A pn junction at 300 (K) of cadmium telluride (CdTe) has the following properties p-side N₁ =10¹7 (cm³) a Hn=H₂=900 cm ²

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A Pn Junction At 300 K Of Cadmium Telluride Cdte Has The Following Properties P Side N 10 7 Cm A Hn H 900 Cm 1 (207 KiB) Viewed 63 times

A pn junction at 300 (K) of cadmium telluride (CdTe) has the following properties p-side N₁ =10¹7 (cm³) a Hn=H₂=900 cm ² V.S 2 n-side N₁ =10¹5 (cm³) H₁==1050 cm V.S 2 cm Hp = μ₁=100 V.s cm V.S Hp = μ₁₂=90 L₁ = 2.856×10-³ (cm) L = 3.085×10-³ (cm) L₂=9.032×104 (cm) L = 9.521×10+ (cm) CdTe has a bandgap of 1.58 (eV) and has an intrinsic carrier concentration of n₁ =1.13×10³ (cm³). Assume the bandgap does not change with temperature. The refractive index for CdTe is 2.95. If a pn junction is fabricated from the materials in the table above to make a laser diode with resonant cavity length of 100 (um), then: a) What are the diffusion coefficients for electrons, De (or Dn), and for holes, Dh (or Dp), used in calculating the reverse saturation current for this device? (6 points) b) What is the wavelength of light coming out of the laser diode based on its energy band diagram? (6 points) c) What cavity mode is closest to the wavelength you found in part b) above? (5 points) d) If the loss coefficient for this laser diode is α = 5×10³ (m³¹), then what is the threshold gain if the radiation confinement factor is r = 0.4? (8 points)