Recombination: Depletion. Auger, and Tunnelling
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1 Recombination: Depletion Region, Bulk, Radiative, Auger, and Tunnelling Ch 140 Lecture Notes #13 Prepared by David Gleason
2 We assume: Review of Depletion Region Recombination Flat Quantum Fermi Levels Requires that the fastest recombination rate is slow with respect to diffusion k n k p = σν There are an even distribution of traps where σ does not depend on x w 0 0 U(x)dx U(x)dx This leads to: U total = πktw q(v bi + V app ) U max pp
3 Review of Depletion Region Recombination (cont.) We also have U max = 1 N V 2 Tσνn i exp( qv app /2kT) and J D/R =qu total So J DR = πktw 4(V + V ) N Tσνn i exp( qv app /2kT) bi app
4 Quasi-Neutral Region The Quasi-neutral region is defined as a region of the semiconductor with an uneven distribution of carriers in a region of flat bands As pictured, the holes will diffuse away from w into the bulk where they will recombine w w h + h + h + h + h + h + Quasi Quasi-Neutral Region E f,n E f,p
5 Bulk Recombination At steady state using Fick s 1st Law δp(x) δ ( ) = R(Δx) D(Δx) = 0 δt δc Flux = D o (x,t) o δ(x) and Fick s 2nd Law δc o δt δ 2 C = D o 2 C o (x,t) = D o (x,t) o δx 2 We have δp(x) δt = 0 = p(x) p o (x) τ p D p δ 2 p(x) δx 2
6 Bulk Recombination (cont.) To solve this we must first establish some boundary conditions 1. p( w ) = p o exp( qv q app /kt) 2. n( w ) p(w) = n 2 i exp( qv app /kt) 3. p( ) = p o Solving for p(x) yields p(x) = p o + p [ o exp( qv app /kt) 1]exp x w L p Where L p = D p τ p is the diffusion length
7 The bulk recombination current can be determined by J BR =qflux where the flux here is for all carriers at any point in the flat band region This is solved easiest at w since there there is no electron movement to consider. At other values of x J BR =-q flux holes +q flux electrons D δp(x) ) At x=w this simplifies to J BR = qflux holes = qd p δx Solving this and evaluating at x=w recognizing g that the last term simplifies ( exp x w ) L 1 p We have = qd p po exp qv app J BR ep L p kt 1
8 From J BR = qd p p o L p qv app exp kt 11 n o p p = n 2 i p o = n 2 i n 2 i We can substitute o p p i p o n o N D To get J BR = qd 2 pn i L p N D exp qv app kt 1 This is the bulk region recombination
9 Radiative Recombination Assume a perfect semiconductor crystal No surface state recombination No depletion region recombination (σ is very small) No bulk recombination (L p is very big) Generate carriers through light absorption or thermal excitation Carriers diffuse until finally they recombine in the inverse of the absorption reaction e - e - k r k r hν h + h + Light is emitted with hν = E g
10 Radiative Recombination (cont.) This process has been ignored until now because for indirect band gap semiconductors the carrier lifetime due to radiative recombination is really long. 99.9% of bulk recombination in Si and Ge will occur across trap states For direct gap semiconductors, including GaAs and porous Si, radiative recombination is more competitive Leads to LEDs, lasers, ect.
11 Radiative Recombination Current Rate of electron recombination given by δn = npk r k r k r k δt r At equilibrium 0 = n o p o k r k r kr = 2 ni k r This expression can be plugged into the rate equation away from equilibrium to give δn 2 = k r np n i δt ( ) ( ) And finally J 2 r = qk r np n i
12 Determination of k r from the absorption spectra Indirect semiconductors can not be made pure enough to emit, so k r must be calculated from the absorption spectra At equilibrium in a perfect sample, the rate of thermal absorption must equal the rate of radiative recombination because they are inverse processes The thermal absorption is given by the overlap between the blackbody curve at temperature T and the absorption spectra at temperature T
13 Determination of k r from the absorption spectra Si absorption spectra GaAs absorption spectra 300K Blackbody Photons absorbed by Si Photons absorbed by GaAs E g GaAs E g Si Note that even though Si has a lower E g than GaAs, less light is absorbed due to the shape of the absorption spectra caused by the indirect band gap of Si. At equilibrium, the amount absorbed is equal to that emitted through radiative recombination, so we can calculate k r, which is sometimes called B, and has units of cm 4 s -1.
14 Auger Recombination Pronounced or Occurs at very high injection or doping conditions This is a 3 body process whereby two majority carriers collide One looses energy E g and combines with a minority carrier The other gains energy E g, which it subsequently looses through thermalization
15 Auger Recombination (cont.) For n-type τ A = 1 G n n 2 G n = recombination rate 1 Auger lifetime e - e - +E g For p-type Auger lifetime τ A = G p p 2 -E g h + heat G p = recombination rate p G p 2x10-31 cm 6 /s for Si at room temperature The dependence is on n or p, and therefore depends on the doping or excitation level h + h + e - -E g +E g heat
16 Tunneling Current Tunneling is only important at high high dopant densities and low temperatures The tunneling probability is given by T tun = exp 8πw 2qm * V + V e( ) 1 2 3h [ ] bi w 1 where N D The tunneling probability is temperature independent, and since most other currents (thermionic emission) are highly temperature dependent it is only seen at low temps Ratio of tunneling current to thermionic current for Si-Au barrier taken from Sze p. 264
17 Summary of Recombination Bulk J BR = qd n 2 p i exp qv app 1 L p N D kt Depletion Region Thermionic Radiative Auger τ A = 1 G n n 2 J DR = πktw 4(V bi + V app ) N Tσνn i exp( qv app /2kT) J th = A * T 2 exp qφ b exp qv app 1 kt kt J r = qk ( 2 r np n ) i Tunneling J T exp( αφ b / N ) D pp
18 Summary of Recombination (cont.) Bulk A=1, depends N D J o proportional to exp(-e g /kt) Depletion Region A=2 Thermionic A=1, does not depend on N D J o proportional to exp(-qφ b /kt) Radiative Insignificant for indirect gap semiconductors, Strictly depends on excess carriers Auger Only at really high carrier concentrations Tunneling Only significant at low T and high N D or N A Constant with temperature
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