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**Unformatted text preview: **k BT &
# (c) δnB δpC δpE
X’=0 X=0 X=xB X’’=0 13.39 Consider an N- Al0.3Ga0.7As—intrinsic GaAs abrupt heterojunction. Assume the AlGaAs is doped to ND = 3 × 1018 cm- 3 and has a thickness of 35.0 nm. Let φBn = 0.89 V, and assume that ΔEC = 0.24 eV. (a) Calculate Voff and (b) calculate ns for VG = 0. (a) Voff will the the applied (negative) bias which just depletes the entire AlGaAs region: ea 2 N D
φ Bn − Voff − ΔEc =
2ε s ea 2 N D
Voff = φ Bn −
− ΔEc
2ε s
= 0.89 − 1.6 × 10 −19 ( 3.5 × 10 −8 )2 3 × 10 24
− 0.24 = −2.07V
2 ⋅ 12.2 ⋅ 8.854 × 10 −12 (b) For the inversion density, we can use the approximation from (13.68) as ε N (VG − Voff ) 12.2 ⋅ 8.854 × 10 −12 ⋅ 2.07
ns =
=
= 3.25 × 1012 cm −3 −19
−9
e(d + Δd )
1.6 × 10 ( 35 + 8 )10 This is grossly too high. 13.41 Cons...

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