The Panasonic Lumix DMC-FZ1000 has, according to the specs, a MOS image sensor - while cameras usually have CMOS (or CCD, for that matter).

Is there any difference between MOS and CMOS? Or is this a simple marketing trick?


4 Answers 4


MOS means Metal Oxide Semiconductor. It is a specific way of device formation on silicon wafer. See this link.

MOS, unqualified, usually means using a single doping technique to either create n-channel (typically) or p-channel FETs (field effect transistors). When ICs are manufactured this way, the costs are reduced. The disadvantage is that a certain amount of power is consumed if the device is powered but actually not doing any useful work. n-MOS has the advantage of reduced chip area, and if the specification is true, they may be using NMOS exclusively for more pixel density.

CMOS means complementary MOS, when both n-channel and p-channel FETs are created (and so it requires at least two doping pass in manufacturing). The effect is increased cost, but n-FET and p-FET transistors together allow for creation of static CMOS logic gates. These consume very little power when not toggling (there is static power consumption only because of some leakage current), and so these day CMOS, and consequently static logic cells are used almost exclusively in low-power applications, where battery lifetime is critical.

That having been said, I can imagine a nice sensor circuit that uses only NMOS devices, and handles power-related issues in a separate circuit.

  • \$\begingroup\$ Sorry but this answer is incorrect, in this regard MOS and CMOS are equivalent \$\endgroup\$
    – clabacchio
    Commented Oct 1, 2016 at 17:13
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    \$\begingroup\$ If you want to start an argument, state your facts (and make sure you know your facts right before). \$\endgroup\$
    – TFuto
    Commented Oct 1, 2016 at 17:26
  • \$\begingroup\$ This is a good answer, but what do you mean in the last paragraph by handling power-related issues in a separate circuit? In the context of MOS vs CMOS, the power issue is the fact that CMOS uses less power MOS. Using NMOS-only circuity consumes more power, so how does a separate circuit resolve that? \$\endgroup\$
    – scottbb
    Commented Oct 2, 2016 at 18:28
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    \$\begingroup\$ You can create a power-management circuit that puts the NMOS chip into sleep (basically disconnecting the VDD input). You probably want to implement that logic (counters, state machines, etc.) using CMOS logic, since that will be constantly on. \$\endgroup\$
    – TFuto
    Commented Oct 2, 2016 at 18:49
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    \$\begingroup\$ Thanks for answering this very technical question! I've considered to buy the Lumix FZ1000, but then I've settled for a Canon (which uses CMOS). The camera arrives today... which makes me very happy. :) \$\endgroup\$
    – user258532
    Commented Oct 3, 2016 at 8:44

From Wikipedia:

"The Live MOS sensor is a brand name of an NMOS Image sensor used by Panasonic, Olympus and Leica in their Four Thirds System DSLR manufactured since 2006. (Olympus E-330, Panasonic Lumix DMC-L1 and Leica Digilux 3).

A reviewer claims that the sensor can achieve the same image quality as CCD-based sensors while keeping energy consumption down to CMOS levels."

After further research it appears that NMOS sensors use only nMOS transistors and CMOS sensors use both nMOS and pMOS transistors. CMOS allows for better power dissipation and more transistors.


It appears that CMOS and MOS(NMOS) are very similar technologies while CCD is something completely different.

  • \$\begingroup\$ The article quoted, though it does show an image of an image sensor, does not seem to be very specific to image sensors. Also, it is kind of odd NMOS shows up in image sensors again - I would have assumed that depletion-load NMOS IC technology is mostly extinct since the late 1980s... \$\endgroup\$ Commented Aug 6, 2020 at 1:35

Shortly put, there is no difference. Generally CMOS is a subset of MOS, but in sensors' context it represent the same thing. I can't find any evidence of the sensor being different from other CMOS sensor, instead it seems it may have the same sensor as Sony's RX10.

MOS sensors are made by photo transistors which allow the direct reading of the photo-generated charge. Conversely, CCD sensors store the charge which is then shifted to other cells where it is read.

What makes me think there is no difference between MOS and CMOS (in imaging sensors) is that the actual phototransistor must be either pMOS or nMOS, and there may or may not be its complementary to form a CMOS pair.

I'd say it's not even marketing, just slightly different terms to indicate the same technology (MOSFET).

For more in-depth discussion you can refer to this question on EE.SE.


A digital imaging chip has a flat surface that is covered with photosites in a grid pattern. During the exposure, the surface of this image sensor is bombarded by photons. The number of photon hits in any given location on the surface of the chip is proportional to the scene’s brightness. The job of the photodiode is to covert photon hits into electrons. Each photon hit induces an electrical charge, the greater the number of hits, the greater the charge. The charge is temporally stored in the photosite.

The CCD (Charged-coupled devise), at the end of the exposure, moves these charges out of each photosite, row by row into a holding site called a transfer register. Each charge is now read out and converted from a charge to a proportional voltage. Next this voltage sent to a separate chip that coverts this analog signal into a digital signal.

The CMOS (Complementary metal–oxide–semiconductor) imaging chip does not store the charge, some processing is preformed and then the charge is sent to another part of the chip or to an adjacent chip for further processing.

The CCD reads out data from each photosite pixel by pixel then row by row. To do so required elaborate timing signals and other processing functions. Best if these are handed off to a separate chip. Because the CCD required adjacent circuit chips, the net result is more power consumption (bigger battery). The CMOS is a more complex chip as most functions are built in. The CCD must move charges around and transfer them to adjacent chips, this takes time to complete. The CMOS gains speed by handling the process internally.

The CCD is more efficient as to sensitivity. The CMOS with onsite processing takes up photosite space. The CMOS is less efficient as to sensitivity. The CMOS chip likely contains millions of converters and amplifiers. Each will be slightly different as to efficiency; this induces a fixed pattern noise. The CCD is more susceptible to “blooming”, this is a leakage of the charge to an adjacent photosite. Blooding is seen as streaks in high brightness areas of the image whereby the CMOS is resistant to blooming. These chips are evolving as is the software. Tomorrows chips will be smaller, more efficient. The goal is always, to make a faithful image.

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    \$\begingroup\$ Thank you for your well-written and clear answer, but I'm sorry to say that my question was about the difference between MOS and CMOS... \$\endgroup\$
    – user258532
    Commented Sep 30, 2016 at 15:50
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    \$\begingroup\$ Again, another brain dump vaguely related to the topic but which completely fails to answer the question. You clearly know a lot about photography and can write well, but you would be a much more valuable contributor to this site if you concentrated on making your posts actually answer what people are asking. \$\endgroup\$
    – Philip Kendall
    Commented Sep 30, 2016 at 16:50
  • \$\begingroup\$ @PhilipKendall -- Teaching is not about just filling the bucket, it’s more about lighting the way. \$\endgroup\$ Commented Sep 30, 2016 at 17:32
  • \$\begingroup\$ Pushing 80 – more than 55 years in the industry, maybe I do go off target – in other words – I talk too much. I generally I think of myself as a good educator in this narrow filed. \$\endgroup\$ Commented Sep 30, 2016 at 19:00
  • \$\begingroup\$ It's a great idea to share your vast knowledge, e.g. by posting a question and answering it yourself - like I've done with photo.stackexchange.com/questions/83178/… . In this instance, though, it doesn't seem like you've understood my question. \$\endgroup\$
    – user258532
    Commented Oct 3, 2016 at 8:38

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