A semiconductor doped Si wafer with the thickness of 525m. Carrier mobility is 15.0

m2/(Vs). Carrier concentration is 323 m10 
n . Carrier charge is 1.6×10-19 C. The resistance
is measured between two circular electrodes with the radius of 1mm each, which are attached
on the opposite sides of the wafer. Assume uniform current flow between the electrodes. That is the resistance?

To determine the resistance between the two circular electrodes attached to the opposite sides of the semiconductor-doped Si wafer, we can use the formula derived from Ohm's law:

\[ R = \frac{L}{\sigma \cdot A} \]

where:
- \( R \) is the resistance.
- \( L \) is the thickness of the wafer (distance between the two electrodes).
- \( \sigma \) is the electrical conductivity of the material.
- \( A \) is the cross-sectional area through which the current flows.

### Step 1: Calculate the electrical conductivity (\(\sigma\))

The electrical conductivity \(\sigma\) is given by:

\[ \sigma = q \cdot n \cdot \mu \]

where:
- \( q \) is the carrier charge (\(1.6 \times 10^{-19} \, \text{C}\)).
- \( n \) is the carrier concentration (\(10^{23} \, \text{m}^{-3}\)).
- \( \mu \) is the carrier mobility (\(15 \, \text{m}^2/(\text{V} \cdot \text{s})\)).

Substituting the given values:

\[
\sigma = (1.6 \times 10^{-19} \, \text{C}) \times (10^{23} \, \text{m}^{-3}) \times (15 \, \text{m}^2/(\text{V} \cdot \text{s})) 
\]

\[
\sigma = 2.4 \times 10^{6} \, \text{S/m}
\]

### Step 2: Determine the cross-sectional area (\(A\))

The cross-sectional area \(A\) is the area of the circular electrode, which can be calculated using the formula for the area of a circle:

\[
A = \pi \times r^2
\]

where \( r = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \).

\[
A = \pi \times (1 \times 10^{-3} \, \text{m})^2 
\]

\[
A = \pi \times 10^{-6} \, \text{m}^2
\]

\[
A \approx 3.14 \times 10^{-6} \, \text{m}^2
\]

### Step 3: Calculate the resistance (\(R\))

Now that we have \(\sigma\) and \(A\), we can calculate the resistance \(R\):

\[
R = \frac{L}{\sigma \cdot A}
\]

where \( L = 525 \, \mu\text{m} = 525 \times 10^{-6} \, \text{m} \).

Substituting the values:

\[
R = \frac{525 \times 10^{-6} \, \text{m}}{2.4 \times 10^{6} \, \text{S/m} \times 3.14 \times 10^{-6} \, \text{m}^2}
\]

\[
R \approx \frac{525 \times 10^{-6}}{7.536 \times 10^0} \, \Omega
\]

\[
R \approx 6.97 \times 10^{-5} \, \Omega
\]

So, the resistance \(R\) is approximately \(6.97 \times 10^{-5}\) ohms.

Would you like more details or have any questions? Here are some related questions:

1. How does doping concentration affect the resistance of a semiconductor?
2. What is the relationship between carrier mobility and conductivity?
3. How would the resistance change if the thickness of the wafer were doubled?
4. What are some common methods to measure the resistance of a semiconductor wafer?
5. How does temperature influence the carrier mobility in semiconductors?

**Tip:** When dealing with small resistances like this, consider potential sources of error in measurement, such as contact resistance at the electrodes.

Learn how to calculate the resistance between two circular electrodes attached to a semiconductor-doped Si wafer. This problem involves applying Ohm's Law and understanding semiconductor physics. Detailed step-by-step solution included.

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