Microelectronics: Circuit Analysis and Design
Microelectronics: Circuit Analysis and Design
4th Edition
ISBN: 9780073380643
Author: Donald A. Neamen
Publisher: McGraw-Hill Companies, The
Question
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Chapter 1, Problem 1.41P

(a)

To determine

The current ID1, ID2, VD1, VD2 for different values of IS1 and IS2 .

(a)

Expert Solution
Check Mark

Answer to Problem 1.41P

  1. For IS1=IS2=1013A,

      ID1=1mA ID2=1mA VD1=0.59867V VD2=0.59867V

  2. For IS1=5×1014A,IS2=5×1013A

      ID1=1mA ID2=1mA VD1=0.61669V VD2=0.55683V

Explanation of Solution

Given:

The value of current is,

  1. IS1=IS2=1013A
  2. IS1=5×1014A,IS2=5×1013A

The given circuit is shown below.

  Microelectronics: Circuit Analysis and Design, Chapter 1, Problem 1.41P , additional homework tip  1

Concept Used:

The diode voltage is,

  VD=VTln(IDIS)

Calculation:

  1. For, IS1=IS2=1013A
  2. From circuit it is clear that, ID1=ID2=1×103A

      VD1=VTln( I D1 I S1 )=(0.026)ln( 1× 10 3 10 13 )=0.59867VVD2=VTln( I D2 I S2 )=(0.026)ln( 1× 10 3 10 13 )=0.59867V

  3. For, IS1=5×1014A;IS2=5×1013A

From circuit it is clear that, ID1=ID2=1×103A

  VD1=VTln( I D1 I S1 )=(0.026)ln( 1× 10 3 5× 10 14 )=0.61669VVD2=VTln( I D2 I S2 )=(0.026)ln( 1× 10 3 5× 10 13 )=0.55683V

(b)

To determine

The current ID1, ID2, VD1, VD2 for different values of IS1 and IS2 .

(b)

Expert Solution
Check Mark

Answer to Problem 1.41P

  1. For IS1=IS2=1013A

    ID1=0.5mA ID2=0.5mA VD1=0.58065V VD2=0.58065V

  2. For IS1=5×1014A,IS2=5×1013A

    ID1=0.09091mAID2=0.9091mA VD1=0.5543V VD2=0.5543V

Explanation of Solution

Given:

The value of current is,

  1. IS1=IS2=1013A
  2. IS1=5×1014A,IS2=5×1013A

The given circuit is shown below.

  Microelectronics: Circuit Analysis and Design, Chapter 1, Problem 1.41P , additional homework tip  2

Concept Used:

The diode voltage is

  VD=VTln(IDIS)

Calculation:

  1. For IS1=IS2=1013A
  2. From circuit it is clear that, VD1=VD2 and ID1=ID2=i2=0.5×103A

    VD1=VTln( I D1 I S1 )=(0.026)ln( 0.5× 10 3 10 13 )=0.58065VVD2=VTln( I D2 I S2 )=(0.026)ln( 0.5× 10 3 10 13 )=0.58065V

  3. For, IS1=5×1014A,IS2=5×1013A

From circuit it is clear that, VD1=VD2 and ID1=ID2=i2=0.5×103A

Since, IS1IS2=ID1ID2=0.10ID1=0.1ID2 ,

Therefore,

  ID1+ID2=1.1ID2ID2=i1.1=1× 10 31.1=0.9091×103A

So,

  ID1=0.09091×103

  VD1=VTln( I D1 I S1 )=(0.026)ln( 0.09091× 10 3 5× 10 14 )=0.5543VVD2=VTln( I D2 I S2 )=(0.026)ln( 0.9091× 10 3 5× 10 13 )=0.5543V

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Chapter 1 Solutions

Microelectronics: Circuit Analysis and Design

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