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7.20 (a) The peak electric field in a reverse-biased silicon pn junction is |E|= 3 × 105 V/cm. The doping concentrations are N = 4 × 10¹5 cm³ and N. = 4 × 107 cm³. Find the magnitude of the reverse-biased voltage. (b) Repeat part (a) for N₁ = 4 × 10 16 cm³ and N₁ = 4 × 10¹7 cm³. (c) Repeat part (a) for Nd = N₁ = 4 × 10¹7 cm³.

7.20 (a) The peak electric field in a reverse-biased silicon pn junction is |E|= 3 × 105 V/cm. The doping concentrations are N = 4 × 10¹5 cm³ and N. = 4 × 107 cm³. Find the magnitude of the reverse-biased voltage. (b) Repeat part (a) for N₁ = 4 × 10 16 cm³ and N₁ = 4 × 10¹7 cm³. (c) Repeat part (a) for Nd = N₁ = 4 × 10¹7 cm³.

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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7.20 (a) The peak electric field in a reverse-biased silicon pn junction is |E|= 3 × 105 V/cm. The doping concentrations are N = 4 × 10¹5 cm³ and N. = 4 × 107 cm³. Find the magnitude of the reverse-biased voltage. (b) Repeat part (a) for N₁ = 4 × 10 16 cm³ and N₁ = 4 × 10¹7 cm³. (c) Repeat part (a) for Nd = N₁ = 4 × 10¹7 cm³.
Transcribed Image Text:7.20 (a) The peak electric field in a reverse-biased silicon pn junction is |E|= 3 × 105 V/cm. The doping concentrations are N = 4 × 10¹5 cm³ and N. = 4 × 107 cm³. Find the magnitude of the reverse-biased voltage. (b) Repeat part (a) for N₁ = 4 × 10 16 cm³ and N₁ = 4 × 10¹7 cm³. (c) Repeat part (a) for Nd = N₁ = 4 × 10¹7 cm³.
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