Math Problem Statement

An intrinsic Si wafer is doped with 5 Γ— 10^16 /𝑐𝑐 phosphorous atoms and 6 Γ— 10^16 / 𝑐𝑐 boron atoms. Assume that at 300K, the intrinsic carrier concentration in Si is 10^10 /cc and all the dopant atoms are ionized. What is the location of the Fermi level with respect to the conduction band edge at 300K? (Assume the effective density of states at the conduction band edge, 𝑁𝐢 = 2.82 Γ— 10^19 /𝑐𝑐). 

a) 0.2 eV    b) 0.56 eV    c) 0.16 eV    d) 0.92 eV 

Solution

To determine the location of the Fermi level (EFE_F) with respect to the conduction band edge (ECE_C) for a doped intrinsic Si wafer at 300K, we can follow these steps:

Step 1: Find the net doping concentration

Given:

  • Phosphorus atoms (donors) = ND=5Γ—1016 cmβˆ’3N_D = 5 \times 10^{16} \, \text{cm}^{-3}
  • Boron atoms (acceptors) = NA=6Γ—1016 cmβˆ’3N_A = 6 \times 10^{16} \, \text{cm}^{-3}

The net doping concentration can be calculated as: NDβˆ’NA=5Γ—1016βˆ’6Γ—1016=βˆ’1Γ—1016 cmβˆ’3N_D - N_A = 5 \times 10^{16} - 6 \times 10^{16} = -1 \times 10^{16} \, \text{cm}^{-3} Since NA>NDN_A > N_D, this is a p-type semiconductor with a net acceptor concentration of 1016 cmβˆ’310^{16} \, \text{cm}^{-3}.

Step 2: Use the formula for the hole concentration

In a p-type semiconductor, the hole concentration (pp) is approximately equal to the net acceptor concentration (NAβˆ’NDN_A - N_D): pβ‰ˆNAβˆ’ND=1016 cmβˆ’3p \approx N_A - N_D = 10^{16} \, \text{cm}^{-3}

Step 3: Calculate the electron concentration

The electron concentration nn in a p-type semiconductor can be found using the mass action law: nβ‹…p=ni2n \cdot p = n_i^2 where ni=1010 cmβˆ’3n_i = 10^{10} \, \text{cm}^{-3} is the intrinsic carrier concentration.

Thus, the electron concentration nn is: n=ni2p=(1010)21016=104 cmβˆ’3n = \frac{n_i^2}{p} = \frac{(10^{10})^2}{10^{16}} = 10^4 \, \text{cm}^{-3}

Step 4: Use the formula for the Fermi level

The Fermi level with respect to the conduction band edge (EFβˆ’ECE_F - E_C) in a p-type semiconductor is given by: EFβˆ’EC=kTln⁑(nNC)E_F - E_C = kT \ln\left(\frac{n}{N_C}\right) where:

  • kT=0.0259 eVkT = 0.0259 \, \text{eV} at 300K
  • NC=2.82Γ—1019 cmβˆ’3N_C = 2.82 \times 10^{19} \, \text{cm}^{-3}
  • n=104 cmβˆ’3n = 10^4 \, \text{cm}^{-3}

Substituting these values into the equation: EFβˆ’EC=0.0259ln⁑(1042.82Γ—1019)E_F - E_C = 0.0259 \ln\left(\frac{10^4}{2.82 \times 10^{19}}\right) EFβˆ’EC=0.0259ln⁑(3.55Γ—10βˆ’16)E_F - E_C = 0.0259 \ln(3.55 \times 10^{-16}) EFβˆ’EC=0.0259Γ—(βˆ’35.57)E_F - E_C = 0.0259 \times (-35.57) EFβˆ’ECβ‰ˆβˆ’0.92 eVE_F - E_C \approx -0.92 \, \text{eV}

Step 5: Conclusion

Thus, the Fermi level is approximately 0.92 eV below the conduction band edge.

The correct answer is: d) 0.92 eV

Would you like more details on any step, or have any other questions?


Here are 5 related questions to expand on this topic:

  1. How does temperature affect the intrinsic carrier concentration in a semiconductor?
  2. What happens to the Fermi level position when the doping level increases?
  3. How does the effective density of states at the conduction band (NCN_C) change with temperature?
  4. How would the Fermi level shift in an n-type semiconductor under similar conditions?
  5. What are the key differences between intrinsic and extrinsic semiconductors?

Tip: Always ensure to check whether a semiconductor is n-type or p-type by comparing donor and acceptor concentrations before calculating the Fermi level.

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Math Problem Analysis

Mathematical Concepts

Semiconductor Physics
Carrier Concentration
Fermi Level

Formulas

Net doping concentration: N_D - N_A
Electron concentration: n = ni^2 / p
Fermi level: E_F - E_C = kT * ln(n / N_C)

Theorems

Mass Action Law
Boltzmann Approximation for Carrier Concentration

Suitable Grade Level

Undergraduate (Physics/Electrical Engineering)