Post Comments (Atom)
According to the announced data, it has an readout noise of 6.3e- in ERS mode and 19.6e- in GS mode. So, where's the "excellent performance in low light conditions"? It may be an oversimplified view, but readout noise is an important criterion for low-light applications...
ReplyDeleteFully agree. I've just quoted e2v site on this.
DeleteHi, can you tell me where you got your data from? I can't find anything about the readout noise in the datasheet. Thanks!
ReplyDeleteFrom dynamic range and Qsat numbers
DeleteAny videos of e2v sensors quality in low light?
ReplyDeleteIt never ceases to amaze me how these brilliant sensor designers are able to extract 66dB dynamic range out of a 10 bit ADC!
ReplyDeleteIf only they could turn their attention to perpetual motion machines, we'd lick the environmental crises within a decade!
This should not be a problem. An ideal 10b ADC has 1/sqrt(12) LSB quantization noise, which is about 0.29 LSB. So, ideally, a 10b ADC can reach over 71dB of DR in image sensor applications.
DeleteThis is different from the ADCs in signal processing applications where the full scale signal is sinewave rather than DC like in image sensors.
You use high gain for the noise floor and low gain for the maximum full well, then you get it: ))
DeleteVladimir, je suis assez skeptical in this noise estimation of ADC. Since this formula is true for communications system where the sampling rate is higher than signal rate. So the quantization noise energy can be statistically calculated in such way. But for image, the spatial frequency could be as high as half of the sampling rate. This noise level is true only when you want to distinguish a large uniform patch in an image, but it's certainly wrong if you want to distinguish high spatial resolution textures. The phenomena should be clearly visible on MTF test chart.
Delete-yang ni
Hi Yang, it depends on the nature of the ADC noise. Quite often, the ADC quantization noise is mostly limited by device noise and looks random. Even if it is not, pixel noise might have a dithering effect and make it more random.
DeleteYes, of course. But in this case, the noise can not be estimated by using 1/sqrt(12)LSB. That is what I would like to say. Thanks !
Delete-yang ni
The ADC noise value does not change its value, and still can be estimated as 1/sqrt(12), in the ideal ADC case. The pixel noise is just added to the ADC quantization noise.
DeleteBut, doing digital CDS, shouldn't she ADC quantization noise increase as well? Then 71dB DR with 10bit ADC should not be possible anymore? Thank's for this interesting post!
DeleteI'm not sure that they are doing a digital CDS. If yes, you are right, the ADC noise rises by 3dB.
DeleteVladimir... when was the last time you saw a spec of an "ideal ADC"? When I look at all the specs, they all have ENOB's that are 1 or 2 bits under the total bits. In other words, a 14bit ADC has ENOB of ~12.x bits. At least that is honest and gives a good idea of what kind of PRECISION can be expected from the device.
DeleteHowever, even if it WERE true that the on-chip e2v ADC was that much better than anything coming from the amateurs at Analog Devices, Linear Technologies or Texas Instruments, it's still disingenuous at best...
Imagine your physician ordering a blood test for you and he had to know with a precision of part in 2000 (66db) of the phosphorus level in your blood. The lab says "No problem, our equipment is precise to one part in 3500!"
And the blood test comes back with 5,500 on a scale of 1 to 100,000 with a little note at the bottom that claims " * measurement uncertainty +/- 50 counts".
in this scenario, what would be the value of the 0.29 LSB noise? and would you trust the test to keep going to this lab or betting your life on the results?
I recall the resolution vs ENOB issue has already been touched in the comments a couple of months ago. Please look at Intersil ADCs, such as these:
Delete8b physical, 7.99 ENOB:
http://www.intersil.com/content/dam/Intersil/documents/isla/isla118p50.pdf
10b physical, 9.8 ENOB:
http://www.intersil.com/content/dam/Intersil/documents/kad5/kad5610p.pdf
12b physical, 11.42 ENOB:
http://www.intersil.com/content/dam/Intersil/documents/isla/isla222s.pdf
It much depends on the definition of enob. In image sensors, INL can be quite relaxed as the pixel signal is quite nonlinear. Also, good DNL is required only at dark signal. So, the formula Vladimir listed above is a good approximation in image sensors.
DeleteI agree, it has been covered to death.
DeleteBut pointing to high quality stand alone ADCs with ENOB's that a just a shade under the physical number of bits does not in any indicate that an on-chip e2v ADC can come even close.
And it certainly does not at all show that the on-chip ADC can provide 11 Bits of DR.
E2v has put out some pretty impressive products recently
DeleteAfter some years of experience in the field you will understand why a 10 bit ADC can give you 11bit imaging DR. It was a shock for me to read about 14bit single-ended column ADs when I graduated. I've been always told by my professors that with single ended circuits you get max 9bits. But my professors were wrong. I've measured myself imaging DR 1 bit higher than ADC resolution.
DeleteCan you explain mathematically the reason of getting 1 bit higher than the ADC resolution?
DeleteMaybe i am wrong, but i totally disagree with the 10bit ADC can achieve 71 dB dynamic range story.
DeleteLet's first clarify the definition. The normal DR definition is the distance between the minimal detectable signal and the maximum allowed input signal. e.g. in sine wave input test case: from noise floor (total integrated in band noise power) to the largest sine wave (signal power = V^2/8 ) that one can apply. The largest sine wave amplitude is smaller than full scale input and normally chosen to be the one that gives jsut SNRmax or a few dB more when the SNRmax drops *dB( *: 0,1,3?).
The ADC dynamic range statement is strange. If you say noise floor==0.29LSB or LSB/12^0.5. I think you measured with sine wave input. But what is the max allowed input amplitude?. Rail to rail square wave? I don’t see how can one get 71dB output of 10 bit ADC. The DR theoretical limitation of this method is about 6.02*B+1.76 + *dB.
For image sensor ADC, our input is static, than we should use other way to define our minimal detectable signal and max input signal. I don't think one can detect something smaller than 1 LSB, thus I do think the minimal detectable signal is just 1 LSB and the max input level is the full input range. Therefore. The DR for image sensor ADC is as simple as 2^N. Unless you do oversampling…
Peng
See here:
Deletehttps://en.wikipedia.org/wiki/Quantization_%28signal_processing%29
"When the quantization step size is small (relative to the variation in the signal being measured), it is relatively simple to show that the mean squared error produced by such a rounding operation will be approximately Δ²/12.
Mean squared error is also called the quantization noise power."
Hi Peng. You are asking a tough and ambiguous question about the "minimal detectable signal." It boils down to the definition of what is detection. Is it enough just to say that the signal is present or not? If yes, an ideal ADC in a noiseless system can detect a signal well below LSB, such as 0.001 LSB or less. Here's how:
DeleteSay, we have an ideal 10b ADC, and set our image sensor analog black level at exactly 0.4999 LSB. Then, as long as there is no signal, the ADC output would be 0. If there is a signal of 0.001 LSB, the signal + black level would be just above 0.5 LSB and the ADC output would be 1. So, we can clearly see a difference between 0 and 1 and detect the signal of 0.001 LSB level.
Does this answer on your question?
Hi Vladimir, glad to read you comments. But sorry i could not agree with you. To my understand, the detectable means information can be identified from unknown "noise or black" background. Therefore, the example you give is not valid, the 0.4999LSB black level is known and given, therefore the 0.001 LSB can indeed be detected. I would consider this example as an known reference (offset) + compactor, i.e. 1-bit ADC. The 1-bit ADC is nonlinear system thus no DR.
DeleteMy example: Assuming that the difference (information) between reset and signal is 0.3 LSB. We should be very lucky to see a +1 in ADC output digital code compared with 0 LSB input or black background. Therefore, It is hard to say this particular piece of information is detectable if we only sample it once.
The only way I know that can get higher accuracy is rely on ADC oversampling and noise of the information itself (noise has to be significantly detectable by ADC and statistic plays a role here).
Peng
Hi Peng, let me clarify what I mean by the analog black level. This is the level that is intentionally added to the signal in the analog domain before the ADC to make sure that ADC does not start from zero when the signal is absolutely black. It's added for many reasons, primarily because of the non-linearity.
DeleteIndeed, if the real, noisy sensor ADC starts from zero, the noise fluctuations would add positive counts, while the negative ones are cut at zero. So, if one averages the ADC output in time, the zero signal would be non-zero. This would cause a low light non-linearity.
To prevent it, most image sensor ADCs black level is set to higher than zero, for example, at 22 or 24 LSB.
What I'm saying is that in our ideal noiseless experiment, one can set the analog black level at 0.4999 analog value at the ADC input. It's set by a bias at the ADC analog input, so the black level does not have ADC quantization limitations and can be set at any level in our noiseless discussion.
Does this answer on your question?
The detectability is associated with a probability. In the example of Vladimir, the detection probability should be slightly slightly larger than 50%. :)-
DeleteHi Vladimir, IMHO this is a misleading discussion. It started for a real system with significant readout noise and has now become a theoretical discussion aside the concrete application of ADCs in image sensors... At least I agree with you regarding the noiseless system you mentioned.
DeleteHi Vladimir,
DeleteYou can change the black level to any level, but for ADC this is called offset, for comparator this called reference, isn’t it? If your offset/reference is 0.0499LSB, for a signal of 0.001 LSB, yes you can get a digital +1.
But you will get a +1 for all analog input between 0.001 LSB and 1.000 LSB, there is no information there, you claim that you detected a analog input signal of 0.001 LSB. but i can also say, sorry, your input was 0.5LSB or 0.9LSB.
Meaning that you will never know if the ADC analog input between 0.001 or 1.000 LSB.
This to my understand can not be called detectable of 0.001 LSB signal/information.
So, back to my conclusion,
Nyquist ADC itself can never output more than SNRmax=6.02*N+1.76 dB/ a few dB more for DR depend on definition.
I do agree that the light to digital (pixel) conversion can have DR>>SNRmax. But it is correct only because of probability/statistic same as other ADC designer called oversampling.
Peng
Hi Peng, it's perfectly OK to reject my definition of detection. There is no commonly accepted definition for that.
DeleteMost signal processing textbooks present ADC as an adder of signal and quantization noise. This way, the whole systems becomes linear and easy to analyse. In a strict sense, the quantization act is non-linear. But if one goes in non-linear system analysis, everything becomes much more complex. Offset is not offset, gain is not gain anymore and all other definitions become blurred too. Once you go that way, you are pretty much on your own and can come to any conclusion.
I agree with Vlad on this.
DeleteHi Vladimir,
DeleteI think we are talking about linear part of the ADC input range. i.e. the linear part of the pixel output. It can be modeled as a linear system.
My main point is that the definition of DR for image sensor is different for ADC. that makes maybe people confuse.
For a 10-bit general purpose ADC, two input + two conversion, e.g. one reset, one signal. There no way that one can get better resolution than 10-bit or 6x dB.
For a pixel quantized with 10-bit ADC, yes, one can claim that the detectablilty of electrons (light) is smaller than the quantization limit of the 10-bit ADC. Because we assume that the same amount of light hit the pixel and that will be quantized many times (oversampling), the averaging in digital domain might give us a sign that there was a small electrons/light input.
What i am trying to explain/understand here is:
The definition bridge from general purpose ADC to image sensor ADC world.
ADC is still ADC, but inherent "oversampling" principle with image sensor is hidden!
Peng
Why the Qmax changes from Rs mode to Gs mode?
ReplyDeleteDoes any1 have Any videos or still pictures of the E2V low light samples, if they claim very good low light capability, why not show any examples when offering sensors for thousands of dollars, it boggles the imagination? Or maybe im not looking in the right place, does anyone have any links to samples? Thank you in advance.
ReplyDeleteThis particular WVGA sensor shouldn't be too pricey as far as I know. At most several hundred bucks :-)
DeleteIt looks expensive then !
ReplyDeleteAre you kidding here or it's the true price ballpark?
Any price list?
No, it was a rough estimation based on the EV76C570, also a member of e2v's Sapphire product family. It's sold for a quite reasonable price...
DeleteI though thier high end sensors were like $1500 to $2500
DeleteSNR=signal-to-noise ratio. Noise is RMS, NOT max value. Same with ADC quantization noise. Max quantization noise is LSB while the RMS of the quantization noise is LSB/sqrt(12). Simply draw the sawtooth of the quantization error and calculate its rms value. Many imagers from Sony, Samsung and others have DR in bits higher than ADC resolution in bits.
ReplyDelete