Assignment 6. Solution: Assumptions - Momentum is conserved, light holes are ignored. Diagram: a) Using Eq a Verdeyen,

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1 Assignment 6 Solution: Assumptions - Momentum is conserved, light holes are ignored. Diagram: a) Using Eq..4.5a Verdeyen, ΔE c = E 2 E c = Using Eq..4.5b Verdeyen, ΔE v = E v E = b) Using Eq.2.9 Verdeyen, m h 0.55 m o (hν E m e + m g ) = (0.05) = ev h 0.55 m m 0 m e 0.067m 0 (hν E m e + m g ) = (0.05) = ev h 0.55 m m 0 N = 3 3π 2 [2mΔE c 2 ħ 2 ] = m 3 = cm 3 This is the minimum number of electron-hole pairs required to achieve optical gain.

2 Solution: (a) Using (b) E = hν = hc λ From the diagram given in the problem, (c) We know that E = hδv Δν FWHM = h E = 0 g(ν) dv =, hc λ = E = hc.476ev = 0.84μm h 2 2 mev = Hz = 69 cm 8

3 From part (a), c = λν ν 0 = Hz Condition for threshold: Δν Base = Hz γ 0 = σng(ν 0 ) g(ν 0 ) = 2 Δν Base = s R R 2 exp[(γ 0 α)2l g ] = γ 0 = α + 2l g ln R R 2 = μm μm ln (0.3)(0.3) = 27.7 cm γ 0 = λ 0 2 2πn 2 (n e Inverted carrier density, τ ) g(ν 0) = (0.84μm)2 2π(3.6) 2 ( n e ns ) = 27.7 cm (d) n e = 27.7 cm cm 2 = cm 3 P = [(βn e 2 + n τ ) (hν 0)] Volume = [(β(0 6 ) ns ) (h(0.84 μm))] ( )μm3 = 369 mw Solution: 9

4 Assumption: the given diagram is drawn to scale. Total energy gap, E = E g + (F n E c ) + (E v F p ) =.43 ev + (0.052)eV Using the plot given above, we can read the loss coefficient off it which corresponds to an energy of ev from E g. Photon energy, α 235 cm E = hν = =.482 ev Solution: (a) 0

5 For the given cavity, (b) ( 2nd τ p = c ) μm exp( 2αdR R 2 ) = ( c ) exp( 2 2cm 400μm ( )) =.35 0 s FSR for the given cavity, (c) Condition for threshold: (d), (e) FSR = c 2nd = GHz Δλ λ = Δν Δν Δλ = ν ν λ = GHz 850nm = 2.5 A 0 R R 2 exp[(γ th α s ) 2d] = γ th = α s + 2d ln ( ) = 2 + R R 2 2 (400μm) ln ( ) = 6.54 cm Using Eq..4.5c Verdeyen, For gain, we need, E g < [hν = E 2 E ] < F n F p For quantum well lasers, the density of states in the energy interval de is (Eq..6.6) When T=0, (f) ρ(e)de = 2π 2 (2m ħ 2 ) (2x e ) de F n F n n = ρ(e)de = 2π 2 (2(0.067m 0) ħ 2 ) ( 2(.3) 00A 0 ) de E E = 2π 2 (2(0.067m 0) ħ 2 ) ( 2(.3) 00A 0 ) [F n E ] = 0 8 cm cm 3 F n E = 2π 2 (2(0.067m 0) ħ 2 ) ( 2(.3) = ev = 50meV 00A 0 ) l z Recombination dominates when we have n = p = 2 0 8

6 Given, λ = 54.5 nm and β = cm 3 /s P = [(βn e 2 + n τ ) (hν 0)] p = P = [(βn 2 hν e + n 0 τ )] = W cm 2 Solution: (a) When there is no pumping, as it is given that the populations in (,0) are related by the Boltzmann factor, we have, 2

7 N = ( exp ΔE v kt + exp ΔE v kt ) 0 20 cm 3 = cm 3 (b) Given that absorption coeff. in the absence of pumping is 20 cm. (c) 20 = N σ abs σ abs = 20 cm cm 3 = cm 2 We are given that thermal processes keep populations in (3,2) related by Boltzmann factor, therefore, (d) N 3 = exp ΔE N c /kt = (exp 0.32 ) = kt Using principle of conservation of Atoms, N = N 0 + N + N 2 + N 3 But, we know that N 3 N & N 2 and can be ignored. And we also know that N + N 2 < N 0 & N. Thus, N N 2 = cm 3 when we have optical transparency. (e) P = V ( n 2 τ 2 ) (hν 3,0 ) P V = (2 08 ) 0 9 (hν 3,0 ) = W/cm 2 3

8 Problem 6.2 (a) Spontaneous emission rate can be written as following. R sp = A ρ jnt (v)f C (E 2 )[ f V (E )] R sp = A 2π 2 (2m r ħ 2 )3/2 hv E g + e E 2 F n kt E F p e kt E F p + e kt We want to express everything in terms of photon energy. From (.4.5ab), we have m h E 2 E C = m e + m (hv E g) = m r h m (hv E g) e m e E V E = m e + m (hv E g) = m r h m (hv E g) h m r = m e m h m e + m h Let s assume the energy reference point is the valence band maximum point, meaning E V = 0, E C = E g. E 2 = m r m e (hv E g) + E g E = m r m h (hv E g) Substitute these two into the first equation, then we can plot it. Spontaneous emission rate hv ev Blue curve corresponds to T=77K. Orange corresponds to T=300K. First, high temperature has higher peak intensity in the plot. This can be understood easily. Because the spontaneous emission results from the carrier recombination. For higher temperature, it means that the carrier distribution tail will go to higher energy in conduction, while hole will go lower into valence band. So we have more carriers to recombine and higher intensity. Higher temperature s peak intensity is at higher energy. This is similar behavior. Because higher temperature moves electron distribution upward in band Solution by Kevin Lee Rsp hv, 77,.4, 0 Rsp hv, 300,.4, 0

9 diagram, while hole distribution moves downward. Therefore, the maximum emission has higher energy. (b) To plot the gain spectrum, we can use the formula in the class. λ2 γ 0 (hv) = A 8πn 2 [h ρ jnt(hv)][f c (E 2 ) f v (E )] The same strategy can be used here to replace E and E 2. Gain coefficent m ρ jnt (hv) = 2π 2 (2m r ħ 2 )3/2 hv E g hv ev 0 hv, 300, 0.7, hv, 300,.4, 0 0 hv, 300,.5, 0.05 ) First case is the blue curve. This case means that the both quasi-electron and hole Fermi levels are in the middle of the band gap. Therefore, there is no population inversion, we have negative gain, which means absorption. 2) Second case is the orange curve. Here the quasi-fermi levels of electron and hole are at the band edge. This is the threshold point that the system is going to have gain. 3) Third case is green curve. Now quasi-fermi level of electron is above conduction band minimum, which means excess electrons in the conduction band. Quasi-Fermi of hole is also below the valence band maximum, meaning excess hole exists. Therefore, we have population inversion created in the semiconductor. There is gain in the material. (c) Now I am going to plot the quantum well laser gain spectrum. The only difference from the previous problem is that the density of states is different for quantum well

10 system. Let s first observe what the density of states. Its analytical form is the following. ρ 2D jnt = m r πħ 2 Θ[hv (E L g + E n + E p )] z E n = ħ2 2m e (πn z L z ) 2, n z =,2,3 E p = ħ2 2m h (πn z L z ) 2, n z =,2,3 f c (hv) = f v (hv) = + e m r m(hv E g ) E n F n e kt m r m(hv E g +F p ) E p e + e kt

11 There are two cases I can plot this problem for the inversion. Let s say the electron and hole quasi-fermi levels are both sitting where they were, not shifted. So you can see that the peak gain is actually decreased. This is expected. Because the quantum confinement shifts the conduction and valence band edges. If the quasi- Fermi levels stay where they were, effectively, we are having less carrier in the bands. Let s assume that quasi-fermi levels differences are away from the new band edges due to the quantum confinement. So now I shift all the cases in problem (b) with respect to the quantized energies of first hole and electron states. We can see that the maximum gain is actually higher than the bulk case. (d) For D quantum wire, the density of states should be modified as following. ρ D jnt = 2 2m r πl x L y ħ 2 Θ[hv (E g + E n + E p )] If I plot it out, it should look something like the following.

12 So if I redo the two plots I had in the previous problem. So we don t have any gain for quasi-fermi levels are the same with respect to the original band edges. This is also expected. Because quantum confinement is too large, that the quasi-fermi levels are less than the bound states energies. For example, the st electron bound state energy is 0.624eV and st hole bound state energy is 0.eV. They re both larger than the original quasi-fermi levels.

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