SSED - Solved Problems For Chapter 3 [PDF]

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HCMUT – Falculty of Electrical and Electronics Engineering Course: Solid State Electronic Devices (EE2411) – Instructor: Hồ Trung Mỹ Solved problems for Chapter 3 - Energy bands and charge carries in semiconductors Notes: • For an electron: m = 9.1 x 10 –31 kg and q = 1.6 x 10–19 Coulomb, h = 6.62 x 10–34 Joule‐sec. • For silicon at T=300 K, ni = 1.5 × 1010 cm–3. The Boltzmann constant kB = k = 8.61 × 10−5 eV/K. Silicon bandgap energy Eg = 1.12 eV. • For problems about carrier concentrations, we use the aproximation X >> Y when X/Y ≥ 100. • Commonly accepted values of ni at T = 300 K  Silicon (Si) ni = 1.5 x 1010 cm–3  Gallium arsenide (GaAs) ni = 1.8 x 106 cm–3  Germanium (Ge) ni = 2.4 x 1013 cm–3 1. The longest wavelength that can be absorbed by silicon, which has the bandgap of 1.12 eV, is 1100 nm. If the longest wavelength that can be absorbed by another material is 870 nm, then bandgap of this material is (A) 1.416 eV (B) 0.886 eV (C) 0.854 eV (D) 0.706 eV Ans. (A) For light absortion, Eg = hc/λ where the bandgap Eg in eV and the longest wavelength λ in nm. We derive that EgSi × λ1 = EgX × λ2 = hc Or EgX = 1.12 × 1100 /870 = 1.416 eV



2. What is meant by an intrinsic semiconductor? Ans. n = p



All the electrons are originating from thermal excitation from the valence band for an intrinsic semiconductor. Then the concentration of electrons will equal the concentration of holes. This situation is nearly realized in a very pure semiconductor with no donors or acceptors.



3. Specify what semiconductor for each energy band diagram as follows:



Ans.



(a) Intrinsic semiconductor (c) N-type semiconductor



(b) P-type semiconductor (d) Degenerate semiconductor



4. For an intrinsic Semiconductor with a band gap of 0.7 eV, determine the positionof EF at T = 300 K if m*h = 6m*e. (Note: m*h = m*p and m*e = m*n) Ans. Bandgap Eg = 0.7 eV = 0.7 x 1.6 x 10–19V and T = 300 K. Fermi energy for an intrinsic semiconductor



SSED – Solved problems For Chapter 3 – page 1/7



Fermi energy level EF = 0.3847 eV (Note: 1 eV = 1.6 x 10−19J)



5. Calculate the thermal equilibrium concentrations of electrons and holes for a given Fermi energy. Consider silicon at T = 300 K so that NC = 2.8 x 1019 cm–3 and NV = 1.04 x 1019 cm-3. Assume that the Fermi energy is 0.25 eV below the conduction band. If we assume that the bandgap energy of silicon is 1.12 eV, then the Fermi energy will be 0.87 eV above the valence band. Ans. Using Equations for n0 and p0 in terms of the Fermi energy and We have



6. Consider a piece of intrinsic semiconductor. For this sample, the intrinsic carrier concentration ni = 1010 cm–3 at a given temperature T1. (a) What are the electron and hole concentrations n0 and p0 for this intrinsic sample at equilibrium at the same temperature T1? (b) Qualitatively, what happens to these concentrations if temperature is reduced? (c) This material is doped with donor atoms, with ND = 2 × 1015 cm–3. What type is the doped material? At the initial temperature T1, what is the new electron concentration? Remembering that at equilibrium n0p0 = ni2, what is the new hole concentration? Qualitatively compare the hole concentration for the intrinsic material to the hole concentration for the material with donors added. Ans.



(a) n0 = p0 = ni = 1010 cm–3 at the same temperature for the intrinsic material. (b) ni drops with lower temperature, therefore the electron and hole concentrations will also drop. (c) The material is n-type, since the type of doping used is donor atoms. Now n0 = ND = 2 × 1015 cm–3. Using the Law of Mass Action, p0 = ni2/n0 = 1020/(2 × 1015) = 0.5 × 105 cm–3. The hole concentration of the n-type doped material is much less than the hole concentration of the intrinsic material; even as the electron concentration is much higher than that of the intrinsic material.



7. Consider a silicon crystal at room temperature (300 K) doped with arsenic atoms so that ND = 6 × 1016 cm–3. Find the equilibrium electron concentration n0, hole concentration p0, and Fermi level EF with respect to the intrinsic Fermi level Ei and conduction band edge EC. Ans. This is an n-type material, as it is doped with donor atoms. Therefore: n0 ≈ ND = 6 × 1016 cm–3 (1) Then we can use the Law of Mass Action to find the hole concentration: p0 = n2i/n0 = (2.25 × 1020) / (6 × 1016) = 3.75 × 103 cm–3 (2)



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To find the Fermi level with respect to the intrinsic Fermi level, we use the expression that links electron concentration to Ei and ni: (3)



At room temperature kT = 8.61 × 10−5 × 300 ≈ 0.026 eV (4) We will be using this quantity often. Then the separation of the Fermi level and intrinsic Fermi level is, from Eqn. 3:



Drawing the band energy diagram, we can then place the Fermi level in the correct place with respect to the intrinsic Fermi level (middle of bandgap) and also find its separation from the conduction band edge EC.



8. Sketch the position of the Fermi level with respect to the intrinsic Fermi level at equilibrium for following cases: (Assume that the semiconductor has Eg of 1.12 eV and ni of 1010 cm–3 at T = 300 K ) a) Doping semiconductor with donor concentration of 3 x 1015 cm–3 b) Doping semiconductor with acceptor concentration of 5 x 1015 cm–3 Ans. a) Donor concentration of 3 x 1015 cm–3 We have n-type semiconductor with n = ND = 3 x 1015 cm–3 Therefore EF – Ei = kT ln(n/ni) = kT ln(ND/ni) = 0.026 ln(3 x 1015 / 1010) ≈ 0.33 eV



b) Acceptor concentration of 5 x 1015 cm–3 We have p-type semiconductor with p = NA = 5 x 1015cm–3. Therefore Ei – EF = kT ln(p/ni) = kT ln(NA/ni) = 0.026 ln(5 x 1015 / 1010) ≈ 0.34 eV



9. A silicon sample is doped with 1017 arsenic atoms cm–3. The minority carrier concentration at room temperature is (a) 2.25 x 103 cm–3 (b) 1.5 x 103 cm–3 (c) 1.5 x 108 cm–3 (d) 2.25 x 108 cm–3 Ans. (a) Given that ND = 1017 for silicon (As in group V), intrinsic concentration ni = 1.5 x 1010 cm–3. Therefore, the concentration for minority carriers (holes) is given by



SSED – Solved problems For Chapter 3 – page 3/7



10. A sample of Ge has n-type impurity concentration of 2 x 1014 donors/cm3 and p-type impurity concentration of 3 x 1014 acceptor/cm3. Find n and p at room temperature. Ans. Given that ND = 2 x 1014 cm–3, NA = 3 x 1014 cm–3 Then



11. Determine the thermal-equilibrium electron and hole concentrations in silicon at T = 300 K for given doping concentrations. a) Let ND = 1016 cm–3 and NA = 0. b) Let ND = 5 x 1015 cm–3 and NA = 2 x 1015 cm–3. Ans.



a) The majority carrier electron concentration is



The minority carrier hole concentration is found to be



b) The majority carrier electron concentration is The minority carrier hole concentration is found to be



 Quick calculations for the case of NA – ND ⪢ ni: a) ND – NA = 1016 cm–3 ⪢ ni = 1.5 x 1010 cm–3 ⇒ n-type: n0 = ND – NA = 1016 cm–3 and p0 = ni2/n0 = (1.5 x 1010)2/1016 = 2.25 x 104 cm–3 b) ND – NA = 3 x 1015 cm–3 ⪢ ni = 1.5 x 1010 cm–3 ⇒ n-type: n0 = ND – NA = 3 x 1015 cm–3 and p0 = ni2/n0 = (1.5 x 1010)2/(3 x 1015) = 7.5 x 104 cm–3



12. Consider a region of Si at room temperature. For each of the following cases, calculate the equilibrium electron and hole concentrations (n and p). Assume that the dopants are fully ionized. a) Intrinsic material (ND = NA =0) b) ND = 1.5 ×1013 cm–3 NA =0 NA =0 c) ND = 1.5 ×1017 cm–3 d) ND = 0 NA =1.5 ×1017 cm–3 e) ND =1.00 ×1017 cm–3 NA =2.5 ×1017 cm–3 Ans. a) For intrinsic material (Si), we know that n = p = ni = 1.5 x 1010 cm–3



b) First, let's ask what to expect. The donor density is 1000 times the intrinsic density. All of the donor electrons will go in the conduction band, overwhelming the number of intrinsic carriers that were there. So we expect n ≈ ND = 1.5 x 1013 cm–3 We would determine the hole density from np = ni2, so p = ni2/n = (1.5 x 1010)2 / (1.5 x 1013) = 1.5 x 107 cm–3 Note that doping the semiconductor n­type means that we now have more electrons in the conduction band, and we have fewer holes in the valence band than for the intrinsic semiconductor.



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c) First, let's ask what to expect. The donor density is 107 times the intrinsic density. All of the donor electrons will go in the conduction band, overwhelming the number of intrinsic carriers that were there. So we expect n ≈ ND = 1.5 x 1017 cm–3 We would determine the hole density from np = ni2, so p = ni2/n = (1.5 x 1010)2 / (1.5 x 1017) = 1.5 x 103 cm–3 d) Again, let's ask what to expect. The acceptor density is 107 times the intrinsic density. All of the acceptor holes will go in the valence band, overwhelming the number of intrinsic carriers that were there. So we expect p ≈ NA = 1.5 x 1017 cm–3 We would determine the electron density from np = ni2, so n = ni2/p = (1.5 x 1010)2 / (1.5 x 1017) = 1.5 x 103 cm–3 Note that doping the semiconductor p­type means that we now have more holes in the valence band, and we have fewer electrons in the conduction band than the intrinsic semiconductor. d) Since the net p­‐type doping is N A − N D = 2. 5 x 1017 − 1 x 1017 = 1 . 5 x 1017 is much greater than the intrinsic density, we expect: p = N A − N D = 1 . 5 x 1017 cm –3 and n = ni2/p = 1.5 x 103 cm –3



13. Indicate whether the following materials, doped with the indicated acceptor and donor concentrations (NA and ND, respectively) are n-type or p-type. For each case, give the equilibrium electron and hole concentrations n0 and p0. Use ni = 1010 cm–3 as the intrinsic carrier concentration a) NA = 4 x 1018 cm–3, no donors (ND = 0). b) NA = 1016 cm–3, ND = 5.1 x 1017 cm–3. c) NA = 1016 cm–3, ND = 8 x 1015 cm–3. d) No acceptors (NA = 0), ND = 5 x 1016 cm–3. Ans. Case



NA – ND (cm–3)



S/C type



n0 (cm–3)



p0(cm–3)



a)



4 x 1018



p



1020/p0 = 25



4 × 1018



b)



1016 – 5.1 x 1017 = –5 x 1017



n



5 ×1017



1020/n0 = 200



c)



1016 – 8 x 1015 = 2 x 1015



p



1020/p0 = 5 x 104



2 x 1015



d)



–5 x 1016



n



5 x 1016



1020/n0 = 2000



14. Consider a piece of uniformly-doped p-type material, with NA = 1 x 1017 cm–3. In this material, ni = 1010 cm–3, μn = 1000 cm2/V.s and μp=500 cm2/V.s. (a) Calculate n0 and p0. (b) Calculate electron and hole conductivities. Note: σn = qμnn; similar for holes. (a) The material is illuminated until an excess carrier concentration δn = δp = 1015 cm–3 is created uniformly throughout the sample. Calculate the new electron and hole conductivities,and compare with the previous case. Which type of carrier conductivity rose more, that of minorities or that of majorities? Ans. (a) p0 = NA = 1017 cm–3 and n0 = ni2/p0 = 1020/1017 = 103 cm–3. (b) σn0 = qμnn0 = 1.6 x 10–19 x1000 x 103 = 1.6 x 10–13 S/cm σp0 = qμpp0 = 1.6 x 10–19 x 500 x 1017 = 8 S/cm (c) σn = qμnn = qμn(n0 + δn) ≈ qμnδn = 1.6 x 10–19 x 1000 x 1015 = 0.16 S/cm >> σn0 σp = qμpp = qμp(p0 + δp) = 1.6 x 10–19 x 500 x (1017 + 1015) = 8.08 S/cm ≈ σp0 Conductivity of minorities rose more. 15. For an n-type semiconductor with the electron concentration of 1019 m–3 and their mobility of 1.6 m2/(V.s), the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored ) is close to: (A) 2 Ωm (B) 4 Ωm (C) 0.2 Ωm (D) 0.4 Ωm Ans. For N-type semiconductor electrons are majority carriers . Conductivity σ ≈ σn = qnµn Resistivity ρ = 1/σ = 1/qnµn = 1/ (1.6 x 10–19 x 1019 x 1.6) = 0.4 Ωm Hence option (D) is correct.



SSED – Solved problems For Chapter 3 – page 5/7



16. An n−type silicon bar 0.1 cm long and 100 μm2 cross-sectional area has a majority carrier concentration of 5 x 1020/m2 and the carrier mobility is 0.13 m2/V-s at 300 K. The resistance of the bar is (A) 106 Ω (B) 104 Ω (C) 10–1 Ω (D) 10–4 Ω Ans. (A)



We have



From above relation we have



Hence option (A) is correct.



17. A Silicon sample A is doped with 1018 atoms/cm3 of boron. Another sample B of identical dimension is doped with 1018 atoms/cm3 phosphorus. The ratio of electron to hole mobility is 3. The ratio of conductivity of the sample A to B is (A) 3 (B) 1/3 (C) 2/3 (D) 3/2 Ans. (B)



Hence option (B) is correct.



18. Design a semiconductor resistor with a specified resistance to handle a given current density. A silicon semiconductor at T = 300 K is initially doped with donors at a concentration of ND = 5 x 1015 cm–3. Acceptors are to be added to form a compensated p-type material. The resistor is to have a resistance of 10 kΩ and handle a current density of 50 A/cm2 when 5 V is applied. Assume that limitation of electric filed E is 100 V/cm. Ans.



For 5 V applied to a 10-k resistor, the total current is I = V / R = 5V/10kΩ = 0.5 mA If the current density is limited to 50 A/cm2, then the cross-sectional area is A = I / J = 0.5 x 10–3 / 50 = 10–5 cm2 The length of the resistor is L = V/ E = 5/100 = 5 x 10–2 cm The conductivity of the semiconductor is σ = L / (RA) = 5 x 10–2 / (104 x 10–5) = 0.50 S/cm



19. A silicon crystal having a cross-sectional area of 0.001 cm2 and a length of 20 µm is connected to its ends to a 20 V battery. At T = 300 K, we want a current of 100 mA in crystal. The concentration of donor atoms to be added is (A) 2.4 x 1013 cm–3 (B) 4.6 x 1013 cm–3 (C) 7.8 x 1014 cm–3 (D) 8.4 x 1014 cm–3 Ans. (B)



SSED – Solved problems For Chapter 3 – page 6/7



20. Six volts is applied across a 2 cm long semiconductor bar. The average drift velocity is 104 cm/s. The electron mobility is (A) 4396 cm2(V.s)–1 (B) 3 x 104 cm2(V.s)–1 (C) 6 x 104 cm2(V.s)–1 (D) 3333 x 104 cm2(V.s)–1 Ans. (D)



21. In a particular semiconductor the donor impurity concentration is Nd = 1014 cm-3. Assume the following parameters,



An electric field of E = 100 V/cm is applied. The electric current density at 300 K is (A) 2.3 A/cm2 (B) 1.6 A/cm2 (C) 9.6 A/cm2 (D) 3.4 A/cm2 Ans. (B)



 The silicon sample with unit cross-sectional area shown below is in thermal equilibrium (for next two questions).



22. The magnitude of the electric field at x = 1 μm is (A) 2.5 kV/cm (B) 5 kV/cm (C) 7.5 kV/cm Ans. (B)



(D) 10 kV/cm



E = V/L = 1V/2µm = 1 V/ (2 x 10–4cm) = 5 kV/cm



23. The magnitude of the electron drift current density at x = 1 μm is (A) 52.7 x 104 A/cm2 (B) 35.1 x 104 A/m2 (C) 21.6 x 104 A/cm2 Ans. (C)



(D) 10.8 x 104 A/cm2



Conductivity σ ≈ σn = qnµn = qNDµn 1.6e-19*2e17*1350*5e3 Electron drift current density J = σE = qNDµnE = 1.6x10–19 x 2x1017 x 1350 x 5x103 = 21.6 x 104 A/cm2



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