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If the conductivity of a new intrinsic semiconductor at 20 ° C is 250 Ω-1 .m-1 and at 100 ° C it is 1100 Ω-1 .m-1 , which one is your bandwidth gap (eV)? Data: k = 86.2 x 10-6 eV / K.

If the conductivity of a new intrinsic semiconductor at 20 ° C is 250 Ω-1 .m-1 and at 100 ° C it is 1100 Ω-1 .m-1 , which one is your bandwidth gap (eV)? Data: k = 86.2 x 10-6 eV / K.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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If the conductivity of a new intrinsic semiconductor at 20 ° C is 250 Ω-1 .m-1 and at 100 ° C it is 1100 Ω-1 .m-1 , which one is your bandwidth gap (eV)? Data: k = 86.2 x 10-6 eV / K.

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= 217 22-¹m-¹. EXEMPLO DE PROBLEMA 17.3 Você precisa caracterizar um novo semicondutor. Se sua condutividade a 20 °C for 250 2-¹ m¹ e a 100 °C for 1.100 2-¹ m¹, qual será seu espaçamen- to entre as bandas de energia, E? SOLUÇÃO Da Equação 17.3, E 1 lnon = Ino 2k T₁ e InoT, =lno Subtraindo (b) de (a), temos OT, Inσ₁ - InoT = In от, E 1 8 2k T₁ Então, E In(0₁/0₂) g = 2k 1/T₁-1/T₂ ou Eg (2k) In(o, /or) 1/T₁-1/T₂ Considerando T, = 20 °C (= 293 K) e T₂ = 100 °C (= 373 K), temos (2x86,2×10 eV/K) In(1.100/250) E 1 K-¹ -K-¹ 373 293 = 0,349 eV. E 1 8 2k T₂ (a) (b). 1 T₂
Transcribed Image Text:= 217 22-¹m-¹. EXEMPLO DE PROBLEMA 17.3 Você precisa caracterizar um novo semicondutor. Se sua condutividade a 20 °C for 250 2-¹ m¹ e a 100 °C for 1.100 2-¹ m¹, qual será seu espaçamen- to entre as bandas de energia, E? SOLUÇÃO Da Equação 17.3, E 1 lnon = Ino 2k T₁ e InoT, =lno Subtraindo (b) de (a), temos OT, Inσ₁ - InoT = In от, E 1 8 2k T₁ Então, E In(0₁/0₂) g = 2k 1/T₁-1/T₂ ou Eg (2k) In(o, /or) 1/T₁-1/T₂ Considerando T, = 20 °C (= 293 K) e T₂ = 100 °C (= 373 K), temos (2x86,2×10 eV/K) In(1.100/250) E 1 K-¹ -K-¹ 373 293 = 0,349 eV. E 1 8 2k T₂ (a) (b). 1 T₂
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