# What is the percent concentration of doping in n and p-type regions of typical solar cell?

I have heard that you need a very small percentage of boron (p-type) and and phosphorus (n-type) in each respective region in order to achieve the desired concentration necessary for the workings of a solar cell. What exactly is the composition of silicon vs. boron (for example 99% silicon vs. 1% boron) in the n and p-type regions of a typical solar cell?

## 1 Answer

Silicon has a density of $$\2.33\ {\rm g/cm^3}\$$. And its atomic weight is about 28. Knowing that 1 atomic mass unit is about $$\1.66\times10^{-24}\ {\rm g}\$$, we can find that there are about $$\5\times10^{22}\$$ silicon atoms per cubic centimeter of material.

If the doping density in a solar cell is about $$\10^{17}\ {\rm cm^{-3}}\$$ (I don't know if this is the most likely value, but it's a common value for fairly strongly doped material in other devices and the first one I came up with in a google search for "solar cell dopant concentration"), then the fraction of silicon atoms replaced by dopants in that material is about $$\2\times10^{-6}\$$, or 2 ppm.

Of course if you have a more accurate figure for the doping concentration in some particular solar cell you're interested in, you can easily calculate the value for that device yourself.

• said another way, at a doping density of 10^17 cm^-3 what percent of the material would be made of dopants and what percent would be made up of silicon? Commented Sep 15, 2020 at 1:16
• @gstudent, 2 ppm is 0.0002% Commented Sep 15, 2020 at 1:40