EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri Second Edition ( 2001 McGraw-Hill) Chapter 5 5.1 Bandgap and photodetection a. Determine the maximum value of the energy gap that a semiconductor, used as a photoconductor, can have if it is to be sensitive to yellow light (600 nm). b. A photodetector whose area is 5 × 10 -2 cm 2 is irradiated with yellow light whose intensity is 2 mW cm -2 . Assuming that each photon generates one electron-hole pair, calculate the number of pairs generated per second. c. From the known energy gap of the semiconductor GaAs (E g = 1.42 eV), calculate the primary wavelength of photons emitted from this crystal as a result of electron-hole recombination. d. Is the above wavelength visible? e. Will a silicon photodetector be sensitive to the radiation from a GaAs laser? Why? Solution a We are given the wavelength λ = 600 nm, therefore we need E ph = h υ = E g so that, E g = hc/ λ = (6.626 × 10 -34 J s)(3.0 × 10 8 m s -1 ) / (600 × 10 -9 m) ∴ E g = 3.31 × 10 -19 J or 2.07 eV b Area A = 5 × 10 -2 cm 2 and light intensity I light = 2 × 10 -3 W/cm 2 . The received power is: P = AI light = (5 × 10 -2 cm 2 )(2 × 10 -3 W/cm 2 ) = 1.0 × 10 -4 W N ph = number of photons arriving per second = P/E ph ∴ N ph = (1.0 × 10 -4 W) / (3.31 × 10 -19 J) = 3.02 × 10 14 Photons s -1 Since the each photon contributes one electron-hole pair (EHP), the number of EHPs is then: N EHP = 3.02 × 10 14 EHP s -1 c For GaAs, E g = 1.42 eV and the corresponding wavelength is λ = hc/E g = (6.626 × 10 -34 J s)(3.0 × 10 8 m s -1 ) / (1.42 eV × 1.602 × 10 -19 J/eV) ∴ λ = 8.74 × 10 -7 m or 874 nm The wavelength of emitted radiation due to electron-hole pair (EHP) recombination is therefore 874 nm. d It is not in the visible region (it is in the infrared). e From Table 5.1 (in the textbook), for Si, E g = 1.10 eV and the corresponding cut-off wavelength is, λ g = hc/E g = (6.626 × 10 -34 J s)(3.0 × 10 8 m s -1 ) / (1.1 eV × 1.602 × 10 -19 J/eV) ∴ λ g = 1.13 × 10 -6 m or 1130 nm 5.1 EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri Since the 874 nm wavelength of the GaAs laser is shorter than the cut-off wavelength of 1130 nm, the Si photodetector can detect the 874 nm radiation (Put differently, the photon energy corresponding to 874 nm, 1.42 eV, is larger than the E g , 1.10 eV, of Si which means that the Si photodetector can indeed detect the 874 nm radiation). 5.2 Minimum conductivity a. Consider the conductivity of a semiconductor, σ = en µ e + ep µ h . Will doping always increase the conductivity? b. Show that the minimum conductivity for Si is obtained when it is p-type doped such that the hole concentration is p m = n i µ e µ h and the corresponding minimum conductivity (maximum resistivity) is σ min = 2en i µ e µ h c. Calculate p m and σ min for Si and compare with intrinsic values. Solution a Doping does not always increase the conductivity. Suppose that we have an intrinsic sample with n = p but the hole drift mobility is smaller. If we dope the material very slightly with p-type then p > n. However, this would decrease the conductivity because it would create more holes with lower mobility at the expense of electrons with higher mobility. Obviously with further doping p increases sufficiently to result in the conductivity increasing with the extent of doping. b To find the minimum conductivity, first consider the mass action law: np = n i 2 isolate n: n = n i 2 /p Now substitute for n in the equation for conductivity: σ = en µ e + ep µ h ∴ σ = en i 2 µ e p + µ h ep To find the value of p that gives minimum conductivity (p m ), differentiate the above equation with respect to p and set it equal to zero: d σ =− en i 2 µ e 2 + µ h e dp p ∴ − en i 2 µ e 2 p m + µ h e = 0 Isolate p m and simplify, 5.2 EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri p m = n i µ e µ h Substituting this expression back into the equation for conductivity will give the minimum conductivity: σ min = en i 2 µ e p m + µ h ep m = en i 2 µ e n i µ e µ h + µ h en i µ e µ h ∴ σ min = en i µ e µ h µ e + en i µ e µ h = en i µ e µ h + en i µ e µ h ∴ σ min = 2en i µ e µ h c From Table 5.1, for Si: µ e = 1350 cm 2 V -1 s -1 , µ h = 450 cm 2 V -1 s -1 and n i = 1.45 × 10 10 cm -3 . Substituting into the equations for p m and σ min : p m = n i µ e µ h = 1.45 ×10 10 cm −3 () 1350 cm 2 V −1 s −1 450 cm 2 V −1 s −1 = 2.51 × 10 10 cm -3 σ min = 2en i µ e µ h ∴ σ min = 2 1.602 ×10 −19 C () 1.45 ×10 10 cm −3 () 1350 cm 2 V −1 s −1 ( ) 450 cm 2 V −1 s −1 ( ) ∴ σ min = 3.62 × 10 -6 Ω -1 cm -1 The corresponding maximum resistivity is: ρ max = 1 / σ min = 2.76 × 10 5 Ω cm The intrinsic value corresponding to p m is simply n i (= 1.45 × 10 10 cm -3 ). Comparing it to p m : p m n i = 2.51 ×10 10 cm −3 1.45 ×10 10 cm −3 =1.73 The intrinsic conductivity is: σ int = en i ( µ e + µ h ) ∴ σ int = (1.602 × 10 -19 C)(1.45 × 10 10 cm -3 )(1350 cm 2 V -1 s -1 + 450 cm 2 V -1 s -1 ) ∴ σ int = 4.18 × 10 -6 Ω -1 cm -1 Comparing this value to the minimum conductivity: σ int σ min = 3.62 ×10 −6 W −1 cm − 1 4.18 ×10 −6 W -1 cm -1 = 0.866 Sufficient p-type doping that increases the hole concentration by 73% decreases the conductivity by 15% to its minimum value. 5.3 Compensation doping in Si a. A Si wafer has been doped n-type with 10 17 As atoms cm -3 . 1. Calculate the conductivity of the sample at 27 °C. 5.3 EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri 2. Where is the Fermi level in this sample at 27 °C with respect to the Fermi level (E Fi ) in intrinsic Si? 3. Calculate the conductivity of the sample at 127 °C. b. The above n-type Si sample is further doped with 9 × 10 16 boron atoms (p-type dopant) per centimeter cubed. 1. Calculate the conductivity of the sample at 27 °C. 2. Where is the Fermi level in this sample with respect to the Fermi level in the sample in (a) at 27 °C? Is this an n-type or p-type Si? Solution a Given temperature T = 27 °C = 300 K, concentration of donors N d = 10 17 cm -3 , and drift mobility µ e ≈ 800 cm 2 V -1 s -1 (from Figure 5Q3-1). At room temperature the electron concentration n = N d >> p (hole concentration). 50 100 1000 2000 10 15 10 16 10 17 10 18 10 19 10 20 -3 Electrons Holes Dopant Concentration, cm Figure 5Q3-1 The variation of the drift mobility with dopant concentration in Si for electrons and holes at 300 K. (1) The conductivity of the sample is: σ = eN d µ e ≈ (1.602 × 10 -19 C)(10 17 cm -3 )(800 cm 2 V -1 s -1 ) = 12.8 Ω -1 cm -1 (2) In intrinsic Si, E F = E Fi , n i = N c exp[−(E c − E Fi )/kT] (1) In doped Si, n = N d , E F = E Fn , n = N d = N c exp[−(E c − E Fn )/kT] (2) Eqn. (2) divided by Eqn. (1) gives, N d = exp E Fn − E Fi (3) n i kT ∴ ln N d n i = E Fn − E Fi kT 5.4 EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri ∴ ∆E F = E Fn − E Fi = kT ln(N d /n i ) (4) Substituting we find ( n i = 1.45 × 10 10 cm -3 from Table 5.1 in the textbook), ∆ E F = (8.617 × 10 -5 eV/K)(300 K)ln[(10 17 cm -3 )/ (1.45 × 10 10 cm -3 )] ∴ ∆ E F = 0.407 eV above E fi 10 100 1000 10000 50000 70 100 80 0 Ge N d =10 13 Si µ L ∝T –1.5 N d =10 14 N d =10 16 N d =10 17 N d =10 18 N d =10 19 µ Ι ∝T 1.5 Temperature (K) Figure 5Q3-2 Log-log plot for drift mobility versus temperature for n-type Ge and n-type Si samples. Various donor concentrations for Si are shown, N d are in cm -3 . The upper right insert is the simple theory for lattice limited mobility whereas the lower left inset is the simple theory for impurity scattering limited mobility. (3) At T i = 127 °C = 400 K, µ e ≈ 450 cm 2 V -1 s -1 (from Figure 5Q3-2). The semiconductor is still n- type (check that N d >> n i at 400 K), then σ = eN d µ e ≈ (1.602 × 10 -19 C)(10 17 cm -3 )(450 cm 2 V -1 s -1 ) = 7.21 Ω -1 cm -1 b The sample is further doped with N a = 9 × 10 16 cm -3 = 0.9 × 10 17 cm -3 acceptors. Due to compensation, the net effect is still an n-type semiconductor but with an electron concentration given by, n = N d − N a = 10 17 cm -3 − 0.9 × 10 17 cm -3 = 1 × 10 16 cm -3 (>> n i ) We note that the electron scattering now occurs from N a + N d (1.9 × 10 17 cm -3 ) number of ionized centers so that µ e ≈ 700 cm 2 V -1 s -1 (Figure 5Q3-1). (1) σ = eN d µ e ≈ (1.602 × 10 -19 C)(10 16 cm -3 )(700 cm 2 V -1 s -1 ) = 1.12 Ω -1 cm -1 (2) Using Eqn. (3) with n = N d − N a we have N d − N a n i = exp ′ E Fn − E Fi kT so that ∆E′ Φ = E Fn ′ − E Fi = (0.02586 eV)ln[(10 16 cm -3 ) / (1.45 × 10 10 cm -3 )] ∴ ∆ E′ Φ = 0.348 eV above E Fi The Fermi level from (a) and (b) has shifted “down” by an amount 0.059 eV. Since the energy is still above the Fermi level, this an n-type Si. 5.5 EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri 5.4 Temperature dependence of conductivity An n-type Si sample has been doped with 10 15 phosphorus atoms cm -3 . The donor energy level for P in Si is 0.045 eV below the conduction band edge energy. a. Calculate the room temperature conductivity of the sample. b. Estimate the temperature above which the sample behaves as if intrinsic. c. Estimate to within 20% the lowest temperature above which all the donors are ionized. d. Sketch schematically the dependence of the electron concentration in the conduction band on the temperature as log( n) versus 1/T, and mark the various important regions and critical temperatures. For each region draw an energy band diagram that clearly shows from where the electrons are excited into the conduction band. e. Sketch schematically the dependence of the conductivity on the temperature as log( σ ) versus 1/T and mark the various critical temperatures and other relevant information. Solution Si Ge GaAs 0°C200°C400°C600°C 27°C 10 18 1 1.5 2 2.5 3 3.5 4 10 15 10 12 10 3 10 6 10 9 1.45×10 10 cm -3 2.4×10 13 cm -3 2.1×10 6 cm -3 1000/ T (1/K) Figure 5Q4-1 The temperature dependence of the intrinsic concentration. a The conductivity at room temperature T = 300 K is ( µ e = 1350 × 10 -4 m 2 V -1 s -1 can be found in Table 5.1 in the textbook): σ = eN d µ e ∴ σ = (1.602 × 10 -19 C)(1 × 10 21 m -3 )(1350 × 10 -4 m 2 V -1 s -1 ) = 21.6 Ω -1 m -1 b At T = T i , the intrinsic concentration n i = N d = 1 × 10 15 cm -3 . From Figure 5Q4-1, the graph of n i (T) vs. 1/T, we have: 5.6 EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri 1000 / T i = 1.9 K -1 ∴ T i = 1000 / (1.9 K -1 ) = 526 K or 253 °C c The ionization region ends at T = T s when all donors have been ionized, i.e. when n = N d . From Example 5.7, at T = T s : n = N d = 1 2 N c N d 1 2 exp −∆E 2kT s ∴ T s = −∆ E 2k ln N d 1 2 N c N d = − ∆ E 2k ln 2N d N c ∴ T s = ∆ E k ln N c 2N d Take N c = 2.8 × 10 19 cm -3 at 300 K from Table 5.1 (in the textbook), and the difference between the donor energy level and the conduction band energy is ∆E = 0.045 eV. Therefore our first approximation to T s is: T s = ∆E k ln N c 2N d = 0.045 eV () 1.602 × 10 − 19 J/eV ( ) 1.381×10 −23 J/K () ln 2.8 ×10 19 cm −3 () 210 15 cm −3 () = 54.68 K Find the new N c at this temperature, N c ′: ′ N c = N c T s 300 3 2 = 2.8 × 10 19 cm −3 () 54.68 K 300 K 3 2 = 2.179 × 10 18 cm -3 Find a better approximation for T s by using this new N c ′: T ′ s = ∆E k ln ′ N c 2N d = 0.045 eV () 1.602 × 10 − 19 J/eV ( ) 1.381×10 −23 J/K () ln 2.179 ×10 18 cm −3 () 210 15 cm −3 () = 74.64 K ∴ ′ ′ N c = N c ′ T s 300 3 2 = 2.8 × 10 19 cm −3 () 74.64 K 300 K 3 2 = 3.475 × 10 18 cm -3 A better approximation to T s is: T ′ ′ s = ∆E k ln ′ ′ N c 2N = 0.045 eV () 1.602 × 10 − 19 J/eV ( ) 1.381×10 −23 J/K () ln 3.475 ×10 18 cm −3 () 210 15 cm −3 () = 69.97 K d ∴ ′ ′ ′ N c = N c ′ ′ T s 300 3 2 = 2.8 × 10 19 cm −3 () 69.97 K 300 K 3 2 = 3.154 × 10 18 cm -3 5.7 EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri ∴ ′ ′ ′ T s = ∆E k ln ′ ′ ′ N c 2N d = 0.045 eV () 1.602 × 10 − 19 J/eV ( ) 1.381×10 −23 J/K () ln 3.154 ×10 18 cm −3 () 210 15 cm −3 () = 70.89 K We can see that the change in T s is very small, and for all practical purposes we can consider the calculation as converged. Therefore T s = 70.9 K = −202.1 °C. d and e See Figures 5Q4-2 and 5Q4-3. ln(n) 1/ T T i T s Intrinsic Extrinsic Ionization n i (T) ln(N d ) slope = –∆E/2k slope = –E g /2k Figure 5Q4-2 The temperature dependence of the electron concentration in an n-type semiconductor. log(n) INTRINSIC EXTRINSIC IONIZATION log( σ ) log( µ ) µ ∝T –3/2 µ ∝T 3/2 Lattice scattering Impurity scattering 1/ T Low TemperatureHigh Temperature T Metal Semiconductor Figure 5Q4-3 Schematic illustration of the temperature dependence of electrical conductivity for a doped (n-type) semiconductor. 5.5 GaAs Ga has a valency of III and As has V. When Ga and As atoms are brought together to form the GaAs crystal, as depicted in Figure 5Q5-1, the 3 valence electrons in each Ga and the 5 valence electrons in each As are all shared to form four covalent bonds per atom. In the GaAs crystal with some 10 23 or so equal numbers of Ga and As atoms, we have an average of four valence electrons per atom, whether Ga or As, so we would expect the bonding to be similar to that in the Si crystal: four bonds per atom. The crystal structure, however, is not that of diamond but rather that of zinc blende (Chapter 1 of the textbook). 5.8 EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri a. What is the average number of valence electrons per atom for a pair of Ga and As atoms and in the GaAs crystal? b. What will happen if Se or Te, from Group VI, are substituted for an As atom in the GaAs crystal? c. What will happen if Zn or Cd, from Group II, are substituted for a Ga atom in the GaAs crystal? d. What will happen if Si, from Group IV, is substituted for an As atom in the GaAs crystal? e. What will happen if Si, from Group IV, is substituted for a Ga atom in the GaAs crystal? What do you think amphoteric dopant means? f. Based on the above discussion ,what do you think the crystal structures of the III-V compound semiconductors AlAs, GaP, InAs, InP, and InSb will be? As GaGa As As GaAsGa As Ga Ga As As GaAsGa AsGa Ga atom (Valency III) As atom (Valency V) Figure 5Q5-1 The GaAs crystal structure in two dimensions. Average number of valence electrons per atom is four. Each Ga atom covalently bonds with four neighboring As atoms and vice versa. Solution As GaGa As As GaAsGa As Ga Ga As As GaAsGa As Ga Ga atom (Valency III) As atom (Valency V) ψ hyb orbitals As ion core (+5e) Valence electron ψ hyb orbitals Ga ion core (+3e) Valence electron Explanation of bonding in GaAs: The one s and three p orbitals hybridize to form 4 ψ hyb orbitals. In As there are 5 valence electrons. One ψ hyb has two paired electrons and 3 ψ hyb have 1 electron each as shown. In Ga there are 3 electrons so one ψ hyb is empty. This empty ψ hyb of Ga can overlap the full ψ hyb of As. The overlapped orbital, the bonding orbital, then has two paired electrons. This is a bond between Ga and As even though the electrons come from As (this type of bonding is called dative bonding). It is a bond because the electrons in the overlapped orbital are shared by both As and Ga. The other 3 ψ hyb of As can overlap 3 ψ hyb of neighboring Ga to form "normal bonds". 5.9 EE145 Spring 2002 Homework 5 Solution Prof. Ali Shakouri Repeating this in three dimensions generates the GaAs crystal where each atom bonds to four neighboring atoms as shown. Because all the bonding orbitals are full, the valence band formed from these orbitals is also full. The crystal structure is reminiscent of that of Si. GaAs is a semiconductor. a The average number of valence electrons is 4 electrons per atom. b Se or Te replacing As will have one additional electron that cannot be involved in any of the four bonds. Hence Se and Te will act as a donor. c Zn or Cd replacing Ga will have one less electron than the substituted Ga atom. This creates a hole in a bond. Zn and Cd will act as acceptors. d The Si atom has 1 less electron than the As atom and when it substitutes for an As atom in GaAs there is a "hole" in one of the four bonds. This creates a hole, or the Si atom acts as an acceptor. e The Si atom has 1 more electron than the Ga atom and when it substitutes for a Ga atom in GaAs there is an additional electron that cannot enter any of the four bonds and is therefore donated into the CB (given sufficiently large temperature). Si substituting for Ga therefore acts as a donor. f All these compounds (AlAs, GaP, InAs, InP, InSb) are compounds of III elements and V elements so they will follow the example of GaAs. 5.10 . N c exp[−(E c − E Fn )/kT] (2) Eqn. (2) divided by Eqn. (1) gives, N d = exp E Fn − E Fi (3) n i kT ∴ ln N d n i = E Fn − E Fi kT 5.4 EE145 Spring 2002 Homework 5. whereas the lower left inset is the simple theory for impurity scattering limited mobility. (3) At T i = 127 °C = 400 K, µ e ≈ 450 cm 2 V -1 s -1 (from Figure 5Q3-2). The semiconductor. ≈ (1.602 × 10 -19 C)(10 16 cm -3 )(700 cm 2 V -1 s -1 ) = 1.12 Ω -1 cm -1 (2) Using Eqn. (3) with n = N d − N a we have N d − N a n i = exp ′ E Fn − E Fi kT so that