When we use Cz-crystalline silicon wafers, what is the maximal doping density that can be processed using for example ionic diffusion ?
Article High efficiency PERL cells on CZ P-type crystalline silicon ...
https://link.springer.com/content/pdf/10.1007/s12633-019-00353-9.pdf
Doping concentration gradually decrease with the depth of the bulk Si. Usually within 0.5 micron depth from the surface the maximum doping concentration can be achieved in diffusion process is around 10^19-10^20 /cm3 for boron doping and 10^20-10^21/cm3 for phosphorus doping.
Gourab Das
I didn't think it was possible to dope crystalline silicon so strongly knowing that it has a density of states of about 10^19 cm-3. Thank you for this very useful information.
Just talking about the density of dopant atoms, for the relevant maximum number one has to compare with the atomic density of silicon crystals, which is 4.99 × 1022 cm–3. Thus, the densities mentioned by Gourab Das are already rather high but still possible – for still speaking about "doped silicon" and not about an alloy or a mixed semiconductor.
Talking about the consequences of such high doping levels (usually called "heavy doping") for the electrical (and optical) properties of silicon is a different story, with "band-gap narrowing", "Burstein-Moss effect", or "Mott transition" coming into play – but that wasn't the question, right?
Jan-Martin Wagner
Thank you for your precious contribution. A density of 4.99 × 1022 cm-3 seems to me to be so high for crystalline silicon, can you provide me with some documents on this? So I can use them as references.
In my simulation work, I use a doping density of 1015 cm-3 for a 200 µm n-doped crystalline silicon layer. I could therefore use a doping density of 1017cm-3 without affecting the electrical and optical properties of silicon. This will allow me to improve the efficiency of my device by 10%.
You are welcome! The number density of Si atoms can easily be calculated "from scratch" (therefore you don't need any reference documents): Take the lattice parameter of silicon, a = 0.543 nm (can be found anywhere), and remember that silicon has a diamond lattice structure so that 8 Si atoms belong to each unit cube (this you can easily figure out yourself). All there is left to do is to (i) find the number of unit cubes per cm3, and (ii) multiply this number by 8 to find the total number of Si atoms per cm3.
(Well, using a = 0.543 nm you will find 4.99678... × 1022 atoms per cm3, which would be rounded to 5 × 1022, of course, but with the high-accuracy value a = 5.431,020,511 Å available from https://physics.nist.gov/cgi-bin/cuu/Value?asil one ends up with the first number I gave you.)
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