Camera sensor size: noise & depth of field trade-offs

[Class A]

As usual, https://www.cambridgeincolour.com/tu...ensor-size.htm is a very nice tutorial. I have two issues with it though:

1. While the observations regarding pixel size are true, they are a bit misleading because if you keep the sensor size constant then different pixel sizes just result in different pixel densities. Pixel density, however, only has small secondary effects on noise, the primary one being sensor size.

2. I don't understand the following "Another important result is that if depth of field is the limiting factor, the required exposure time increases with sensor size for the same sensitivity."
Why? Given the same shutter speed and DOF (aperture size), the same amount of light will hit both small and big sensor. While the light/area ratio is worse for the big sensor, it also has the larger area to make up for it.

So why would you need to increase the exposure time (or sensitivity) to get an equivalent image?

[McQ]

Thanks for the feedback. Please see below for some sensor size comments:

1. Yes, sensor size is the dominant factor when considering image noise. When viewed at 100% though, pixel density corresponds closely with perceived noise. In a fixed print size, such as 8x10 inches, then pixel density will not correspond as closely with perceived noise, as you point out. However, this all assumes negligible gaps between pixels on the sensor; if these are at all present, then increasing pixel density will increase apparent noise since a greater fraction of sensor real estate is dedicated to the electronics between the pixels, as opposed to the pixels themselves. This effect is becoming less important though, since the microlenses are improving and the gaps between pixels are being reduced in the latest models. Canon even claims "gapless" pixels in their latest 50D I believe. However, real-world tests have shown that the 50D has a little higher noise (at 100%) than the 40D.

2. They key here is the phrase "for the same sensitivity". Another way of putting your comment is that, in theory, one ought to be able to use a higher ISO on the larger sensor to achieve the same shutter speed (and not have any more or less noise than the smaller sensor). This will likely negate any resolution advantage though, but that does not matter if the goal is to get an equivalent image. However, in practice I think it's a lot more complicated. I do not think medium format backs quite have 6-7X lower noise at the same ISO (and print size) than a Canon 40D, for example. Overall though, yes, it's true that *if you vary the sensitivity*, then you ought to be able to get an equivalent image -- everything else being equal.

The reason for this is that the total flux of light across both the smaller and larger sensors are equal for the same depth of field. The larger sensor therefore has to increase the sensitivity since this light is spread out over a greater area. However, if both the smaller and larger sensors had the same number of megapixels, then one could use the same ISO/sensitivity -- it's just that that's rarely the case. I could add a qualifier to the statement by saying "for the same pixel density". Perhaps the more generalized example would be in the case of traditional film photography; here you definitely need higher sensitivity film if you intend to achieve both the same exposure time and depth of field...it's just that the larger format's image does not need to be enlarged as much for the same print size -- effectively canceling out any increase in noise.

[Class A]

Dear McQ,

many thanks for your reply.

In a fixed print size, such as 8x10 inches, then pixel density will not correspond as closely with perceived noise, as you point out.

Why do you say "as closely"?
Should the perceived noise not be identical (assuming gaps are negligible or accounted for by microlenses)?

At pentaxforums an apparently knowledgeable member said the following:

"as pixel sizes shrink, light gathering technology (microlenses) and quantum efficiency both are typically improved." which results actually in less noise from higher density sensors.

Another member stated:
"
* In the first order, noise is a function of sensor surface only.
* Second order effect in favor of a high pixel count is: The read-out noise is minimized.
* Second order effect in favor of a low pixel count is: The area sensitive to light is maximized.
* So, the combined second order effects favor a specific pixel density which is technology dependent (and may currently be at about 3-4 µm).
"
His explanation for the reduction of read-out noise for higher density sensor was that read-out-noise per pixel is relatively constant and if you are able to combine many small pixels to emulate a larger pixel, you actually reduce the noise level. (Read-out noise obviously being not as important as shot noise).

Overall though, yes, it's true that *if you vary the sensitivity*, then you ought to be able to get an equivalent image -- everything else being equal.

I agree w.r.t. the fact that you can cancel out a shorter shutter speed with a higher sensitivity.

What I still don't understand is why you would need to increase the sensitivity for a larger sensor. (See towards the end of the posting for how to potentially resolve this conundrum.)

The reason for this is that the total flux of light across both the smaller and larger sensors are equal for the same depth of field. The larger sensor therefore has to increase the sensitivity since this light is spread out over a greater area.

It is true that the light is spread out over a greater area for the larger sensor. However, it also has a larger area to make up for fact that it receives less intensity per square-mm. The light collected by the whole area is exactly the same in both cases.

Assuming that we are comparing using the same output size in each case, using the same aperture size and shutter speed, the available light to the sensor will be the same. Hence, the output image will be a representation of the same amount of light. It doesn't really matter to what area the light was spread out in between (small or large sensor). This obviously neglects engineering difficulties but I'd like to concentrate on first order effects first.

However, if both the smaller and larger sensors had the same number of megapixels, then one could use the same ISO/sensitivity

Why does the number of megapixels should play a role? If the larger sensor has more, it will just have a different pixel density, the latter not being relevant (as a first order effect) for overall sensitivity or noise.

Again, I'm assuming that the comparison will be made with an image of the same size and pixel density in both cases.

Perhaps the more generalized example would be in the case of traditional film photography; here you definitely need higher sensitivity film if you intend to achieve both the same exposure time and depth of field.

My rationalisation of this effect is that sensitivity must be measured as light gathering effectiveness per square-mm but not just in terms of the material, but also considering its usage context. Both larger sensor and larger film area have a worse light gathering effectiveness per square-mm (since the receive less light per square-mm), but that doesn't mean they actually need to be made of more sensitive material to produce an equivalent image since their total amount of light collected is the same (due to the larger area).

If this is true than using the same film material in two cameras, one with with a format of area A and the other with area 2 * A, then if the film is specified as ISO 100 in the first case then it would be specified as ISO 200 in the second case. Is that true?

Comparing ISO values without conversion than makes sense when regarding the same areas on a negative / sensor with respect to a usage context (small vs large format), but not w.r.t. to final equivalent images.

I therefore believe it should be possible to use the same sensor technology in both small and large sensor cases (just more of it in the large case) and obtain equivalent images (when using the same shutter speed and aperture size). The fact that the larger sensor would be regarded as requiring a higher sensitivity would be just an effect of measuring sensitivity not in terms of total light collected but in terms of light collected per square-mm.

Hope this makes sense.

When you just enlarge the film area its se..it's just that the larger format's image does not need to be enlarged as much for the same print size -- effectively canceling out any increase in noise.

Yes, that's why pixel density is irrelevant (in terms of first order effects). And it is also the explanation why the light represented in both final images should be the same even if you don't use different (physically more sensitive) film in the larger format (just more of it).

[rc53]

2. I don't understand the following "Another important result is that if depth of field is the limiting factor, the required exposure time increases with sensor size for the same sensitivity."
Why? Given the same shutter speed and DOF (aperture size), the same amount of light will hit both small and big sensor. While the light/area ratio is worse for the big sensor, it also has the larger area to make up for it.

So why would you need to increase the exposure time (or sensitivity) to get an equivalent image?

This puzzled me for a long time. I understand it to mean: a larger sensor needs a longer focal length lens for the same field of view as a smaller sensor. But this longer lens will need a smaller aperture for the same depth of field. Therefore, the larger sensor needs a longer exposure - a longer shutter speed - for an image equivalent to that from the smaller sensor. Alternatively, the larger sensor needs a higher ISO for the same shutter speed [with the smaller aperture].

Am I right?

Bertie

[Class A]

Hi Berti,

I understand it to mean: a larger sensor needs a longer focal length lens for the same field of view as a smaller sensor.

Correct.

But this longer lens will need a smaller aperture for the same depth of field.

More precise language is helpful in this case. The longer lens needs a higher f-ratio (or aperture value), say 8 instead of 5.6. This is necessary to result in the *same* aperture size, i.e., the actual opening diameter.
The same aperture size will cause the same DOF and the same amount of light. Note that McQ writes above:
"The reason for this is that the total flux of light across both the smaller and larger sensors are equal for the same depth of field."

Therefore, the larger sensor needs a longer exposure - a longer shutter speed - for an image equivalent to that from the smaller sensor. Alternatively, the larger sensor needs a higher ISO for the same shutter speed [with the smaller aperture].

McQ agrees with you that this is necessary but not because of a smaller aperture but because the sensor area is larger and hence the light gets spread over a larger area, loosing intensity (per square-mm).

This is were I get lost, because a larger sensor manages to capture the same flux of light and although it receives less of it per square-mm, it has accordingly more area to exactly make up for the local intensity loss.

[rc53]

This is were I get lost, because a larger sensor manages to capture the same flux of light and although it receives less of it per square-mm, it has accordingly more area to exactly make up for the local intensity loss.

I find this all rather confusing. F stop values seem to be both absolute and relative at the same time. Absoulte because of the 'amount' of light passing, and relative because of the relation between focal length and aperture size.

And the smaller sensor can be expected to have a lens which is physically smaller, as there seems little point having an image circle much greater than the diagonal of the sensor. So if the small sensor lens is put on a large sensor camera [assuming that this is physically possible] presumably only part of the larger sensor gets illuminated.

We seem to have several variables; sensor size, absolute aperture size and relative aperture size; and sensor sensitivity - only the shutter speed is the same between sensors.

And the use of the term 'exposure time' is confusing. Does it mean shutter speed, the time for which the sensor is exposed, or does it mean exposure value, appropriate combinations of aperture and shutter speed which will give the same 'quantity' of light on the sensor?

Perhaps McQ could explain further?

Bertie

[McQ]

This thread is getting quite long and potentially hard to follow for many readers. I'm going to try and keep this as focused on the original question as possible in the reply.

Why do you say "as closely"?
Should the perceived noise not be identical (assuming gaps are negligible or accounted for by microlenses)?

The phrase "as closely" is intended to address just that: gaps and microlens efficiency. We're mincing words here though. The sensor sizes article/tutorial is not intended to give a precise recipe for estimating noise as a function of sensor size, pixel density and sensor technology -- it is instead aimed at giving a general overview of the first order effects and their relevance to the camera user. Second order effects can in some situations, as mentioned, substantially mitigate the first order effects, depending on the sensor technology. The explanation of other second order effects you listed below is helpful at illuminating some of these. However, first order effects are first order for a reason: in the vast majority of situations they are the most important considerations to keep in mind for the average (or even technically oriented) camera user. Further, the second order effects are often not camera-universal, so they can become very hard to judge depending on the brand or generation of the camera and what technology it incorporates.

What I still don't understand is why you would need to increase the sensitivity for a larger sensor.

It is true that the light is spread out over a greater area for the larger sensor. However, it also has a larger area to make up for fact that it receives less intensity per square-mm. The light collected by the whole area is exactly the same in both cases.

Assuming that we are comparing using the same output size in each case, using the same aperture size and shutter speed, the available light to the sensor will be the same. Hence, the output image will be a representation of the same amount of light. It doesn't really matter to what area the light was spread out in between (small or large sensor).

This is true: the net amount of light collected over the entire sensor area is the same for both the large and small sensor when keeping depth of field the same (the larger sensor needs to use a larger f-number for its aperture). Since the light is spread out over a larger area for the larger sensor, the flux per square mm decreases. This means that to get the same exposure as with the smaller sensor, you will need to increase sensitivity. The necessary sensitivity is determined by the flux of light, not the total amount summed over the entire sensor.

An example is that if I were to take a panoramic image comprised of 2 photos, I would effectively be increasing the sensor area by let's say 1.7X (accounting for overlap). This means that a total of 1.7X more light is reaching the sensor (assuming the same exposure for both shots). This does not, however, mean that I can therefore decrease the sensitivity since there's more total light. The flux of light per unit sensor area has remained the same, and therefore the sensitivity needs to remain the same for a given exposure.

Why does the number of megapixels should play a role? If the larger sensor has more, it will just have a different pixel density, the latter not being relevant (as a first order effect) for overall sensitivity or noise.

This is true, I got side-tracked onto second order stuff. It should be the same to first order (when viewed at the same absolute print size as opposed to 100% on the screen).

My rationalisation of this effect is that sensitivity must be measured as light gathering effectiveness per square-mm but not just in terms of the material, but also considering its usage context. Both larger sensor and larger film area have a worse light gathering effectiveness per square-mm (since the receive less light per square-mm), but that doesn't mean they actually need to be made of more sensitive material to produce an equivalent image since their total amount of light collected is the same (due to the larger area).

If this is true than using the same film material in two cameras, one with with a format of area A and the other with area 2 * A, then if the film is specified as ISO 100 in the first case then it would be specified as ISO 200 in the second case. Is that true?

A lower light gathering effectiveness per mm will indeed require higher sensitivity for the same exposure. As with the panorama example, both will be ISO 100 since they are the same sensitivity film; it is independent of total area.

I therefore believe it should be possible to use the same sensor technology in both small and large sensor cases (just more of it in the large case) and obtain equivalent images (when using the same shutter speed and aperture size). The fact that the larger sensor would be regarded as requiring a higher sensitivity would be just an effect of measuring sensitivity not in terms of total light collected but in terms of light collected per square-mm.

Overall, I think we both agree on the central point that started this thread: that exposure times do not necessarily have to be any longer with a larger sensor if the aim is achieving an equivalent image. By equivalent, we both mean that when viewed at the same print size, that the perceived noise and depth of field are the same. The only point of disagreement was whether the sensitivity had to increase. I believe that it does, but that the requirement for less enlargement in the larger sensor effectively cancels out any noise increase from higher sensitivity. Again though, this is all first order effects, but that's really all the photographer need be concerned with 99% of the time...

[McQ]

I find this all rather confusing. F stop values seem to be both absolute and relative at the same time. Absolute because of the 'amount' of light passing, and relative because of the relation between focal length and aperture size.

Yes, it does get a bit confusing, especially because larger f-numbers correspond to smaller apertures. The f-number or f-stop is quite useful because it is in many ways a universal description of the lens aperture. A longer focal length lens at f/8 will have a larger physical aperture size (in absolute dimensions, such as mm's) than shorter focal length with the same f/8 f-number. However, when it comes to diffraction, f/8 is universal in describing the size of the airy disc. In addition, both the long and short lens will have the same total depth of field if the subject is magnified the same amount in both shots. This is in part why f-numbers are so useful.

And the smaller sensor can be expected to have a lens which is physically smaller, as there seems little point having an image circle much greater than the diagonal of the sensor. So if the small sensor lens is put on a large sensor camera [assuming that this is physically possible] presumably only part of the larger sensor gets illuminated.

This is true, if it is physically possible. There's a number of factors which can prevent this from being possible, such as the back focus distance and mounting mechanism.

We seem to have several variables; sensor size, absolute aperture size and relative aperture size; and sensor sensitivity - only the shutter speed is the same between sensors.

And the use of the term 'exposure time' is confusing. Does it mean shutter speed, the time for which the sensor is exposed, or does it mean exposure value, appropriate combinations of aperture and shutter speed which will give the same 'quantity' of light on the sensor?

Perhaps McQ could explain further?

We have, to my knowledge, been using "shutter speed" and "exposure time" synonymously in this thread. However, when I use the phrase "the same exposure" or "equivalent exposure", I mean that the aperture, shutter speed and sensitivity are all adjusted so that a middle gray tone on one sensor is more or less mapped to the equivalent middle gray on the other sensor/aperture/shutter/ISO combination.

[Class A]

Thanks McQ, for your reply.

We're mincing words here though.

I agree, but I also think that the topic requires precise observations and descriptions, and this sometimes requires mincing of words.

I also fully agree that the tutorial should concentrate on first order effects.

I welcome the fact that we are in agreement about most issues, because I'd like to arrive at an understanding that is consistent and in agreement with knowledgeable people.

Since the light is spread out over a larger area for the larger sensor, the flux per square mm decreases. This means that to get the same exposure as with the smaller sensor, you will need to increase sensitivity.

I believe, whether or not sensitivity needs to be increased depends on what you are comparing:

If you want to achieve equivalence for a given sensor area size then you are correct. But I don't think that this is the most useful way to do the comparison, i.e., not the most appropriate notion of equivalence.

Here are a number of equivalence notions (assuming identical incoming total flux of light in each case):

1. Per Pixel:
In this case a sensor with higher pixel density (smaller pixels) needs higher sensitivity per pixel to make up for the lower light flux per pixel (smaller receiving area).
However, when comparing identical output sizes, we agreed that pixel density has no first order effect on required sensitivity. If each pixel were more sensitive, the total flux of light represented in the output image would be higher.

2. Per square mm sensor area:
In this case a larger sensor, which receives the same flux of light as a smaller sensor (with identical pixel density), needs higher sensitivity to make up for the lower flux per square mm (same receiving area but the light is more spread out over a larger area).
However, when comparing identical output sizes, the overall captured light flux is the same for both small and larger sensors. If the larger sensor were more sensitive, the the total flux of light represented in the output image would be higher.

3. With respect to the output image:
Since both small and large sensors receive the same flux of light there is no need for either of them to be more sensitive in order to achieve an output image which represents the same flux of light.
Cutting out any analogue to digital conversion, assuming our output image is just a projection of light, it should be clear that either enlarging the image from the small sensor area or decreasing the image from the large sensor to an identical size yields the same image intensity. Hence no amplification (higher sensitivity) is required.

BTW, the illustration with the projected image also illustrates the irrelevance of pixel density. It doesn't matter how fine a grid is that is placed within the projection path. Discounting, of course, the (second order) effect of the relative proportion of the grid material which blocks light.

I can agree with that statement "a larger sensor requires more sensitivity" if you are comparing output images that do not have the same size. If the output image of the larger sensor is larger by the factor LargeSensorArea/SmallSensorArea then, obviously, the sensitivity of the large sensor needs to be higher by the factor LargeSensorArea/SmallSensorArea.

An example is that if I were to take a panoramic image comprised of 2 photos, I would effectively be increasing the sensor area by let's say 1.7X (accounting for overlap). This means that a total of 1.7X more light is reaching the sensor (assuming the same exposure for both shots). This does not, however, mean that I can therefore decrease the sensitivity since there's more total light. The flux of light per unit sensor area has remained the same, and therefore the sensitivity needs to remain the same for a given exposure.

Yes, but I don't understand what point you are trying to make. The output image is 1.7X as wide, 1.7X more flux of light has been captured, so of course the sensitivity can stay the same.

If the output image is 1.7X as wide but the same flux of light has been captured (this is the case when comparing a small to a large sensor system, when the sensor area ratio is 1.7) then the larger sensor needs to be more sensitive.

If the output image is not wider and the same flux of light has been captured (this is the case when comparing a small to a large sensor system and equivalence is defined on the output image, not on per square mm sensor area) then the larger sensor does not need to be more sensitive.

The only point of disagreement was whether the sensitivity had to increase. I believe that it does, but that the requirement for less enlargement in the larger sensor effectively cancels out any noise increase from higher sensitivity.

I agree w.r.t. to the cancellation, but I still don't get my head around your postulated need for increased sensitivity. Perhaps I'm using the wrong notion of sensitivity?

For the purpose of this discussion, I regard sensitivity as factor in the equation PixelCharge = Sensitivity * NumberOfPhotonsCapturedByPixel. PixelCharge then gets A/D converted into a number, i.e., a numerical pixel value. In order to achieve the same total sum over all pixel values for two sensor readouts where the second time the total flux of light was 1/2 of that of the first time, I need 2X the Sensitivity to make up for the loss in light.

With this definition of sensitivity, sensor size is irrelevant for the total amount of flux and noise represented in the final image. Assuming the larger sensor has the same pixel density, i.e., F times more pixels if F = LargerSensorArea/SmallerSensorArea then the total sum for the larger sensor is decreased by the worse light intensity per square mm (X 1/F) but is increased by the larger number of pixels (X F). This is the cancellation I'm talking of and which you seem to refer to as well when talking about no increase of noise regarding the same output size.

Now I only need to understand your notion of sensitivity which requires a higher sensitivity for the larger sensor. I'm very much looking forward to your response.

[Class A]

Perhaps the more generalized example would be in the case of traditional film photography; here you definitely need higher sensitivity film if you intend to achieve both the same exposure time and depth of field...

I have recently read the same statement somewhere else and someone made a comment which I believe to be true.

Film photographers do want the larger format and the same exposure / mm2. They are not contend only with a larger negative, which when scaled down would have the same exposure / mm2 as a smaller negative. They want to have their cake and eat it too. That, I believe, why they have to increase sensitivity (use higher ISO) film.

If they were only concerned with "equivalent images" (i.e., same representation of light captured when rescaled to the same output size) then they wouldn't have to increase ISO. Increasing ISO leads to an image that represents more light then the image done with the smaller negative format.

It would be great if someone could confirm this. A consequence would be that one could state that FF sensors do not need to have higher ISO capabilities in order to produce images which are equivalent to those from APS-C sensors.