The current I D 1 , I D 2 , V D 1 , V D 2 for different values of I S 1 and I S 2 .

Question

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Answer 1

Question

Chapter 1, Problem 1.41P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 1.41P

For IS1=IS2=10−13 A,
ID1=1 mA ID2=1 mA VD1= 0.59867V VD2=0.59867V
For IS1=5×10−14 A,IS2=5×10−13A
ID1=1 mA ID2=1 mA VD1= 0.61669V VD2=0.55683V

Explanation of Solution

Given:

The value of current is,

IS1=IS2=10−13 A
IS1=5×10−14 A, IS2=5×10−13A

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:

For, IS1=IS2=10−13 A

From circuit it is clear that, ID1=ID2=1×10−3 A

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

For, IS1=5×10−14 A; IS2=5×10−13A

From circuit it is clear that, ID1=ID2=1×10−3 A

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 )=0.55683 V

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 1.41P

For IS1=IS2=10−13 A
ID1=0.5 mA ID2=0.5 mA VD1=0.58065 V VD2=0.58065 V
For IS1=5×10−14 A, IS2=5×10−13A
ID1= 0.09091 mAID2=0.9091 mA VD1= 0.5543 V VD2= 0.5543 V

Explanation of Solution

Given:

The value of current is,

IS1=IS2=10−13 A
IS1=5×10−14 A, IS2=5×10−13A

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:

For IS1=IS2=10−13 A

From circuit it is clear that, VD1=VD2 and ID1=ID2=i2=0.5×10−3 A

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

For, IS1=5×10−14 A, IS2=5×10−13A

From circuit it is clear that, VD1=VD2 and ID1=ID2=i2=0.5×10−3 A

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

Therefore,

ID1+ID2=1.1ID2ID2=i1.1=1× 10 −31.1=0.9091×10−3 A

So,

ID1=0.09091×10−3

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 )=0.5543 V

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Answer 2

Question

Chapter 1, Problem 1.41P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 1.41P

For IS1=IS2=10−13 A,
ID1=1 mA ID2=1 mA VD1= 0.59867V VD2=0.59867V
For IS1=5×10−14 A,IS2=5×10−13A
ID1=1 mA ID2=1 mA VD1= 0.61669V VD2=0.55683V

Explanation of Solution

Given:

The value of current is,

IS1=IS2=10−13 A
IS1=5×10−14 A, IS2=5×10−13A

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:

For, IS1=IS2=10−13 A

From circuit it is clear that, ID1=ID2=1×10−3 A

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

For, IS1=5×10−14 A; IS2=5×10−13A

From circuit it is clear that, ID1=ID2=1×10−3 A

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 )=0.55683 V

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 1.41P

For IS1=IS2=10−13 A
ID1=0.5 mA ID2=0.5 mA VD1=0.58065 V VD2=0.58065 V
For IS1=5×10−14 A, IS2=5×10−13A
ID1= 0.09091 mAID2=0.9091 mA VD1= 0.5543 V VD2= 0.5543 V

Explanation of Solution

Given:

The value of current is,

IS1=IS2=10−13 A
IS1=5×10−14 A, IS2=5×10−13A

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:

For IS1=IS2=10−13 A

From circuit it is clear that, VD1=VD2 and ID1=ID2=i2=0.5×10−3 A

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

For, IS1=5×10−14 A, IS2=5×10−13A

From circuit it is clear that, VD1=VD2 and ID1=ID2=i2=0.5×10−3 A

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

Therefore,

ID1+ID2=1.1ID2ID2=i1.1=1× 10 −31.1=0.9091×10−3 A

So,

ID1=0.09091×10−3

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 )=0.5543 V

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Answer 3

Question

Chapter 1, Problem 1.41P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 1.41P

For IS1=IS2=10−13 A,
ID1=1 mA ID2=1 mA VD1= 0.59867V VD2=0.59867V
For IS1=5×10−14 A,IS2=5×10−13A
ID1=1 mA ID2=1 mA VD1= 0.61669V VD2=0.55683V

Explanation of Solution

Given:

The value of current is,

IS1=IS2=10−13 A
IS1=5×10−14 A, IS2=5×10−13A

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:

For, IS1=IS2=10−13 A

From circuit it is clear that, ID1=ID2=1×10−3 A

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

For, IS1=5×10−14 A; IS2=5×10−13A

From circuit it is clear that, ID1=ID2=1×10−3 A

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 )=0.55683 V

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 1.41P

For IS1=IS2=10−13 A
ID1=0.5 mA ID2=0.5 mA VD1=0.58065 V VD2=0.58065 V
For IS1=5×10−14 A, IS2=5×10−13A
ID1= 0.09091 mAID2=0.9091 mA VD1= 0.5543 V VD2= 0.5543 V

Explanation of Solution

Given:

The value of current is,

IS1=IS2=10−13 A
IS1=5×10−14 A, IS2=5×10−13A

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:

For IS1=IS2=10−13 A

From circuit it is clear that, VD1=VD2 and ID1=ID2=i2=0.5×10−3 A

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

For, IS1=5×10−14 A, IS2=5×10−13A

From circuit it is clear that, VD1=VD2 and ID1=ID2=i2=0.5×10−3 A

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

Therefore,

ID1+ID2=1.1ID2ID2=i1.1=1× 10 −31.1=0.9091×10−3 A

So,

ID1=0.09091×10−3

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 )=0.5543 V

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Answer 4

Question

Chapter 1, Problem 1.41P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 1.41P

For IS1=IS2=10−13 A,
ID1=1 mA ID2=1 mA VD1= 0.59867V VD2=0.59867V
For IS1=5×10−14 A,IS2=5×10−13A
ID1=1 mA ID2=1 mA VD1= 0.61669V VD2=0.55683V

Explanation of Solution

Given:

The value of current is,

IS1=IS2=10−13 A
IS1=5×10−14 A, IS2=5×10−13A

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:

For, IS1=IS2=10−13 A

From circuit it is clear that, ID1=ID2=1×10−3 A

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

For, IS1=5×10−14 A; IS2=5×10−13A

From circuit it is clear that, ID1=ID2=1×10−3 A

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 )=0.55683 V

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 1.41P

For IS1=IS2=10−13 A
ID1=0.5 mA ID2=0.5 mA VD1=0.58065 V VD2=0.58065 V
For IS1=5×10−14 A, IS2=5×10−13A
ID1= 0.09091 mAID2=0.9091 mA VD1= 0.5543 V VD2= 0.5543 V

Explanation of Solution

Given:

The value of current is,

IS1=IS2=10−13 A
IS1=5×10−14 A, IS2=5×10−13A

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:

For IS1=IS2=10−13 A

From circuit it is clear that, VD1=VD2 and ID1=ID2=i2=0.5×10−3 A

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

For, IS1=5×10−14 A, IS2=5×10−13A

From circuit it is clear that, VD1=VD2 and ID1=ID2=i2=0.5×10−3 A

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

Therefore,

ID1+ID2=1.1ID2ID2=i1.1=1× 10 −31.1=0.9091×10−3 A

So,

ID1=0.09091×10−3

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 )=0.5543 V

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Answer 5

Chapter 1, Problem 1.41P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 1.41P

For IS1=IS2=10−13 A,
ID1=1 mA ID2=1 mA VD1= 0.59867V VD2=0.59867V
For IS1=5×10−14 A,IS2=5×10−13A
ID1=1 mA ID2=1 mA VD1= 0.61669V VD2=0.55683V

Explanation of Solution

Given:

The value of current is,

IS1=IS2=10−13 A
IS1=5×10−14 A, IS2=5×10−13A

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:

For, IS1=IS2=10−13 A

From circuit it is clear that, ID1=ID2=1×10−3 A

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

For, IS1=5×10−14 A; IS2=5×10−13A

From circuit it is clear that, ID1=ID2=1×10−3 A

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 )=0.55683 V

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 1.41P

For IS1=IS2=10−13 A
ID1=0.5 mA ID2=0.5 mA VD1=0.58065 V VD2=0.58065 V
For IS1=5×10−14 A, IS2=5×10−13A
ID1= 0.09091 mAID2=0.9091 mA VD1= 0.5543 V VD2= 0.5543 V

Explanation of Solution

Given:

The value of current is,

IS1=IS2=10−13 A
IS1=5×10−14 A, IS2=5×10−13A

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:

For IS1=IS2=10−13 A

From circuit it is clear that, VD1=VD2 and ID1=ID2=i2=0.5×10−3 A

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

For, IS1=5×10−14 A, IS2=5×10−13A

From circuit it is clear that, VD1=VD2 and ID1=ID2=i2=0.5×10−3 A

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

Therefore,

ID1+ID2=1.1ID2ID2=i1.1=1× 10 −31.1=0.9091×10−3 A

So,

ID1=0.09091×10−3

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 )=0.5543 V

Answer 6

Chapter 1, Problem 1.41P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 1.41P

For IS1=IS2=10−13 A,
ID1=1 mA ID2=1 mA VD1= 0.59867V VD2=0.59867V
For IS1=5×10−14 A,IS2=5×10−13A
ID1=1 mA ID2=1 mA VD1= 0.61669V VD2=0.55683V

Explanation of Solution

Given:

The value of current is,

IS1=IS2=10−13 A
IS1=5×10−14 A, IS2=5×10−13A

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:

For, IS1=IS2=10−13 A

From circuit it is clear that, ID1=ID2=1×10−3 A

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

For, IS1=5×10−14 A; IS2=5×10−13A

From circuit it is clear that, ID1=ID2=1×10−3 A

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 )=0.55683 V

Answer 7

Chapter 1, Problem 1.41P

(a)

To determine

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

Answer 8

(a)

Expert Solution

Answer to Problem 1.41P

For IS1=IS2=10−13 A,
ID1=1 mA ID2=1 mA VD1= 0.59867V VD2=0.59867V
For IS1=5×10−14 A,IS2=5×10−13A
ID1=1 mA ID2=1 mA VD1= 0.61669V VD2=0.55683V

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The value of current is,

IS1=IS2=10−13 A
IS1=5×10−14 A, IS2=5×10−13A

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:

For, IS1=IS2=10−13 A

From circuit it is clear that, ID1=ID2=1×10−3 A

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

For, IS1=5×10−14 A; IS2=5×10−13A

From circuit it is clear that, ID1=ID2=1×10−3 A

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 )=0.55683 V

Answer 13

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 1.41P

For IS1=IS2=10−13 A
ID1=0.5 mA ID2=0.5 mA VD1=0.58065 V VD2=0.58065 V
For IS1=5×10−14 A, IS2=5×10−13A
ID1= 0.09091 mAID2=0.9091 mA VD1= 0.5543 V VD2= 0.5543 V

Explanation of Solution

Given:

The value of current is,

IS1=IS2=10−13 A
IS1=5×10−14 A, IS2=5×10−13A

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:

For IS1=IS2=10−13 A

From circuit it is clear that, VD1=VD2 and ID1=ID2=i2=0.5×10−3 A

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

For, IS1=5×10−14 A, IS2=5×10−13A

From circuit it is clear that, VD1=VD2 and ID1=ID2=i2=0.5×10−3 A

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

Therefore,

ID1+ID2=1.1ID2ID2=i1.1=1× 10 −31.1=0.9091×10−3 A

So,

ID1=0.09091×10−3

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 )=0.5543 V

Answer 14

(b)

To determine

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

Answer 15

(b)

Expert Solution

Answer to Problem 1.41P

For IS1=IS2=10−13 A
ID1=0.5 mA ID2=0.5 mA VD1=0.58065 V VD2=0.58065 V
For IS1=5×10−14 A, IS2=5×10−13A
ID1= 0.09091 mAID2=0.9091 mA VD1= 0.5543 V VD2= 0.5543 V