N-type silicon (Si) has been produced by doping with phosphorus (P). The charge-carrier population based on electrical conductivity measurments is determined to be 3.091*1017 carriers/cm3 at room temperature.
Calculate the doping level of P (how much P has been added to the Si). Express your answer in units of g P / kg Si.
What is the majority charge carrier in the material described above?
p-type n-type electrons holes
majority charge carrier are electrons.
Answer for doping level =6.82*10^-3 gP
electrons
Where you find it anonyous?
To calculate the doping level of phosphorus (P) in the silicon (Si), we need to understand that each phosphorus atom contributes one free electron to the silicon crystal structure.
1. Convert the given charge-carrier population of 3.091*10^17 carriers/cm^3 to carriers/m^3:
- Given: 3.091*10^17 carriers/cm^3
- Conversion: (3.091*10^17 carriers/cm^3) * (1 m^3 / 1e6 cm^3) = 3.091*10^11 carriers/m^3
2. Determine the volume of silicon (Si) needed to obtain this carrier population:
- As the concentration is given in carriers/m^3, we can assume the volume of silicon is equal to 1m^3.
- Therefore, the number of phosphorus atoms added would be 3.091*10^11.
3. Determine the molar mass of phosphorus (P) and the molar mass of silicon (Si):
- The molar mass of phosphorus (P): 30.974 g/mol
- The molar mass of silicon (Si): 28.086 g/mol
4. Convert the number of phosphorus atoms to grams:
- Grams of P = (Number of phosphorus atoms) * (Molar mass of P)
- Grams of P = (3.091*10^11) * (30.974 g/mol)
5. Determine the number of silicon atoms in 1 kg (1000 g) of silicon:
- Number of silicon atoms = (1 kg / Molar mass of Si) * (6.022*10^23 atoms/mol)
- Number of silicon atoms = (1000 g / 28.086 g/mol) * (6.022*10^23 atoms/mol)
6. Calculate the doping level of P:
- Doping level of P = (Grams of P) / (Number of silicon atoms) * 1000 (to get g P / kg Si)
To determine the majority charge carrier, we need to look at the type of doping. In N-type silicon, the majority charge carrier is electrons.
Please note that the exact values of molar masses and other constants used in the calculation might vary slightly based on the specific reference source.