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

Want to see more full solutions like this?

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

how to measure ripple factor and peak input and peak output voltage for this circuits ?

Draw the outputs Q1, Q2 AND X in the system given in the figure CLK- CLK D Q1 Q2 X U O a Q1 Q2 HAL LO L1 MUX SO X

can you go into more detail about how you solved the derivative? and when was the t=1sec used? what if it was t=30sec what would be different then in the calculation?

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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