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1 ELEC425/1-2012 1. Assignment 3. Assignment 3. ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ). Spectral widths a) Suppose that the frequency spectrum of a radiation emitted from a source has a central frequency and a spectral width . The spectrum of this radiation in terms of wavelength will have a central wavelength and a spectral width . Clearly, = c/ . Since and , using = c/ , show that the line width and hence the coherence length lc are 0 2. = = 0. 0 c and 20. l c = c t =.. b) Calculate for a lasing emission from a He-Ne laser that has = nm and GHz. Solution. a). d c = 2. d . c = . d . = 2 = . d . The negative sign means that if increases by d then decreases by d . The spectral width, or are much smaller than the emission wavelength (or the central wavelength, ) or the emission frequency (or the central frequency), respectively.

2 The negative sign is omitted since and the intervals centered on and , respectively. and are positive quantities. ELEC425/1-2012 2. Assignment 3. 0 2. = = 0. 0 c The coherent length lc is determined by the temporal coherent time t which is determined by the frequency width and hence by . 1 2 2. lc = c t = c = c 0 = 0. c . b) = nm and GHz 20. = . c ( 10 9 ) 2. = 10 9 pm 3 10 8. Einstein coefficients and critical photon concentration. (h ) is the energy of the electromagnetic radiation per unit volume per unit frequency due to photons with energy h = E -E . Suppose that there are nph photons per unit volume. Each has an energy h . The frequency range of emission is . Then, n ph h . (h ) =.. Consider the Ar ion laser system. Given that the emission wavelength is at 488 nm and the linewidth in the output spectrum is about 5 10 Hz between half intensity points, estimate the photon concentration necessary to achieve more stimulated emission than spontaneous emission.

3 Solution. For stimulated photon emission to exceed photon absorption the population inversion should be reached, N > N . R21 (stim ) c3 N. = (h ) = 2 > 1. R21 (spon) 8 h 3. N1. ELEC425/1-2012 3. Assignment 3. c3. (h ) > 1. 8 h 3. 8 h 3 8 h (h ) > = 3. c3 . 8 10 34. (h ) > = 10 13 J s m 3. (. 488 10 9 3. ). N ph h . (h ) .. (h ) . n ph = N ph =. h . 10 13 2 5 109 488 10 9. n ph = 34 8. = 1015 photons/m 3. 10 3 10. The obtained critical photon concentration for stimulated emission just exceeds spontaneous emission in the absence of any photon losses. It does not represent the photon concentration for laser operation. In practice, the photon concentration is much greater during laser operation. Fabry-Perot optical resonator. a) Consider an idealized He-Ne laser optical cavity. Taking L = m, R = , calculate the separation of the modes and the spectral width following Example b) Consider a semiconductor Fabry-Perot optical cavity of length 200 micron with end-mirrors that have a reflectance of If the semiconductor refractive index is , calculate the cavity mode nearest to the free space wavelength of 1300 nm.

4 Calculate the separation of the modes and the spectral width following Example Solution. a) separation of the modes is , m = f =. c =. (3 108 ) = 3 108 Hz 2L 2 0 .5. The finesse is R1 / 2 / 2. F= = = 1 R 1 ELEC425/1-2012 4. Assignment 3. and each mode width, spectral width, is f 3 108. m = = = 105 Hz F b) the cavity mode nearest to the emission wavelength 1300 nm is m=. 2L. =. (. 2 200 10 6. = ). / n 10 6 / m=1138. the separation of the modes is m = f =. c/n =. (. 3 108 / ). = 1011 Hz 2 L 2 200 10 6 ( ). The finesse is R 1 / 2 / 2. F= = = 1 R 1 0 .8. Each mode width or the mode spectral width is f 1011. m = = = 1010 Hz F Population inversion in a GaAs laser diode Consider the energy diagram of a forward biased GaAs laser diode as in Figure which results in EFn-EFp=Eg. ELEC425/1-2012 5.

5 Assignment 3. p+ n+. Ec e EFn A Ec Eg Ev B. EFp Ev h+. V. The energy band diagram of a degenerately doped p-n with with a sufficiently large forward bias to just cause population inversion where A and B overlap. 1999 Kasap, Optoelectronics (Prentice Hall). Figure GaAs laser diode, energy diagram. Estimate the minimum carrier concentration n = p for population inversion in GaAs at 300 K. The intrinsic carrier concentration in GaAs is of the order of 10 . cm . Assume for simplicity that n = ni exp[(EFn EFi ) / (k BT )] and p = ni exp[(EFi EFn ) / (k BT )]. (Note: The analysis will only be an order of magnitude as the above equations do not hold in degenerate semiconductors. A better approach is to use the Joyce- Dixon equations as can be found in advanced textbooks, applied for degeneracies of EF-EC 8kT).

6 Solution. The potential barrier from Ec (n-side) to Ec (p-side) is Ec and in valence band Ec=Ev(p-side)-Ev(n-side). To reach population inversion, the Fermi level EFp must be at least 1/2 Ec below Ev(p-side) or 1/2 Ec above Ev (n-side) and EFn must be at least 1/2 Ec above Ec (n- side), Figure Thus population inversion occurs when, ELEC425/1-2012 6. Assignment 3. EFn-EFp=[Ec(n-side)+1/2 Ec]-[Ev(n-side)+1/2 Ec]=Eg If EFi is Fermi level in intrinsic material, then for n=p EFn-EFi= EFi-EFp Substituting for EFp=EFn-Eg EFn-EFi= EFi-(EFn-Eg). 2 EFn-2 EFi=Eg Assuming that EFn-EFi=Eg/2= eV. Minimum carrier concentration is n = ni exp[(E Fn E Fi ) / (k B T )] = 10 7 exp[ / ] = 1019 cm -3. n >> ni this is a degenerate doping. Threshold current and power output from a laser diode. a) Consider the rate equations and their results in Section It takes t = nL/c second for photons to cross the laser cavity length L, where n is the refractive index.

7 If Nph is the coherent radiation photon concentration, then only half of the photons, (1/2)Nph, in the cavity would be moving towards the output face of the crystal at any instant. Given that the active layer has a length L, width W and thickness d, show that the coherent optical output power and intensity are hc 2 N ph dW hc 2 N ph . P0 = (1 R ) and I = (1 R ). 2 n 2n . where R is the reflectance of the semiconductor crystal face. b) If is the attenuation coefficient for the coherent radiation within the semiconductor active layer due to various loss processes such as scattering and R is the reflectance of the crystal ends then the total attenuation coefficient t is, 1 1 . t = + ln . 2L R 2 . Consider a double heterostructure InGaAsP semiconductor laser operating at 1310. nm. The cavity length L 60 m, width W 10 m, and d m.

8 The refractive index n The loss coefficient 10 cm . Find t, ph. c) For the above device, threshold current density Jth 500 A cm and sp 10 ps. What is the threshold electron concentration? Calculate the lasing optical power and intensity when the current is 5 mA. Solution. ELEC425/1-2012 7. Assignment 3. a) P =Energy flow per unit time in cavity towards face Transmittance hc 1 hc 1 . 2 N ph (dWL ) 2 N ph (dWL ) . P0 = Transmittance = (1 R ). t nL / c .. hc 2 N ph dW . P0 = (1 R ). 2 n . I=(Optical power)/Area P0 hc N ph . 2. I= = (1 R ). Wd 2n . where R is the reflectance of the crystal face. b) Consider one round trip through the cavity. The length L is traversed twice and there is one reflectance at each end. The overall attenuation of the coherent radiation after one-round trip is R Rexp[- (2L)].

9 Where R is the reflectance of the crystal end. Equivalently, this reduction can be represented as an effective or a total loss coefficient t such that after one round trip, the reduction factor is exp[- t(2L)]. Equating these two expressions R Rexp[- (2L)] = exp[- t(2L)] and rearranging, t = +1/(2L)ln(1/R ). The reflectance is 2 2. n 1 1 . R= = = n + 1 + 1 . The total loss coefficient is 1 1 1 1 . t = + ln 2 = 1000m 1 + 6. ln 2 . = 104 m -1. 2L R 2 60 10 . The average time for a photon to be lost from the cavity due to transmission through the end-faces, scattering and absorption in the semiconductor is n ph = = = ps c t (3 10 )( 104 ). 8. Coherent radiation is lost from the cavity after, on average, ps. c) From ELEC425/1-2012 8. Assignment 3. nth ed J th =. sp threshold concentration is nth =.

10 J th sp =. (500 10 )(10 10 ). 4 12. 1021 m -3 or 1015 cm-3. ed ( 10 19 )( 10 6 ). from given current of 5 mA the current density is J=I/(WL). J= = 106 A m -2. (10 10 6 )(60 10 6 ). The coherent radiation photon concentration is ph 10 12. N ph (J J th ) 19 6. (833 500) 104 1017 1019 photons m -3. ed ( 10 )( 10 ). The optical power is hc 2 N ph dW . P0 = (1 R ). 2n . P0 =. ( ). 2. 10 34 3 108 1019 10 6 10 10 6. (1 ) 10-4 W or mW. 9. 2 1310 10. Intensity= Optical power/area 10 3. I= 6 6. = 212 106 W/m 2 or 212 W/mm2. ( 10 )(10 10 ). This intensity is right at the crystal face over the optical cavity cross section. As the beam diverges, the intensity decreases away from the laser diode. InGaAsP-InP Laser Consider an InGaAsP-InP laser diode which has an optical cavity of length 250. microns.