Dear Yifan Meng,

I think the color of the Al2O3 film is due to non-stoichiometry. It is known that Al2O3 was already prepared in the form of fine powder with a black color due to non-stoichiometry. The film will probably lost its color after annealing in air. Adding oxygen in the chamber during deposition would help to have a transparent film. Changing the sputtering parameters can also to lead to more or less colored films due to a more or less reduction of the oxide. For the influence of sputtering conditions with iron oxides you can have a look to :

Oudrhiri-Hassani et al. Appl Surf Sci (2008) and Mauvernay et al. Thin Solid Films (2007). These publications are available on research gate on my account.

I think also that the problem of non-stoichiometry will not be changed by replacing sapphire by a ceramic target.

Best regards and good work!

Philippe.

Dear Yifan Meng,

I think the color of the Al2O3 film is due to non-stoichiometry. It is known that Al2O3 was already prepared in the form of fine powder with a black color due to non-stoichiometry. The film will probably lost its color after annealing in air. Adding oxygen in the chamber during deposition would help to have a transparent film. Changing the sputtering parameters can also to lead to more or less colored films due to a more or less reduction of the oxide. For the influence of sputtering conditions with iron oxides you can have a look to :

Oudrhiri-Hassani et al. Appl Surf Sci (2008) and Mauvernay et al. Thin Solid Films (2007). These publications are available on research gate on my account.

I think also that the problem of non-stoichiometry will not be changed by replacing sapphire by a ceramic target.

Best regards and good work!

Philippe.

As additional comment, you can also sputter pure Al in reactive mode (Ar+O2), you should have good transparency of your Al2O3 film.

Regards,

Emile HAYE

Thin transparent materials have interfirence colors, like soup bubbls or  thin oil film colors on water. Color depends on thickness and dielectric constant.

I concur with Philippe and the Emile that the cause of the coloring is due to the fact that you should add O2 to the Ar. Under all sputter conditions the target, whether this is sapphire or Al2O3, will loose some oxygen due to the impact of Ar ions and adding 1-2% oxygen to the argon gas will help to remediate this. Sub-stoichiometric Al2O3 will show a yellow or brownish colour depending on thickness and amount and oxygen lost (the delta in Al2O3-d). When the sputtering is done right, thin layers of Al2O3 (

The band gap of Al2O3is 7.0 and 7.6 eV for the amorphous and crystalline phases, respectively. Thus the stoichiometric phase is completely transparent in the visible range and it should look like this:

https://search.yahoo.com/search;_ylc=X3oDMTFiN25laTRvBF9TAzIwMjM1MzgwNzUEaXRjAzEEc2VjA3NyY2hfcWEEc2xrA3NyY2h3ZWI-?p=aluminum+oxide%2C+sapphire&fr=yfp-hrmob&fr2=p%3Afp%2Cm%3Asb&.tsrc=yfp-hrmob&fp=1&toggle=1&cop=mss&ei=UTF-8

The yellowish color is very indicative of a non-stoichiometric phase where there is excess metal (lack of oxygen) as a result of sputtering from the Al2O3 using Argon gas only. Consequently, there is loss of oxygen. The yellow color center is at ~ 600 nm, which is equivalent to ~ 2 eV. By considering the Bandgap of crystalline Al2O3 of 7.6 eV and the zero eV metallic Bandgap of Al, then the Bandgap of your non-stoichiometric film can be considered as an engineered Bandgap approximately equal to:

Eg(film) = xEg(excess Al) + (1-x)Eg(Al2O3) = zero + (1-x)Eg(Al2O3) = (1-x)Eg(Al2O3)

Thus,

2 eV = (1-x)(7.6) => x = 1 - (2/7.6) = 0.74

which means that the stoichiometric percent is only 0.26 and the rest is metal aluminum. In other words, your film composition is approximately:

Al: 0.74 + 0.26*(2/5) = 0.84 = 84%

O: 0.26*(3/5) = 0.16 = 16%

Therefore, your non-stoichiometric film is approximately Al0.86O0.16 , while the stoichiometric phase Al2O3 is equivalent to Al0.4O0.6

Too much aluminum in the film as a result of oxygen loss during sputtering

Dear Philippe Tailhades :

Thanks for your answer, however according to my recent experiments,adding 3%-10% oxygen into vaccum during deposition for 2 hours can deposit nothing on the substrate. Four-probe measurements show that the surface resistance of the sample was equal to pure Si substrate, and we can not find any films on Si substrate by step profiler.I hope you can give me some advices and many thanks to you.

Yifan Meng

Dear Emile Haye 

     Thanks for your answer, I had tried to use reactive RF sputtering method to fabricate Al2O3 thin film, here are my parameters:

Substrate temperature:350℃

     Target:Al(99.99%)

     Gas:Ar and O2

     O2 partial pressure: varied from 2% to 15%

     Total pressure:varied from 0.1 to 0.6 Pa

     Substrate:Si(100) with thickness of 0.5mm

     However, I can't detect any films in substrate by four-probe and step profiler, I'd like to ask that is it all right about my parameters? Thanks a lot!

    Yifan Meng

You mention two cases where you say you did not see a film. However if you made about 25 nm of a transparent film maybe:

• it is not visible by eye
• the probes of the four point probe just puncture the small layer and still contact the silicon (and anyway finite contact resistances should have no influence on this measurement)
• it was somehow not so clear to detect with profilometry

How do you determine layer thickness with profilometry anyway? Because it can only measure a height difference (step), so it cannot measure a continuous film. Do you pattern the film before profilometry? And what kind of profilometry? AFM?

I think ellipsometry would in this case be a more useful method to determine the presence of a thin transparent layer.

I concur with Pieter, you probably deposited a thin transparant Al2O3 layer which is difficult to measure wit profilometry. Try using ellipsometry or inteferometry to measure the optical layers.