# Thermo-optic effect

1. Jun 14, 2006

### DePurpereWolf

I would like to know more about the thermo-optic effect. I've got the following down:
The thermo-optic effect is the thermal modulation of the refractive index of a material. The refractive index of a material can be modulated as a function of its thermo-optic coefficient α.
$$n(\DELTA T)=n_0 + \alpha \cdot \DELTA T .$$
A tunable fabry perot structure can operate by change in its refractive index as function of applied heat.

But my question is what makes this thermo-optic effect. Let's say it is a thin film of amorphous Silicon, how does this change the refractive index when it is heated? I do not think it is just linear to the change in size. Does anything change internally? I would appreciate it if anybody would have more info on this.

2. Jun 14, 2006

### Gokul43201

Staff Emeritus
It need not be stricly linear, but I think the thermal expansion is responsible. The primary reason for a change in dielectric constant would be the change in lattice parameter. Another reason would be a change in the crystal structure (at a phase transition), where a sharper change in dielectric properties is observed.

In any case, do not take my word on this. I'm sort of extrapolating from the dielectric properties of gases!

Last edited: Jun 14, 2006
3. Jun 15, 2006

### DePurpereWolf

Off course I cannot disagree on this that it is the lattice constant change that creates this refractive index change, I just think it's hard to believe.

If it's just a thin layer of a-Si on a wafer that is getting thicker, this thin layer is jused in DBR and fabry-perot etalons in which the thickness of the layer is very important. Why don't they just call it optical path change, or correlate it to the lattice parameter change.

Both refractive index and layer thickness are important parameters in the equations for reflectivity in Distributed Bragg Reflectors (DBRs) and filter wavelengths in fabry-perot etalons. When these scientists speak of tunable thin film filters, they use the term tunable refractive index by thermo-optic effect of the medium (a-Si in this case) and are silent about the layer height change.

Having a tunable optical filter utilizing the thermo-optic effect means there are multiple thin layers of alternating amorpheus silicon (Hydrogen 'doped') and silicon nitride. Having these alternating thin layers prevents the free lattice paramter increase of the a-Si as otherwise it might delaminate. Because of the thin alternating structure the heat will first only result in stress change in the medium.

I just don't believe it is as easy as the lattice paramter change :yuck:

P.S in an amorpheus silicon you can't really say lattice parameter, but just interpred it as the average bond distance between the Si atoms.

Here is a reference you can check: http://www.aegis-semi.com/technology/publications/Aegis_OFC_Mar2003_main.pdf

4. Jun 15, 2006

### Gokul43201

Staff Emeritus
Oh wait, I'm not saying that the effect is a change in the path length. No, that's not what I meant.

What I should have said is that the change in lattice parameter changes the electron density and the shape of the lattice potential. In turn, this will affect the dielectric constant.

This is not saying that this is the only thing respoonsible for the effect. In a semiconductor, the conduction electron density is a stronger function of temperature than just size effects. The electron density is important because it determines the size of screening, which in turn determines the value of the (Thomas-Fermi, for instance) dielectric function.

Also, look at the typical sizes of the two effects (thermo-optic coefficient, and thermal expansion coefficient). They are both of the order of 10-6 to 10-5 per K. Of course, that's not proof of anything - just supporting evidence, perhaps.

Why don't you just spend an hour in the library (or the internet, if you have journal access) and simply hunt down the theory? I'm interested enough, that if you don't do it by the weekend and settle this question, I might just take a look.

Last edited: Jun 15, 2006
5. Jun 15, 2006

### DePurpereWolf

I just understood that if the lattice parameter changes, the whole volume per atom has to change, and thus the layer expands. If it is related to the coefficient of expansion it is related to the stress induced by that. If that is the case than heating is indeed the best way to apply the stress, but you could possible get the same thing by (mechanical)pressure. If it is not the expansion of the material, but more like electrons going in the valence band or change in the energy-gap or what not, than it is more of a quantum phenomena, not a 'mechanical' one.
I will, I'm just not near a library now (friday again), i do not have journal access. There are some interesting IOP and IEEE journals that might be about the issue, and I've read some (that I could access), but it just doen't clarify the mechanism that changes the refractive index.

6. Jun 15, 2006

### Gokul43201

Staff Emeritus
For what it's worth, I'm almost positive that strain affects the thermo-optic coefficient. In any case, I'll look more into this tomorrow night or saturday.

7. Jun 15, 2006

### ZapperZ

Staff Emeritus
I don't quite understand the difficulty here. ANY change in the lattice will cause a change in the phonon spectrum. This is what couples to any photon going through the material, and thus, would affect optical property (dispersion, refractive index, transmission, reflectivity etc...). Such a change can be accomplished via stress/strain/mechanical pressure/heat/structural phase transition/etc.

Now, for a semiconductor/dielectric, depending on the energy gap, you can see a difference in, for example, the reflectivity, of the material if you heat it up can cause a significant enough of a change in the charge carrier density. However, if the heating effect is small enough that it doesn't change the phonon spectrum, I'm not so sure as to the extent that such additional charge carriers can affect the refractive index.

Zz.

8. Jun 15, 2006

### Dr Transport

The thermooptic constant for many materials is less than 0.0001%, so the index of refraction chnages very little over normal operating temperatures, the carrier density has a much larger affect on index than temperture change.....

9. Jun 16, 2006

### DePurpereWolf

From the reference given in post #3 the following:
300C times 0.36x10^-3 is about 0.1, 0.1 over 3.6 is 0.03, 3% change.
There is no talk of applying a current through the a-Si and seeing a change in refractive index.
I also am not certain if a-Si is conductive or not.

10. Dec 1, 2008

### Energy Recrui

hi,
The Poly(vinylidene) fluoride (PVDF) thin films with a high content of Î²-phase were prepared by controlling heat-treatment temperature using casting from the poled solvents. The crystallite microstructure of thin films was depicted by the techniques of X-ray diffraction and FTIR. The results showed that heat treatment was favorable for inducing the Î²- and Î³-phase formation of PVDF. The Î² phase films were obtained with heat treatment at temperatures ranging from 60Â°C to 120Â°C and annealing at 120Â°C after casting from DMF. The thermo-optical effect of Î² phase PVDF films was investigated using a spectroscopic ellipsometer. At temperatures ranging from 20Â°C to 100Â°C, the refractive index of PVDF was negatively correlated with the temperature between 350 and 1500 nm. The value of the t.o. coefficient of PVDF films was calculated at all temperatures. The maximum value of the t.o. coefficient was about 3.3 Ã 10-4/Â°C at the ascending stage of temperature and 3.0 Ã 10-4/Â°C at the descending stage of temperature. Therefore, it is possible to use the thermo-optic effect of the Î² phase PVDF for long wavelength infrared imaging.