1. Current in a wire. If a current of 1 A is flowing in a conductor
wire, find the number of electrons that pass through its cross section
each second. Get solution
2. Current in a copper wire. If a copper wire of cross-sectional area 1 mm2 is carrying a current of 1 A, find the average drift velocity of the conduction electrons in the wire. The density of conduction electrons in copper is nc = 8.45 × 1028 el-m−3. Get solution
3. Electric field in an aluminum wire. An aluminum wire of conductivity σ = 3.82 × 107 S-m−1 is carrying a uniform current of density 100 A-cm−2. (a) Find the electric field in the wire. (b) Find the electric potential difference per meter of wire. Get solution
4. Silver versus porcelain. The conductivity of silver is σ = 6.17 × 107 S-m−1 and that of porcelain is about σ ≃ 10−14 S-m−1. (a) If a voltage of 1 V is applied across a silver plate of 1 cm thickness, find the current density through it. Assume a uniform electric field. (b) Repeat part (a) for a porcelain plate of same thickness and compare your results. Get solution
5. Electron mobility in copper. The average drift velocity of free electrons in a metal is proportional to the applied electric field approximately given by...where μe is a proportionality factor called the mobility of the electron. Since vd = μeE, the mobility describes how strongly the motion of an electron is influenced by an applied electric field. The conductivity of copper is σ = 5.8 × 107 S-m−1 at room temperature and is due to the mobility of electrons that are free (one per atom) to move under the influence of an electric field. (a) Find the electron mobility in copper at room temperature and compare it with the mobility values of pure silicon and germanium (see Problem). (b) Find the average drift velocity of the electrons in the direction of the current flow in a copper wire of 1 mm diameter carrying a current of 1 A.ProblemIntrinsic semiconductor. A semiconductor is said to be intrinsic (pure) when it is free of any dopant impurity atoms. The only mobile carriers are those caused by thermal excitation, a process that creates an equal number of holes and electrons. The intrinsic charge carrier concentration Ni is a strong function of temperature. Under most conditions, it is given by...where Nc and Nv are related to the density of allowed states near the edges of the conduction and valence bands, Eg is the energy gap between the conduction and the valence bands, kB is Boltzmann’s constant kB ≃ 1.38 × 10−23 J-K−1, and T is the temperature in K. The conductivity of a semiconductor sample is given by...where |qe| ≃ 1.6 × 10−19 C; μe and μp are the electron and the hole mobilities, which are both functions of temperature; and Ne and Np are the electron and hole concentrations, which, in the case of a pure semiconductor, are both equal to the intrinsic charge carrier concentration; that is, Ne = Np = Ni . (a) For a bulk sample of pure silicon (Si) at 300K, Nc = 2.8 × 1019 cm−3, Nv = 1.04 × 1019 cm−3, Eg = 1.124 eV (where 1 eV ≃ 1.602 × 10−19 J), μe = 1450 cm2-(V-s)−1, and μp = 450 cm2-(V-s)−1. Calculate the intrinsic carrier concentration (in cm−3) and the conductivity (in S-m−1) of the Si sample at 300K. (b) Repeat part (a) for a bulk sample of pure germanium (Ge). For Ge at 300K, Nc = 1.04 × 1019 cm−3, Nv = 6.0 × 1018 cm−3, Eg = 0.66 eV, μe = 3900 cm2-(V-s)−1, and μp = 1900 cm2-(V-s)−1. Get solution
6. Resistance of a short, small-diameter conductor. A technique has been developed to measure the dc resistance at room temperature of conductors that are short in length and small in diameter.31 To demonstrate this system, the resistance of commercial bare copper wire (σ = 5.85 × 107 S-m−1 at 20◦C) was measured in four different diameters. The manufacturer’s reported diameters for these wires were 25.4 μm, 20.3 μm, 12.7 μm, and 7.6 μm.The resistances per unit length of these wires were measured to be 0.340, 0.527, 1.22, and 2.56, all in Ω-cm−1. (a) Calculate the actual diameter of each wire using the measured resistance values. (b) Find the percentage of error in the reported diameter values by comparing them to the calculated values. (Note that the uniformity and the surface condition of each wire were examined with a scanning electron microscope. All of the wires had uniform diameters, and only the 7.6-μm diameter wire showed significant surface cracking and roughness. So a larger error for the 7.6-μm diameter wire is most likely due to its degraded surface condition.) Get solution
7. Resistance of a copper wire. Consider a copper wire of 1 mm diameter. Find its resistance per unit length if the wire temperature is (a) −20◦C, (b) 20◦C, and (c) 60◦C. For copper, the conductivity is σ = 5.8 × 107 S-m−1 at 20◦C. Get solution
8. Resistance of a copper wire. Consider a copper wire at 20◦C. To what temperature must it be heated in order to double its resistance? Get solution
9. Intrinsic semiconductor. A semiconductor is said to be intrinsic (pure) when it is free of any dopant impurity atoms. The only mobile carriers are those caused by thermal excitation, a process that creates an equal number of holes and electrons. The intrinsic charge carrier concentration Ni is a strong function of temperature. Under most conditions, it is given by...where Nc and Nv are related to the density of allowed states near the edges of the conduction and valence bands, Eg is the energy gap between the conduction and the valence bands, kB is Boltzmann’s constant kB ≃ 1.38 × 10−23 J-K−1, and T is the temperature in K. The conductivity of a semiconductor sample is given by...where |qe| ≃ 1.6 × 10−19 C; μe and μp are the electron and the hole mobilities, which are both functions of temperature; and Ne and Np are the electron and hole concentrations, which, in the case of a pure semiconductor, are both equal to the intrinsic charge carrier concentration; that is, Ne = Np = Ni . (a) For a bulk sample of pure silicon (Si) at 300K, Nc = 2.8 × 1019 cm−3, Nv = 1.04 × 1019 cm−3, Eg = 1.124 eV (where 1 eV ≃ 1.602 × 10−19 J), μe = 1450 cm2-(V-s)−1, and μp = 450 cm2-(V-s)−1. Calculate the intrinsic carrier concentration (in cm−3) and the conductivity (in S-m−1) of the Si sample at 300K. (b) Repeat part (a) for a bulk sample of pure germanium (Ge). For Ge at 300K, Nc = 1.04 × 1019 cm−3, Nv = 6.0 × 1018 cm−3, Eg = 0.66 eV, μe = 3900 cm2-(V-s)−1, and μp = 1900 cm2-(V-s)−1. Get solution
10. Extrinsic semiconductor. Semiconductors in which conduction results primarily from carriers contributed by impurity atoms are said to be extrinsic (impure). The impurity atoms, which are intentionally introduced to change the charge carrier concentration, are called dopant atoms. In doped semiconductors, only one of the components of the conductivity expression in Problem is generally significant because of the very large ratio between the two carrier densities Ne and Np , the product of which is always equal to the square of the intrinsic charge carrier concentration, that is, NeNp = N2i . In addition, the mobility values of the charge carriers vary with different doping levels. (a) Phosphorus donor atoms with a concentration of Nd = 1016 cm−3 are added uniformly to a pure sample of Si (which is called n-type Si since the electrons of the phosphorus atoms are the majority carriers, that is, Ne≃ Nd ≫ Np ). Find the conductivity of the n-type Si sample at 300K. For the mobility of majority charge carriers, use μe = 1194 cm2-(V-s)−1. (b) Repeat part (a) if boron acceptor atoms with a concentration of Na = 1016 cm−3 are added to a pure sample of Si (which is called p-type Si since the holes are now the majority carriers, that is, Np ≃ Na ≫ Ne). For the mobility of majority charge carriers, use μp = 444 cm2-(V-s)−1. (c) Compare your results in parts (a) and (b) with the conductivity of the intrinsic Si and comment on the differences.ProblemIntrinsic semiconductor. A semiconductor is said to be intrinsic (pure) when it is free of any dopant impurity atoms. The only mobile carriers are those caused by thermal excitation, a process that creates an equal number of holes and electrons. The intrinsic charge carrier concentration Ni is a strong function of temperature. Under most conditions, it is given by...where Nc and Nv are related to the density of allowed states near the edges of the conduction and valence bands, Eg is the energy gap between the conduction and the valence bands, kB is Boltzmann’s constant kB ≃ 1.38 × 10−23 J-K−1, and T is the temperature in K. The conductivity of a semiconductor sample is given by...where |qe| ≃ 1.6 × 10−19 C; μe and μp are the electron and the hole mobilities, which are both functions of temperature; and Ne and Np are the electron and hole concentrations, which, in the case of a pure semiconductor, are both equal to the intrinsic charge carrier concentration; that is, Ne = Np = Ni . (a) For a bulk sample of pure silicon (Si) at 300K, Nc = 2.8 × 1019 cm−3, Nv = 1.04 × 1019 cm−3, Eg = 1.124 eV (where 1 eV ≃ 1.602 × 10−19 J), μe = 1450 cm2-(V-s)−1, and μp = 450 cm2-(V-s)−1. Calculate the intrinsic carrier concentration (in cm−3) and the conductivity (in S-m−1) of the Si sample at 300K. (b) Repeat part (a) for a bulk sample of pure germanium (Ge). For Ge at 300K, Nc = 1.04 × 1019 cm−3, Nv = 6.0 × 1018 cm−3, Eg = 0.66 eV, μe = 3900 cm2-(V-s)−1, and μp = 1900 cm2-(V-s)−1. Get solution
11. A silicon resistor. A silicon bar 1 mm long and 0.01 mm2 in cross-sectional area is doped with Nd = 1017 cm−3 arsenic (As) atoms. Find the resistance of the bar and compare with the resistance of the same bar made of pure silicon (see Problem). For the mobility of the free arsenic electrons, use μe = 731 cm2-(V-s)−1.ProblemIntrinsic semiconductor. A semiconductor is said to be intrinsic (pure) when it is free of any dopant impurity atoms. The only mobile carriers are those caused by thermal excitation, a process that creates an equal number of holes and electrons. The intrinsic charge carrier concentration Ni is a strong function of temperature. Under most conditions, it is given by...where Nc and Nv are related to the density of allowed states near the edges of the conduction and valence bands, Eg is the energy gap between the conduction and the valence bands, kB is Boltzmann’s constant kB ≃ 1.38 × 10−23 J-K−1, and T is the temperature in K. The conductivity of a semiconductor sample is given by...where |qe| ≃ 1.6 × 10−19 C; μe and μp are the electron and the hole mobilities, which are both functions of temperature; and Ne and Np are the electron and hole concentrations, which, in the case of a pure semiconductor, are both equal to the intrinsic charge carrier concentration; that is, Ne = Np = Ni . (a) For a bulk sample of pure silicon (Si) at 300K, Nc = 2.8 × 1019 cm−3, Nv = 1.04 × 1019 cm−3, Eg = 1.124 eV (where 1 eV ≃ 1.602 × 10−19 J), μe = 1450 cm2-(V-s)−1, and μp = 450 cm2-(V-s)−1. Calculate the intrinsic carrier concentration (in cm−3) and the conductivity (in S-m−1) of the Si sample at 300K. (b) Repeat part (a) for a bulk sample of pure germanium (Ge). For Ge at 300K, Nc = 1.04 × 1019 cm−3, Nv = 6.0 × 1018 cm−3, Eg = 0.66 eV, μe = 3900 cm2-(V-s)−1, and μp = 1900 cm2-(V-s)−1. Get solution
12. A silicon resistor. A sample of p-type silicon is 8 μm long and has a cross-sectional area of 2.5 μm × 2.5 μm. If the measurements show that the average hole concentration and the resistance of the sample are Nd = 1016 cm−3 and R = 18 kΩ, respectively, find the hole mobility μp . Get solution
13. Sheet resistance Rsq. Consider the resistance of a uniformly doped n-type Si layer of length l , width w, and thickness t , as shown in Figure. This resistor can be divided into square sheets of dimension w on each side, as shown. The resistance of any one of these square sheets is called the sheet resistance, denoted by the symbol Rsq, in units of Ω-(sq)−1. (a) Show that the sheet resistance is given by...(b) Show that the total resistance of the Si layer is given by...(c) Calculate the total resistance of a Si layer that has a length of 50 μm, width of 5 μm and a sheet resistance of 150Ω-(sq)−1.Figure Sheet resistance. Problem.... Get solution
14. Integrated-circuit (IC) resistor. Consider an integrated-circuit (IC) resistor of length l , width w, and thickness t that is made of Si doped with 1016 cm−3 phosphorus atoms (i.e., Ne ≃2.5 × 1016 cm−3). Given t = 2 μm, find the aspect ratio, w/l , such that the resistance of the IC resistor is 10 kΩ at 300K. Take μe = 1000 cm2-(V-s)−1. Get solution
15. An ion-implanted IC resistor. An ion-implanted n-type resistor layer with an average doping concentration Nd = 4 × 1017 cm−3 is designed with 1 μm thickness and 2 μm width to provide a resistance of 3 kΩ for an IC chip. Find the sheet resistance and the required length of the resistor. Assume the electron mobility as μe = 450 cm2-(V-s)−1. Get solution
16. Diffused IC resistor. An IC resistor is frequently fabricated by diffusing a thin layer of p-type impurity into an n-type isolation island, as shown in Figure. If contacts are made near the two ends of the p-type region and a voltage is applied, a current will flow parallel to the surface in this region. It is not possible to use R = l/(σA), however, to calculate the resistance of this region because the impurity concentration in it is not uniform. The impurity concentration resulting from the diffusion process is maximum near the surface (x = 0) and decreases as one moves in the x direction. (a) Show that the sheet resistance of the p-type layer is given by...(b) Assuming the conductivity of the p-type layer decreases linearly from σ0 at the surface (x = 0) to σ1 ≪ σ0 at the interface (x = t ) with the n-type wafer, find the sheet resistance, Rsq. (c) Repeat part (b) if the conductivity decreases exponentially from σ0 at x = 0 to σ1 ≪ σ0 at x = t .Figure Diffused IC resistor. Problem.... Get solution
17. Resistance of a copper-coated steel wire. A stainless steel wire (σ = 1.11 × 106 S-m−1) 3 mm in diameter is to be coated with copper (σ = 5.8 × 107 S-m−1) in order to reduce its resistance per unit length by 50%. Find the thickness of the copper coating needed to achieve this goal. Get solution
18. Resistance of an aluminum conductor, steel-reinforced (ACSR) wire. A power utility company uses a steel-reinforced aluminum conductor (ACSR) wire of 3 cm diameter as an extra-high-voltage (EHV) transmission line. It is made of aluminum (σ = 3.82× 107 S-m−1) with an inner core of stainless steel (σ = 1.11 × 106 S-m−1) along the center axis such that the steel content of the ACSR wire is only 10%. (a) Find the resistance per kilometer of the wire at 20◦C. (b) If a current of 1000 A flows in the wire, find the current that flows in each metal. (Neglect the changes in the conductivity values due to changes in temperature because of the current flow.) (c) If the ACSR wire between two EHV towers is 300 m long, find the voltage drop between the two towers. (d) Repeat parts (a), (b), and (c) if the steel content of the ACSR wire is increased to 25%. Get solution
19. Conductivity of lunar soil. A simulated version of the lunar soil obtained from the Apollo 11 site on the moon is investigated as a possible material to be used for electrical insulation in high-voltage power systems in space.32 A sample of this soil is placed as an insulator between the electrodes of a parallel-plate capacitor with 1 mm separation and 10 cm2 area, and dc voltages are applied across the electrodes. Leakage currents of 10 nA, 20 nA, 30 nA, 40 nA, and 50 nA are recorded for electric field strengths of 2.5 kV-(mm)−1, 3 kV-(mm)−1, 3.6 kV-(mm)−1, 4 kV-(mm)−1, and 4.2 kV-(mm)−1, respectively. (a) Using Ohm’s law, find the conductivity of the lunar soil in each case. Note that the conductivity of this soil varies with the applied field. (b) Using a curve-fitting technique, the conductivity of the soil is approximated as...where σ is in S-(mm)−1 and E is in kV-(mm)−1. Using this expression, find the value of σ and its percentage difference from its corresponding value found in part (a) for each case. Get solution
20. Resistance of a semicircular ring. The ends of a semicircular conductor ring of rectangular cross section are connected to a dc battery, as shown in Figure. (a) Write the total resistance R of the conductor using the result of Example. (b) Find R by assuming the conductor to be a straight conductor with a length l = π(a + b)/2. (c) Using σ = 7.4 × 104 S-m−1 (graphite) and b = 1.5a = 10t = 3 cm, find R using both expressions obtained in parts (a) and (b) and compare the results. (d) Show that for a ≫ (b − a), the expression of part (a) reduces to the expression of part (b).Figure Semicircular ring. Problem....ExampleCurved bar. The resistance of a rectangular block of metal was discussed in this section and found to be given by R = l/(σA), where σ is the conductivity, l is the length, and A is the cross-sectional area. As a more complicated geometry, consider the resistor in the form of a curved bar, which is made of the rectangular block bent to form the arc of a circle, with the two end surfaces remaining flat, as shown in Figure. The edges are coated with perfectly conducting material so that they constitute electrodes at uniform potential. Find the resistance R between the electrodes.Figure A curved-bar resistor. (a) A rectangular bar bent into a circular arc of rectangular cross section.... Get solution
21. Resistance of a hemispherical conductor. A perfectly conducting hemispherical conductor is buried in the earth to achieve a good ground connection (see Figure). If the diameter of the conductor is 25 cm and the conductivity of the ground is 10−4 S-m−1, find the resistance of the conductor to distant points in the ground.Figure Buried spherical electrode. The electrode is made of perfectly conducting material and is thus an equipotential surface.... Get solution
22. Resistance of a toroidal conductor. Consider a segment of a toroidal (doughnut-shaped) resistor with a horizontal cross section, as shown in Figure. Show that the resistance between the flat ends having a circular cross section is given byFigure Toroidal conductor. Problem....... Get solution
23. Leakage resistance. A 2-mm-diameter copper wire is enclosed in an insulating sheath of 4-mm outside diameter. If the wire is buried in a highly conducting ground, what is the leakage resistance per kilometer of the sheath to the ground? The sheath can be assumed to have a conductivity of 10−8 S-m−1. Get solution
24. Pipeline resistance. Two parallel steel pipelines have centers spaced 10 m apart. The pipes are half buried in the ground, as shown in Figure. The diameter of the pipes is 1 m. The ground in which the pipes lie is marshy soil (σ ≃ 10−2 S-m−1). Find the resistance per kilometer between the two pipes.Figure Pipeline resistance. Problem.... Get solution
25. Energy dissipation in lossy coaxial cable. Consider the coaxial cable shown in Figure, with inner conductor radius of a, outer conductor radius of b, and filled with two different conductive materials, having the same thickness, w1 = w2 = (b − a)/2. Available in hand are two different types of lossy material, with conductivities of σ and 3σ. (a) Assuming a constant voltage Vab to be applied between the inner and outer conductors, which of the two lossy materials should be used as the inner one in order to minimize losses (per unit length)? In other words, is it better to have σ1 = σ, σ2 = 3σ or vice versa? By what factor does the loss per unit length increase if the two materials are flipped? (b) Determine the values that w1 and w2 must have in order for the loss per unit length to be the same regardless of the position of the materials. In other words, find values of w1 and w2 (in terms of a and b) such that the loss per unit length is the same for (i) σ1 = σ and σ2 = 3σ, or (ii) σ1 = 3σ and σ2 = σ.Figure Lossy coaxial cable. Problem.... Get solution
26. Ground current. A vertical lightning rod discharges 105 amperes into the ground. Find the voltage produced by the discharge between two points 1 m apart on a radial line if the point nearest the rod is at a distance of (a) 3 m, and (b) 10 m from the rod. Assume the ground to consist of a 20-cm layer of conducting soil with σ ≃ 10−2 S-m−1. Get solution
27. Inhomogeneous medium. Consider an inhomogeneous medium in which both ∊ and σ are functions of position. Show that a steady current J flowing though such a medium would necessarily establish a charge distribution given by ρ = −[∇∊− (∊/σ )∇σ] • ∇ϕ. Note that the field arising from the flow of current can still be derived from a scalar potential, but the potential no longer satisfies Laplace’s equation, since ρ ≠ 0. Get solution
28. Leaky capacitor. Consider the parallel-plate capacitor with spacing 2d and plate area A, as shown in Figure. The region between the plates is filled with two lossy dielectric slabs, each of thickness d and with parameters ∊1, σ1, and ∊2, σ2, respectively. A potential V0 is applied across the plates. (a) Find the steady-state value of the electric field in each of the two materials. (b) Find an expression for the surface charge density at the interface between the two materials.Figure Leaky capacitor. Problem.... Get solution
29. Lossy dielectric between spherical electrodes. Consider a spherical capacitor with the region between the inner and outer electrodes filled with lossy material with permittivity ∊0 and conductivity...where b is the radius of the outer electrode and σ0 is a constant. The inner electrode (with radius a) is held at a constant potential of V0 while the outer electrode (with radius b >a) is at zero potential. (a) Determine the charge Q on the inner and outer electrodes, and the charge distribution ρ(r) in the region between the electrodes. (b) Determine the total space-distributed charge in the region between the electrodes. Get solution
30. Inhomogeneous medium. A spherical electrode of radius a is surrounded by a concentric spherical shell of radius b. The space between the two conductors is filled with material whose conductivity varies linearly with distance from the common center of the spheres (i.e., σ = Kr, where K is a proportionality constant). If a potential difference of V volts is maintained between the spheres, with the inner electrode grounded (i.e., at zero potential), what is the electric potential at a distance r (a r b) from the center? Get solution
31. Leakage resistance. A very long wire of radius a is suspended horizontally near the bottom of a deep lake. Assume the lake to have a plane bottom that is a very good conductor. The wire is parallel to the bottom and is at a height h ≫ a above it. If the conductivity of the water is σ, find the resistance (per unit length) between the wire and the planar bottom of the lake. Get solution
32. Two conducting spheres. Two metallic conducting spheres of radii a1 and a2 are buried deep in poorly conducting ground of conductivity σ and permittivity ∊. The distance b between the spheres is much larger than both a1 and a2. Determine the resistance between the two spheres. Get solution
2. Current in a copper wire. If a copper wire of cross-sectional area 1 mm2 is carrying a current of 1 A, find the average drift velocity of the conduction electrons in the wire. The density of conduction electrons in copper is nc = 8.45 × 1028 el-m−3. Get solution
3. Electric field in an aluminum wire. An aluminum wire of conductivity σ = 3.82 × 107 S-m−1 is carrying a uniform current of density 100 A-cm−2. (a) Find the electric field in the wire. (b) Find the electric potential difference per meter of wire. Get solution
4. Silver versus porcelain. The conductivity of silver is σ = 6.17 × 107 S-m−1 and that of porcelain is about σ ≃ 10−14 S-m−1. (a) If a voltage of 1 V is applied across a silver plate of 1 cm thickness, find the current density through it. Assume a uniform electric field. (b) Repeat part (a) for a porcelain plate of same thickness and compare your results. Get solution
5. Electron mobility in copper. The average drift velocity of free electrons in a metal is proportional to the applied electric field approximately given by...where μe is a proportionality factor called the mobility of the electron. Since vd = μeE, the mobility describes how strongly the motion of an electron is influenced by an applied electric field. The conductivity of copper is σ = 5.8 × 107 S-m−1 at room temperature and is due to the mobility of electrons that are free (one per atom) to move under the influence of an electric field. (a) Find the electron mobility in copper at room temperature and compare it with the mobility values of pure silicon and germanium (see Problem). (b) Find the average drift velocity of the electrons in the direction of the current flow in a copper wire of 1 mm diameter carrying a current of 1 A.ProblemIntrinsic semiconductor. A semiconductor is said to be intrinsic (pure) when it is free of any dopant impurity atoms. The only mobile carriers are those caused by thermal excitation, a process that creates an equal number of holes and electrons. The intrinsic charge carrier concentration Ni is a strong function of temperature. Under most conditions, it is given by...where Nc and Nv are related to the density of allowed states near the edges of the conduction and valence bands, Eg is the energy gap between the conduction and the valence bands, kB is Boltzmann’s constant kB ≃ 1.38 × 10−23 J-K−1, and T is the temperature in K. The conductivity of a semiconductor sample is given by...where |qe| ≃ 1.6 × 10−19 C; μe and μp are the electron and the hole mobilities, which are both functions of temperature; and Ne and Np are the electron and hole concentrations, which, in the case of a pure semiconductor, are both equal to the intrinsic charge carrier concentration; that is, Ne = Np = Ni . (a) For a bulk sample of pure silicon (Si) at 300K, Nc = 2.8 × 1019 cm−3, Nv = 1.04 × 1019 cm−3, Eg = 1.124 eV (where 1 eV ≃ 1.602 × 10−19 J), μe = 1450 cm2-(V-s)−1, and μp = 450 cm2-(V-s)−1. Calculate the intrinsic carrier concentration (in cm−3) and the conductivity (in S-m−1) of the Si sample at 300K. (b) Repeat part (a) for a bulk sample of pure germanium (Ge). For Ge at 300K, Nc = 1.04 × 1019 cm−3, Nv = 6.0 × 1018 cm−3, Eg = 0.66 eV, μe = 3900 cm2-(V-s)−1, and μp = 1900 cm2-(V-s)−1. Get solution
6. Resistance of a short, small-diameter conductor. A technique has been developed to measure the dc resistance at room temperature of conductors that are short in length and small in diameter.31 To demonstrate this system, the resistance of commercial bare copper wire (σ = 5.85 × 107 S-m−1 at 20◦C) was measured in four different diameters. The manufacturer’s reported diameters for these wires were 25.4 μm, 20.3 μm, 12.7 μm, and 7.6 μm.The resistances per unit length of these wires were measured to be 0.340, 0.527, 1.22, and 2.56, all in Ω-cm−1. (a) Calculate the actual diameter of each wire using the measured resistance values. (b) Find the percentage of error in the reported diameter values by comparing them to the calculated values. (Note that the uniformity and the surface condition of each wire were examined with a scanning electron microscope. All of the wires had uniform diameters, and only the 7.6-μm diameter wire showed significant surface cracking and roughness. So a larger error for the 7.6-μm diameter wire is most likely due to its degraded surface condition.) Get solution
7. Resistance of a copper wire. Consider a copper wire of 1 mm diameter. Find its resistance per unit length if the wire temperature is (a) −20◦C, (b) 20◦C, and (c) 60◦C. For copper, the conductivity is σ = 5.8 × 107 S-m−1 at 20◦C. Get solution
8. Resistance of a copper wire. Consider a copper wire at 20◦C. To what temperature must it be heated in order to double its resistance? Get solution
9. Intrinsic semiconductor. A semiconductor is said to be intrinsic (pure) when it is free of any dopant impurity atoms. The only mobile carriers are those caused by thermal excitation, a process that creates an equal number of holes and electrons. The intrinsic charge carrier concentration Ni is a strong function of temperature. Under most conditions, it is given by...where Nc and Nv are related to the density of allowed states near the edges of the conduction and valence bands, Eg is the energy gap between the conduction and the valence bands, kB is Boltzmann’s constant kB ≃ 1.38 × 10−23 J-K−1, and T is the temperature in K. The conductivity of a semiconductor sample is given by...where |qe| ≃ 1.6 × 10−19 C; μe and μp are the electron and the hole mobilities, which are both functions of temperature; and Ne and Np are the electron and hole concentrations, which, in the case of a pure semiconductor, are both equal to the intrinsic charge carrier concentration; that is, Ne = Np = Ni . (a) For a bulk sample of pure silicon (Si) at 300K, Nc = 2.8 × 1019 cm−3, Nv = 1.04 × 1019 cm−3, Eg = 1.124 eV (where 1 eV ≃ 1.602 × 10−19 J), μe = 1450 cm2-(V-s)−1, and μp = 450 cm2-(V-s)−1. Calculate the intrinsic carrier concentration (in cm−3) and the conductivity (in S-m−1) of the Si sample at 300K. (b) Repeat part (a) for a bulk sample of pure germanium (Ge). For Ge at 300K, Nc = 1.04 × 1019 cm−3, Nv = 6.0 × 1018 cm−3, Eg = 0.66 eV, μe = 3900 cm2-(V-s)−1, and μp = 1900 cm2-(V-s)−1. Get solution
10. Extrinsic semiconductor. Semiconductors in which conduction results primarily from carriers contributed by impurity atoms are said to be extrinsic (impure). The impurity atoms, which are intentionally introduced to change the charge carrier concentration, are called dopant atoms. In doped semiconductors, only one of the components of the conductivity expression in Problem is generally significant because of the very large ratio between the two carrier densities Ne and Np , the product of which is always equal to the square of the intrinsic charge carrier concentration, that is, NeNp = N2i . In addition, the mobility values of the charge carriers vary with different doping levels. (a) Phosphorus donor atoms with a concentration of Nd = 1016 cm−3 are added uniformly to a pure sample of Si (which is called n-type Si since the electrons of the phosphorus atoms are the majority carriers, that is, Ne≃ Nd ≫ Np ). Find the conductivity of the n-type Si sample at 300K. For the mobility of majority charge carriers, use μe = 1194 cm2-(V-s)−1. (b) Repeat part (a) if boron acceptor atoms with a concentration of Na = 1016 cm−3 are added to a pure sample of Si (which is called p-type Si since the holes are now the majority carriers, that is, Np ≃ Na ≫ Ne). For the mobility of majority charge carriers, use μp = 444 cm2-(V-s)−1. (c) Compare your results in parts (a) and (b) with the conductivity of the intrinsic Si and comment on the differences.ProblemIntrinsic semiconductor. A semiconductor is said to be intrinsic (pure) when it is free of any dopant impurity atoms. The only mobile carriers are those caused by thermal excitation, a process that creates an equal number of holes and electrons. The intrinsic charge carrier concentration Ni is a strong function of temperature. Under most conditions, it is given by...where Nc and Nv are related to the density of allowed states near the edges of the conduction and valence bands, Eg is the energy gap between the conduction and the valence bands, kB is Boltzmann’s constant kB ≃ 1.38 × 10−23 J-K−1, and T is the temperature in K. The conductivity of a semiconductor sample is given by...where |qe| ≃ 1.6 × 10−19 C; μe and μp are the electron and the hole mobilities, which are both functions of temperature; and Ne and Np are the electron and hole concentrations, which, in the case of a pure semiconductor, are both equal to the intrinsic charge carrier concentration; that is, Ne = Np = Ni . (a) For a bulk sample of pure silicon (Si) at 300K, Nc = 2.8 × 1019 cm−3, Nv = 1.04 × 1019 cm−3, Eg = 1.124 eV (where 1 eV ≃ 1.602 × 10−19 J), μe = 1450 cm2-(V-s)−1, and μp = 450 cm2-(V-s)−1. Calculate the intrinsic carrier concentration (in cm−3) and the conductivity (in S-m−1) of the Si sample at 300K. (b) Repeat part (a) for a bulk sample of pure germanium (Ge). For Ge at 300K, Nc = 1.04 × 1019 cm−3, Nv = 6.0 × 1018 cm−3, Eg = 0.66 eV, μe = 3900 cm2-(V-s)−1, and μp = 1900 cm2-(V-s)−1. Get solution
11. A silicon resistor. A silicon bar 1 mm long and 0.01 mm2 in cross-sectional area is doped with Nd = 1017 cm−3 arsenic (As) atoms. Find the resistance of the bar and compare with the resistance of the same bar made of pure silicon (see Problem). For the mobility of the free arsenic electrons, use μe = 731 cm2-(V-s)−1.ProblemIntrinsic semiconductor. A semiconductor is said to be intrinsic (pure) when it is free of any dopant impurity atoms. The only mobile carriers are those caused by thermal excitation, a process that creates an equal number of holes and electrons. The intrinsic charge carrier concentration Ni is a strong function of temperature. Under most conditions, it is given by...where Nc and Nv are related to the density of allowed states near the edges of the conduction and valence bands, Eg is the energy gap between the conduction and the valence bands, kB is Boltzmann’s constant kB ≃ 1.38 × 10−23 J-K−1, and T is the temperature in K. The conductivity of a semiconductor sample is given by...where |qe| ≃ 1.6 × 10−19 C; μe and μp are the electron and the hole mobilities, which are both functions of temperature; and Ne and Np are the electron and hole concentrations, which, in the case of a pure semiconductor, are both equal to the intrinsic charge carrier concentration; that is, Ne = Np = Ni . (a) For a bulk sample of pure silicon (Si) at 300K, Nc = 2.8 × 1019 cm−3, Nv = 1.04 × 1019 cm−3, Eg = 1.124 eV (where 1 eV ≃ 1.602 × 10−19 J), μe = 1450 cm2-(V-s)−1, and μp = 450 cm2-(V-s)−1. Calculate the intrinsic carrier concentration (in cm−3) and the conductivity (in S-m−1) of the Si sample at 300K. (b) Repeat part (a) for a bulk sample of pure germanium (Ge). For Ge at 300K, Nc = 1.04 × 1019 cm−3, Nv = 6.0 × 1018 cm−3, Eg = 0.66 eV, μe = 3900 cm2-(V-s)−1, and μp = 1900 cm2-(V-s)−1. Get solution
12. A silicon resistor. A sample of p-type silicon is 8 μm long and has a cross-sectional area of 2.5 μm × 2.5 μm. If the measurements show that the average hole concentration and the resistance of the sample are Nd = 1016 cm−3 and R = 18 kΩ, respectively, find the hole mobility μp . Get solution
13. Sheet resistance Rsq. Consider the resistance of a uniformly doped n-type Si layer of length l , width w, and thickness t , as shown in Figure. This resistor can be divided into square sheets of dimension w on each side, as shown. The resistance of any one of these square sheets is called the sheet resistance, denoted by the symbol Rsq, in units of Ω-(sq)−1. (a) Show that the sheet resistance is given by...(b) Show that the total resistance of the Si layer is given by...(c) Calculate the total resistance of a Si layer that has a length of 50 μm, width of 5 μm and a sheet resistance of 150Ω-(sq)−1.Figure Sheet resistance. Problem.... Get solution
14. Integrated-circuit (IC) resistor. Consider an integrated-circuit (IC) resistor of length l , width w, and thickness t that is made of Si doped with 1016 cm−3 phosphorus atoms (i.e., Ne ≃2.5 × 1016 cm−3). Given t = 2 μm, find the aspect ratio, w/l , such that the resistance of the IC resistor is 10 kΩ at 300K. Take μe = 1000 cm2-(V-s)−1. Get solution
15. An ion-implanted IC resistor. An ion-implanted n-type resistor layer with an average doping concentration Nd = 4 × 1017 cm−3 is designed with 1 μm thickness and 2 μm width to provide a resistance of 3 kΩ for an IC chip. Find the sheet resistance and the required length of the resistor. Assume the electron mobility as μe = 450 cm2-(V-s)−1. Get solution
16. Diffused IC resistor. An IC resistor is frequently fabricated by diffusing a thin layer of p-type impurity into an n-type isolation island, as shown in Figure. If contacts are made near the two ends of the p-type region and a voltage is applied, a current will flow parallel to the surface in this region. It is not possible to use R = l/(σA), however, to calculate the resistance of this region because the impurity concentration in it is not uniform. The impurity concentration resulting from the diffusion process is maximum near the surface (x = 0) and decreases as one moves in the x direction. (a) Show that the sheet resistance of the p-type layer is given by...(b) Assuming the conductivity of the p-type layer decreases linearly from σ0 at the surface (x = 0) to σ1 ≪ σ0 at the interface (x = t ) with the n-type wafer, find the sheet resistance, Rsq. (c) Repeat part (b) if the conductivity decreases exponentially from σ0 at x = 0 to σ1 ≪ σ0 at x = t .Figure Diffused IC resistor. Problem.... Get solution
17. Resistance of a copper-coated steel wire. A stainless steel wire (σ = 1.11 × 106 S-m−1) 3 mm in diameter is to be coated with copper (σ = 5.8 × 107 S-m−1) in order to reduce its resistance per unit length by 50%. Find the thickness of the copper coating needed to achieve this goal. Get solution
18. Resistance of an aluminum conductor, steel-reinforced (ACSR) wire. A power utility company uses a steel-reinforced aluminum conductor (ACSR) wire of 3 cm diameter as an extra-high-voltage (EHV) transmission line. It is made of aluminum (σ = 3.82× 107 S-m−1) with an inner core of stainless steel (σ = 1.11 × 106 S-m−1) along the center axis such that the steel content of the ACSR wire is only 10%. (a) Find the resistance per kilometer of the wire at 20◦C. (b) If a current of 1000 A flows in the wire, find the current that flows in each metal. (Neglect the changes in the conductivity values due to changes in temperature because of the current flow.) (c) If the ACSR wire between two EHV towers is 300 m long, find the voltage drop between the two towers. (d) Repeat parts (a), (b), and (c) if the steel content of the ACSR wire is increased to 25%. Get solution
19. Conductivity of lunar soil. A simulated version of the lunar soil obtained from the Apollo 11 site on the moon is investigated as a possible material to be used for electrical insulation in high-voltage power systems in space.32 A sample of this soil is placed as an insulator between the electrodes of a parallel-plate capacitor with 1 mm separation and 10 cm2 area, and dc voltages are applied across the electrodes. Leakage currents of 10 nA, 20 nA, 30 nA, 40 nA, and 50 nA are recorded for electric field strengths of 2.5 kV-(mm)−1, 3 kV-(mm)−1, 3.6 kV-(mm)−1, 4 kV-(mm)−1, and 4.2 kV-(mm)−1, respectively. (a) Using Ohm’s law, find the conductivity of the lunar soil in each case. Note that the conductivity of this soil varies with the applied field. (b) Using a curve-fitting technique, the conductivity of the soil is approximated as...where σ is in S-(mm)−1 and E is in kV-(mm)−1. Using this expression, find the value of σ and its percentage difference from its corresponding value found in part (a) for each case. Get solution
20. Resistance of a semicircular ring. The ends of a semicircular conductor ring of rectangular cross section are connected to a dc battery, as shown in Figure. (a) Write the total resistance R of the conductor using the result of Example. (b) Find R by assuming the conductor to be a straight conductor with a length l = π(a + b)/2. (c) Using σ = 7.4 × 104 S-m−1 (graphite) and b = 1.5a = 10t = 3 cm, find R using both expressions obtained in parts (a) and (b) and compare the results. (d) Show that for a ≫ (b − a), the expression of part (a) reduces to the expression of part (b).Figure Semicircular ring. Problem....ExampleCurved bar. The resistance of a rectangular block of metal was discussed in this section and found to be given by R = l/(σA), where σ is the conductivity, l is the length, and A is the cross-sectional area. As a more complicated geometry, consider the resistor in the form of a curved bar, which is made of the rectangular block bent to form the arc of a circle, with the two end surfaces remaining flat, as shown in Figure. The edges are coated with perfectly conducting material so that they constitute electrodes at uniform potential. Find the resistance R between the electrodes.Figure A curved-bar resistor. (a) A rectangular bar bent into a circular arc of rectangular cross section.... Get solution
21. Resistance of a hemispherical conductor. A perfectly conducting hemispherical conductor is buried in the earth to achieve a good ground connection (see Figure). If the diameter of the conductor is 25 cm and the conductivity of the ground is 10−4 S-m−1, find the resistance of the conductor to distant points in the ground.Figure Buried spherical electrode. The electrode is made of perfectly conducting material and is thus an equipotential surface.... Get solution
22. Resistance of a toroidal conductor. Consider a segment of a toroidal (doughnut-shaped) resistor with a horizontal cross section, as shown in Figure. Show that the resistance between the flat ends having a circular cross section is given byFigure Toroidal conductor. Problem....... Get solution
23. Leakage resistance. A 2-mm-diameter copper wire is enclosed in an insulating sheath of 4-mm outside diameter. If the wire is buried in a highly conducting ground, what is the leakage resistance per kilometer of the sheath to the ground? The sheath can be assumed to have a conductivity of 10−8 S-m−1. Get solution
24. Pipeline resistance. Two parallel steel pipelines have centers spaced 10 m apart. The pipes are half buried in the ground, as shown in Figure. The diameter of the pipes is 1 m. The ground in which the pipes lie is marshy soil (σ ≃ 10−2 S-m−1). Find the resistance per kilometer between the two pipes.Figure Pipeline resistance. Problem.... Get solution
25. Energy dissipation in lossy coaxial cable. Consider the coaxial cable shown in Figure, with inner conductor radius of a, outer conductor radius of b, and filled with two different conductive materials, having the same thickness, w1 = w2 = (b − a)/2. Available in hand are two different types of lossy material, with conductivities of σ and 3σ. (a) Assuming a constant voltage Vab to be applied between the inner and outer conductors, which of the two lossy materials should be used as the inner one in order to minimize losses (per unit length)? In other words, is it better to have σ1 = σ, σ2 = 3σ or vice versa? By what factor does the loss per unit length increase if the two materials are flipped? (b) Determine the values that w1 and w2 must have in order for the loss per unit length to be the same regardless of the position of the materials. In other words, find values of w1 and w2 (in terms of a and b) such that the loss per unit length is the same for (i) σ1 = σ and σ2 = 3σ, or (ii) σ1 = 3σ and σ2 = σ.Figure Lossy coaxial cable. Problem.... Get solution
26. Ground current. A vertical lightning rod discharges 105 amperes into the ground. Find the voltage produced by the discharge between two points 1 m apart on a radial line if the point nearest the rod is at a distance of (a) 3 m, and (b) 10 m from the rod. Assume the ground to consist of a 20-cm layer of conducting soil with σ ≃ 10−2 S-m−1. Get solution
27. Inhomogeneous medium. Consider an inhomogeneous medium in which both ∊ and σ are functions of position. Show that a steady current J flowing though such a medium would necessarily establish a charge distribution given by ρ = −[∇∊− (∊/σ )∇σ] • ∇ϕ. Note that the field arising from the flow of current can still be derived from a scalar potential, but the potential no longer satisfies Laplace’s equation, since ρ ≠ 0. Get solution
28. Leaky capacitor. Consider the parallel-plate capacitor with spacing 2d and plate area A, as shown in Figure. The region between the plates is filled with two lossy dielectric slabs, each of thickness d and with parameters ∊1, σ1, and ∊2, σ2, respectively. A potential V0 is applied across the plates. (a) Find the steady-state value of the electric field in each of the two materials. (b) Find an expression for the surface charge density at the interface between the two materials.Figure Leaky capacitor. Problem.... Get solution
29. Lossy dielectric between spherical electrodes. Consider a spherical capacitor with the region between the inner and outer electrodes filled with lossy material with permittivity ∊0 and conductivity...where b is the radius of the outer electrode and σ0 is a constant. The inner electrode (with radius a) is held at a constant potential of V0 while the outer electrode (with radius b >a) is at zero potential. (a) Determine the charge Q on the inner and outer electrodes, and the charge distribution ρ(r) in the region between the electrodes. (b) Determine the total space-distributed charge in the region between the electrodes. Get solution
30. Inhomogeneous medium. A spherical electrode of radius a is surrounded by a concentric spherical shell of radius b. The space between the two conductors is filled with material whose conductivity varies linearly with distance from the common center of the spheres (i.e., σ = Kr, where K is a proportionality constant). If a potential difference of V volts is maintained between the spheres, with the inner electrode grounded (i.e., at zero potential), what is the electric potential at a distance r (a r b) from the center? Get solution
31. Leakage resistance. A very long wire of radius a is suspended horizontally near the bottom of a deep lake. Assume the lake to have a plane bottom that is a very good conductor. The wire is parallel to the bottom and is at a height h ≫ a above it. If the conductivity of the water is σ, find the resistance (per unit length) between the wire and the planar bottom of the lake. Get solution
32. Two conducting spheres. Two metallic conducting spheres of radii a1 and a2 are buried deep in poorly conducting ground of conductivity σ and permittivity ∊. The distance b between the spheres is much larger than both a1 and a2. Determine the resistance between the two spheres. Get solution
