EE 529 Semiconductor Optoelectronics – Semiconductor Basics
EE529 Semiconductor Optoelectronics
Semiconductor Basics 1. Semiconductor materials 2. Electron and hole distribution 3. Electron-hole generation and recombination 4. p-n junction
Reading: Liu, Chapter 12, Sec. 13.1, 13.5 Reference: Bhattacharya, Sec. 2.1-2.2, 2.5-2.6, 4.2
Lih Y. Lin EE 529 Semiconductor Optoelectronics – Semiconductor Basics Energy Bands in Semiconductors
Origin: Periodic lattice structure in the crystal. E-k diagram details the band structure. k: Electron wave vector k = 2π/λ Indirect bandgap Si
Direct bandgap Near the band edge: 22k = + EEc * 2me 22k = − EEv * 2mh * meh, : Effective masses of electrons/holes
Lih Y. Lin 2 EE 529 Semiconductor Optoelectronics – Semiconductor Basics Semiconductor Materials
Elementary: e.g., Si and Ge
Binary
Ternary
Quaternary
Lih Y. Lin
EE 529 Semiconductor Optoelectronics – Semiconductor Basics III-V Compound Semiconductors
Solid curve: Direct bandgap Dashed curve: Indirect bandgap
AlxGa1-xAs closely lattice matched to GaAs
In1-xGaxAsyP1-y lattice matched to InP for 0 y 1 ≤ ≤ and x = 0.47y
Matching lattice constant is important when depositing one semiconductor on another. Lih Y. Lin 4 EE 529 Semiconductor Optoelectronics – Semiconductor Basics Electron-Hole Generation How do electrons get to the conduction band (and leave holes in the valence band)?
Free electrons and holes can be generated by: • Thermal excitation (thermal equilibrium) • Optical excitation (quasi-equilibrium) • Current injection (quasi-equilibrium)
It’s important to understand the carrier (electrons and holes) distribution as a function of energy. Lih Y. Lin 5 EE 529 Semiconductor Optoelectronics – Semiconductor Basics Carrier Distribution in the Energy Band
Concentration of electrons (holes) versus energy in the conduction (valence) band =
Density of states (density of allowed energy levels) Analogy: Density of available office spaces
X
Probability of occupancy (probability that each of these levels is occupied) Analogy: People’s desire of occupying these spaces
A skyscraper without elevators
Lih Y. Lin 6 EE 529 Semiconductor Optoelectronics – Semiconductor Basics Probability of Occupancy
Fermi-Dirac function 1 f (E) = E − E f exp +1 kBT f(E) = probability of occupancy by an electron 1 - f(E) = probability of occupancy by a hole (in valence band) (Electrons like to sink to the bottom, holes like to float to the top.)
Lih Y. Lin 7 EE 529 Semiconductor Optoelectronics – Semiconductor Basics Density of States ― Bulk Material k 2 ρ=()k π2 (2m* ) 3/2 ρ=()Ee ( EE − )1/2 CC2π23 (2m* ) 3/2 ρ=()Eh ( EE − )1/2 VV2π23
Lih Y. Lin 8 EE 529 Semiconductor Optoelectronics – Semiconductor Basics Density of States ― Quantum Wells
Green laser diode
Nature Photonics 3, 432 - 434 (2009) Discrete energy levels 2 2 (qπ/ d ) E = , q =1, 2, ... q 2m Density of states m* e >+ 2 , EECq E1 π ρ=C ()E d1 <+ 0, EECq E1 Lih Y. Lin 9 EE 529 Semiconductor Optoelectronics – Semiconductor Basics Quantum Wires and Quantum Dots
Quantum wires
(2 /dd )( m* / 2π ) 12 e , EE>+ E + E ρ= 1/2 Cq12 q C ()E ()EE−−−Cq E12 E q 0, otherwise
(Gudiksen et al., Nature, 2002)
Quantum dots
1 ρc (E) = 2δ(E − Ec − Eq1 − Eq2 − Eq3 ) d1d2d3
Lih Y. Lin 10 EE 529 Semiconductor Optoelectronics – Semiconductor Basics Thermal-Equilibrium Carrier Concentration
Bulk Semiconductor ∞ EEFC− n0 =ρ=∫ CC()() E f E dE N () T F1/2 (a) (b) (c) (d) E c kTB g(E) ∝ (E–E )1/2 E c E E Ev EE− =ρ−= VF E +χ [1–f(E)] p0 ∫ VV( E )[1 f ( E )] dE N ( T ) F1/2 c −∞ kTB CB For = AreaArea == ∫ nnEE (E ) d dEE = n n electrons n (E) Ec E Ec
E EF F
E p (E) v Ev E Area = p For holes VB
0 We can simplify this more … fE) g(E) nE(E) or pE(E) (a) Energy band diagram. (b) Density of states (number of states per unit energy per unit volume). (c) Fermi-Dirac probability function (probability of occupancy of a state). (d) The product of g(E) and f(E) is the energy density of electrons in the CB (number of electrons per unit energy per unit volume). The area under nE(E) vs. E is the electron concentration. Lih Y. Lin 11 EE 529 Semiconductor Optoelectronics – Semiconductor Basics Carrier Concentration and Mass Action Law
−≥ For non-degenerate semiconductors, (ECFB−≥ E ) / kT 3.6 (EFVB E ) / kT 3.6
()EECF− nN0 =C exp − kTB * 2 3/2 NC=2(2 π mkT eB / h ) : Effective density of states at the conduction band edge
()EEFV− pN0 =V exp − kTB * 2 3/2 NV=2(2 π mkT hB / h ) : Effective density of states at the valence band edge
The location of the Fermi level energy EF is the key. 3 2πkT 3/2 Eg = B ** −=2 Mass action law: np00 42 (mmeh) expni h kTB → Knowing one carrier concentration, you can determine the other (no matter intrinsic or extrinsic)
Lih Y. Lin 12 EE 529 Semiconductor Optoelectronics – Semiconductor Basics Intrinsic and Extrinsic Semiconductor
EECV+ 1 NC Intrinsic: EFi = − kTB ln 22 NV n-type semiconductor p-type semiconductor
Electron energy Electron Energy B atom sites every 106 Si atoms
Ec x Distance into crystal CB + As E h+ – c e ~0.05 eV B– E – – – d As+ As+ As+ As+ B B B B– Ea ~0.05 eV h+ Ev VB
Ev x i 6 i
Lih Y. Lin 13 EE 529 Semiconductor Optoelectronics – Semiconductor Basics Exercise: Fermi Levels in Semiconductors
a) Where is the Fermi level of intrinsic bulk Si at room temperature? b) What kind of dopants can make it n-type? 16 -3 c) If the donor concentration Nd is 10 cm , where will the Fermi level be? 17 -3 d) If the wafer is compensation-doped with boron (Na = 2 x 10 cm ), where will the Fermi level be?
Lih Y. Lin 14 EE 529 Semiconductor Optoelectronics – Semiconductor Basics Quasi Equilibrium ― What happens to the Fermi levels during photon absorption 1 fEC ()= Probability of occupancy for electrons: EE− expFc +1 kTB 1 fE()= Probability of occupancy for holes: V EE− expFv +1 kTB
How to calculate the quasi Fermi levels? Lih Y. Lin 15 EE 529 Semiconductor Optoelectronics – Semiconductor Basics Quasi-Fermi Levels
The figure below shows positions of quasi-Fermi levels as a function of photo- generated electron-hole pair density. The semiconductor is n-type GaAs with 15 -3 ND = 10 cm .
Q: Why does εfp decrease gradually with increasing density while εfn shows a sudden increase?
Lih Y. Lin 16 EE 529 Semiconductor Optoelectronics – Semiconductor BasicsExercise: Carrier Lifetime and Internal Quantum Efficiency in GaAs 18− 3 A n-type (n0 =5 × 10 m ) GaAs is under optical excitation generating excess carrier
concentration Nnn=−=−00 p p. It has the followsing recombination coefficients: 51− −−17 3 1 −−42 6 1 A =5 × 10 s , B =8 × 10 m s , and CC=+=×eh C 5 10 m s . Assume that
CCCeh= = /2. (1) Find the range of N where each of the three different recombination processes (Shockley-Read, bimolecular, Auger) dominates. (2) Plot the spontaneous 18 26− 3 carrier lifetime τs as a function of N for 10≤≤N 10 m . (3) Assume only the bimolecular recombination process is radiative, plot the internal quantum efficiency vs N.
Lih Y. Lin 17 EE 529 Semiconductor Optoelectronics – Semiconductor Basics Bimolecular Recombination and Steady-State Concentration
Rate of recombination R= Bnp ()1 3 cm⋅ sec In thermal equilibrium, generation = recombination
G= Bn p ()1 3 0 00 cm⋅ sec With electron-hole injection (by external current or photon) Net radiative recombination rate N R= Bnp −= G0 τrad 1 τ=rad BN()++ n00 p
In steady sate: dN N =−=G 0 dt τrad
→ Determines Ν if G is known, and therefore the quasi-Fermi levels.
Lih Y. Lin 18 EE 529 Semiconductor Optoelectronics – Semiconductor Basics Exercise: Carrier Injection at Steady State
Electron-hole pairs are injected into n-type GaAs at a rate G =1023 /cm 3 s . At room 6 -3 temperature GaAs has the following parameters: Eg = 1.42 eV, ni = 2.33× 10 cm , 17 -3 18 -3 NC =4.35 × 10 cm , and NV =9.41 × 10 cm . The thermal equilibrium concentration 16 -3 of electrons is n0 = 10 cm . Assume bimolecular recombination process dominates and the coefficient for bimolecular recombination B =8 × 10−−11 cm 3 s 1 . Determine:
(a) The thermal equilibrium concentration of holes p0. (b) The steady-state excess carrier concentration Ν.
(c) The recombination lifetime τrad . (d) The separation between the quasi-Fermi levels EFc – EFp.
Lih Y. Lin 19 EE 529 Semiconductor Optoelectronics – Semiconductor Basics The p-n Junction
p n Build-in field B- As+ + eN W eNap W h (a) dn E0 =−=− e– εε
M M Metallurgical Junction Built-in potential E (x) Neutral p-region Eo Neutral n-region –Wp 0 Wn 1 x V=−+ EW() W (e) 002 np (b) –E o kT NN = B ad V(x) ln 2 eni M V Space charge region o Wp Wn (f) log(n), log(p) Depletion widths p po x n no ε PE(x) 21Na n (c) WVn = 0 i eV o eNdd N+ N a
pno Hole PE(x) npo 21ε N x = d ρ x = 0 WVp 0 net x (g) eN N+ N M ad a
Electron PE(x) WWW0 =np + eNd –Wp –eV x (d) o 2ε NN+ W ad n = V0 e NNad -eNa
Lih Y. Lin 20
EE 529 Semiconductor Optoelectronics – Semiconductor Basics Biased p-n Junction
Lih Y. Lin 21 EE 529 Semiconductor Optoelectronics – Semiconductor Basics p-n Junction Under Forward Bias
Law of the junction Excess minority carrier concentration Diffusion length eV x ' pp(0)= exp nn0 ∆=∆−pxnn( ') p (0)expLDhe,= he ,, τ he kTB Lh Dhe, : Diffusion coefficient eV x '' nnpp(0)= 0 exp ∆=∆nx( '') n (0)exp τ : Minority carrier lifetime kT pn he, B Le
Lih Y. Lin 22 EE 529 Semiconductor Optoelectronics – Semiconductor Basics Current in Forward-biased p-n Junction J
p-region SCL n-region
J = Jelec + Jhole Total current Majority carrier diffusion The total current and drift current anywhere in the device is Jhole constant. Just outside the depletion region it is due Jelec Minority carrier diffusion current to the diffusion of minority carriers. x W –Wp n
Diffusion current Recombination current 2 eDhi n eV eninWp W eV J D, hole = exp− 1 J = + exp LN kT recom hd B 22ττehkT B 2 eDei n eV J D, elec = exp− 1 LNea kTB A more accurate result:
JJdiff= D,, hole + J D elec eV eDhe eD 2 eV JJ= exp− 1 =+−ni exp 1 recom r0 LNhd LN ea kTB 2kTB eV JJdiff= so exp− 1 kTB Lih Y. Lin 23 EE 529 Semiconductor Optoelectronics – Semiconductor Basics I-V Characteristics of a p-n Junction
I I = Io[exp(eV/ηkBT) − 1]
mA Reverse I-V characteristics of a pn junction (the positive and V negative current axes have Shockley equation different scales)
nA Space charge layer generation.
eV II= 0 exp− 1 ηkTB η: Diode ideality factor η = 1: Diffusion controlled η = 2: SCL recombination controlled
Lih Y. Lin 24 EE 529 Semiconductor Optoelectronics – Semiconductor Basics p-n Junction under Reverse Bias
Current due to thermally Total reverse current generated EHP: J≅+ JJ eWn rev s gen J = i gen eDhe eD2 eWn i τg =++ni LNhd LN ea τg τg : Mean thermal generation time
Lih Y. Lin 25 EE 529 Semiconductor Optoelectronics – Semiconductor Basics Exercise: GaAs p-n Junction
• Device parameters: – Cross sectional area A = 1 mm2. 23 -3 – Na (p-side doping) = Nd (n-side doping) = 10 m . – Coefficient of recombination Β = 7.21 x 10 -16 m3s-1 12 -3 – ni = 1.8 x 10 m – εr = 13.2 2 -1 -1 – µh (in the n-side) = 250 cm V s , µe (in the p-side) = 5000 cm2V-1s-1 – Carrier recombination time in the depletion region = 10 ns • What is the I-V characteristics? Calculate the diffusion current and the recombination current under 1V bias.
Lih Y. Lin 26