The area and dimensions (width and length) of the given capacitor which is fabricated using MOS technology.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 5, Problem 5.1P

To determine

The area and dimensions (width and length) of the given capacitor which is fabricated using MOS technology.

Expert Solution & Answer

Answer to Problem 5.1P

For, tox1=1nmArea=2.89×10−10 m2W=L=17μmFor, tox2=10nmArea=0.289×10−10 m2W=L=5.37μm

Explanation of Solution

Given:

( t OX ) 1 =1nm ( t OX ) 2 =10nm C=10pF

Calculation:

Case 1:

tox1=1nm

Although formula oxide capacitance which is the quotient of permittivity of the silicon dioxide and oxide thickness is given by

Cox=εoxtoxCox=3.9ε0toxplugging valuesCox=3.9×8.854×10−121×10−9Cox=34.53×10−3 F/m2

Although,

C=Cox⋅⋅W⋅Lwhere, W= channel width, L = channel length, C= total capacitanceHence,W⋅L=CCox⋅so,W⋅L=10×10−1234.43×10−3W⋅L=2.89×10−10 m2

Now, take square root from W · L in order to get channel width and channel length.

W=L=2.89×10−10W=L=1.70×10−5W=L=17 μm

Case 2:

tox2=10nm

Although formula oxide capacitance which is the quotient of permittivity of the silicon dioxide and oxide thickness is given by:

Cox=εoxtoxCox=3.9ε0toxplugging valuesCox=3.9×8.854×10−1210×10−9Cox=3.453×10−3 F/m2

Although,

C=Cox⋅⋅W⋅Lwhere, W= channel width, L = channel length, C= total capacitanceHence,W⋅L=CCox⋅so,W⋅L=10×10−123.443×10−3W⋅L=0.289×10−10 m2

Now take square root from W · L in order to get channel width and channel length

W=L=0.289×10−10W=L=0.537×10−5W=L=5.37 μm

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

Find the Thevenin Equivalent of the circuit below, show all steps;

Find the Thevenin Equivalent Circuit of the following and find the current through R_L, show all steps;

Find the Thevenin Equivalent Circuit of the circuit below and the current through R_L , show all steps;

Answer 2

Question

Chapter 5, Problem 5.1P

To determine

The area and dimensions (width and length) of the given capacitor which is fabricated using MOS technology.

Expert Solution & Answer

Answer to Problem 5.1P

For, tox1=1nmArea=2.89×10−10 m2W=L=17μmFor, tox2=10nmArea=0.289×10−10 m2W=L=5.37μm

Explanation of Solution

Given:

( t OX ) 1 =1nm ( t OX ) 2 =10nm C=10pF

Calculation:

Case 1:

tox1=1nm

Although formula oxide capacitance which is the quotient of permittivity of the silicon dioxide and oxide thickness is given by

Cox=εoxtoxCox=3.9ε0toxplugging valuesCox=3.9×8.854×10−121×10−9Cox=34.53×10−3 F/m2

Although,

C=Cox⋅⋅W⋅Lwhere, W= channel width, L = channel length, C= total capacitanceHence,W⋅L=CCox⋅so,W⋅L=10×10−1234.43×10−3W⋅L=2.89×10−10 m2

Now, take square root from W · L in order to get channel width and channel length.

W=L=2.89×10−10W=L=1.70×10−5W=L=17 μm

Case 2:

tox2=10nm

Although formula oxide capacitance which is the quotient of permittivity of the silicon dioxide and oxide thickness is given by:

Cox=εoxtoxCox=3.9ε0toxplugging valuesCox=3.9×8.854×10−1210×10−9Cox=3.453×10−3 F/m2

Although,

C=Cox⋅⋅W⋅Lwhere, W= channel width, L = channel length, C= total capacitanceHence,W⋅L=CCox⋅so,W⋅L=10×10−123.443×10−3W⋅L=0.289×10−10 m2

Now take square root from W · L in order to get channel width and channel length

W=L=0.289×10−10W=L=0.537×10−5W=L=5.37 μm

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 5, Problem 5.1P

To determine

The area and dimensions (width and length) of the given capacitor which is fabricated using MOS technology.

Expert Solution & Answer

Answer to Problem 5.1P

For, tox1=1nmArea=2.89×10−10 m2W=L=17μmFor, tox2=10nmArea=0.289×10−10 m2W=L=5.37μm

Explanation of Solution

Given:

( t OX ) 1 =1nm ( t OX ) 2 =10nm C=10pF

Calculation:

Case 1:

tox1=1nm

Although formula oxide capacitance which is the quotient of permittivity of the silicon dioxide and oxide thickness is given by

Cox=εoxtoxCox=3.9ε0toxplugging valuesCox=3.9×8.854×10−121×10−9Cox=34.53×10−3 F/m2

Although,

C=Cox⋅⋅W⋅Lwhere, W= channel width, L = channel length, C= total capacitanceHence,W⋅L=CCox⋅so,W⋅L=10×10−1234.43×10−3W⋅L=2.89×10−10 m2

Now, take square root from W · L in order to get channel width and channel length.

W=L=2.89×10−10W=L=1.70×10−5W=L=17 μm

Case 2:

tox2=10nm

Although formula oxide capacitance which is the quotient of permittivity of the silicon dioxide and oxide thickness is given by:

Cox=εoxtoxCox=3.9ε0toxplugging valuesCox=3.9×8.854×10−1210×10−9Cox=3.453×10−3 F/m2

Although,

C=Cox⋅⋅W⋅Lwhere, W= channel width, L = channel length, C= total capacitanceHence,W⋅L=CCox⋅so,W⋅L=10×10−123.443×10−3W⋅L=0.289×10−10 m2

Now take square root from W · L in order to get channel width and channel length

W=L=0.289×10−10W=L=0.537×10−5W=L=5.37 μm

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 5, Problem 5.1P

To determine

The area and dimensions (width and length) of the given capacitor which is fabricated using MOS technology.

Expert Solution & Answer

Answer to Problem 5.1P

For, tox1=1nmArea=2.89×10−10 m2W=L=17μmFor, tox2=10nmArea=0.289×10−10 m2W=L=5.37μm

Explanation of Solution

Given:

( t OX ) 1 =1nm ( t OX ) 2 =10nm C=10pF

Calculation:

Case 1:

tox1=1nm

Although formula oxide capacitance which is the quotient of permittivity of the silicon dioxide and oxide thickness is given by

Cox=εoxtoxCox=3.9ε0toxplugging valuesCox=3.9×8.854×10−121×10−9Cox=34.53×10−3 F/m2

Although,

C=Cox⋅⋅W⋅Lwhere, W= channel width, L = channel length, C= total capacitanceHence,W⋅L=CCox⋅so,W⋅L=10×10−1234.43×10−3W⋅L=2.89×10−10 m2

Now, take square root from W · L in order to get channel width and channel length.

W=L=2.89×10−10W=L=1.70×10−5W=L=17 μm

Case 2:

tox2=10nm

Although formula oxide capacitance which is the quotient of permittivity of the silicon dioxide and oxide thickness is given by:

Cox=εoxtoxCox=3.9ε0toxplugging valuesCox=3.9×8.854×10−1210×10−9Cox=3.453×10−3 F/m2

Although,

C=Cox⋅⋅W⋅Lwhere, W= channel width, L = channel length, C= total capacitanceHence,W⋅L=CCox⋅so,W⋅L=10×10−123.443×10−3W⋅L=0.289×10−10 m2

Now take square root from W · L in order to get channel width and channel length

W=L=0.289×10−10W=L=0.537×10−5W=L=5.37 μm

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 5, Problem 5.1P

To determine

The area and dimensions (width and length) of the given capacitor which is fabricated using MOS technology.

Expert Solution & Answer

Answer to Problem 5.1P

For, tox1=1nmArea=2.89×10−10 m2W=L=17μmFor, tox2=10nmArea=0.289×10−10 m2W=L=5.37μm

Explanation of Solution

Given:

( t OX ) 1 =1nm ( t OX ) 2 =10nm C=10pF

Calculation:

Case 1:

tox1=1nm

Although formula oxide capacitance which is the quotient of permittivity of the silicon dioxide and oxide thickness is given by

Cox=εoxtoxCox=3.9ε0toxplugging valuesCox=3.9×8.854×10−121×10−9Cox=34.53×10−3 F/m2

Although,

C=Cox⋅⋅W⋅Lwhere, W= channel width, L = channel length, C= total capacitanceHence,W⋅L=CCox⋅so,W⋅L=10×10−1234.43×10−3W⋅L=2.89×10−10 m2

Now, take square root from W · L in order to get channel width and channel length.

W=L=2.89×10−10W=L=1.70×10−5W=L=17 μm

Case 2:

tox2=10nm

Although formula oxide capacitance which is the quotient of permittivity of the silicon dioxide and oxide thickness is given by:

Cox=εoxtoxCox=3.9ε0toxplugging valuesCox=3.9×8.854×10−1210×10−9Cox=3.453×10−3 F/m2

Although,

C=Cox⋅⋅W⋅Lwhere, W= channel width, L = channel length, C= total capacitanceHence,W⋅L=CCox⋅so,W⋅L=10×10−123.443×10−3W⋅L=0.289×10−10 m2

Now take square root from W · L in order to get channel width and channel length

W=L=0.289×10−10W=L=0.537×10−5W=L=5.37 μm

Answer 6

Chapter 5, Problem 5.1P

To determine

The area and dimensions (width and length) of the given capacitor which is fabricated using MOS technology.

Expert Solution & Answer

Answer to Problem 5.1P

For, tox1=1nmArea=2.89×10−10 m2W=L=17μmFor, tox2=10nmArea=0.289×10−10 m2W=L=5.37μm

Explanation of Solution

Given:

( t OX ) 1 =1nm ( t OX ) 2 =10nm C=10pF

Calculation:

Case 1:

tox1=1nm

Although formula oxide capacitance which is the quotient of permittivity of the silicon dioxide and oxide thickness is given by

Cox=εoxtoxCox=3.9ε0toxplugging valuesCox=3.9×8.854×10−121×10−9Cox=34.53×10−3 F/m2

Although,

C=Cox⋅⋅W⋅Lwhere, W= channel width, L = channel length, C= total capacitanceHence,W⋅L=CCox⋅so,W⋅L=10×10−1234.43×10−3W⋅L=2.89×10−10 m2

Now, take square root from W · L in order to get channel width and channel length.

W=L=2.89×10−10W=L=1.70×10−5W=L=17 μm

Case 2:

tox2=10nm

Although formula oxide capacitance which is the quotient of permittivity of the silicon dioxide and oxide thickness is given by:

Cox=εoxtoxCox=3.9ε0toxplugging valuesCox=3.9×8.854×10−1210×10−9Cox=3.453×10−3 F/m2

Although,

C=Cox⋅⋅W⋅Lwhere, W= channel width, L = channel length, C= total capacitanceHence,W⋅L=CCox⋅so,W⋅L=10×10−123.443×10−3W⋅L=0.289×10−10 m2

Now take square root from W · L in order to get channel width and channel length

W=L=0.289×10−10W=L=0.537×10−5W=L=5.37 μm

Answer 7

Chapter 5, Problem 5.1P

To determine

The area and dimensions (width and length) of the given capacitor which is fabricated using MOS technology.

Answer 8

Expert Solution & Answer

Answer to Problem 5.1P

For, tox1=1nmArea=2.89×10−10 m2W=L=17μmFor, tox2=10nmArea=0.289×10−10 m2W=L=5.37μm

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

( t OX ) 1 =1nm ( t OX ) 2 =10nm C=10pF

Calculation:

Case 1:

tox1=1nm

Although formula oxide capacitance which is the quotient of permittivity of the silicon dioxide and oxide thickness is given by

Cox=εoxtoxCox=3.9ε0toxplugging valuesCox=3.9×8.854×10−121×10−9Cox=34.53×10−3 F/m2

Although,

C=Cox⋅⋅W⋅Lwhere, W= channel width, L = channel length, C= total capacitanceHence,W⋅L=CCox⋅so,W⋅L=10×10−1234.43×10−3W⋅L=2.89×10−10 m2

Now, take square root from W · L in order to get channel width and channel length.

W=L=2.89×10−10W=L=1.70×10−5W=L=17 μm

Case 2:

tox2=10nm

Although formula oxide capacitance which is the quotient of permittivity of the silicon dioxide and oxide thickness is given by:

Cox=εoxtoxCox=3.9ε0toxplugging valuesCox=3.9×8.854×10−1210×10−9Cox=3.453×10−3 F/m2

Although,

C=Cox⋅⋅W⋅Lwhere, W= channel width, L = channel length, C= total capacitanceHence,W⋅L=CCox⋅so,W⋅L=10×10−123.443×10−3W⋅L=0.289×10−10 m2

Now take square root from W · L in order to get channel width and channel length

W=L=0.289×10−10W=L=0.537×10−5W=L=5.37 μm

Answer 12

Ch. 5.1 - Prob. 5.1E Ch. 5.1 - Prob. D5.3E Ch. 5.2 - Prob. 5.4E Ch. 5.2 - Prob. 5.5E Ch. 5.3 - Prob. D5.9E Ch. 5.3 - Prob. D5.10E Ch. 5.3 - Prob. 5.11E Ch. 5.3 - Prob. 5.12E Ch. 5.3 - Prob. D5.13E Ch. 5.3 - Prob. D5.14E

The area and dimensions (width and length) of the given capacitor which is fabricated using MOS technology.

The area and dimensions (width and length) of the given capacitor which is fabricated using MOS technology.

Answer to Problem 5.1P

Explanation of Solution

Want to see more full solutions like this?

Chapter 5 Solutions