Refraction of IR "light" in CCD sensors SiO2 layer

[Balint Kovats]

Dear friends I am new at this forum thank you for accepting my application first of all.

My question is that I don't understand the optics/physics behind the reason why Si-based CCDs are not sensitive for IR-light (above 1000-1100 nm) if on the top of the p-type Si there is a SiO2 layer which has a refractive index of 1.449 at 1100 nm wavelength. Taking into consideration that EM waves entering into a new medium because of their frequency is constant their wavelenght and speed will change can we calculate in this way: wavelength = speed of light in vacuum/frequency >> frequency = speed of light in vacuum/wavelength in our case f = 3*10^8[m/s]/1100*10^-9[m] = 2,73*10^14 [1/s] and as written in this post (https://www.physicsforums.com/threads/changing-wavelength-of-light-ps-hello.1833/) new wavelength in the new medium = speed of light in vacuum/frequency*refraction index >> new wavelength in the new medium = 3*10^8[m/s]/2,73*10^14[1/s]*1,449 = 7,58*10^-7 [m] = 758 [nm] There we go: if the Si is perfectly coated with SiO2 layer why isn't the incoming IR-light impacting the surface of the Si at 758 nm wavelength and so it should generate photoelectric effects and the CCD should be sensitive for that range too. Can anyone explain this? Is there something wrong with my knowledge about CCD manufacturing or the basic optical principles?

Thank you very much

 

[rbelli1]

This has to do with the critical wavelength of the photoelectric effect.

http://hamamatsu.magnet.fsu.edu/articles/quantumefficiency.html

BoB

 

[Drakkith]

This has to do with the critical wavelength of the photoelectric effect.

http://hamamatsu.magnet.fsu.edu/articles/quantumefficiency.html

BoB

Whoa! 1100 nm light penetrates 7600 μm into the silicon! That's most of the way through the chip!

Can anyone explain this? Is there something wrong with my knowledge about CCD manufacturing or the basic optical principles?

From the article linked by rbelli1:

In cases where the photon energy is greater than the band gap energy, an electron has a high probability of being excited into the conduction band, thus becoming mobile. This interaction is also known as the photoelectric effect, and is dependent upon a critical wavelength above which photons have insufficient energy to excite or promote an electron positioned in the valence band and produce an electron-hole pair. When photons exceed the critical wavelength (usually beyond 1100 nanometers), band gap energy is greater than the intrinsic photon energy, and photons pass completely through the silicon substrate. Table 1 lists the depth (in microns) at which 90 percent of incident photons are absorbed by a typical CCD.

Note that the energy per photon is not wavelength dependent, but frequency dependent. This is a requirement of energy conservation. Otherwise light would gain or lose energy as it passed through mediums of different refractive indices.

 

[rbelli1]

That's most of the way through the chip!

That's all the way through a handful of chips. It seems that for silicon something on the order of a microchip's thickness is essentially transparent.

BoB

 

[Balint Kovats]

Thank you for the great article and explanation rbelli1 and Drakkith. What do you think is there any way to increase somehow the energy of a single photon or a beam of photons higher than the band gap energy to make them energized enough to cause photoelectric effect in Si?

 

[Drakkith]

What do you think is there any way to increase somehow the energy of a single photon or a beam of photons higher than the band gap energy to make them energized enough to cause photoelectric effect in Si?

Not that I know of. We usually have to choose a different material with a different bandgap energy if we want to detect IR beyond 1100 nm.

 

[rbelli1]

You could use a phosphor coating and increase your illuminance to induce two photon fluorescence. Are you in control of the light source? Can your subject withstand the additional infrared irradiance?

BoB

 

[Balint Kovats]

Yes I am in control of the source and the subject can withstand the IR radiance, though inducing two photon fluoresence does change the energy of a single photon?

 

[Balint Kovats]

And to get back to the original question: on the surface of the Si photodiode the SiO2 coating should change the visible light's wavelength too based on the above mentioned optical equations. Is there an other medium between the photodiode and the SiO2 layers in which the light's wavelength changes back for example air or vacuum?

 

[rbelli1]

The changes are all cumulative. The wavelength in the silicon is the same no matter how many different materials it passes through. The intensity may be altered due to absorption.

BoB