Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

For an ideal p-n junction rectifier with a sharp boundary between its two semiconducting sides, the current I is related to the potential difference V across the rectifier by $l = l_{0} (e^{e V / k T} - 1)$ , where $l_{0}$ , which depends on the materials but not on I or V, is called the reverse saturation current. The potential difference V is positive if the rectifier is forward-biased and negative if it is back-biased. (a) Verify that this expression predicts the behavior of a junction rectifier by graphing I versus V from to $- 12 V to + 0.12 V$ . Take $T = 300 K$ and $l_{0} = 5.0 n A$ . (b) For the same temperature, calculate the ratio of the current for a 0.50 V forward bias to the current for a 0.50 V back bias.

The graph plotted for the given expression predicts the behavior of a junction rectifier.
The ratio of the current for a forward bias to the current for a 0.50 V backward bias is $2.5 \times 10^{8}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

The graph voltage values range from $- 0.12 V to + 0.12 V, for T = 300 K and l_{0} = 5 n A$
Voltage of the forward bias, $V = + 5.0 V$
Voltage of the backward bias, $V = - 5.0 V$

Step 2: Understanding the concept of current in a p-n junction rectifier

When the positive terminal of the battery is connected to the p-type and the negative terminal of the battery to the n-type semiconductor, the junction is said to be forward biased. If negative is connected with p and positive with n-type semiconductor, then the junction is said to be reverse biased. The current flows through the p-n junction when it is forward biased.

Formula:

The given current equation, $l = l_{0} (e^{e V / k T} - 1)$ (i)

Here e is the elementary charge, V is the volume, T is the absolute temperature and $l_{0}$ is the initial value of current.

Step 3: a) Calculation to plot the current graph

The vertical axis represents the current in nanoamperes and the horizontal axis represents the voltage.

Hence, the graph plotted for the given expression predicts the behavior of a junction rectifier.

Step 4: b) Calculation of the ratio of the current in the forward bias to the backward bias

Using the given data and equation (i), the ratio of the current for a 0.50V forward bias to the current for a backward bias is given by:

$\frac{l |_{V = + 0.50 V}}{l |_{_{V = - 0.50 V}}} = \frac{l_{0} e x p [(\frac{(+ 0.50 e V)}{(8.62 \times 10^{- 5} e V / K) (300 K)}) - 1]}{l_{0} e x p [(\frac{(- 0.50 e V)}{(8.62 \times 10^{- 5} e V / K) (300 K)}) - 1]} = 2.5 \times 10^{8}$

Hence, the ratio value is $2.5 \times 10^{8}$ .

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of current in a p-n junction rectifier

Step 3: a) Calculation to plot the current graph

Step 4: b) Calculation of the ratio of the current in the forward bias to the backward bias

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of current in a p-n junction rectifier

Step 3: a) Calculation to plot the current graph

Step 4: b) Calculation of the ratio of the current in the forward bias to the backward bias

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