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The University of Western Australia School of Electrical, Electronic and Computer Engineering J.H 2006 1 TS 4 Solutions ______________________________________________________________________________________ Problems 1. (Pierret 2.2) Using the energy band model for a semiconductor indicate how one visualises a) an electron,b) a hole, c) donor sites, d) acceptor sites, e) an intrinsic semiconductor, f) an n-type semiconductor, g) a p-type semiconductor Solution Donor Level eg Phosphorus doped Acceptor Level eg Boron doped - Electron E D E Hole A + n - type p-type Intrinsic CB EG VB 2. Explain in general terms how we might produce an n-type and a p-type semiconductor from an ultra-pure Gr IV semiconductor. Solution: Si Si Si Si Substitute - Si P Si Si Si Si The University of Western Australia School of Electrical, Electronic and Computer Engineering J.H 2006 2 TS 4 Solutions ______________________________________________________________________________________ 3. (Callister Eg 19.2) Phosphorus is added to high purity silicon to give a 23 3 concentration of 10 /m charge carriers at room temperature. a) is this material n or p type. b) calculate the room temperature conductivity of this material, assuming that electron and hole mobilities are the same as for the intrinsic material. Solution a) P Group V Si Group IV n-type material b) Conductivity = neµn (µn is the same as for an intrinsic material) µn = 0.15 m2/V-s = 1500 cm2/V-s σ = 1023. 1.6 x 10-19. 0.15 = 2400 (Ω-m)-1 4. (Callister 19.27) Will each of the following elements act as a donor or an acceptor when added to the indicated semiconducting material? Assume that the impurity elements are substitutional. (Note for example, GaAs is a II-V semiconductor and Cd is a Group IIb. Cd will substitute for Ga, and being an electron short will give rise to p-type conduction) Solution Impurity Semiconductor Al Si Acceptor P Ge Donor Cd GaAs Acceptor (Cd replaces Ga - 1 el short) S AlP Donor (S subs for P - 1 el extra) Sb ZnSe Acceptor (Sb subs for Se 1 el short) In CdTe Donor (In subs for Cd, 1 extra el) 5. (Callister 19.32) The following electrical characteristics have been determined for both intrinsic and p-type gallium antimonide (GaSb) at room temperature: The University of Western Australia School of Electrical, Electronic and Computer Engineering J.H 2006 3 TS 4 Solutions ______________________________________________________________________________________ σ (Ω-m)-1 n (m-3) p (m-3) Intrinsic 8.9 x 104 8.7 x 1023 8.7 x 1023 Extrinsic 2.3 x 105 7.6 x 1022 1.0 x 1025 Calculate the electron and hole mobilities. Intrinsic: σ = n|e|µe + pqµh 4 To start: 8.9 x 10 = 8.7 x 1023 . 1.6 x 10-19 µe + 8.7 x 1023 . 1.6 x 10-19 µp So we need µe and µh Finding µh: Extrinsic σ = pqµh 2.3x 105 = 1 x 1025.1.6x10-19µh µh=0.1438 m2/V-s Therefore µe=0.50 m2/V-s 6. Callister Eg19.1 For intrinsic silicon the room temperature conductivity is 4 -4 -1 x 10 (Ω-m) ; the electron and hole mobilities are respectively 0.14 and 0.048 m2/V-s. Compute the electron and hole concentration at room temperature. Solution a) σ = 4 x 10-4 (Ω-m)-1 µe = 0.14 m2/V-s µh = 0.048 m2/V-s σ = neµe + pqµh 4 x 10-4 = n . 1.6 x 10-19 (0.14 + 0.048) n=p = 1.33 x 1016 /m3 7. (Pierret pge 127 (5)) Which diagram shown below best shows the position of electrons at 0 K in an n-type material? Solution Only (b) and (c) show a donor level . AT 0oK there will be no ionisation of donors all electrons will be in the donor level so the answer is (b) The University of Western Australia School of Electrical, Electronic and Computer Engineering J.H 2006 4 TS 4 Solutions ______________________________________________________________________________________ 8. Using energy band diagrams, explain the mechanisms of conduction in an intrinsic semiconductor, an extrinsic p-type material and an extrinsic n-type semiconductor. Which mechanism is dominant in each case? Solution Extrinsic p-type Major mechanism Minor mechanism Many holes Small number of electrons CB CB ! EA VB ++++++ VB + Intrinsic Mechanism Applied field E - + The two mechanisms that occur in the same crystal and at the same time, with the extrinsic generation producing the majority hole carriers and the intrinsic producing the minority electrons. Extrinsic n-type Major mechanism Minor mechanism Many electrons Small number of holes The University of Western Australia School of Electrical, Electronic and Computer Engineering J.H 2006 5 TS 4 Solutions ______________________________________________________________________________________ 9. Pierret pge 127 The illustration below is likely to depict the situation inside an n-type semiconductor at approximately what temperature? a) T<100K b) T about room temperature c) T> 400K d) T about 0 K Solution 0K no ionisation takes place Room Temp More ionisation occurs >400K fully ionised So the answer is (a) 1 10. The Fermi-Dirac distribution function is given by : f(E) = E ! Ef $ 1+ exp" # kT % Show that at 0oK all states below Ef are filled and all states above Ef are empty. Solution 1 f(E) = "E ! E f % 1+ exp $ # kT ' & 1 1 At T=0K f(E) = ! + $ ' 1+ ( ' 0 1+ exp # & "0 % ie at 0K all the states above the Fermi level are empty 1 1 Say E<Ef f(E) = "! % ( 1 ( 1 1+ exp $ ' #0 & ie at 0K all the states below the Fermi level are full 11. Calculate the number of holes in the valence band of pure Ge at 300oK, given Eg=0.7eV and mp*=m. Solution !E N p = N v exp f kT 3 " 2 ! m * kT % 2 N v = 2$ p with 2 ' # h & The University of Western Australia School of Electrical, Electronic and Computer Engineering J.H 2006 6 TS 4 Solutions ______________________________________________________________________________________ 3 3 " 2 ! m* kT % 2 " -31 -23 %2 N v = 2$ p ' = 2$ 2 ! 9.1x10 .1.38x10 300 ' $ ' $ h 2 ' (34 2 (6.62x10 ) # & # & = 2.51 x 1025 states/m3 !E f !0.35x1.6x10!19 N p = N v exp = 2.51x1025 exp kT 1.38x10-23.300 = 3.35x1019 /m3 12. In questions 3 and 4, a simplified distribution function is used to obtain carrier densities i.e Maxwell-Boltzmann. In what region of space would this be invalid? Solution This approximation is valid far from the Fermi level as the “1” becomes insignificant in comparison with the exponential term. This approximation would be invalid close to the Fermi level. 13. (Pierret 2.4) At T>0 K what is the probability of an electron state being occupied if it is located at the Fermi level? Solution By definition, te Fermi level is the “half filled” level ie there is a 50% chance of finding an electron at the Fermi level. 1 1 1 1 f(E) = = = = "E ! E f % " 0 % 1 +1 2 1+ exp $ ' 1 + exp $ ' # kT & # kT & The University of Western Australia School of Electrical, Electronic and Computer Engineering J.H 2006 7 TS 4 Solutions ______________________________________________________________________________________ 14. A silicon chip is doped with one phosphorus atom per 108 host atoms. Assume that the device operates at a temperature where complete impurity ionisation occurs, and calculate the resistivity and conductivity. Silicon has an atomic density of 5 1022 3 atoms/cm and an electron mobility of 1500 cm2/volt sec. Solution 1/108 = 10-8 P atoms per silicon. Complete Ionisation Si has 5 x 1022 atoms per cm3 µ = 15 cm2/V-s No of P atoms = 10-8 x 5x1022 /cm3 = 5 x1014 atoms/cm3 Complete ionisation: so we have n = 5 x1014 /cm3 σ = neµ = 5 x1014x1.6x10-19 = 1500=0.12 (Ω-cm)-1 ρ=8.33 Ω-cm 15. If a field of 500 mV/mm is applied across the device of Q21, calculate the current density J flowing through the device. E=500V/mm = 0.5 V/mm = 5 V/cm J=σE = 0.12.5 = 0.6A/cm2 16. Explain how minority carriers are thermally generated in doped semiconductors and the precise meaning of the terms n>>p and also p>>n. Solution These terms refer to minority and majority carriers in an extrinsic semiconductor. Let us take a p-type material. There will be many holes left behind in the VB from the promotion of electrons from the VB to the acceptor level - this is the main mechanisn of carrier promotion. There is however a finite probability that some electrons will have enough energy to bridge the large gap from the VB to the CB giving us some conducting electrons - minor mechanism. As a result there are many holes and a few electrons p>>n. Let us take a n-type material. There will be many electrons in the CB from the promotion of electrons from the donor level to the CB - this is the main mechanisn of carrier promotion. There is however a finite probability that some electrons will have enough energy to bridge the large gap from the VB to the CB giving us some conducting holes - minor mechanism. As a result there are many electrons and a few holes n>>p 17. Show that a donor impurity level in an n-type semiconductor is approximately 95% ionised when the fermi level lies 3kT below ED. The University of Western Australia School of Electrical, Electronic and Computer Engineering J.H 2006 8 TS 4 Solutions ______________________________________________________________________________________ Solution Ef = 3kT below ED 1 Ef =ED - 3kT f(E D ) = " E D ! E D ! 3kT % = 0.048 1+ exp $ ' # kT & There is a 5% chance of finding an electron in the donor level ie a 95% chance of NOT finding an electron. This means that the donor level is 95% ionised 18. (Neaman 3.1) Calculate the intrinsic carrier concentration ni at T=200, 400 and 600K for Si and Ge. !(E G - E f ) ! Ef kT kT 3 np = Nc Nv e e 3 " 2 ! m* kT % 2 " 2 ! m*kT % 2 p !E G N v = 2$ $ 2 ' ' N c = 2$ e ' $ 2 ' = Nc Nv e kT = n2 i # h & # h & There are large discrepancies for values of effective mass (mainly because they are experimentally measured) in these results, we have used: mp = 0.16m and me = 0.98m. As long as you have the right order of magnitude that is ok. T(K) Eg (Si) Nv Nc Eg/kT ni (m-3) ni (cm-3) 200 1.12eV 8.74x 1023 1.325x1025 64.9 2.74 x 1010 2.74 x 104 400 1.12eV 2.47x1024 3.75x1025 32.46 8.6x1017 8.6x1011 600 1.12eV 4.54x1024 6.89x1025 21.6 3.53x1020 3.53x1014 Another (easier) way to calculate is to group the terms together: !(E G - E f ) ! Ef np = Nc Nv e kT e kT 3 3 # 2 " m* kT & 2 # 2 " m* kT & 2 !E G p = 2% % 2 ( ( 2% % 2 e ( e kT ( $ h ' $ h ' G !E # T & 3 kT = N VN C% ( e $ 300 ' NC300K = 1.04x1019 /cm3 NV300K = 2.8x1019 /cm3 The University of Western Australia School of Electrical, Electronic and Computer Engineering J.H 2006 9 TS 4 Solutions ______________________________________________________________________________________ T(K) Eg (Si) Eg/kT ni (cm-3) 200 1.12eV 64.9 7.5 x 104 400 1.12eV 32.46 2.35x1012 600 1.12eV 21.6 9.84x1014 NC300K = 6.0x1018 /cm3 NV300K = 1.04x1019 /cm3 T(K) Eg (Ge) Eg/kT ni (cm-3) 200 0.66eV 38.26 2.1 x 1010 400 0.66eV 19.13 8.5x1014 600 0.66eV 12.75 3.8x1016 19. (Neaman 3.6) The carrier effective masses in a semiconductor are m * = 0.62m 0 and m * = 1.4m 0 . Determine the position of the intrinsic Fermi level with e p respect to the centre of the bandgap at T=300K. EG 3 m* p Ef = + kT ln * 2 4 me EG 3 m* p (1.38x10!23) " 1.4 % Ef ! = kT ln * = 0.75 (300)ln $ = 0.0158eV 2 4 me 1.6x10!19 # 0.62 ' & So it is 0.0158e above midgap and therefore an n-type material. 20. (Neaman 3.13) The electron concentration in Si at T=300K is.n 0 = 5 ! 104 cm"3 a) Determine p. Is this an n or p-type material? b) Determine the position of the Fermi level with respect to the intrinsic Fermi level. ni = 1.4 x 1010 /cm3 2 i ( n 2 = n 0p 0 = 5 ! 104 p0 = 1.4x1010 ) p0 = 3.92x1015 / cm3 Therefore this is a p type material. The University of Western Australia School of Electrical, Electronic and Computer Engineering J.H 2006 10 TS 4 Solutions ______________________________________________________________________________________ kT n 1.38x10 !23 " 5x10 4 % E f ! E fi = ln = 0.5 !19 300ln $ ' = !0.32eV 2 p 1.6x10 $ 3.92x1015 ' # & As expected the Fermi level is below the intrinsic Fermi level (midgap). Note that if you used ni = 1.5 x 1010 /cm3 then the difference in energy levels would be: 0.3266eV 21. (Neaman 3.15) a) If Ec - EF =0.25eV in GaAs at T=400K, calculate the values of n and p b) Assuming that n doesn’t change from part (a) determine Ec - EF at 300K !(E G ! E f ) N e = Nc exp kT From Neaman: GaAs: Egap =1.42eV; mh = 0.48m, mn = 0.067m So calculating: 1) 3 3 " 2 ! m*kT % 2 " % N c = 2$ e ' = 2$ 2 ! 0.067mk400 ' 2 = 6.7x1023 $ ' # h2 & # h2 & !(0.25) N e = 6.7x10 23 exp = 6.7x10 23 • 7.13x10 !4 = 4.81x1020 / m 3 = 4.8x1014 / cm3 kT 2) 3 ! T $2 # & " 300 % Or use: NC300 = 4.7x1017/cm3: NC400 = 4.7x1017 = 7.24x1017 /cm3 So !(0.25) N e = 6.7x10 23 exp = 7.24x1017 • 7.13x10 !4 = 5.16x1014 / cm3 kT 22. (Neaman 3.22) The Fermi level in n-type Si at T=300K is 245meV below the conduction band and 200meV below the donor level. Determine the probability of finding an electron a) in the donor level and b) in a state in the conduction band kT above the conduction band edge. The University of Western Australia School of Electrical, Electronic and Computer Engineering J.H 2006 11 TS 4 Solutions ______________________________________________________________________________________ EC - Efn = 0.245eV ED - EEfn = 0.2eV so therefore EC - ED = 0.045eV Ec E 0.245eV D 0.2eV E F We can use FD or a MB approximation since we are quite far from the Fermi level: f(E) = 1 = 1 = 4.42x10 !4 E - Ef 0.2 (1 + e kT ) (1 + e kT ) b) E-EC = kT=0.0259eV, E-EF = 0.245+kT f(E) = 1 = 1 = 2.87x10 !5 E - Ef 0.245+.0259 (1 + e kT ) (1 + e kT ) 23. (Neaman 3.28) A GaAs device is doped with a donor concentration of 3 x1015 cm-3. For the device to operate properly the intrinsic carrier concentration must remain at less than5% of the total electron concentration. What is the maximum temperature at which the device may operate? ND = 3x1015 /cm3 so ni = 0.05x3 1015 /m3= 1.5x1014 /m3 !E G np = Nc Nv e kT = n2 i G !E " T % 3 kT n2 28 i = 2.25x10 = NV300 NC300 $ ' e # 300 & !1.42 " T %3 2.25x1028 = 7x1018 4.7x1017 $ ' e kT # 300 & !1.42 " T % 3 kT 6.8x10 !9 =$ ' e # 300 & using trial and error: T=500K: 2.3x10-14 < 6.8x 10-9 T=1000K: 2.62x10-6 > 6.8x 10-9 T=800K: 2.19x10-8> 6.8x 10-9 T=700K 7.75x10-10< 6.8x 10-9 The University of Western Australia School of Electrical, Electronic and Computer Engineering J.H 2006 12 TS 4 Solutions ______________________________________________________________________________________ T=750K 4.57x10-9< 6.8x 10-9 etc Final answer is about 762K 24. (Neaman 3.31) Silicon at T=300K is doped with acceptor atoms at a concentration of NA = 7 x 1015 cm-3 a) Determine EF - EV b) Calculate the concentration of additional acceptor atoms that must be added to move the Fermi level a distance kT closer to the valence band edge. a) !E f N p = N v exp kT At 300K, Si has NV 1.04 x 1019 / cm3: therefore "E f 7 ! 1015 = 1.04 ! 1019 exp 0.0259 E f = 0.189eV b) The full form of this is !(E f ! E v ) N p = Nv exp kT So the energy gap would be Ef-Ev = 0.189-0.0.259=0.1633eV "0.1633 N p = 1.04 ! 1019 exp 0.0259 N p = 1.90 ! 1016 #Np = 1.90 ! 1016 " 7 ! 1015 = 1.2 ! 1016 / cm3

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