From what I understand, the lower the EV value, the more exposed (bright) the image is.

  1. EV decreases as aperture decreases. More light gets in with lower apertures.
  2. EV decreases as shutter speed increases. More light gets in with a higher shutter speed.
  3. EV increases as ISO increases. This does not make sense to me because as ISO increases, the sensor is more sensitive to light, exposing the image more. I would expect EV to decrease as ISO increases.

I understand that the EV calculation taking ISO into consideration is:

$$\log_2\left({N^2\over t}\right) + \log_2\left({S\over 100}\right)$$

Where \$N\$ is the F-stop, \$t\$ is the shutter speed, and \$S\$ is the ISO.

I guess my question is, why is the second part of that equation having the do with the ISO added to the EV instead of subtracted, since a greater ISO makes the sensor more sensitive to light?

I am new to this calculation so any insight is greatly appreciated.

  • \$\begingroup\$ My assumptions were wrong and the way I was calculating EV was also wrong. Details are in the comments of the accepted answer below. Not sure if this question should be closed but it was certainly informative to me. Thanks, all. \$\endgroup\$
    – Chuck C.
    Commented Jan 10, 2013 at 17:10

3 Answers 3


You're fundamentally right, because the equation you've given has the sign reversed. The adjustment should go on the left side of the equals sign, not the right side; if you want to move it to the right side, it needs to be subtracted. Or, to answer the title question: it doesn't.

But, I think part of the confusion comes from looking at what EV is meant for in reverse as well.

You say:

the lower the EV value, the more exposed (bright) the image is

This is looking at it backwards. A better way to put it would be: The lower the EV value, the more you need to use more-light-gathering and more-sensitive camera settings to get a proper exposure.

The goal is to get images with the same brightness ("correct exposure") regardless of the actual level of light in the scene. When we're out in the world, our eyes and brain adjust and normalize the light levels — sometimes we're aware that it's very dark or very bright, but overall, we just adjust. We can't adjust when looking at a fixed photograph, though: the exposure is whatever it is.

So, exposure value is a tool for helping to get this right. In modern cameras, of course, it's all done with automatic metering (even in manual mode, we usually follow what the meter says as a guideline). But if the meter weren't built in, you could use readings of the "exposure value" of the scene (or, just estimate, once you're familiar enough) to set the camera settings so that you get a nicely exposed image.

A higher exposure value represents a brighter scene. At ISO 100, bright sunlight is around EV 16, an overcast day more like EV 12, a home interior at night is something like EV 5, and starlight around negative 6 EV. From this, you can figure out what shutter speed, aperture, and ISO you'll need.

Although one might look at the right side of the equation and say that a certain aperture and shutter speed "produce" a certain EV, it's more normal to look at the EV of a scene and then decide what shutter and aperture are required to get the correct exposure. (Where "correct" means "basically average" — you're free to expose in a higher or lower key if you prefer!)

The "EV compensation" adjustment on modern cameras is meant to work with this. By dialing in negative –1 EV, you're telling the camera "Your meter says the EV is N, but please calculate exposure settings as if it were a one stop darker". Or by dialing in +1 EV, you tell it that you want the scene exposed one stop more brightly. This can be used to correct for metering errors or simply for artistic preference.

So, where does ISO come into this? Simply put, each stop of ISO (that is, each doubling, from 100 to 200 or from 200 to 400 or from 400 to 800, and so on) shifts the scale by 1. (\$\log_2({S\over100})\$ is just a fancy way of saying that.)

You can also rearrange the formula into:


if you like — it's another way of looking at the same thing, and in fact a mathematical way of expressing two slightly different definitions of the word "exposure", as explained in this other answer.

Or, you can write it out with all the terms separate, which I think is most clear of all:

$$\text{EV} = \log_2\left({N^2}\right) + \log_2\left({1\over t}\right) - \log_2\left({100\over S}\right)$$

Because that's really how one is meant to think about it in the real world. Or rather, one is meant to think:

$$\text{EV} = \text{aperture} + \text{shutter} - \text{ISO}$$

and not worry about all of the logarithms.

  • \$\begingroup\$ Ok, for your example, when I have: F/16, 1/100s, ISO 100, I get an EV of 14.6438561897747, which is what you got. But when I make a change from F/16 to F/18, the EV changes to 14.9837061926594. F/18 allows less light in than F/16, so the image will be darker, so why does the EV increase? If I change to F/8, allowing in more light, the EV is 12.6438561897747. I don't understand why the EV would become lower as I allow in more light. Does that make sense? Thanks for your help. \$\endgroup\$
    – Chuck C.
    Commented Jan 10, 2013 at 16:17
  • \$\begingroup\$ Yeah, I didn't catch that right away but it quickly became clear. Answer edited. \$\endgroup\$
    – mattdm
    Commented Jan 10, 2013 at 16:26
  • \$\begingroup\$ The update to your answer has been very helpful. I just have one more question if you don't mind. These exposures have the same EV of 14.64: F/4, 1/100s, ISO 1600 and F/8, 1/100s, ISO 400 I don't understand that. I would expect the second exposure to be darker because the aperture is allowing less light in and the ISO setting is less sensitive. I just don't get it. Thanks for your patience. \$\endgroup\$
    – Chuck C.
    Commented Jan 10, 2013 at 16:48
  • \$\begingroup\$ @ChuckC — I think your expectation is right and your math is wrong. :) (More specifically, your calculation is fine but you didn't reverse the sign for ISO.) \$\endgroup\$
    – mattdm
    Commented Jan 10, 2013 at 16:57
  • \$\begingroup\$ As in aside, using the hundredths place in EV calculations is excess precision, because it exceeds the likely real-world precision of the aperture, shutter speed, and ISO numbers you're inputing by a good measure, and also because even in a theoretical world it's a negligible difference. \$\endgroup\$
    – mattdm
    Commented Jan 10, 2013 at 16:59

I think you got it backwards, EV increases when brightness increases.

See the wikipedia article for details

  • \$\begingroup\$ I am sure it is something I am not understanding, but then why does EV go down as I increase shutter speed and decrease aperture? Don't those things increase brightness? \$\endgroup\$
    – Chuck C.
    Commented Jan 10, 2013 at 15:28
  • \$\begingroup\$ @ChuckC. - EV is the brightness of the scene, full sunlight is 15 EV and require a shutter speed of 1/500 at f/8 ISO 100, indoor home lighting is EV 5 to 7, EV 7 is 15-7=8 stops darker and so require 1/2 shutter at the same aperture and ISO \$\endgroup\$
    – Nir
    Commented Jan 10, 2013 at 15:38

"EV increases as ISO increases. This does not make sense to me..."

Let's put this in plain English: Higher ISO film is more sensitive to light. In digitalese, a higher ISO setting makes the sensor more sensitive. If you are not careful you can overexpose your shots easily.

"A high Exposure Value(EV)" means that the film will appear to have been dosed with more light. Let's take two shots of an object – both with the same aperture and shutter speed settings – one on 100 ISO film and one on 800 ISO film. You will, of course, agree that the 800 ISO film will "appear" to be more exposed, when in reality they got the exact same amount of light. When you are comparing various ISO settings, you will have to agree "EV increases as ISO increases."


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