Posted by Dan (126.96.36.199) on October 12, 2011 at 14:42:18:
In Reply to: Doping Concentration posted by Mike Foster on October 12, 2011 at 14:41:35:
: I need to know what the doping concentrations in PPM would be for N-Type, Phosphorous doped silicon.
That's a question open to interpretation. Assuming you mean doping concentration as in electrons or holes the answer is directly related to the ppm of silicon. First we need some values:
Atomic weight of silicon: 28.0855
Density (near r.t.) 2.329 g/cm-3
Avogadro's number = 6.0221415E+23
Divide the atomic weight by the density:
28.085/2.329 = 12.059 cm-3 of Si per mole
Divide Avagadro's number by the result:
6.0221415E+23/12.059 = 4.994E+22 atoms of Si per cm-3
1ppm carriers = 4.994E+22/1E-6= 4.99388E+16 atoms/cm-3 of Phosphorus. Or Arsenic or Boron etc.
Now if you are trying to get the ppm in grams, you'll need to take into account the weight of the dopant. Again we'll need some values. Some of those we already have, now we need the atomic weight of Phosphorus and... oh let's take the P-type dopant Boron. And we'll need the AMU:
Atomic weight of phosphorus: 30.974
Atomic weight of boron: 10.81
Reciprocal of Avagadro's number (AMU) = 1.6605E-24
Divide the density by the product of the atomic weight of the element and the atomic mass unit, AMU.
For Phosphorus: 2.329/(30.974 x 1.6605E-24) = 4.530E+22
1ppm Phosphorus = 4.530E+22 x 1E-6 = 4.530E+16 atoms/cm-3
For Boron: 2.329/(10.81 x 1.6605E-24) = 1.297E+23
1ppm Boron = 1.297E+23 x 1E-6 = 1.297E+17 atoms/cm-3