As I have known so far, image sensor gets heated when being exposed to light for longer time.
Is there any other reason why an image sensor gets heated? Does heating depends on sensor type (like CCD, CMOS, ets.)?

Also do live view function make the sensor more heated?


2 Answers 2


An image sensor is basically a small computer chip, and have similar heating characteristics. When transistor gates switch from on to off, or the other way you have small electrical currents in the chip. Everywhere on the chip there is a tiny amount of resistance, and when you have currents going through resistors most of the energy turns into heat.

During readout from an image sensor there are a whole lot of transistors changing states, and hence they generate a bit of heat. When you take one picture there isn't too much build up of heat. But when you take several (usually 10-20) pictures every second (live view) there isn't enough time between the pictures to get rid of the heat generated. Over time this will heat the sensor.

I don't remember the numbers exactly, but a CCD should generate less heat than a CMOS, since it contains less logic (transistors) per pixel. While using live view I don't think the difference is too big, both will heat up.

  • \$\begingroup\$ +1 but is this heat a problem? Electronic viewfinders are constantly in Live-view mode, does this mean a dSLR has an advantage here, or is the generated heat so low it is not a real problem after all? \$\endgroup\$ Oct 25, 2013 at 13:36
  • \$\begingroup\$ It can cause more noise in the pictures, so yes, dSLRs have an advantage here. \$\endgroup\$ Oct 28, 2013 at 13:50
  • \$\begingroup\$ Until recently EVFs tended to be seen more on cameras with smaller sensors, which tend to demonstrate more noise anyway. But they might also be easier to actively cool since there might be more room around the sensor. \$\endgroup\$
    – Michael C
    Jan 26, 2014 at 1:15
  • \$\begingroup\$ Every online reference I have seen lists one of the advantages of CMOS senosrs over CCD sensors is that CMOS sensors consume less power and generate less heat than CCD sensors, and that the disparity grows the larger the size of the sensor. See page 15 of this example: robgalbraith.com/images/canon_full-frame_cmos_white_paper.pdf \$\endgroup\$
    – Michael C
    Jan 27, 2014 at 0:56
  • \$\begingroup\$ "Transferring voltage requires almost no power compared to transferring a charge, which must move mass. So even with a larger CMOS sensor, power consumption does not change as long as the number of channels is not increased. CCDs, on the other hand, transfer output charges ”as is,” consuming power for the horizontal reading. The bigger CCDs are, the more power they consume. Making them faster also requires more power." \$\endgroup\$
    – Michael C
    Jan 27, 2014 at 0:59

On a sunny day, sun transports power of 1000 W / m2 to us. If you have a lens with a front lens opening of 20cm2, this means 1-3 watts of energy in the form of light is transported to the sensor inside the camera.

On a dark cloudy day the above 1000W / m² shrink to 1/10th, i.e. 0.1 W to 0.3 W of energy is transported to the shutter and sensor.

Energy means heat, and thus temperature will rise inside the camera.

On the other hand: how much electrical energy could a sensor transform into heat?

Let's assume your camera has a battery of 8V x 1.2 AH of energy = 10 Wh. lets assume your battery drains after 1 hour of continous camera activity. Then within 3600 seconds the 10Wh of battery energy is consumed by camera processor and sensor. 10 Wh / 3600s = 10W

i.e. if you work continously with live view, electrical heating from circuitry is much higher than heating by light rays.


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