Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 43, Problem 49P
To determine

The magnitude of the current in the diode when the battery is reversed.

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A pn junction diode and a Schottky diode have equal cross-sectional areas and have forward-bias currents of 0.5 mA. The reverse-saturation current of the Schottky diode is  The difference in forward-bias voltages between the two diodes is 0.30 V. Determine the reverse-saturation current of the pn junction diode.  
What fraction of the electrons in a good conductor have energies between 0.90 EF and EF at T = 0?
A diode is an electronic device that let the current flow in only one direction, that is the current through the diode is I=V/R (R is the diode resistance, a constant) if V > 0 and I=0 if V≤0. The diode is excited by an alternative voltage V(t) = cos(wot). (a) Sketch the diode response I(t) as a function of t. Indicate clearly the minimum values +to for which I(t) drops to zero. (b) Calculate the Fourier transform F(w) of the diode response. Hint: because the function is periodic, you only have to calculate the integral between ±to. (c) Schematically plot F(w) versus w. (d) A Lock-in amplifier is an electronic setup that can measure the response at integer frequency of the excitation frequency. What would a Lock-in measure at f = 2fo? (wo = 2nfo).

Chapter 43 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

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