We are bombarding Silicon surface with Kaufmann ion source of current density 450 uA/cm^2. Then what will be the heat produced and the temperature of the bombarded surface during that bombardment so that we can calculate the diffusivity of the silicon.
As mentioned by Mr. Nazmabadi, the deposit energy by ion is roughly collect. and the ion energy is very low so we can recognize the deposit energy on sueface of Sillicon.
Then the surface temperature is calculated by Q=C（Ts-T∞）.
Probably you should take into account the profile of beam typically Gussian.
Mohammad Nazmabadi,
what are the specifications of the equation P/A=VG, P= power, A area, V voltage, G for current density . Do I have the right capture or not.?
These answers have not provided enough information to actually calculate the temperature rise. Also you have not identified the time period. The power per unit area calculation is correct but you have to appreciate what happens to that energy. There are essentially three pathways for that energy and they are all linked to the temperature. They are an increase in the temperature of the surface region (the previous responses did not identify a thickness in their calculations), the thermal diffusion from the surface to be back (assuming it is mounted on a block of some sort) and the thermal radiation from the surface. If it is in a steady state then the sum of these three components will equal the power incident due to the ion beam energy and the current per unit area. As material diffusion is VERY sensitive to temperature, a reliable estimate of the temperature taking these three into account (either in a dynamic or steady state mode) is essential to provide you with an accurate estimate of diffusion.
John O''Connor sir,
can you refer me some paper from where I can calculate the temperature?
No because it is basic physics. Perhaps we can communicate directly rather than through this forum. Email me with a little more detail directly.
I agree with Dr. O'Connor, but would like to add the surface sputtering effect like another component of primary ion energy transfer.
I don't know any paper where the temperature of ion irradiated surface is calculated. It looks essentially more easy to measure this temperature by an IR thermo-sensor for remote temperature measurement.
Yuriy, technically you are correct but the fraction of energy leaving in this manner is a small fraction and not significant when compared to the uncertainties associated with the other components. The peak in the sputter distribution is say about 10eV, and assume that the sputter coefficient is say 5, then for each 2keV ion coming in 50eV is lost to sputtering. Thus the energy lost to sputtering is a couple of percent. The errors associated with the other components can easily be 10% so normally the sputtering component can be ignored. Having said that, I have to agree that if you want to be thorough and comprehensive in your treatment then sputtering has to be included.
Thank you for your contribution.
Dr. O'connor and Dr.Kudriavtsev, I absolutely agree with you but as a simple estimation, our suggestion was not so far. 2keV ion range is very very short it regognized on only surface and let's say the experimental system is in vacumm, then the 3 thermal tranfer are not so large problem what I want to say irradiation and convection are negligeble I think.
In my opinion, inprotant things is what and/ or how quesioner wont to know?
Sorry for bother you.
Assuming the sample is thin the dominant thermal loss is conduction through the sample however it does not take too much input power (or temperature increase) for radiation to become the second most important form of energy loss or even the most important. Naturally convection plays no role as it has to be in a vacuum. The issue of the range of the 2keV particle is not the issue as diffusion quickly moves the thermal energy over a much greater range.
Hi Basanta,
If I multiply your current density, times your maximum beam energy, I get 9 kW/m^2. How big is this? Well it's about 9x the power density (1000 watts per square mater) of sunlight reaching the earth's surface. So if you can imaging a piece of silicon getting hot if it was sitting in direct sunshine, plan on it getting 9x hotter! If you provide the incident beam's spot size, then it would be possible to estimate the total power and temperature. Otherwise, you can't get too many more quantitative results...
John,
It wont get 9 times hotter as in that temperature regime (ie 9 times room temperature) the dominant energy loss process is by radiation rather than conduction so the power radiated goes as T^4. It will get square root of 3 times hotter if that is the only heat loss process - however it is not.
Dear John Anthony Notte Sir,
The beam spot size is 12.56 cm^2 (pi*2cm*2cm) in our case. but not all the ion beam incident over the substrate because sample holder is rectangular and beam size is circular so few part cross that sample holder. 12cm ^2 is the area of the sample holder. only 12 Cm^2 area comes to be irradiated.
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