85 04 02 028 077 e en 16 10 19 09 1016

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ion, as Now, the electric field in this region is given by the ionized impurities in the δ- doped layer, as and from this, we find the thickness of the GaAlAs layer is ϕ bi = 0.85 + 0.4 − 0.2 − 0.28 = 0.77 . E= eNδ 1.6 × 10 −19 ⋅ 0.9 × 1016 = = 1.2 × 10 7 V / m ε 13.2 ⋅ 8.854 × 10 −12 d = 1.5 × 10 −9 + 0.77 = 64 nm 1.2 × 10 7 2. Consider a GaAs/AlGaAs npn heterojunction bipolar transistor at 300K with ND,E = 1019 cm- 3, NA,B = 1019 cm- 3, and ND,C = 1017 cm- 3. The junction area is 15 mm2. If the base width of the transistor is measured to be 600 nm, what is β? For your calculations, you may assume that the diffusion constants and recombination times are De = 5.2 cm2/s, Dh,E = 1.9 cm2/s, Dh,C = 11 cm2/s, τe = 0.9 µs, τh...
View Full Document

This document was uploaded on 01/28/2014.

Ask a homework question - tutors are online