Question

A rectangular wafer of pure silicon, with resistivity \(\rho=2300 \Omega \mathrm{m},\) measures \(2.00 \mathrm{~cm}\) by \(3.00 \mathrm{~cm}\) by \(0.010 \mathrm{~cm}\) Find the maximum resistance of this rectangular wafer between any two faces.

Short answer

Answer: The maximum resistance of the rectangular wafer between any two faces is 3450 Ω.

Step-by-step solution

Understand the resistance formula for a cylinder

The formula we will use for calculating the resistance \(R\) of a cylindrical conductor with resistivity \(\rho\), length \(L\), and cross-sectional area \(A\) is given by: \(R = \frac{\rho L}{A}\)

Calculate the three possible pairs of opposite faces

We have three possible pairs of opposite faces, each with different dimensions: 1. Pair A: Faces measuring \(2.00\,\text{cm} \times 3.00\,\text{cm}\) 2. Pair B: Faces measuring \(0.010\,\text{cm} \times 2.00\,\text{cm}\) 3. Pair C: Faces measuring \(0.010\,\text{cm} \times 3.00\,\text{cm}\) For each of these pairs, we will calculate the length \(L\) and cross-sectional area \(A\) of the corresponding rectangular cyliner, then use the resistance formula to find the resistance.

Calculate the resistance across Pair A

For Pair A, the length \(L\) is equal to the thickness of the wafer: \(0.010\,\text{cm}\), and the cross-sectional area A is equal to the face's area: \(2.00\,\text{cm} \times 3.00\,\text{cm}\): \(L_A = 0.010\,\text{cm}\) \(A_A = 2.00\,\text{cm} \times 3.00\,\text{cm}=6.00\,\text{cm}^2\) Now, convert L and A to meters: \(L_A = 1.0 \times 10^{-4}\,\text{m}\) \(A_A = 6.0 \times 10^{-4}\,\text{m}^2\) Using the resistance formula, we find the resistance across Pair A: \(R_A = \frac{\rho L_A}{A_A} = \frac{2300 \Omega\,\text{m} \times 1.0 \times 10^{-4}\,\text{m}}{6.0 \times 10^{-4}\,\text{m}^2} = 38.33\,\Omega\)

Calculate the resistance across Pair B

For Pair B, the length \(L\) is equal to the longer side of the wafer: \(3.00\,\text{cm}\), and the cross-sectional area A is equal to the face's area: \(0.010\,\text{cm} \times 2.00\,\text{cm}\): \(L_B = 3.00\,\text{cm}\) \(A_B = 0.010\,\text{cm} \times 2.00\,\text{cm}=0.020\,\text{cm}^2\) Now, convert L and A to meters: \(L_B = 3.0 \times 10^{-2}\,\text{m}\) \(A_B = 2.0 \times 10^{-6}\,\text{m}^2\) Using the resistance formula, we find the resistance across Pair B: \(R_B = \frac{\rho L_B}{A_B} = \frac{2300 \Omega\,\text{m} \times 3.0 \times 10^{-2}\,\text{m}}{2.0 \times 10^{-6}\,\text{m}^2} = 3450\,\Omega\)

Calculate the resistance across Pair C

For Pair C, the length \(L\) is equal to the shorter side of the wafer: \(2.00\,\text{cm}\), and the cross-sectional area A is equal to the face's area: \(0.010\,\text{cm} \times 3.00\,\text{cm}\): \(L_C = 2.00\,\text{cm}\) \(A_C = 0.010\,\text{cm} \times 3.00\,\text{cm}=0.030\,\text{cm}^2\) Now, convert L and A to meters: \(L_C = 2.0 \times 10^{-2}\,\text{m}\) \(A_C = 3.0 \times 10^{-6}\,\text{m}^2\) Using the resistance formula, we find the resistance across Pair C: \(R_C = \frac{\rho L_C}{A_C} = \frac{2300 \Omega\,\text{m} \times 2.0 \times 10^{-2}\,\text{m}}{3.0 \times 10^{-6}\,\text{m}^2} = 1533.33\,\Omega\)

Determine the maximum resistance

We now compare the resistances across the three pairs of opposite faces to find the maximum resistance: - Across Pair A: \(R_A = 38.33\,\Omega\) - Across Pair B: \(R_B = 3450\,\Omega\) - Across Pair C: \(R_C = 1533.33\,\Omega\) Since \(R_B > R_A\) and \(R_B > R_C\), the maximum resistance is across Pair B. Therefore, the maximum resistance of the rectangular wafer between any two faces is \(R_{\text{max}} = 3450\,\Omega\).

Key concepts

Resistivity

Resistivity, represented by the Greek letter rho \( \rho \), is a fundamental property of materials that quantifies how strongly a material opposes the flow of electric current. It is measured in ohm-meters \( \Omega\cdot\mathrm{m} \) and depends on the material's nature and temperature. A higher resistivity means the material is less conductive, and a lower resistivity indicates a better conductor. When calculating the resistance of an object, such as the rectangular wafer in the exercise, the resistivity is crucial as it helps to determine how much resistance the material inherently provides against the electrical current.

Understanding resistivity is not just about memorizing a formula; it's about appreciating the material's characteristics. When students need to find the resistance of a given sample, they should remember that the resistivity value is intrinsic to the sample's material, in this case, silicon, and use it in conjunction with the sample's geometry to compute the resistance.

Ohm's Law

Ohm's Law is a fundamental principle in the field of electronics and electrical engineering, stating that the current through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance between them. It is concisely represented by the equation \( V = IR \), where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance.

This law is not explicitly used in the resistance calculation of the silicon wafer, but it underpins much of the reasoning about electrical circuits. For students who are practicing these calculations, understanding Ohm's Law is crucial because it helps them relate voltage, current, and resistance in any electric circuit. It's also instrumental when analyzing circuit problems where you might be given two quantities (like voltage and current) and asked to solve for the third (resistance in this case).

Conductivity

Conductivity is the reciprocal of resistivity and represents a material's ability to conduct electric current. It is denoted usually by the Greek letter sigma \( \sigma \) and is measured in siemens per meter \( S/m \). When students encounter problems involving resistivity, it's essential also to understand conductivity since it describes the same phenomenon but from an opposite perspective. High conductivity means low resistivity and vice versa.

The exercise on the wafer's resistance involves resistivity, but implicitly it's about the silicon's ability to conduct electricity. By knowing the resistivity, you can also find the conductivity by taking the inverse \( \sigma = 1/\rho \). This could provide additional insights into how materials behave in an electrical circuit and is particularly important when comparing materials--a highly conductive material will allow current to pass through easier than a material with low conductivity. For practical applications, materials with high conductivity are often used for wires and other components where low resistance is desired.

Rectangular Wafer Resistance

When measuring the resistance of a rectangular wafer, it's critical to consider the physical dimensions -- the length, width, and thickness -- of the material. The resistance across any two faces of the wafer will be affected by the size and shape of the wafer as well as its resistivity. As demonstrated in the problem, calculating the resistance requires knowledge of the formula \( R = \frac{\rho L}{A} \), and correctly identifying the length \( L \) and the cross-sectional area \( A \) for each potential path through the wafer.

The thickest dimension is typically considered the length through which current travels, and the opposing faces’ area is the cross-sectional area. As shown in the solution, the path that provides the highest resistance is the one with the longest length and smallest cross-sectional area, aligning with the intuition that longer paths and narrower channels increase resistance. For students, visualizing the current's path and correctly applying the formula are keys to solving such problems and will greatly improve their understanding of physical principles that govern electrical resistance.