Problem 2(a)Consider a gallium arsenide pn junction at T = 300 Kwith doping concentrations of Na = 1 X 10¹8 cm-³ andNd = 1 X 10¹6 cm-³.Determine the total space charge width-5A) W = 6.04 × 107 cm= 4.33 × 10-5 cmB) W =-5C) W = 2.78 x 107 cm-54.51 x 10 cmD) WE) WF) W===-53.76 × 10 cm3.26 x 10-5 cm Problem 2(b)Consider a gallium arsenide pn junction at T = 300 Kwith doping concentrations of Na = 1 X 10¹8 cm-³ andNd = 1 X 10¹6 cm-³.Use the space charge width from 2(a) anddetermine the magnitude of the maximum electricfieldA) |Emax = 5.92 x 104 V/cmEmax 4.13 x 104 V/cm=Emax 4.99 × 104 V/cmEmax = 3.11 x 104 V/cmEmax = 3.45 x 104 V/cmEmax = 2.10 x 104 V/cmB)C)D)E)F)=

Answered: Consider a gallium arsenide pn junction…

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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Problem 2(a)
Consider a gallium arsenide pn junction at T = 300 K
with doping concentrations of Na = 1 X 10¹8 cm-³ and
Nd = 1 X 10¹6 cm-³.
Determine the total space charge width
-5
A) W = 6.04 × 107 cm
= 4.33 × 10-5 cm
B) W =
-5
C) W = 2.78 x 107 cm
-5
4.51 x 10 cm
D) W
E) W
F) W
=
=
=
-5
3.76 × 10 cm
3.26 x 10-5 cm

Problem 2(b)
Consider a gallium arsenide pn junction at T = 300 K
with doping concentrations of Na = 1 X 10¹8 cm-³ and
Nd = 1 X 10¹6 cm-³.
Use the space charge width from 2(a) and
determine the magnitude of the maximum electric
field
A) |Emax = 5.92 x 104 V/cm
Emax 4.13 x 104 V/cm
=
Emax 4.99 × 104 V/cm
Emax = 3.11 x 104 V/cm
Emax = 3.45 x 104 V/cm
Emax = 2.10 x 104 V/cm
B)
C)
D)
E)
F)
=

Expert Solution

Step 1: Define total space charge & electric field.

Electric field and total space charge are strongly connected. The entire space charge generates the electric field. The electric field is greater the more total space charge there is.
Pn junction functioning is significantly influenced by the electric field in the depletion zone. By preventing the diffusion of majority carriers over the junction, it also raises the possibility of a current flow barrier.
In a pn junction, the ionised acceptor and donor impurities in the depletion area produce the total space charge. In the depletion area, the electric field is directed from the p-side to the n-side.

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