EE 472 Solutions to some chapter 4 problems

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1 EE 472 Solutions to some chapter 4 problems 4.4. Erbium doped fiber amplifier An EDFA is pumped at 1480 nm. N1 and N2 are the concentrations of Er 3+ at the levels E 1 and E 2 respectively as shown in Figure 4.7(b). N 1 + N 2 = N, the concentration of Er 3+. At transparency (no net absorption nor gain), g = 0, and N 1 = N 1T, and N 2 = N 2T. Show that N 2T = Nσ ab (ν)/(σ ab (ν) + σ em (ν)) where σ ab (ν) and σ em (ν) are the absorption and emission cross sections which depend on frequency. The 1480 nm pump generates N 2T of Er 3+ at level E 2. The spontaneous decay time from E 2 to E 1 is τ sp. The pump must accumulate the Er 3+ ions at E 2 before they can decay. Show that the minimum pump power needed for transparency is about P pump ALhν p Nσ ab (ν)/τ sp (σ ab (ν) + σ em (ν)) where ν p is the pump frequency, A the pump core area, and L is the fiber length. Calculate the minimum pump power needed for an EDFA with N = cm 3, τ sp = 10 ms, a core diameter of 4 µm, a fiber length of 7 m, and the transparency at 1550 nm where emission and absorption cross sections are cm 2 and cm 2. What power is required to pump all Er 3+ ions to E 2? The transparency condition is σ em N 2T σ ab N 1T = 0 where N 1T + N 2T = N. Rewriting the transparency condition by eliminating N 1T yields σ em N 2T σ ab (N N 2T ) = 0 (σ em + σ ab )N 2T σ ab N = 0 and solving for N 2T gives N 2T = Nσ ab σ em + σ ab The total pump energy reuired to raise N 2T ions to E 2 is (N 2T )(hν p ) (volume). Since the ions will spontaneously decay from E 2 to E 1 with lifetime τ sp, the above energy needs to be supplied at a minimum rate of 1/τ sp. Thus the minimum pump power to achieve N 2T is P pump = V N 2T (hν p )/τ sp = ALhν pnσ ab (ν) τ sp (σ ab (ν) + σ em (ν)) Using the numbers given, A = π( cm) 2, L = 7 m, hν p = ev (1480 nm), N = cm 3, τ sp = 10 2 s, σ ab = cm 2, and σ em = cm 2, we find the

2 minimum pump power for this EDFA to reach transparency is P pump = 5.1 mw. To find the minimum power needed to pump all Er ions to E 2, substitute N for N 2T in the above formula for P pump which then results in 11.8 mw Erbium doped fiber amplifier A 2 m long EDFA has an Er concentration of cm 3 and is pumped at 1480 nm. Determine the small signal gain g at 1530 nm and 1570 nm for three cases corresponding to full population inversion 100%, 80% and 60% inversion. The small signal gain is g = σ em N 2 σ ab N 1 = N(σ em N 2 /N σ ab N 1 /N) where N = N 1 + N 2 is the total Er 3+ concentration. Let x = N 2 /N, the extent of pumping into E 2 ; 1 x = N 1 /N. g = N(σ em x σ ab (1 x)) The cross-sections can be read from Fig. 4.8 in the text. λ (nm) σ ab (10 21 cm 2 ) σ em (10 21 cm 2 ) x g (m 1 ) g (m 1 ) 100% % % x G G 100% 43.4dB 8.7dB 80% 26.1dB 3.5dB 60% 8.7dB 1.7dB Gaussian laser beam A high power (35 mw) commercial HeNe laser emits at nm. The emitting mode is TEM 00 with M 2 = 1.1. The beam diameter is 1.23 mm (between the 1/e 2 points), and the beam divergence is 0.66 mrad. Assume that the output beam is Gaussian (see ch. 1.1), and find the beam diameter at a distance of 100 m. Equation (1.11.1) (corrected) in Chapter 1.1 gives the width of a real beam at distance

3 z as ( ) zλm 2 2 2w r = 2w 0r 1 + πw 2 0r ( ) 2 (100 m)(632.8 nm)(1.1) 2w r = (1.23 mm) 1 + π(1.23 mm/2) 2 1/2 1/2 = 72 mm Because M 2 > 1, the beam is diverging faster than a pure Gaussian beam. 4.x. Laser linewidth Describe two physical mechanisms that produce homogeneous broadening in a laser and two physical mechanisms that produce inhomogeneous broadening. Homogeneous broadening occurs when every active atom in the laser is subject to the same linewidth broadening mechanism; the centre frequency for every atom is the same. For example, lifetime broadening is due to the uncertainty principle, E h/τ, where τ is the lifetime of the electron in the excited state. Another example is collisional broadening in solid state (crystalline) lasers where interactions with phonons shift the photon s energy. Inhomogeneous broadening occurs when the active atoms have their centre frequencies shifted by differing amounts. One example is doppler broadening where the emitted photon is red- or blue-shifted depending on the speed and direction of the emitting atom. Another example is amorphous broadening that occurs in solid state (glass) lasers. In a glass, each active atom has a somewhat different local environment leading to differing centre frequencies Photon cavity lifetime and total attenuation A semiconductor Fabry-Perot (FP) optical cavity that has length L, end mirrors with reflectances R 1 and R 2, and an attenuation coefficient inside the cavity of α s. The total attenuation coefficient for the optical cavity is denoted α t. Calculate the photon cavity lifetime for an In 0.60 Ga 0.40 As 0.85 P 0.15 FP optical cavity that has a refractive index n = 3.7, L = 300 µm, and α s = 20 cm 1. Since each end of the optical cavity is a cleaved semiconductor, R 1 = R 2 = (n 1) 2 (n + 1) 2 = 0.33 The total attenuation coefficient is α t = α s 1 2L ln(r 1R 2 ) = 57 cm 1

4 The photon cavity lifetime depends on α t. τ ph = n cα t = 2.1 ps Semiconductor lasers and temperature A Fabry-Perot semiconductor laser diode operates at 1550 nm. The active region is a III-V quaternary semiconductor alloy of InGaAsP (but quite close to being InGaAs). The LD has a cavity length of 250 µm. The refractive index of InGaAsP is approximately 3.60 with dn/ K 1, and the linear thermal expansion coefficient is K 1. The bandgap of InGaAs follows the Varshi equation E g = E g0 AT 2 /(B + T ) with E g0 = ev, A = ev/k, and B = 301K. Find the shift in the optical gain curve and the emission wavelength for a given mode per unit temperature change. Which effect is dominant? Temperature affects the emission wavelength in several ways, by changing the index of refraction, by changing the length of the optical cavity, and by changing the bandgap. For a given mode m, λ m = 2nL m dλ m = 2 m [ L dn ] + ndl = 2L m [ ] dn + nα For the laser diode described, the mode number nearest to 1550 nm is m = 2Ln λ = m = 1161 dλ m = 2L m [ K K 1] = nm/k Notice that the shift because of the changing index of refraction is larger than the shift because of the length of the optical cavity. The emitted photon energy follows the bandgap. The wavelength shift because of changes to E g can be obtained by differentiating the Varshi equation. dλ m = d ( ) hc E g = hc de g Eg 2 = hc ( 2AT Eg 2 B + T AT 2 ) = 0.76 nm/k (B + T ) 2 For the given laser, the change in the bandgap dominates the shift in output wavelength.

5 4.30. Threshold current and power output A double heterostructure InGaAsP semiconductor laser operates at 1310 nm. The cavity length is 100 µm, the width is 10 µm, and the depth is 0.1 µm. The refractive index is n = 3.7. The loss coefficient α s is 20 cm 1 and the direct recombination coefficient B is m 3 /s. Assume that the optical confinement factor is 1. Find the threshold gain g th, carrier concentration n th, current density J th, and the threshold current I th. Find the output optical power at I = 1.3I th, and the external slope efficiency η slope. What would I th be if Γ were 0.3 (a more realistic value)? The cleaved ends of the semiconductor result in R 1 = R 2 = (n 1) 2 /(n + 1) 2 = The threshold gain is (assuming Γ = 1) g th = α t = α s 1 2L ln(r 1R 2 ) = 131 cm 1 The threshold carrier concentration can be read from Fig. 4.48b; n th = cm 3. The radiative lifetime is τ r = 1/(Bn th ) = 6.7 ns. The threshold current density is J th = en thd τ r = 360 A/cm 2 and the threshold current is I th = wlj th = 3.6 ma The photon cavity lifetime is τ ph = n/(cα t ) = 0.94 ps. The optical output power at I = 1.3I th is [ hc 2 ] τ ph (1 R) P o = (I I th ) = (0.25 W/A)(I I th ) = 0.27 mw 2enλL The slope efficiency is (above threshold) η slope = dp o di = 0.25 W/A We now repeat the problem for Γ = 0.3. The threshold gain is now g th = α t /Γ = 437 cm 1 resulting in n th = cm 3. The new radiative lifetime is τ r = 5.6 ns. The threshold current density becomes J th = 566 A/cm 2, and the threshold current is I th = 5.7 ma Laser diode efficiencies A commercial InGaAsP FP laser diode (Mitsubishi ML720 and 725 series) operates at 1310 nm. The threshold current is 5 ma. At I = 20 ma, the output optical power is 5 mw

6 and the voltage across the diode is 1.1 V. Calculate the external quantum efficiency (QE), external differential QE, power conversion efficiency, and the slope efficiency of the diode. What is the forward diode current that gives an output optical power of 2 mw? η slope = η PCE = P o IV = P o (I I th ) = (5 mw) (20 ma 5 ma) (5 mw) (20 ma)(1.1 V) = 23% = 0.33 W/A η EQE = P o/(hν) (I/e) = (5 mw)/(0.95 ev) (20 ma)/e = 26% η EDQE = P o/(hν) (I I th )/e = P o/(hν) ( I/e) = (5 mw)/(0.95 ev) (15 ma)/e = 35% For P o = 2 mw, I I th = P o /η slope = 6 ma so I = 11 ma.