# Are silicon microbolometers inherently more expensive than conventional CMOS light sensors?

We're finally starting to see practical thermal imaging sensors (microbolometers) entering the consumer market. However, they are still vastly more expensive than comparable visible imaging sensors. The most basic 384x288 17µm pixel (i.e., 32mm2) thermal imagers run about $500, whereas$500 will get a 6000x4000 2µm pixel (i.e., 96mm2) CMOS sensor ... plus 5-axis sensor stabilization and more.

My question: Assuming the same economies of scale were applied as are already used for conventional CMOS sensors, is the manufacturing process for silicon microbolometers inherently more expensive? Or in the limit is it still just some (similar) number of photolithographic steps?

To elaborate: Thermal cameras look for radiation with wavelengths between 7-14µm, whereas visible light is in the range 0.4-0.7µm. Based on the physics alone, at the diffraction limit microbolometer pixels will have an order of magnitude greater surface area. Apparently commercial sensors are at the diffraction limit for both visible light (at 1 micron pixels) and thermal light (at 17 micron pixels). So, to make it fair, we might compare a 1" 24Mpx visible sensor with a 1" 300kpx thermal sensor.

Both sensors can be made from silicon using a CMOS process. The structure of microbolometers looks a little trickier than state-of-the-art visible spectrum CMOS sensors, requiring a thermal bridge for each pixel as well as vacuum encapsulation of the sensor. But I know little of large-scale manufacturing processes, so are these variables significant in the limit on a per-unit basis?

• I would think the older VGA CMOS cameras with 25um pixels would be good for far infrared with the IR block filter replaced with a daylight blocking filter. – Tony Stewart Sunnyskyguy EE75 Jan 13 '17 at 20:22
• @TonyStewart.EEsince'75 - Depends on your definition of "far infrared." For light around 1 micron, sure. But thermal radiation is around 10 microns. – feetwet Jan 13 '17 at 20:24
• I define FIR as 5~25um get an old Sony cam and try the mod then SIR is 0.7 to 5um – Tony Stewart Sunnyskyguy EE75 Jan 13 '17 at 20:31
• Poor question: you're including the thermal-IR lens, then ignoring its cost. Even if the sensor array costs zero, FLIR camera prices remain high. Choose: a single-crystal germanium lens? Zinc Selenide? Silicon lens? Chalcogenide-glass? Single element fixed-focus? Zoom lens? – wbeaty Feb 6 '17 at 20:11
• @wbeaty - True: For this question I am only interested in, and talking about, the IC. I'll remove references to the lenses. – feetwet Feb 6 '17 at 20:29

Microbolometers are inherently more expensive than other silicon ICs specifically because of their required 3-dimensional structure. In order to suspend the sensor pixels on thermal bridges the sensor layer has to be put on a substrate that is subsequently etched out from underneath the ~17-micron panels. Many procedures that are commonly used on CMOS to do something quick (such as fairly harsh cleaning steps, CMP, etc.) have to be replaced by more complex, slow alternatives, and even those have a significant enough failure rate that more chips have to be scrapped. For example, immersing the wafers into liquids or any process that involves a flow of some medium over them has to be done extremely carefully and slowly. Surface tension is a huge problem when it comes time to get the liquid etching reagent out from under those pixels without breaking them: You can't blow it off with pressure or boil it off with heat.

To give an idea of how sensitive they are: Here is an electron micrograph of an array that was hit with compressed air to remove dust:

The 3-D fabrication problems with microbolometers are comparable to the ones involved in making DLP chips, which remained relatively expensive even in large-scale production for consumer devices.

(Microbolometers are made that use only amorphous silicon, but for performance a vanadium oxide sensor is preferred. Adding VO necessitates a separate and more expensive fabrication line because it is a hazardous substance.)

Your "problem" (not your question) is: "How can I see LWIR (~ 7-14µm)" utilizing a Technology (in the Future) that would permit lower cost Sensors if "the same economies of scale were applied as are already used for conventional CMOS sensors.".

This image was produced using an uncooled ("hot") T2SL MWIR (3-5 μm) Detector. It has better contrast than a high quality LWIR Image.

Using SWIR permits seeing through common glass (and using conventional lenses) but only very hot objects (Engines, Fire, etc.) are readily discernable without reflected light which is required to see anything that isn't very hot. Using LWIR is better for exact measurement of temperature but requires expensive Optics and unless you're using Microbolometers you'll need cooling.

MWIR cameras are employed when the primary goal is to obtain high-quality images rather than focusing on temperature measurements and mobility.

The MWIR band of the spectrum is the region where the thermal contrast is higher due to blackbody physics; while in the LWIR band there is quite more radiation emitted from terrestrial objects compared to the MWIR band, the amount of radiation varies less with temperature (see Planck’s curves): this is why MWIR images generally provide better contrast than LWIR.

For example, the emissive peak of hot engines and exhaust gasses occurs in the MWIR band, so these cameras are especially sensitive to vehicles and aircraft.

Instead of confining your choice to microbolometers you need to look at QWIP, Type-II Strained Lattice (T2SL), or even Cooled LWIR all of which are more 'similar' to CMOS than a microbolometer; and thus have a better future potential for scaling (assuming enough interest in seeing LWIR radiation).

• Note: This question was transferred from Photo.SE (where this text appears: "Update: This question now answered here." created by an edit from the OP) from this question: photo.stackexchange.com/questions/86172/… - I mention this to inform the Reader that this is a "Photography Question" turned "Electronics Question", the presumption being that the OP wants a 'good photo' from reasonably priced Equipment rather than the ability to measure temperature with a Photo precisely. – Rob Mar 19 '17 at 18:04

The answer is in economies of scale. CMOS imaging sensors get produced much more than infrared sensors. Therfore CMOS imaging sensors get more R&D funding, more companies producing them and a bigger supply chain ect. Infared sensors are starting to get cheaper and adopted more widely.

MEMS are not inherently more difficult to manufacture and sometimes use the same processes, and can even use older cheaper lithography equipment because the size is bigger.

• The question explicitly asks you to assume "the same economies of scale were applied as are already used for conventional CMOS sensors...." Yes, it would seem that a 17-micron pixel is easier to manufacture than a 1-micron pixel. But on a microbolometer there is a 3-D structure supporting each pixel not needed on visible-light sensors, and the entire sensor has to be vacuum sealed. The question is whether those make the process significantly more expensive in the limit. Or, perhaps photolithography is already done in a vacuum and the thermal bridge is not a special feature? – feetwet Jan 13 '17 at 18:50