Chapter 5. Second Edition ( 2001 McGraw-Hill) 5.1 Bandgap and photodetection. Solution

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1 Chapter Bandgap and photodetetion Seond Edition ( 001 MGraw-Hill) a. Determine the maximum value of the energy gap that a semiondutor, used as a photoondutor, an have if it is to be sensitive to yellow light (600 nm). b. A photodetetor whose area is m is irradiated with yellow light whose intensity is mw m -. suming that eah photon generates one eletron-hole pair, alulate the number of pairs generated per seond.. From the known energy gap of the semiondutor (E g = 1.4 ev), alulate the primary wavelength of photons emitted from this rystal as a result of eletron-hole reombination. d. Is the above wavelength visible? e. Will a silion photodetetor be sensitive to the radiation from a laser? Why? Solution a b We are given the wavelength λ = 600 nm, therefore we need E ph = hυ = E g so that, E g = h/λ = ( J s)( m s -1 ) / ( m) E g = J or.07 ev Area A = m and light intensity I light = 10 - W/m. The reeived power is: P = AI light = ( m )( 10 - W/m ) = W N ph = number of photons arriving per seond = P/E ph N ph = ( W) / ( J) = Photons s -1 Sine the eah photon ontributes one eletron-hole pair (EHP), the number of EHPs is then: N EHP = EHP s -1 For, E g = 1.4 ev and the orresponding wavelength is λ = h/e g = ( J s)( m s -1 ) / (1.4 ev J/eV) λ = m or 874 nm The wavelength of emitted radiation due to eletron-hole pair (EHP) reombinatios therefore 874 nm. d e is, It is not in the visible region (it is in the infrared). From Table 5.1 (in the textbook), for Si, E g = 1.10 ev and the orresponding ut-off wavelength λ g = h/e g = ( J s)( m s -1 ) / (1.1 ev J/eV) λ g = m or 110 nm 5.1

2 Sine the 874 nm wavelength of the laser is shorter than the ut-off wavelength of 110 nm, the Si photodetetor an detet the 874 nm radiation (Put differently, the photon energy orresponding to 874 nm, 1.4 ev, is larger than the E g, 1.10 ev, of Si whih means that the Si photodetetor andeed detet the 874 nm radiation). 5. Minimum ondutivity a. Consider the ondutivity of a semiondutor, σ = en + ep. Will doping always inrease the ondutivity? b. Show that the minimum ondutivity for Si is obtained whet is p-type doped suh that the hole onentratios p m = and the orresponding minimum ondutivity (maximum resistivity) is σ min = e. Calulate p m and σ min for Si and ompare with intrinsi values. Solution a Doping does not always inrease the ondutivity. Suppose that we have antrinsi 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 derease the ondutivity beause it would reate more holes with lower mobility at the expense of eletrons with higher mobility. Obviously with further doping p inreases suffiiently to result in the ondutivity inreasing with the extent of doping. b To find the minimum ondutivity, first onsider the mass ation law: np = isolate n: n = /p Now substitute for n the equation for ondutivity: σ = en + ep σ = en i p + ep To find the value of p that gives minimum ondutivity (p m ), differentiate the above equation with respet to p and set it equal to zero: Isolate p m and simplify, dσ en = i + e dp p en i + µ p h e = 0 m 5.

3 p m = Substituting this expression bak into the equation for ondutivity will give the minimum ondutivity: σ min = en i + ep m = en i µ + en e i p m σ min = e + e = e + e σ min = e From Table 5.1, for Si: = 150 m V -1 s -1, = 450 m V -1 s -1 and = m -. Substituting into the equations for p m and σ min : p m = ( m ) 150 m V 1 s 1 = m V 1 s 1 = m - σ min = e σ min = ( C)1.45 ( m ) ( 150 m V 1 s 1 )( 450 m V 1 s 1 ) σ min = Ω -1 m -1 The orresponding maximum resistivity is: ρ max = 1 / σ min = Ω m The intrinsi value orresponding to p m is simply (= m - ). Comparing it to p m : p m = m m =1.7 The intrinsi ondutivity is: σ int = e ( + ) σ int = ( C)( m - )(150 m V -1 s m V -1 s -1 ) σ int = Ω -1 m -1 Comparing this value to the minimum ondutivity: σ int σ min = W 1 m W -1 m -1 = Suffiient p-type doping that inreases the hole onentration by 7% dereases the ondutivity by 15% to its minimum value. 5. Compensation doping in Si a. A Si wafer has been doped n-type with atoms m Calulate the ondutivity of the sample at 7 C. 5.

4 . Where is the Fermi level in this sample at 7 C with respet to the Fermi level (E Fi ) intrinsi Si?. Calulate the ondutivity of the sample at 17 C. b. The above n-type Si sample is further doped with boron atoms (p-type dopant) per entimeter ubed. 1. Calulate the ondutivity of the sample at 7 C.. Where is the Fermi level in this sample with respet to the Fermi level in the sample in (a) at 7 C? Is this an n-type or p-type Si? Solution a Given temperature T = 7 C = 00 K, onentration of donors N d = m -, and drift mobility 800 m V -1 s -1 (from Figure 5Q-1). At room temperature the eletron onentration n = N d >> p (hole onentration) Holes Eletrons Dopant Conentration, m Figure 5Q-1 The variation of the drift mobility with dopant onentration Si for eletrons and holes at 00 K. - (1) The ondutivity of the sample is: σ = en d ( C)(10 17 m - )(800 m V -1 s -1 ) = 1.8 Ω -1 m -1 () Intrinsi Si, E F = E Fi, = N exp[ (E E Fi )/kt] (1) In doped Si, n = N d, E F = E Fn, n = N d = N exp[ (E E Fn )/kt] () Eqn. () divided by Eqn. (1) gives, N d = exp E E Fn Fi kt ln N d = E Fn E Fi kt 5.4 ()

5 E F = E Fn E Fi = kt ln(n d / ) (4) Substituting we find ( = m - from Table 5.1 in the textbook), E F = ( ev/k)(00 K)ln[(10 17 m - )/ ( m - )] E F = ev above E fi µ L T 1.5 Ge N d =10 14 N N d =10 1 d = N d = N d =10 18 N d =10 19 Si µ Ι T Temperature (K) Figure 5Q- Log-log plot for drift mobility versus temperature for n-type Ge and n-type Si samples. Various donor onentrations for Si are shown, N d are in m -. The upper right insert is the simple theory for lattie limited mobility whereas the lower left inset is the simple theory for impurity sattering limited mobility. () At T i = 17 C = 400 K, 450 m V -1 s -1 (from Figure 5Q-). The semiondutor is still n- type (hek that N d >> at 400 K), then σ = en d ( C)(10 17 m - )(450 m V -1 s -1 ) = 7.1 Ω -1 m -1 b The sample is further doped with N a = m - = m - aeptors. Due to ompensation, the net effet is still an n-type semiondutor but with an eletron onentration given by, n = N d N a = m m - = m - (>> ) We note that the eletron sattering now ours from N a + N d ( m - ) number of ionized enters so that 700 m V -1 s -1 (Figure 5Q-1). (1) σ = en d ( C)(10 16 m - )(700 m V -1 s -1 ) = 1.1 Ω -1 m -1 () Using Eqn. () with n = N d N a we have N d N a = exp E Fn E Fi kt so that E Φ = E Fn E Fi = ( ev)ln[(10 16 m - ) / ( m - )] E Φ = 0.48 ev above E Fi The Fermi level from (a) and (b) has shifted down by an amount ev. Sine the energy is still above the Fermi level, this an n-type Si. 5.5

6 5.4 Temperature dependene of ondutivity An n-type Si sample has been doped with phosphorus atoms m -. The donor energy level for P in Si is ev below the ondution band edge energy. a. Calulate the room temperature ondutivity of the sample. b. Estimate the temperature above whih the sample behaves as if intrinsi.. Estimate to within 0% the lowest temperature above whih all the donors are ionized. d. Sketh shematially the dependene of the eletron onentration the ondution band on the temperature as log(n) versus 1/T, and mark the various important regions and ritial temperatures. For eah region draw an energy band diagram that learly shows from where the eletrons are exited into the ondution band. e. Sketh shematially the dependene of the ondutivity on the temperature as log(σ) versus 1/T and mark the various ritial temperatures and other relevant information. Solution 600 C 400 C 00 C 7 C 0 C m Ge m - Si m /T (1/K) Figure 5Q4-1 The temperature dependene of the intrinsi onentration. a The ondutivity at room temperature T = 00 K is ( = m V -1 s -1 an be found in Table 5.1 in the textbook): σ = en d σ = ( C)( m - )( m V -1 s -1 ) = 1.6 Ω -1 m -1 b At T = T i, the intrinsi onentration = N d = m -. From Figure 5Q4-1, the graph of (T) vs. 1/T, we have: 5.6

7 1000 / T i = 1.9 K -1 T i = 1000 / (1.9 K -1 ) = 56 K or 5 C The ionization region ends at T = T s when all donors have beeonized, i.e. when n = N d. From Example 5.7, at T = T s : n = N d = 1 N N E d exp kt s E E T s = = N k ln d N 1 N N k ln d d N E T s = k ln N N d 1 Take N = m - at 00 K from Table 5.1 (in the textbook), and the differene between the donor energy level and the ondution band energy is E = ev. Therefore our first approximation to T s is: E ( ev) J/eV T s = k ln N = N d ( J/K)ln m m Find the new N at this temperature, N : T s N = N 00 = ( m ) K 00 K Find a better approximation for T s by using this new N : = K = m - E ( ev) J/eV T s = = N k ln N d ( J/K)ln m m T N = N s = A better approximation to T s is: ( m ) K 00 K = K = m - E ( ev) J/eV T s = = N k ln N ( J/K)ln m d m T N = N s K = m 00 K = K = m - 5.7

8 E ( ev) J/eV T s = = N k ln N d ( J/K)ln m m = K We an see that the hange in T s is very small, and for all pratial purposes we an onsider the alulation as onverged. Therefore T s = 70.9 K = 0.1 C. d and e See Figures 5Q4- and 5Q4-. ln(n) Intrinsi slope = E g /k Extrinsi T s Ionization ln(n d ) slope = E/k T i (T) 1/T Figure 5Q4- The temperature dependene of the eletron onentration an n-type semiondutor. log(n) log(σ) Semiondutor INTRINSIC Metal T EXTRINSIC Lattie sattering log(µ) High Temperature µ T / µ T / Impurity sattering IONIZATION 1/T Low Temperature Figure 5Q4- Shemati illustration of the temperature dependene of eletrial ondutivity for a doped (n-type) semiondutor. 5.5 has a valeny of III and has V. When and atoms are brought together to form the rystal, as depited in Figure 5Q5-1, the valene eletrons in eah and the 5 valene eletrons in eah are all shared to form four ovalent bonds per atom. In the rystal with some 10 or so equal numbers of and atoms, we have an average of four valene eletrons per atom, whether or, so we would expet the bonding to be similar to that in the Si rystal: four bonds per atom. The rystal struture, however, is not that of diamond but rather that of zin blende (Chapter 1 of the textbook). 5.8

9 a. What is the average number of valene eletrons per atom for a pair of and atoms and in the rystal? b. What will happef Se or Te, from Group VI, are substituted for an atom in the rystal?. What will happef Zn or Cd, from Group II, are substituted for a atom in the rystal? d. What will happef Si, from Group IV, is substituted for an atom in the rystal? e. What will happef Si, from Group IV, is substituted for a atom in the rystal? What do you think amphoteri dopant means? f. Based on the above disussion,what do you think the rystal strutures of the III-V ompound semiondutors Al, P, In, InP, and InSb will be? atom (Valeny III) atom (Valeny V) Figure 5Q5-1 The rystal struture in two dimensions. Average number of valene eletrons per atom is four. Eah atom ovalently bonds with four neighboring atoms and vie versa. Solution atom (Valeny V) atom (Valeny III) Valene eletron ψ hyb orbitals ψ hyb orbitals Valene eletron ion ore (+5e) ion ore (+e) Explanation of bonding in : The one s and three p orbitals hybridize to form 4 ψ hyb orbitals. In there are 5 valene eletrons. One ψ hyb has two paired eletrons and ψ hyb have 1 eletron eah as shown. In there are eletrons so one ψ hyb is empty. This empty ψ hyb of an overlap the full ψ hyb of. The overlapped orbital, the bonding orbital, then has two paired eletrons. This is a bond between and even though the eletrons ome from (this type of bonding is alled dative bonding). It is a bond beause the eletrons in the overlapped orbital are shared by both and. The other ψ hyb of an overlap ψ hyb of neighboring to form "normal bonds". 5.9

10 Repeating this in three dimensions generates the rystal where eah atom bonds to four neighboring atoms as shown. Beause all the bonding orbitals are full, the valene band formed from these orbitals is also full. The rystal struture is reminisent of that of Si. is a semiondutor. a The average number of valene eletrons is 4 eletrons per atom. b Se or Te replaing will have one additional eletron that annot be involved in any of the four bonds. Hene Se and Te will at as a donor. Zn or Cd replaing will have one less eletron than the substituted atom. This reates a hole in a bond. Zn and Cd will at as aeptors. d The Si atom has 1 less eletron than the atom and whet substitutes for an atom in there is a "hole" in one of the four bonds. This reates a hole, or the Si atom ats as an aeptor. e The Si atom has 1 more eletron than the atom and whet substitutes for a atom in there is an additional eletron that annot enter any of the four bonds and is therefore donated into the CB (given suffiiently large temperature). Si substituting for therefore ats as a donor. f All these ompounds (Al, P, In, InP, InSb) are ompounds of III elements and V elements so they will follow the example of. 5.10