Picking up from this answer and this question, What exactly is ETTR? How it may reduce the image noise? And how is it difference from film to digital sensors?

In the answer linked above, what are the 5 stops and is it related to ETTR?

In real life how can I apply this technique when I'm shooting?

  • \$\begingroup\$ The question of the meaning of a stop in this context is answered under What is one “stop”? \$\endgroup\$
    – mattdm
    May 1, 2012 at 22:14
  • \$\begingroup\$ @mattdm I understand what a stop means however the answer linked in the question mentioned "5 stop range", is this a standard range for tone brightness? \$\endgroup\$
    – K''
    May 2, 2012 at 18:42
  • \$\begingroup\$ Oh, I see the confusion. That number comes from a quote from the Luminous Landscape ETTR article, and 5 stops was chosen as a reasonable number to represent the total dynamic range of a DSLR of the time the article was written. You can work the same calculation with any other arbitrary number for total stops. Five is just the example. \$\endgroup\$
    – mattdm
    May 2, 2012 at 19:13
  • \$\begingroup\$ @mattdm oh okay that makes much more sense, thanks \$\endgroup\$
    – K''
    May 2, 2012 at 19:29

6 Answers 6


"Expose to the right" means record the brightest image you can and then reduce the brightness in post to achieve the desired level.

The word "right" comes from the histogram, where conventionally brightness increases left to right, thus increasing brightness shifts the whole histogram to the right.

ETTR helps reduce noise simply by capturing more light, which reduces photon noise, and gives a better signal to [electrical] noise ratio (by virtue of a bigger signal). The reason high ISO photos look noisy is due to low levels of light and amplifying a weak signal.

The technique works provided you don't increase the exposure to the point where it hits the maximum possible value and gets cut off, as this will result in a loss of information (known as clipping/blowing the highlights). Typically this is seen as an area of the image (usually sky) which has gone pure white.

In principle the technique works for film, certainly exposing the left and then having to push your image when printing will increase grain. However film has a different cutoff characteristic, as highlights gently roll off rather than hitting a hard limit.

Here's an experiment I did to demonstrate the effect (and rebuff a blog article which claimed ETTR didn't work):

Here's the camera metered exposure:

Here I've used ETTR and increased the camera meter's exposure by 1 stop using a longer exposure:

Finally, to show the difference here's the standard exposure with the ETTR image offset in the centre:

The reduction in noise is visible, particularly in the purple patch in the bottom left.

  • 3
    \$\begingroup\$ +1, especially for providing a nice example and for stressing the issue with clipped highlights, an important practical consideration. \$\endgroup\$
    – mattdm
    May 1, 2012 at 20:37

To be short ETTR is a smart usage of two fact:

  1. There is more information in the high light (the right of the level curve) than in the low light (the left of the level curve). This is due to the fact that capter has linear response to the light intensity while human perception is rather log (what you perceive as twice brighter is in fact not twice the amount of light but much more)

  2. The noise is present everywhere but what you perceive is the ratio noise over signal: if the signal is big you cannot see the noise, if the signal is of the same order or smaller than the noise you will see noise. So the more you collect light the bigger is your signal and the smaller is the noise perception

When overexposing your image (and in particular a globally dark image) you are using the right part of level curve for storing your image rather than the left one. Doing that you have two advantages (1) more information (more distinct tones) and (2) by collecting more light you increase the signal/noise ratio (so get less visible noise)

In post-treatment you can then correct your level and get the tone you want.

Back to film camera (I get the B&W picture which is equivalent to the color one but easier to figured out) each grain has a threshold (a number of photon) above which it will turn black and bellow which it will stay white (and be washed out in the film processing) the "noise" was the size of the grain which was related to the sensitivity.

  • \$\begingroup\$ +1 I liked "what you perceive as twice brighter is in fact not twice the amount of light but much more" \$\endgroup\$
    – K''
    May 2, 2012 at 18:23
  • 1
    \$\begingroup\$ "more information" is slightly misleading. There are the same number of bits for the right half of the histogram as there are for the left half aren't there? \$\endgroup\$
    – Joe
    May 3, 2012 at 6:59
  • \$\begingroup\$ @Joe you are right. However your perception act as "compressing" the right part and "inflating" the left part of the histogram, so there is more tones in the bright lights \$\endgroup\$
    – floqui
    May 3, 2012 at 9:11

There are those who think ETTR is folklore, not fact. Ctein (who has multiple decades of experience and is a master printmaker) has written that ti's all bull. (link: http://theonlinephotographer.typepad.com/the_online_photographer/2011/10/expose-to-the-right-is-a-bunch-of-bull.html) I'd suggest at least looking at his commentary.

Me? I respect Ctein a lot, but I tend to expose towards the right a bit (typically about 3/4 of a stop of compensation), depending on the subject. At its worst, ETTR seems to be placebo, not harmful. Whether it's really helpful? Not everyone agrees about it..

  • 4
    \$\begingroup\$ Before getting too riled up by the inflammatory title of the linked article, note that this paragraph summarizes the key point: These days, noise is really not a big source of image quality loss[....] Cameras and sensors are so much better. Clipped highlights, as Mike and I discussed last week, haven't gone away. It's still a big issue when trying to get real quality in a digital photograph. The argument is that blown pixels are a bigger real-world problem than noise in most situations. \$\endgroup\$
    – mattdm
    May 1, 2012 at 22:50

The answers you cite contain the information you want. It may not be "accessible" enough without reading and re and re-re reading. I'll try to summarise what was said in those references and in many other places, but do note that this is a summary and lots of detail are available elsewhere.

A digital camera sensor tends to produce an output that is linearly related to light level. this does not have to be the case, and here may be advantages in doing otherwise, but that's the norm so far.

With a linear sensor, if you halve the brightness you halve the numerical "reading" or light level. If the 'reading' is 4000 at 100% of sensor max light level capability, then it will be 2000 at 50% of sensor max level,
and it will be 1000 at 25% of max
500 at 12.5% of max
250 at 6.25% of max
125 at 3.125% OF MAX
62 AT ...

BUT each halving of light level is equivalent to one stop, or one EV level. It's far more intuitive to think in EV units but it can be equally expressed in stops.

So the first "stop" of sensor range has a certain EV of actual brightness at the top of this range and 1 EV less at the bottom, and the sensor has max reading of 4000 and minimum of 2000 and there are 2000 "counts" across this or EV level.
Areas in the image which are one EV level less bright than maximum brightness = the second stop / EV level in the image and have light levels from 1000 to 2000 and a 1000 range
The third stop has light levels from 500 to 1000 and a 500 range
The fourth stop has light levels from 250 to 500 and a 250 range

This means that the first stop of exposure has many numeric values between its top and bottom levels. Noise of a given magnitude that is a certain percentage of its range will be an increasing percentage of the range of a stop as light level falls. eg say noise was +/- 5 units relative to the sensors 4000:1 dynamic range.
In the top stop noise is 5/2000 = 1/400 = 0.25% of the range.
In the 2nd stop noise is 5/1000 = 0.5% .
By the time we are down to the 8th stop the dynamic range available
= 4000 /(2 x 2 x 2 x 2 x 2 x 2 x 2 x 2) ~+ 16 sensor steps, and the 5 units of noise are 5/16 or about 31% of the range. ie at the op end of brightness a given level of noise may have little effect but as brightness falls the noise double for every 1 stop decrease and the % that the noise is of signal variation doubles.

Translating this into practice - take a highish ISO photo where the image is starting to get noisy. Now look in the shadow areas - you will find that they are far more affected - in about inverse proportion to their brightness.

So - EV levels that are close to the top of the sensors maximum light handling level are less noise affected. It does not matter about what the light level is as long as it can be corrected in due course. Rather, we push all brightness levels up until the brightest level is almost clipping. This allows the lower levels to have as much sensor variation as possible.

Note that 5 stops was just a convenient range to consider - this effect of right shifting matters right across the range.

Film tends to have a logarithmic response to light so comoresses a wider variation of levels into a lower effective range.

  • \$\begingroup\$ I would compute sensor DR stops a bit differently. A/D converters are binary devices, and can only encode, at most, as much information as their bit depth. Since, in binary, every additional digit is a doubling of the numeric space of all previous digits, modern cameras are effectively limited to 14 stops, or 2^14 levels. Its extremely difficult in reality to actually achieve 14 stops of dynamic range, however, given the necessary overhead of converting an electronic charge into ADU (analog-to-digital units). Maximum saturation is usually less than 2^14, so real-world performance... \$\endgroup\$
    – jrista
    May 1, 2012 at 20:58
  • \$\begingroup\$ ...is usually limited to around 13 stops of dynamic range or less (assuming a very forgiving method of computing dynamic range...many would dispute even that much is actually possible, and offer that 10-11 stops is all we can really get in reality with more conservative methods.) The binary nature of an ADC also leads to every additional bit adding almost twice as many possible luminance levels as the previous, so a 15-bit sensor would offer about 32000 levels vs. the approximage 16000 of a 14-bit sensor. \$\endgroup\$
    – jrista
    May 1, 2012 at 20:59
  • \$\begingroup\$ The dynamic range of the best modern camera systems slightly exceeds the number of bits in the ADC. This apparent impossibility is well cobered in prior stack exchange answer and relates in par the ability to "dither" an ADC output to beyond he number of bits provided if the signal and measurement systems are able to support such accuracy. Rushing out, else more ... \$\endgroup\$ May 1, 2012 at 23:21

I thought it's worth adding this quote, from a whitepaper from Adobe, as it is an explanation from the company that makes the most popular software for processing photos and especially converting RAW data to images.

You may be tempted to underexpose images to avoid blowing out the highlights, but if you do, you’re wasting a lot of the bits the camera can capture, and you’re running a significant risk of introducing noise in the midtones and shadows. If you underexpose in an attempt to hold highlight detail, and then find that you have to open up the shadows in the raw conversion, you have to spread those 64 levels in the darkest stop over a wider tonal range, which exaggerates noise and invites posterization.

Correct exposure is at least as important with digital capture as it is with film, but in the digital realm, correct exposure means keeping the highlights as close as possible to blowing out, without actually doing so. Some photographers refer to this concept as “Expose to the Right” because you want to make sure that your highlights fall as close to the right side of the histogram as possible.


One thing that is important to realize is that digital and film photography are utterly different with respect to dealing with sensitivity, and on top of that, different sensor types are different as well.

For negative film exposure, your film sensitivity is implemented by the size of individual grains. While the grains become quite more visible with underexposure (since they overlap less), the film choice fundamentally determines both spatial resolution and the ability to represent different luminosity.

Also film is really, really, inert on its own. If no light falls on it, you can "expose" it for months (namely just keep it in-camera or in-cartridge) without change before handing it off to development

Digital sensors are quite different. The size of photocells is fixed (though you might combine severals in post-processing to reduce noise somewhat) and the concept of "charge wells" means that the resulting voltage is pretty much proportional to the arriving light energy. Sensors these days are either considerably smaller than typical film sensor and/or quite more sensitive. A major factor regarding the sensitivity particularly with smaller sensor or high resolution sensor is photon counts: the number of photons registering for each pixel can be so small that the statistical variation of their numbers is a significant source of image noise: photon noise.

Then there is analog amplification and subsequent quantization.

ISO on digital sensors will be used for determining "correct exposure" and for influencing the analog amplification (a process audio engineers know as "gain staging" before quantization).

To what degree? Some sensor types let whole ISO stops influence the analog amplification while fractional ISO stops just affect metering and processing (so ISO160, ISO200, ISO250 might all be using the same analog/quantization setup but meter with +1/3EV, 0EV, and -1/3EV of correction and then compensate the result digitally).

There are also "ISO invariant" sensors like Sony Exmor that don't change anything in the analog and quantization paths: an ISO200 image underexposed by 4 stops contains the same data as a properly exposed ISO3200 image on those sensors, it is just interpreted differently. It also means that it's almost impossible to blow highlights at higher ISO values with those sensors at least in the raw files.

While not all sensors have complete ISO invariance, larger sensors with potentially larger photosites often still have good digitisation reserves and consequently resilience against blown highlights so that overexposed higher ISO images tend to be quite comparable in quality (at least when working with raw files) to "properly" exposed lower ISO images, so dialing in positive exposure compensation or flash compensation can yield better shadow resolution.

So "expose to the right" will have quite different reserves depending on the sensor used and the ISO setting, with larger sensors and larger ISO values often having larger reserves for getting more light into the camera as "average" metering would.


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