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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 42 / Problem 33E

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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Chapter 42, Problem 33E

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QUESTION:

Problem 33E

(a) A forward-bias voltage of 15.0 mV produces a positive current of 9.25 mA through a p-njunction at 300 K. What does the positive current become if the forward-bias voltage is reduced to 10.0 mV? (b) For reverse-bias voltages of -15.0 mV and -10.0 mV, what is the reverse-bias negative current?

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QUESTION:

Problem 33E

(a) A forward-bias voltage of 15.0 mV produces a positive current of 9.25 mA through a p-njunction at 300 K. What does the positive current become if the forward-bias voltage is reduced to 10.0 mV? (b) For reverse-bias voltages of -15.0 mV and -10.0 mV, what is the reverse-bias negative current?

ANSWER:

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$The\ current\ depends\ on\ the\ voltage\ across\ the\ diode\ and\ its\ temperature,\ so\ the\ resistance\ also\ depends$

$\ on\ these\ quantities.\$

$The\ current\ is\ I={I}_{s}({e}^{eV/kT}-1)\ \ \ and\ resistance\ is\ R=V/I$

$I=current\ through\ diode\ in\ Amps$

${I}_{0}=The\ diode's\ saturation\ current\ value$

$e=electron\ charge,1.602\times{}{10}^{-19}C$

$T=Temperature\ in\ degree\ kelvin$

$V=applied\ voltage\ in\ volts$

$k=Boltzmann's\ constant,1.380\times{}{10}^{-23}J/K$

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