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My impression is that a smartphone has no physical shutter. Rather, it admits light to the sensor's photo-sites under the control of some on-and-off, or perhaps read-twice-and-subtract, electronic method.

Is that generally accurate? If so, could someone point me to a beginner-level description of how it works?

If the operation is electronic, does that mean there's no inherent fastest or slowest "shutter speed" for a given smartphone?

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  • \$\begingroup\$ What photographic problem are you trying to solve? \$\endgroup\$
    – Michael C
    Commented Nov 6, 2020 at 2:33
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    \$\begingroup\$ My Pixel 4a, for all its strengths, takes fuzzy pictures of grandkids in motion. I'll be aiming to resolve this by better lighting, and/or downloading and experimenting with manual/pro mode camera apps. But I'm wondering if the phone itself might have inherent limitations on "shutter speed". \$\endgroup\$ Commented Nov 6, 2020 at 3:07
  • \$\begingroup\$ Small cameras use CCD sensors, and do normally time the exposure by enabling the senors on-off for the duration. Your image file EXIF data surely shows the shutter speed for each picture. \$\endgroup\$
    – WayneF
    Commented Nov 6, 2020 at 4:37
  • \$\begingroup\$ EXIF data certainly shows the "shutter" speed for a given picture. What I'm looking to find is the upper and lower limits (if any) on a particular smartphone's range of speeds. (The things I've read say nearly all smartphones use CMOS sensors rather than CCD. Would this make a difference in the analysis?) \$\endgroup\$ Commented Nov 8, 2020 at 21:18

2 Answers 2

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CMOS sensors read each line sequentially. How long each line is "on" and "off" determines exposure time. But just as with mechanical focal plane shutters, the total time needed to take an exposure is the same, regardless of the exposure time. It takes any given sensor the same amount of time to read out each line of the sensor regardless of how long each line was "on". This provides a limit to the maximum number of frames that may be captured per second (assuming the exposure time is shorter than the readout time - otherwise the exposure time is the limit on fps).

Since the sensors in smartphones do double duty to not only record images, but to also give an almost real time preview on the phone's screen in lieu of an optical viewfinder, exposures longer than the frame rate of the video feed supplied to the phone's screen remove the ability to provide a constantly updated stream of frames to the screen during exposure.

If the operation is electronic, does that mean there's no inherent fastest or slowest "shutter speed" for a given smartphone?

That depends upon what one means by inherent. There are physical limits to how quickly a line or photosite on a sensor can be switched on and then off. There are also limits to how short an exposure time can be and still allow enough time to collect meaningful information (i.e. collect enough photons to rise above the noise floor for a given sensor). Currently in most typical lighting conditions and scenarios, the latter is reached at a longer exposure time than the limit imposed by the former.


Some background information for those who may not be knowledgeable about how mechanical focal plane shutters and sensor with "electronic shutters" that read the output sequentially line-by-line rather than a sensor with a "global shutter" that reads the entire sensor all-at-once. (Hint: There are a good number of existing questions here that go into great detail about how various types of mechanical and electronic shutters operate.)

So I don’t get of on the wrong foot, could you please address my confusion on the first few sentences? 1. CMOS sensors read each line sequentially. How long each line is "on" and "off" determines exposure time. The first sentence is about reading; the second is about exposure. I’m not grasping how they relate to each other. 2. But just as with mechanical focal plane shutters, the total time needed to take an exposure is the same, regardless of the exposure time. I haven’t been able to parse this.

The sensor can not be collecting photons that accumulate charges in each photosite (exposure) and move those charges off the sensior (readout) at the same time, Just like you can't pour the water out of a rainbucket into a measuring cup and catch rain falling on your yard at the same time. When a sensor is read it is "cleared" all charges are removed. When it is "turned on" it begins accumulated charges created by photons falling on each photosite. When it is "turned off" it stops accumulating charges created by photons falling on each photosite.

The time difference between when each photosite is turned on and turned off is the exposure time. That is, it's the total amount of time the photosite was collecting charges.how long it takes the sensor to sequentially turn on each line on the sensor is fixed. It (usually) starts at the bottom (which is the top of the inverted image) and goes line by line to the top (the bottom of the inverted image).

In much the same way mechanical shutter curtains in modern cameras transit the sensor at a fixed rate. It's the difference between when the first curtain begins opening and the second curtain begins following it in the same direction across the sensor that determines exposure time. At short exposure times the entire sensor is never uncovered at the same time. It may take 4 milliseconds (1/250) for each shutter curtain to transit across the sensor. If the second curtain begins closing 1 millisecond after the first curtain begins opening, the first curtain has only revealed 25% of the sensor.

As the first curtain continues travelling across the face of the sensor the second curtain is "chasing" it 1ms (6mm for a 24mm high FF sensor) behind. When the first curtain finishes opening, revealing the bottom of the sensor to light, the second curtain has already covered up the first 75% of the sensor. 1ms later it finishes closing to completely cover the sensor. The GIF in the linked question and linked video (below) shows how it takes about 2.5ms (1/400) to take exposures of 1/1000, 1/2000, 1/4000, and 1/8000 by decreasing the size of the slit between the two curtains as they transit the sensor (The transit time for the Canon EOS 7D is about 2.5ms).

If a rolling shutter travels from top to bottom, why does this image seem to show skew in the other direction?

The accepted answer includes a link to the entire video.

With long exposure times it's the same thing. Let's say we use 1 second exposure with a camera that has a shutter transit time of 2.5ms (1/400). The first curtain opens and takes 2.5 milliseconds to cross from the top to the bottom of the sensor and uncover the sensor. Exactly one second after the first curtain began opening the second curtain begins to close. It also takes 2.5ms to cross the sensor and cover it back up. Thus from the time the first curtain began to open until the second curtain completely closes is 1,002.5 ms. But no part of the sensor has been uncovered more than 1,000ms.

When the exposure time is shorter than the transit time, the second curtain begins closing before the first curtain is completely open. The total time it takes to make a 1/1000 exposure is the transit time (2.5ms) + the delay between first and second curtain (1.0ms), or 3.5ms. No spot on the sensor is uncovered for more than 1ms, but the photosites on one edge are exposed 2.5ms earlier in time than the photosites on the far edge are exposed.

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  • \$\begingroup\$ Thanks, that's helpful. It will take me a day or two to absorb it more fully, but I hope to be back on the weekend with a semi-coherent followup. \$\endgroup\$ Commented Nov 6, 2020 at 3:24
  • \$\begingroup\$ So I don’t get of on the wrong foot, could you please address my confusion on the first few sentences? 1. CMOS sensors read each line sequentially. How long each line is "on" and "off" determines exposure time. The first sentence is about reading; the second is about exposure. I’m not grasping how they relate to each other. 2. But just as with mechanical focal plane shutters, the total time needed to take an exposure is the same, regardless of the exposure time. I haven’t been able to parse this. \$\endgroup\$ Commented Nov 8, 2020 at 21:43
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    \$\begingroup\$ The sensor can not be collecting photons that accumulate charges in each photosite (exposure) and move those charges off the sensior (readout) at the same time, Just like you can't pour the water out of a rainbucket into a measuring cup and catch rain falling on your yard at the same time. When a sensor is read it is "cleared" all charges are removed. When it is "turned on" it begins accumulated charges created by photons falling on each photosite. When it is "turned off" it stops accumulating charges created by photons falling on each photosite. \$\endgroup\$
    – Michael C
    Commented Nov 9, 2020 at 2:20
  • \$\begingroup\$ The time difference between when each photosite is turned on and turned off is the exposure time. That is, it's the total amount of time the photosite was collecting charges.how long it takes the sensor to sequentially turn on each line on the sensor is fixed. It (usually) starts at the bottom (which is the top of the inverted image) and goes line by line to the top (the bottom of the inverted image). \$\endgroup\$
    – Michael C
    Commented Nov 9, 2020 at 2:25
  • \$\begingroup\$ In much the same way mechanical shutter curtains in modern cameras transit the sensor at a fixed rate. It's the difference between when the first curtain begins opening and the second curtain begins following it in the same direction across the sensor that determines exposure time. At short exposure times the entire sensor is never uncovered at the same time. It may take 4 milliseconds (1/250) for each shutter curtain to transit across the sensor. If the second curtain begins closing 1 millisecond after the first curtain begins opening, the first curtain has only revealed 25% of the sensor. \$\endgroup\$
    – Michael C
    Commented Nov 9, 2020 at 2:27
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(Reason for this second answer: I just noticed that after I submitted my first answer to this question you responded to my comment to the question requesting more information about what photographic purpose led you to ask this question.)

The easiest and most effective way to get non-blurry photos of children moving around as children do when in low light situations is to provide more light. This allows your camera to use a shorter exposure time and/or allows it to expose so that the dim continuous ambient light has very little influence on the shot and the very short duration flash provides almost all of the light captured by the camera. Many flashes also provide "focus assist" lights that help the camera focus faster and more accurately.

The first thing we need to do is get this out of the way: Your smartphone does NOT have a flash. The maker and seller may call it a flash, but it is actually a low intensity continuous LED light that is turned on before the image is taken, left on for a significantly long enough amount of time that it won't "freeze" motion blur, and only turned off after the image has been exposed. It is nowhere near bright enough to allow a short enough exposure time to freeze children running around at play. It's also nowhere near short enough to freeze motion by overpowering the ambient light.

We have a good number of existing questions here at Photography SE that deal with the scenario with which you, and pretty much anyone else with children or grandchildren, are having difficulty. Here are links to a few of them:

What do I need in a camera to take non-blurry photos of a fast-moving child?
How do I best use a pop-up flash for indoor birthday parties?
How to take photos of children in difficult lighting?
How can I take pictures of active children with a DSLR in low light?
What lens should I buy to get better pictures of a one-year old running around a dimly-lit house?
How can I get good pictures of a fast-moving child without high FPS?
How can I get non-blurry photos of a wriggly baby with a point-and-shoot camera?

Notice that none of them even mention attempting to do it with a smartphone.

As one of the answers to that last question regarding point-and-shoot cameras begins:

Well, you are bringing a knife to a gunfight here.

For a very long time, most point-and-shoot cameras were more capable in this area than most smartphones. That's because many point-and-shoot cameras have a true flash. Some, though not as many, can even control external flash, which is where you really need to go to get high quality pictures of moving subjects indoors in dim light.

But don't lose heart just yet. Getting phones and flashes to work together is already beginning to happen! There are already products available that work with a large number of smartphones that can trigger and control sophisticated off-camera flashes! It's revolutionizing the way that phone cameras can be used to get shots they couldn't get before.

For instance, there's this recent article posted at Fstoppers:

Will Off Camera Flash Make The iPhone a Professional Camera?

Here's a PetaPixel article about the Godox A1.

Godox Made an Off-Camera Flash and Transmitter for Smartphones

You can find plenty of other articles about the Godox A1 with a simple internet search. It's a unit that has a small built-in flash, a built-in continuous LED light, and most significantly, it can control the entire Godox 2.4Ghz lighting system! That's all the way from their cheap $60-70 manual flashes to their high end studio lights and battery powered monolights.

Speaking of flashes, here are a few more links to existing questions here at Photography SE:

How do I manage good photos of babies and kids?
Prime lens or flash: which upgrade will most improve baby photos?

There are plenty of others with the [Flash] tag. Some also include the [children] tag. To find question that include both, search for [flash] [children]

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  • \$\begingroup\$ Thank you. Very informative, especially about flash. As it happens, I've already bought a cheapish clip-on light. If it doesn't do the job the Godox A1 will be next in the in the queue. \$\endgroup\$ Commented Nov 17, 2020 at 23:52

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