7 we have n n 4 1018 4 1016 vbi kbt ln a 2

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Unformatted text preview: or Vp 0 − Vbi = VDS − VGS = 5 + 0 = 5.0V Now, the built- in voltage is so that we need Vp0 = 5.886 V. From (13.7), we have ⎛N N ⎞ ⎛ 4 × 1018 ⋅ 4 × 1016 ⎞ Vbi = kBT ln ⎜ a 2 d ⎟ = 0.0259 ln ⎜ = 0.886 ⎝ 2.25 × 10 20 ⎟ ⎠ ⎝ ni ⎠ a= 2ε sVp 0 2 ⋅ 11.7 ⋅ 8.854 × 10 −12 ⋅ 5.886 = = 0.44 µ m . eN d 1.6 × 10 −19 ⋅ 4 × 10 22 (b) (i) From the above, we find that Vp0 = 5.886 V. (ii) From (13.4), we find that Vp = Vbi – Vp0 = - 5.0 V. 13.13 Consider an n- channel GaAs JFE...
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This document was uploaded on 02/19/2014.

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