An intrinsic Si wafer is doped with 5 × 10^16 /𝑐𝑐 phosphorous atoms and 6 × 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, 𝑁𝐶 = 2.82 × 10^19 /𝑐𝑐). 

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

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

### Step 1: Find the net doping concentration
Given:
- Phosphorus atoms (donors) = \(N_D = 5 \times 10^{16} \, \text{cm}^{-3}\)
- Boron atoms (acceptors) = \(N_A = 6 \times 10^{16} \, \text{cm}^{-3}\)

The net doping concentration can be calculated as:
\[
N_D - N_A = 5 \times 10^{16} - 6 \times 10^{16} = -1 \times 10^{16} \, \text{cm}^{-3}
\]
Since \(N_A > N_D\), this is a **p-type** semiconductor with a net acceptor concentration of \(10^{16} \, \text{cm}^{-3}\).

### Step 2: Use the formula for the hole concentration
In a p-type semiconductor, the hole concentration (\(p\)) is approximately equal to the net acceptor concentration (\(N_A - N_D\)):
\[
p \approx N_A - N_D = 10^{16} \, \text{cm}^{-3}
\]

### Step 3: Calculate the electron concentration
The electron concentration \(n\) in a p-type semiconductor can be found using the mass action law:
\[
n \cdot p = n_i^2
\]
where \(n_i = 10^{10} \, \text{cm}^{-3}\) is the intrinsic carrier concentration.

Thus, the electron concentration \(n\) is:
\[
n = \frac{n_i^2}{p} = \frac{(10^{10})^2}{10^{16}} = 10^4 \, \text{cm}^{-3}
\]

### Step 4: Use the formula for the Fermi level
The Fermi level with respect to the conduction band edge (\(E_F - E_C\)) in a p-type semiconductor is given by:
\[
E_F - E_C = kT \ln\left(\frac{n}{N_C}\right)
\]
where:
- \(kT = 0.0259 \, \text{eV}\) at 300K
- \(N_C = 2.82 \times 10^{19} \, \text{cm}^{-3}\)
- \(n = 10^4 \, \text{cm}^{-3}\)

Substituting these values into the equation:
\[
E_F - E_C = 0.0259 \ln\left(\frac{10^4}{2.82 \times 10^{19}}\right)
\]
\[
E_F - E_C = 0.0259 \ln(3.55 \times 10^{-16})
\]
\[
E_F - E_C = 0.0259 \times (-35.57)
\]
\[
E_F - E_C \approx -0.92 \, \text{eV}
\]

### 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?

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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 (\(N_C\)) 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.

An intrinsic Si wafer is doped with 5 × 10^16 /cc phosphorous atoms and 6 × 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, NC = 2.82 × 10^19 /cc).

Learn how to calculate the Fermi level location with respect to the conduction band edge in a p-type Si wafer doped with phosphorous and boron at 300K. This step-by-step solution covers key concepts such as carrier concentration, Fermi level calculation, and semiconductor physics.

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