EECE 363 Quiz 2B 2008 Solutions

# 4 question 3 contd c using the figures shown below

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Unformatted text preview: 5 V ) C j0 ) )⎤ = 0.745 ⎥ ⎥ ⎦ 1 = 1 ⎞2 V = 2. 8 0.65 ⎛ ⎜1 − ⎟ ⎝ 0.745 ⎠ The application of a forward/reverse bias decreases/increases the depletion width, which modifies the junction capacitance. -4- Question 3, Cont’d. c) Using the figures shown below, sketch as a function of distance x: the charge density ρ(x), the electric field E(x), and the voltage V(x) with the ground at the metal contact of the n-type side. Calculate and indicate in your sketch the edges of the depletion region -xp and xn, the charge densities on both sides of the pn junction, the magnitude of the electric field at the metallurgical junction (i.e., x=0), and the voltage drop across the junction. Show all your calculation work on the next page. (10 points) -5- Question 3, Cont’d. [Use this page for the question 3(c).] ( )( ) ) = 3.2 × 10 ρ p = −qN A = − 1.6 × 10 −19 1017 = −1.6 × 10 −2 C/cm 3 ( )( ρ n = qN D = 1.6 × 10 −19 2 × 1016 −3 C/cm 3 From 3(b), V0 = 0.745 V Wdep = 2ε s q ⎛1 1 ⎜ + ⎜N ⎝ A ND ( ⎞ ⎟(V0 − V ) ⎟ ⎠ ) 2 × 1.04 × 10 −12 ⎛ 1 1 ⎞ −6 ⎜ 17 + ⎟(0.745 − 0.65 ) = 8.61× 10 cm = 0.0861 μm −19 16 1.6 × 10 2 × 10 ⎠ ⎝ 10 = NA 1017 xn = Wdep = 17 × 0.0861 = 0.0718 μm N A + ND 10 + 2 × 1016 x p = Wdep -x n = 0.0143 μm E (0 ) = − qN D Vn = 0V , εs (1.6 × 10 )(2 × 10 ) × (0.0718 × 10 ) = −2.21× 10 =− 1.04 × 10 −19 xn 16 −4 −12 V p = −(V0 − V ) = −(0.745 − 0.65 ) = −0.095 V -6- 4 V/cm...
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