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Unformatted text preview: onservation. e.g. Si; thermal energy released. 3. Auger recombination: three-carrier process. Electron takes energy to lattice. Photon (Light) Etrap Band-to-band
EE3161 Semiconductor Devices Sang-Hyun Oh R-G center
13 Auger UNIVERSITY OF MINNESOTA Direct vs. Indirect gap semiconductor
Direct Band Gap (GaAs) Conduction Band Indirect Band Gap (Si) Conduction Band
Phonon-assisted Transition Valence Band
Momentum Valence Band Momentum Recombination process must conserve both energy and momentum. The crystal lattice has so much inertia (momentum) whereas light doesn’t have much. In those materials that require much momentum (and energy) change when an electron goes from “bound” to “free”, light just cannot handle the momentum.
EE3161 Semiconductor Devices Sang-Hyun Oh 14 UNIVERSITY OF MINNESOTA Absorption Cross Section EE3161 Semiconductor Devices Sang-Hyun Oh 15 UNIVERSITY OF MINNESOTA Absorption in Solar Cell
Silicon Eg GaAs Absorption and charge separation in a solar cell
EE3161 Semiconductor Devices Sang-Hyun Oh 16 UNIVERSITY OF MINNESOTA Multi-Junction Solar Cell
Solar illumination Suppose we’re making a three-junction solar cell with GaAs (Eg=1.42 eV), GaInP (Eg=1.9 eV) and InGaAsP (Eg=1.05 eV). How should you arrange the stack (Eg1, Eg2, Eg3 in the image)?
EE3161 Semiconductor Devices Sang-Hyun Oh 17 UNIVERSITY OF MINNESOTA Equilibrium At equilibrium dp0 = −Cn0 p0 + Gthermal dt This must be zero, so that the generation rate G=Cn0p0 This is band-to-band recombination. EE3161 Semiconductor Devices Sang-Hyun Oh 18 UNIVERSITY OF MINNESOTA Low-Level Injection n-type: p-type: ∆ p n0 n ≈ n0 ∆ n p0 p ≈ p0 EE3161 Semiconductor Devices Sang-Hyun Oh 19 UNIVERSITY OF MINNESOTA Non-Equilibrium Assume an excess number of holes Δp in an n-type sample under conditions of low-level injection:
dp = −Cnp + Gthermal = −Cn0 (p0 + ∆p) + Cn0 p0 dt
EE3161 Semiconductor Devices Sang-Hyun Oh 20 UNIVERSITY OF MINNESOTA Non-Equilibrium (cont.)
Since p0 is constant in time,
d ∆p = −Cn0 ∆p dt Deﬁne a recombination time
d ∆p ∆p =− dt τp τp = 1/Cn0 which can be solved to be
∆p(t) = ∆p(0)e−t/τp EE3161 Semiconductor Devices Sang-Hyun Oh 21 UNIVERSITY OF MINNESOTA Redistribution of Excess Carriers p = p0 + ∆p(x, t) Change in carrier concentration = (illumination) ± (current ﬂow) - (recombination)
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