EML4550HW6ans

EML4550HW6ans - HW #6 Homework Solutions Power desired P :=...

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HW #6 Homework Solutions Power desired P 2500W := eV 1.6021764610 19 J := Givens on the photons: Φ .059 W cm 2 := ε av 1.25eV := N ph 4.8 10 17 1 scm 2 := Converter Properties: ε g 1.11eV := μ p 400 cm 2 Vs := μ n 1000 cm 2 := τ p 10 5 s := τ n 10 7 s := N a 10 19 1 cm 3 := N d 10 17 1 cm 3 := T 350K := e c 1.62020910 19 coul := Short Circuit Density: J s J s η co 1r () 1e α l ( ) e n ph ε g γ 4 173eV γ 4 173eV ε g := γ 3.533 = r γ 1 2 γ 1 + 2 := r 0.312 = e α l .368
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Potential number of absorbed photons from figure 5-2: n ph 310 17 sec cm 2 := J s η co 1r () 1 .368 e c n ph := J s 0.013 A cm 2 = Reverse Saturation Density: J o J o p n e c D p L p D p Hole Diffusion Constant: D p μ p kT e := D p 12.133 cm 2 s = Hole Diffusion Length: L p L p D p τ p .5 := L p 0.011cm = P n n n 2.23 10 31 T 3 exp ε g 9.93610 22 1 cm 6 n n N d 2 := n n 510 16 1 cm 3 := p n 9.93610 22 cm 6 n n := p n 1.987 10 6 × cm 3 = J o p n e c D p L
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This note was uploaded on 10/22/2011 for the course EML 4450 taught by Professor Greska during the Fall '06 term at FSU.

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EML4550HW6ans - HW #6 Homework Solutions Power desired P :=...

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