Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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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
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Transcribed Image Text: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)
=
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Transcribed Image Text: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
Check Mark
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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