Answered: The Maxwell-Boltzmann distnbutlOn…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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The Maxwell-Boltzmann distnbutlOn function

Expert Solution

Step 1

Solution-

Given,

$f (E) = e^{- (E - E_{f}) / K T} D_{c} (E) = A {(E - E_{c})}^{1 / 2}$

a) Carrier distribution taken as f(E) in conduction band is directly proportional to the distribution of E,

Distribution of (E) $α {(E - E_{c})}^{\frac{1}{2}} . e^{- (E - E_{f}) / K T} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)$

Differentiate the above eq and equate to zero ,

$\frac{d}{d E} [{(E - E_{c})}^{1 / 2} . e^{- (E - E_{f}) / K T}] = 0 \frac{1}{2} \frac{1}{{(E - E_{c})}^{1 / 2}} e^{- (E - E_{f}) / K T} + {(E - E_{c})}^{1 / 2} [\frac{- 1}{K T}] e^{- (E - E_{f}) / K T} = 0 e^{- (E - E_{f}) / K T} [\frac{1}{2} \frac{1}{\sqrt{E - E_{c}}} + \sqrt{E - E_{c}} [\frac{- 1}{K T}]] = 0 \frac{1}{2} \frac{1}{\sqrt{E - E_{c}}} = - [(\sqrt{E - E_{c}}) (- \frac{1}{K T})] \frac{1}{2} \frac{1}{\sqrt{E - E_{c}}} = + \sqrt{E - E_{c}} (\frac{1}{K T}) \frac{{(E - E_{c})}^{- 1 / 2}}{2} = \frac{{(E - E_{c})}^{1 / 2}}{K T} \frac{K T}{2} = \frac{{(E - E_{c})}^{1 / 2}}{{(E - E_{c})}^{- 1 / 2}} \frac{K T}{2} = {(E - E_{c})}^{1 / 2} . {(E - E_{c})}^{1 / 2} \frac{K T}{2} = E - E_{c} E = E_{c} + \frac{KT}{2}$

b) Conduction band electron concentration is,

n = $\int_{C . B} D_{c} (E) f (E) d E$ Where C.B = conduction band

= $\int_{E_{c}}^{t o p o f C . B} D_{c} (E) f (E) d E$

assume loop of conduction band be infinite,

$n = \int_{E_{c}}^{\infty} D_{c} (E) f (E) d E$

Put the values,

n = $\int_{E_{c}}^{\infty} A {(E - E_{c})}^{1 / 2} . e^{- (E - E_{f}) / K T}$

Assume,

$\frac{E - E_{c}}{K T} = x \Rightarrow E = K T_{x} + E_{c}$

$\frac{1}{K T} d E = d x \Rightarrow d E = d x K T$ where x $\to$ 0 to $\infty$

$n = \int_{0}^{\infty} A {[E_{c} + K T_{x} - E_{c}]}^{1 / 2} . e^{- [K T_{x} + E_{c} - E_{f}] / K T} . K T d x = \int_{0}^{\infty} A \sqrt{K T_{x}} . e^{- (K T_{x} + E_{c} - E_{f}) / K T} . K T d x = A {(K T)}^{3 / 2} e^{- (E_{c} - E_{f}) / K T} \int_{0}^{\infty} \sqrt{x} e^{- K T_{x} / K T} d x = A {(K T)}^{3 / 2} e^{- (E_{c} - E_{f})} \int_{0}^{\infty} \sqrt{x} e^{- x} d x$

By Integrating ,

$n = A {(K T)}^{3 / 2} e^{- (E_{c} - E_{f}) / K T} . \int_{0}^{\infty} {(x)}^{\frac{3}{2} - 1} e^{- x} d x [\int_{0}^{\infty} x^{(\frac{3}{2} - 1)} e^{- x} d x = g a m m a f u n c t i o n (\frac{3}{2})] = A {(K T)}^{3 / 2} e^{- (E_{c} - E_{f}) / K T} g a m m a f u n c (\frac{3}{2}) [g a m m a (\frac{3}{2}) = \frac{\sqrt{π}}{2}]$

$n = \frac{\sqrt{π}}{2} A {(K T)}^{3 / 2} e^{- (E_{c} - E_{f}) / K T}$

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