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L6 ENGR_ECE4243-6243 Handout#1 10042016(Summary L5+HW#4) F. Jain Absorption coefficient expressions for band-to-band direct and excitonic transitions, indirect transitions, band filling and band tailing. The power continuity equation gives power P at a point in semiconductor medium: -∇ P(x,y,z) = W*ħ (Rate of absorption) (1) The absorption coefficient is also defined in terms of probability of absorption P CV when an intensity I() is on for t sec duration on a sample of volume V having bandband transitions. α = ħω * 𝑃𝑐𝑣 𝑡 * 1 (2) 𝐼(𝑣)𝑉 PCV is expressed in terms of integral of one transition from state o in valence band to state m in the conduction band. ∞ Pcv= ∫0 𝑃𝑚𝑜 ∗ ϱ (ℎ𝑣)𝑑𝑣 (3) Here, ϱ is join density of states expressed as: Here, mr is the reduced mass defined below. 3 2 1 2 ϱ (ℎ𝑣) = [V4π(2mr ) (ℎ𝑣 − 𝐸𝑔 ) ]/ℎ3 (3b) 1 1 1 mr me mh (3C) The probability is obtained by solving a time dependent Schrodinger equation having photon interaction as a perturbation Hamiltonian. It is shown [Moss et al.] that 𝑠𝑖𝑛2 (ω−ω𝑚𝑜 )∗𝑡 Pmo(t)=|𝐻𝑚𝑜 |2 * (4) ħ2 (ω−ω𝑚𝑜 )2 Here, is the frequency of radiation and mo is the angular frequency corresponding to energy difference Em-Eo. 2𝑒 2 𝐼(𝑣)(|𝒑𝑚𝑜 |)2 |𝐻𝑚𝑜 |2 = (5) 3m2o 𝑛𝑟 𝜀𝑟 𝑐ω2 Here, pmo is the momentum matrix element which has three components. p x m0 * j m 0, other than k s 0 d x C , other than k kC kV |𝑝𝑚𝑜 |2 = |𝑝𝑥,𝑚𝑜 |2 + |𝑝𝑦,𝑚𝑜 |2 + |𝑝𝑧,𝑚𝑜 |2 Add Eq. 7-9 Using Eqs. 5, 4 and 3, Pcv (t ) 2e 2 I (v) ( mo ) 3m02 nr 0 c 2 2 sin 2 1 m0 t 2 d m0 2 m0 0 (3.6) 𝑝 𝑖𝑠 𝑠𝑞𝑢𝑎𝑟𝑒 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 (6) p mo 2 1 (10) sin 2 x in the integral, so it is taken out. x2 sin 2 1 m 0 t sin 2 ax a sin 2 ax 2 d m 0 =t dx ], or Using the standard integral, dx , [ 2 2 2 2 ( ax ) m0 ax 2 0 0 0 Let us take another look at the integration: 2 1 km t 2 sin km t 2dx t 2 d , let x , d 2 4 t 2 t km 2 t 2 2 sin 2 x t sin 2 x = dx =(t/2)*2*t dx 2 x 2 4 t x 2 ρ is assumed to be varying much slowly than 2e 2 I (v) ( mo ) t Pcv (t ) 3m02 n r 0 c 2 2 p mo 2 (11) E. O. Kane has shown that 2 2 2 𝑚𝑜 𝑃 |𝑝𝑚𝑜 | = 2 ħ (12) Where, 2 𝑃 =( P2 = 1 𝑚𝑒 2 ħ Eg 2𝑚𝑒 − 1 𝑚𝑛 ) 3ħ2 Eg (Eg +Δs0 ) (13) 2(3Eg +2Δs0 ) , if 𝑚𝑒 ≪ 𝑚𝑛 𝐸𝑔 > 2 3 𝛥𝑠𝑜 (14) Equation 14 and 12 gives 𝑚𝑜2 Eg 2 |𝑝𝑚𝑜 | 𝐾=𝐾𝑜 = (15) 2𝑚𝑒 Substitute equation 15 into equation 5 to obtain 𝐻𝑚𝑜 |2 : 𝑚𝑜2 Eg 2𝑒 2 𝐼(𝑣) 2 |𝐻𝑚𝑜 | = 2 ∗[ ] 3mo 𝑛𝑟 𝜀𝑟 𝑐ω2 2𝑚𝑒 Substitute equation 15b into equation 4 to obtain Pmo: 𝑚𝑜2 Eg 2𝑒 2 𝐼(𝑣) 𝑠𝑖𝑛2 (ω−ω𝑚𝑜 )∗𝑡 Pmo(t) = 2 ∗ [ ] * 3mo 𝑛𝑟 𝜀𝑟 𝑐ω2 2𝑚𝑒 ħ2 (ω−ω𝑚𝑜 )2 Substitute Pmo from Eq. 15C in Eq. 3 to obtain PCV 2𝑒 2 𝐼(𝑣)ϱ(ω𝑚𝑜 )πt 𝑚02 Eg 𝑃𝑐𝑣 (𝑡) = ∗ 3𝑚02 𝑛𝑟 𝜀0 𝑐ħ2 ω2 2𝑚𝑒 Subsisting for density of states 2 (15b) (15c) (16A) 4𝜋 𝑃𝑐𝑣 (𝑡) = 3 2𝑒 2 𝐼(𝑣)𝑉[ 3 ](2𝑚𝑟 )2 (ℎ𝑣−Eg )1/2 πt ℎ 3𝑚02 𝑛𝑟 𝜀0 𝑐ħ2 ω2 4π 𝛂= ħω2e2 [ 3 ][2mr ]3/2 ħ [ ] [hv 6m20 nr 𝜀0 𝑐ħω2 − ∗ 𝑚02 Eg (16B) 2𝑚𝑒 1/2 2 1/2 mo Eg Eg ] [ ] 2me (17) 𝛂 = A (h – Eg)1/2 3 2 4 2 1 2e 3 2mv m 2 E 2 o g , Where A= 2 2 6m0 nr 0 c 2me A= 3.38 x 10-7 (nr)-1 (me/mo)1/2 in units of m-1 eV-1/2 (17B) Excitonic transitions: For exciton transitions in direct band gap semiconductor, the absorption confident is derived by Elliott (1957). e , ex AR sinh 1 2 γ=( 𝑅 ħω−Eg 1/2 ) , (Eq. 2 page 164) Here, R is the exciton binding energy Eex. It is simplified to as photon energy approaches Eg or below. 𝛂𝐞𝐱 = AR1/2 π eπγ (eπγ −e−πγ )/2 ≈ 2πAR1/2 1 (1−e−2πγ ) ≈ 2πAR1/2 , at h=Eg. (X, page 165 Eq. 4B) The plot is shown below (Moss et al.) Excitonic peak at (h-Eg)/R =-1. That is, when h is Eex or R below Eg. 3 Case II: When photon energy is equal or less than band gap Eg. h E g , Now we can see from equation (4b): ex h 2 AR 2 2 AE ex2 (6, page 166) We see that Equation (2) expresses both direct band-to-band at energy above band gap and slightly below band gap. Case I: When photon energy is much larger than the band gap ℎ𝑣 ≫ Eg, 1 𝛂𝐞𝐱 = 2A(ħω−Eg )1/2 γπ 2πγ 1 , this simplifies to 𝛂𝐞𝐱 = A(ħω − Eg )1/2 , which is 𝛂𝐞𝐱 = α(ħω) (band-to-band) h Eg Absorption coefficient as a funciton h Absorption with excitonic contribution shown in the form of peak below Eg. (without excitonic contributions). In the case of quantum wells, the excitonic binding energy Eex or R increases. In addition the direct band-to-band absorption coefficient is higher due to increased joint density of states magnitude. 4 Calculation of exciton binding energy in bulk layers: (LED Notes, part II) The binding energy of an exciton ranges from 0.004 eV to 0.04 eV. In general the exciton binding energy Eex is: 4 m rq (1) E ex = 2 2 2 8o r h Where:, mr = reduced mass of the exciton given by 1 1 1 = + (2) m r mh me q = electron charge, εr = dielectric constant of semiconductor or carbon nanotube etc. h = Plank's constant Binding energy can also be viewed as the dissociation energy. The latter being the energy needed to make the electron and hole free, overcoming the Columbic attraction. Quantitatively, Eex = (13.6/r2)*(mr/mo); the unit is in electron Volt. This comes from the fact that 4 q the ionization potential of an hydrogen atom is 13.6 eV = mo2 2 . Semiconductors that have small 8o h dielectric constant has larger exciton binding energy. Bound and free excitons: Excitons are free to move in semiconductor layer or they are bound to certain impurity sites. For example in p-type GaP (dopted with zinc), when electrons are injected they form excitons if GaP is doped with oxygen or nitrogen. In the case of oxygen, Zn-O locations provides sites for excitons to remain localized at those sites as energy levels facilitates exciton formation. Bound exciton at Zn- O site in p-GaP When the electron and hole forming an exciton recombine, this recombination results in photon emission. The probability of photon emission is as high (if not higher) as in the case of direct transitions. The energy of photon hv emitted upon the decay of an exciton at the Zn-O site is: h = E g - 0.3 - Eex . (3) if Eex = 0.04 eV h = 2.24 - 0.3 - .04 = 1.9eV (4) or wavelength of emission is =1.24/h = 0.65 μm or 6500Å. Bound exciton at nitrogen sites in p-GaP An injected electron gets trapped at the nitrogen level, and subsequently binds a hole to form an exciton. The decay of this exciton results in a photon emission. The energy of the photon is h = E g - 0.08 - Eex (6) If Eex = 0.004eV h = 2.24 - .08 - .004 = 2.156eV (7) 1.24 = = 0.575 m = 5750 (8) 2.156 Although GaP is an indirect energy gap material, the electron-hole recombination via excitonic decay is almost like a vertical (direct) transition. This is due to momentum rule relaxation as Heisenberg uncertainty principle (x*p ~h) provides larger latitude in momentum change (p) as x is very small due to bound excitons. 5 Calculation of exciton binding energy in quantum wells: In quantum wells, the heavy and light hole bands split. In unstrained wells and compressive strained wells, the heavy holes are dominant where as in tensile strained quantum wells, light holes are dominant. Exciton binding is higher for smaller width quantum wells than the larger width wells for a given material/semiconductor layer. The electron and hole levels are computed (like HW). Ee1 Ehh1 Fig. 1(a) E = 0 Fig.2(a) Reverse biased p-n diode with MQWs. Change in electro-absorption and exciton peak shift due to E-field: The exciton binding energy as well as absorption plot is a funciton of electric field which is applied by an external radio frequency source. One way to apply is to make a p-n junction diode and reverse biasing it. This way the p and n- layers sandwich the multiple quantum well layers. Electron wave function E’e 1g E’hh 1 Hole wave function Fig. 1(b) Quantum well in the presence of E Electric field moves the electron and holes in Fig. 2(b) Absorption/responsivity as a different positions. This reduces absorption as funciton of applied voltage or electric field. well as excitonic binding energy. The electric field also changes the electron E’e1and hole E’hh1energy levels. The change in electron energy levels and change in exciton binding energy E’ex results in the photon energy value at which the exciton absorption peaks. See Fig. 2(b). Note that the band-to-band absorption due to free carriers is not shown in Fig. 2(b). It is to the right of plots shown here. It varies as 𝛂 = A (h – Eg)1/2. 6 Change in electro-refraction in multiple quantum wells due to E-field: Index of refraction nc is complex when there are losses as photons travel in the medium. nc= nr – i k Where extinction coefficient is which is k = α/4, here alpha is the absorption coefficient. The square root of permittivity is nc. nc= nr – i k = (ri)1/2 The imaginary part of the permittivity is given by: The imaginary part i = 2nr , alpha is absorption coefficient, lambda is the wavelength and nr the real part of index of refraction. i=2nre2/(m2o2)]*(Density of states)*(transition matrix element)*(polarization factor)*(Line shape funciton). Ref. W. Huang, UCONN Ph.D. thesis 1995] Gaussian Line shape function: L(E)=[1/()1/2]*exp[-(ho-h)2/2], =h/(ln2)-1/2,=1.54*10-13 s. The density of states changes with quantum well, quantum wire and dots. Transition matrix element involves wave functions in one, two or three dimensional confinement. The real part r is related to imaginary part i as The imaginary and real and index of refraction are: Use i and relation above to recognize various terms in the following gain expression using excitons: Note gain coefficient g = - (fc + fv -1); here fc and fv are the Fermi distributions for electrons and holes. 2 e2 [| M b|2 2 1/2 | ex (0) |2 g ex ( ) = c L L o n r mo y z l,h (| e ( y )h ( y )dy |2 | e ( z )h ( z )dz |2 ) ex L( E ex ) ( f c + f v - 1)] Exciton Binding Energy in Quantum Wires (OPTIONAL) (Ref. W. Huang, UCONN Ph.D. thesis, 1995) 7 8 Indirect band-to-band transitions: summary (optional) (A) Steps: (h ) PCV (t ) I (h )Vt Definition Initial state ‘0’, intermediate state ‘i’ or ‘i’’, and final state ‘m’. Indirect transition can take place in two ways: 0 i (a) Phonon assisted Direct i m Process (a) is dominant if E 0 E m The probability that an electron makes a transition from the initial ‘0’ to the final ‘m’ state is Pm0 , which is: (B) Pm 0 4 H i0 2 i 0 4 2 H 2 mi sin 2 m 0 p t p 2 2 m0 (2) Eq.2 is obtained using second order perturbation theory. Note that Pi 0 H i0 2 2 sin 2 m 0 i 0 t m0 i 0 2 2(b) 2 2 (Direct transition) Hi 0 2 2 e2 I ( ) Pi 0 3m02 nr 0c 2 2 Hmi is the matrix element which represents electron-phonon interaction. ‘+’ sign represents phonon emission and the ‘-‘ is for phonon absorption. Qualitatively, [here, '1' accounts for phonon emission] 2 and Hmi is proportional to N p V Hmi 2 is proportional to N p 1 V (3) N p is the number of phonons in a particular mode in the crystal of volume V. By definition, 4) 5) N p 1 Np 1 e h q kT 1 1 e 1 hq kT e 1 h q kT e (or the occupancy of the qth mode) 11 h q kT 1 1 1 e E p kT The energy balance equations are: h E m E 0 E p (6) h E m E 0 E f (phonon absorption) (C) PCV is obtained by summing Pm0 over all pairs of initial states (in the valence band) and all final states (in the conduction band). (phonon emission) 9 PCV PmV Pm 0 7) and PmV Pm 0 v Ev d m 0 0 E p E g P 8) PCV (t ) mv c Ec dEc 0 Where E p for phonon absorption and (- E p ) for phonon emission Eq.2, 7 & 8 give: 2 H i 0 H mi 2 PCV (t ) 2 2 E0 h 2 E p Eg E E t c c v c E p dEc 0 Density of states (in volume V) 4 2mc c Ec V h3 12 32 V 4 2mv E g Ev v Ev h3 32 Ec m m is the CB (10) minima Conservation of energy – just like joint density of states c ( E c ) and v ( E c E p ) in the integrand give 12 12 Ec Ec dEc E c2 2 1 2 E Eg 2 p Thus, Eq.9 integral is mc me m m n v 8V 2 m 3 mc mv h6 32 E Eg 2 p 2 Pcv (t ) 2 H i 0 H mi 8V 2 m 3 mc mv 3 2 E p E g 2 and 2 2 6 t h E 0 2 Or for phonon emission 2 H i 0 Pcv (t ) B 1 2 C 2 E t E0 V 1 e p 2 mc mv 3 2 E p E g 2 kT 8V 2 m 3 h6 Combining Eq.1 with Eq.11, we get 12) 10 (9) h C E p E g 2 C E p E g 2 E kT E kT e p 1 1 e p [Phonon Absorption ] [Phonon Emission] Jα C has contribution from the 0 i m route. 13) 2 H i 0 BC 8V 2 m 3 mc mv 3 2 C 2 I ( )V E 0 2 V h6 2 Eg Eg+EP C 2 Substiutin g for H i 0 using Eq.2(b) 2e 2 I ( ) Pi 0 2 2 2 2 3m0 nr 0 c BC 8V 2 M 3 mc mv 3 2 C 2 3 I ( )V V h E0 32 Pi 0 BC M mc mv e 2 or C 2 7 2 6 m0 nr 0 c E 0 2 This is: 2 14) 32 Pmi BV M mc mv e 2 C 6 2 7 m02 nr 0 c E m 2 2 Exciton transitions in indirect gap semiconductors Jα Eg Eg+EP C Band-to band transitions in heavily doped semiconductors: band filling and band tailing: Band filling takes place in low energy gap semiconductors such as InSb where effective mass for electrons and holes is very small as compared with Si and GaAs. 11 Fabry-Perot Cavity (Modulators and Lasers): Cr-Au Contact W = 3 -> 5 um p+ GaAs cap Excited Region Top Contact Stripe P-Al0.3Ga0.7As SiO2 50µm 1 um - 2 um p-Ga0.93As0.07As n - Al0.2Ga0.8As (Cladding)=ND>9.7x1018cm-3 d=0.2um p-Ga0.83As0.17As~0.1µm p-GaAs 1 um - 2 um p - Al0.12Ga0.88As (Active)=1016cm-3 Output n - Al0.2Ga0.8As (Cladding)=ND>9.7x1018cm-3 n-GaAs substrate Au-Ge-Ni contact Laser output p-GaAs Active Layer~0.1 to 0.5µm n-Ga0.7Al0.3As-0.2µm + n - GaAs Substrate (100 150 um) ND>1019cm-3 y Ohmic Contact Fig .48 Distributed feedback (DFB) edgeemitting laser (first order grating). Fig .28. An edge-emitting cavity laser with ridge. La P01 Lg Lp P02 P02 r1 Lg1 r2 Leff rg La+Lp LDBR La LDBR r1 Fig. 55 Cavity formed by one conventional mirror and one distributed Bragg reflector (DBR). Fig. 56 (right panel): In a vertical cavity surface emitting laser (VCSEL) the cavity thickness is multiple of half wavelengths and DBRs are quarter wavelengths with high and low index. Lg1 Fig. 56 VCSEL with DBR reflectors. P01 Condition of oscillations: -jL t1 t 2 e Eo = - 2jL ) E i (1 - r1 r 2 E 2 1 - r1 r 2 e-2j n r + j Gain condition 1 - r1 r 2 e(g- )L = 0 1 1 g = + ln L r1 r 2 Phase condition - j2m e =e -4 jnrL ; L= (g- ) L 2 m 2 nr Fig. 1a. Electric field strength after many passes in a cavity of length L. 12 =0 ECE 4243-6243 10042016 L6: F. Jain HW6B Waveguide Equations for light waves: One dimensional confinement along x-axis. Three slab waveguide structure. Waveguide thickness is d. Waveguide equation is like Schrodinger equation for electrons. 2Ey 5. + 2Ey = 0 2Ey x2 z2 t2 Wave propagating in the z-direction E y (x,z,t) = A ¢ coskx + A ¢ sin kx e j wt βz 18. e o EVEN TE MODES Solution to the Eigenvalue equation: find k and by solving eq. 38 and 39. β n k tank d / 2 = = k n k β 2 2 12 1 0 2 2 2 12 2 0 38. 2 Substitution(15),(23) k 2 = n 22 k 02 β 2 as and (equation 15) =β and (equation 23) Add equations 15 and 23: 2 k 2 + 2 = n22 n12 k 02 , multiplying by d / 2 2 k 2 + 2 2 d 2 = 2 n 22 n12 k0 d 2 2 2 kd d 2 2 k d + = n 2 n1 0 Rewriting Equation 27 as: 2 2 2 2 39. n12 k 02 2 13 14