EECE 363 Quiz 2B 2008 Solutions

EECE 363 Quiz 2B 2008 Solutions

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Unformatted text preview: q = 1.5 × 1010 (1350 + 480 ) 1.6 × 10 −19 = 4.39 × 10 −6 (Ω ⋅ cm) -1 R= 1L 1 0 .5 = = 1.14 × 10 8 = 11.4 MΩ −6 −3 σ A 4.39 × 10 10 b) Assume that the silicon is doped with acceptors while the donor concentration is zero (ND=0 cm-3). Find the acceptor concentration NA if the resistance between the contacts is 10 Ω. (2 points) R= 1 L 1 0. 5 = = 1 0 Ω , or σ = 50 (Ω ⋅ cm) -1 σ A σ 10 −3 σ = pμ p q ≅ N A μ p q = 50 (Ω⋅cm)-1, then NA = 50 50 = = 6.51× 1017 cm-3 −19 μ p q 480 × 1.6 × 10 ( ) c) Which type of silicon (n or p) will exhibit higher conductivity if the dopant concentration is identical? Why? (2 points) The mobility of electrons is higher than that of holes. Since the conductivity is proportional to the mobility, the n-type silicon will result in higher conductivity. -3- 3) A step graded pn-junction silicon diode with a cross section area of 2.5×10-5 cm2 is forward biased with a voltage of 0.65V. The following information is known about the diode: ni=1.5...
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This document was uploaded on 02/16/2014 for the course EECE 363 at University of British Columbia.

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