The intrinsic carrier n i for silicon and the fraction value of atom ionized at T = − 55 ∘ C , T = 0 ∘ C , T = 20 ∘ C , T = 75 ∘ C and T = 125 ∘ C .

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Answer 1

Question

Chapter 3, Problem 3.1P

To determine

The intrinsic carrier ni for silicon and the fraction value of atom ionized at T=−55∘ C , T=0∘ C , T=20∘ C , T=75∘ C and T=125∘ C .

Expert Solution & Answer

Answer to Problem 3.1P

The value of intrinsic carrier ni at T=−55∘ C is 26.84×105 carriers/cm3 , T=0∘ C is 1.523×109 carriers/cm3 , T=20∘ C is 8.595×109 carriers/cm3 , T=75∘ C is 3.7001×1011 carriers/cm3 and T=125∘ C is 4.7227×1012 carriers/cm3 and the fraction of atom ionized at T=−55∘ C is 5.368×10−17 , T=0∘ C is 3.04×10−14 , T=20∘ C is 1.719×10−13 , T=75∘ C is 7.4002×10−12 and T=125∘ C is 9.4455×10−10 .

Explanation of Solution

Calculation:

The intrinsic carrier density is given by

ni=BT32e−Eg2kT …… (1)

Here, ni is the intrinsic carrier density, T is the given temperature, B is the material dependent parameter, B=7.3×1015 cm−3 K−32 , Eg is the bandgap energy, Eg=1.12 eV and k is the Boltzmann’s constant, k=8.62×10−5 eV/K .

The fraction of atom ionized is calculated as,

F=ni5×1022 …… (2)

Here, ni is the concentration, F is the fraction of ionized atom and 5×1022 is the atoms in a silicon crystal.

For −55∘C into K is given by,

−55∘C=−55∘ C+273 K=218 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 218 K for T in equation (1).

ni=(7.3× 10 15)(218)32e( −1.12 2( 8.62× 10 −5 )( 218 ) )=(7.3× 10 15)(3218.73)e( −1.12 3758.32× 10 −5 )=(7.3× 10 15)(3218.73)e−29.8=(23496.74× 10 15)(1.1423× 10 −13)

Solve it further as

ni=26.84×105 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=26.84× 1055× 10 22=5.368×10−17

For 0∘C into K is given by

0∘C=0∘ C+273 K=273 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 273 K for T in equation (1).

ni=(7.3× 10 15)(273)32e( −1.12 2( 8.62× 10 −5 )( 273 ) )=(7.3× 10 15)(4510.7)e( −1.12 4706.2× 10 −5 )=(32928.112)e−23.79677=(23496.74× 10 15)(4.625× 10 −11)

Solve it further as

ni=1.523×109 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=1.523× 1095× 10 22=3.04×10−14

For 20∘C into K is given by

20∘C=20∘ C+273 K=293 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 293 K for T in equation (1).

ni=(7.3× 10 15)(293)32e( −1.12 2( 8.62× 10 −5 )( 293 ) )=(7.3× 10 15)(5015.35)e( −1.12 5051.32× 10 −5 )=(7.3× 10 15)(5015.35)e−22.1724=(36612.071× 10 15)(2.347× 10 −10)

Solve it further as

ni=8.595×109 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=8.595× 1095× 10 22=1.719×10−13

For 75∘C into K is given by,

75∘C=75∘ C+273 K=348 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 348 K for T in equation (1).

ni=(7.3× 10 15)(348)32e( −1.12 2( 8.62× 10 −5 )( 348 ) )=(7.3× 10 15)(6491.85)e( −1.12 5999.52× 10 −5 )=(47390.547× 10 15)e−18.66=(47390.547× 10 15)(7.807× 10 −9)

Solve it further as

ni=3.7001×1011 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=3.7001× 10 115× 10 22=7.4002×10−12

For 125∘C into K is given by,

125∘C=125∘ C+273 K=398 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 398 K for T in equation (1).

ni=(7.3× 10 15)(398)32e( −1.12 2( 8.62× 10 −5 )( 398 ) )=(7.3× 10 15)(7940.07)e( −1.12 6861.52× 10 −5 )=(57962.54796× 10 15)e−16.32291=(57962.54796× 10 15)(8.1479× 10 −8)

Solve it further as

ni=4.7227×1012 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=4.72276× 10 125× 10 22=9.4455×10−10

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Answer 2

Question

Chapter 3, Problem 3.1P

To determine

The intrinsic carrier ni for silicon and the fraction value of atom ionized at T=−55∘ C , T=0∘ C , T=20∘ C , T=75∘ C and T=125∘ C .

Expert Solution & Answer

Answer to Problem 3.1P

The value of intrinsic carrier ni at T=−55∘ C is 26.84×105 carriers/cm3 , T=0∘ C is 1.523×109 carriers/cm3 , T=20∘ C is 8.595×109 carriers/cm3 , T=75∘ C is 3.7001×1011 carriers/cm3 and T=125∘ C is 4.7227×1012 carriers/cm3 and the fraction of atom ionized at T=−55∘ C is 5.368×10−17 , T=0∘ C is 3.04×10−14 , T=20∘ C is 1.719×10−13 , T=75∘ C is 7.4002×10−12 and T=125∘ C is 9.4455×10−10 .

Explanation of Solution

Calculation:

The intrinsic carrier density is given by

ni=BT32e−Eg2kT …… (1)

Here, ni is the intrinsic carrier density, T is the given temperature, B is the material dependent parameter, B=7.3×1015 cm−3 K−32 , Eg is the bandgap energy, Eg=1.12 eV and k is the Boltzmann’s constant, k=8.62×10−5 eV/K .

The fraction of atom ionized is calculated as,

F=ni5×1022 …… (2)

Here, ni is the concentration, F is the fraction of ionized atom and 5×1022 is the atoms in a silicon crystal.

For −55∘C into K is given by,

−55∘C=−55∘ C+273 K=218 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 218 K for T in equation (1).

ni=(7.3× 10 15)(218)32e( −1.12 2( 8.62× 10 −5 )( 218 ) )=(7.3× 10 15)(3218.73)e( −1.12 3758.32× 10 −5 )=(7.3× 10 15)(3218.73)e−29.8=(23496.74× 10 15)(1.1423× 10 −13)

Solve it further as

ni=26.84×105 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=26.84× 1055× 10 22=5.368×10−17

For 0∘C into K is given by

0∘C=0∘ C+273 K=273 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 273 K for T in equation (1).

ni=(7.3× 10 15)(273)32e( −1.12 2( 8.62× 10 −5 )( 273 ) )=(7.3× 10 15)(4510.7)e( −1.12 4706.2× 10 −5 )=(32928.112)e−23.79677=(23496.74× 10 15)(4.625× 10 −11)

Solve it further as

ni=1.523×109 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=1.523× 1095× 10 22=3.04×10−14

For 20∘C into K is given by

20∘C=20∘ C+273 K=293 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 293 K for T in equation (1).

ni=(7.3× 10 15)(293)32e( −1.12 2( 8.62× 10 −5 )( 293 ) )=(7.3× 10 15)(5015.35)e( −1.12 5051.32× 10 −5 )=(7.3× 10 15)(5015.35)e−22.1724=(36612.071× 10 15)(2.347× 10 −10)

Solve it further as

ni=8.595×109 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=8.595× 1095× 10 22=1.719×10−13

For 75∘C into K is given by,

75∘C=75∘ C+273 K=348 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 348 K for T in equation (1).

ni=(7.3× 10 15)(348)32e( −1.12 2( 8.62× 10 −5 )( 348 ) )=(7.3× 10 15)(6491.85)e( −1.12 5999.52× 10 −5 )=(47390.547× 10 15)e−18.66=(47390.547× 10 15)(7.807× 10 −9)

Solve it further as

ni=3.7001×1011 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=3.7001× 10 115× 10 22=7.4002×10−12

For 125∘C into K is given by,

125∘C=125∘ C+273 K=398 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 398 K for T in equation (1).

ni=(7.3× 10 15)(398)32e( −1.12 2( 8.62× 10 −5 )( 398 ) )=(7.3× 10 15)(7940.07)e( −1.12 6861.52× 10 −5 )=(57962.54796× 10 15)e−16.32291=(57962.54796× 10 15)(8.1479× 10 −8)

Solve it further as

ni=4.7227×1012 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=4.72276× 10 125× 10 22=9.4455×10−10

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Answer 3

Question

Chapter 3, Problem 3.1P

To determine

The intrinsic carrier ni for silicon and the fraction value of atom ionized at T=−55∘ C , T=0∘ C , T=20∘ C , T=75∘ C and T=125∘ C .

Expert Solution & Answer

Answer to Problem 3.1P

The value of intrinsic carrier ni at T=−55∘ C is 26.84×105 carriers/cm3 , T=0∘ C is 1.523×109 carriers/cm3 , T=20∘ C is 8.595×109 carriers/cm3 , T=75∘ C is 3.7001×1011 carriers/cm3 and T=125∘ C is 4.7227×1012 carriers/cm3 and the fraction of atom ionized at T=−55∘ C is 5.368×10−17 , T=0∘ C is 3.04×10−14 , T=20∘ C is 1.719×10−13 , T=75∘ C is 7.4002×10−12 and T=125∘ C is 9.4455×10−10 .

Explanation of Solution

Calculation:

The intrinsic carrier density is given by

ni=BT32e−Eg2kT …… (1)

Here, ni is the intrinsic carrier density, T is the given temperature, B is the material dependent parameter, B=7.3×1015 cm−3 K−32 , Eg is the bandgap energy, Eg=1.12 eV and k is the Boltzmann’s constant, k=8.62×10−5 eV/K .

The fraction of atom ionized is calculated as,

F=ni5×1022 …… (2)

Here, ni is the concentration, F is the fraction of ionized atom and 5×1022 is the atoms in a silicon crystal.

For −55∘C into K is given by,

−55∘C=−55∘ C+273 K=218 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 218 K for T in equation (1).

ni=(7.3× 10 15)(218)32e( −1.12 2( 8.62× 10 −5 )( 218 ) )=(7.3× 10 15)(3218.73)e( −1.12 3758.32× 10 −5 )=(7.3× 10 15)(3218.73)e−29.8=(23496.74× 10 15)(1.1423× 10 −13)

Solve it further as

ni=26.84×105 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=26.84× 1055× 10 22=5.368×10−17

For 0∘C into K is given by

0∘C=0∘ C+273 K=273 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 273 K for T in equation (1).

ni=(7.3× 10 15)(273)32e( −1.12 2( 8.62× 10 −5 )( 273 ) )=(7.3× 10 15)(4510.7)e( −1.12 4706.2× 10 −5 )=(32928.112)e−23.79677=(23496.74× 10 15)(4.625× 10 −11)

Solve it further as

ni=1.523×109 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=1.523× 1095× 10 22=3.04×10−14

For 20∘C into K is given by

20∘C=20∘ C+273 K=293 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 293 K for T in equation (1).

ni=(7.3× 10 15)(293)32e( −1.12 2( 8.62× 10 −5 )( 293 ) )=(7.3× 10 15)(5015.35)e( −1.12 5051.32× 10 −5 )=(7.3× 10 15)(5015.35)e−22.1724=(36612.071× 10 15)(2.347× 10 −10)

Solve it further as

ni=8.595×109 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=8.595× 1095× 10 22=1.719×10−13

For 75∘C into K is given by,

75∘C=75∘ C+273 K=348 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 348 K for T in equation (1).

ni=(7.3× 10 15)(348)32e( −1.12 2( 8.62× 10 −5 )( 348 ) )=(7.3× 10 15)(6491.85)e( −1.12 5999.52× 10 −5 )=(47390.547× 10 15)e−18.66=(47390.547× 10 15)(7.807× 10 −9)

Solve it further as

ni=3.7001×1011 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=3.7001× 10 115× 10 22=7.4002×10−12

For 125∘C into K is given by,

125∘C=125∘ C+273 K=398 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 398 K for T in equation (1).

ni=(7.3× 10 15)(398)32e( −1.12 2( 8.62× 10 −5 )( 398 ) )=(7.3× 10 15)(7940.07)e( −1.12 6861.52× 10 −5 )=(57962.54796× 10 15)e−16.32291=(57962.54796× 10 15)(8.1479× 10 −8)

Solve it further as

ni=4.7227×1012 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=4.72276× 10 125× 10 22=9.4455×10−10

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Answer 4

Question

Chapter 3, Problem 3.1P

To determine

The intrinsic carrier ni for silicon and the fraction value of atom ionized at T=−55∘ C , T=0∘ C , T=20∘ C , T=75∘ C and T=125∘ C .

Expert Solution & Answer

Answer to Problem 3.1P

The value of intrinsic carrier ni at T=−55∘ C is 26.84×105 carriers/cm3 , T=0∘ C is 1.523×109 carriers/cm3 , T=20∘ C is 8.595×109 carriers/cm3 , T=75∘ C is 3.7001×1011 carriers/cm3 and T=125∘ C is 4.7227×1012 carriers/cm3 and the fraction of atom ionized at T=−55∘ C is 5.368×10−17 , T=0∘ C is 3.04×10−14 , T=20∘ C is 1.719×10−13 , T=75∘ C is 7.4002×10−12 and T=125∘ C is 9.4455×10−10 .

Explanation of Solution

Calculation:

The intrinsic carrier density is given by

ni=BT32e−Eg2kT …… (1)

Here, ni is the intrinsic carrier density, T is the given temperature, B is the material dependent parameter, B=7.3×1015 cm−3 K−32 , Eg is the bandgap energy, Eg=1.12 eV and k is the Boltzmann’s constant, k=8.62×10−5 eV/K .

The fraction of atom ionized is calculated as,

F=ni5×1022 …… (2)

Here, ni is the concentration, F is the fraction of ionized atom and 5×1022 is the atoms in a silicon crystal.

For −55∘C into K is given by,

−55∘C=−55∘ C+273 K=218 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 218 K for T in equation (1).

ni=(7.3× 10 15)(218)32e( −1.12 2( 8.62× 10 −5 )( 218 ) )=(7.3× 10 15)(3218.73)e( −1.12 3758.32× 10 −5 )=(7.3× 10 15)(3218.73)e−29.8=(23496.74× 10 15)(1.1423× 10 −13)

Solve it further as

ni=26.84×105 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=26.84× 1055× 10 22=5.368×10−17

For 0∘C into K is given by

0∘C=0∘ C+273 K=273 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 273 K for T in equation (1).

ni=(7.3× 10 15)(273)32e( −1.12 2( 8.62× 10 −5 )( 273 ) )=(7.3× 10 15)(4510.7)e( −1.12 4706.2× 10 −5 )=(32928.112)e−23.79677=(23496.74× 10 15)(4.625× 10 −11)

Solve it further as

ni=1.523×109 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=1.523× 1095× 10 22=3.04×10−14

For 20∘C into K is given by

20∘C=20∘ C+273 K=293 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 293 K for T in equation (1).

ni=(7.3× 10 15)(293)32e( −1.12 2( 8.62× 10 −5 )( 293 ) )=(7.3× 10 15)(5015.35)e( −1.12 5051.32× 10 −5 )=(7.3× 10 15)(5015.35)e−22.1724=(36612.071× 10 15)(2.347× 10 −10)

Solve it further as

ni=8.595×109 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=8.595× 1095× 10 22=1.719×10−13

For 75∘C into K is given by,

75∘C=75∘ C+273 K=348 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 348 K for T in equation (1).

ni=(7.3× 10 15)(348)32e( −1.12 2( 8.62× 10 −5 )( 348 ) )=(7.3× 10 15)(6491.85)e( −1.12 5999.52× 10 −5 )=(47390.547× 10 15)e−18.66=(47390.547× 10 15)(7.807× 10 −9)

Solve it further as

ni=3.7001×1011 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=3.7001× 10 115× 10 22=7.4002×10−12

For 125∘C into K is given by,

125∘C=125∘ C+273 K=398 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 398 K for T in equation (1).

ni=(7.3× 10 15)(398)32e( −1.12 2( 8.62× 10 −5 )( 398 ) )=(7.3× 10 15)(7940.07)e( −1.12 6861.52× 10 −5 )=(57962.54796× 10 15)e−16.32291=(57962.54796× 10 15)(8.1479× 10 −8)

Solve it further as

ni=4.7227×1012 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=4.72276× 10 125× 10 22=9.4455×10−10

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Answer 5

Chapter 3, Problem 3.1P

To determine

The intrinsic carrier ni for silicon and the fraction value of atom ionized at T=−55∘ C , T=0∘ C , T=20∘ C , T=75∘ C and T=125∘ C .

Expert Solution & Answer

Answer to Problem 3.1P

The value of intrinsic carrier ni at T=−55∘ C is 26.84×105 carriers/cm3 , T=0∘ C is 1.523×109 carriers/cm3 , T=20∘ C is 8.595×109 carriers/cm3 , T=75∘ C is 3.7001×1011 carriers/cm3 and T=125∘ C is 4.7227×1012 carriers/cm3 and the fraction of atom ionized at T=−55∘ C is 5.368×10−17 , T=0∘ C is 3.04×10−14 , T=20∘ C is 1.719×10−13 , T=75∘ C is 7.4002×10−12 and T=125∘ C is 9.4455×10−10 .

Explanation of Solution

Calculation:

The intrinsic carrier density is given by

ni=BT32e−Eg2kT …… (1)

Here, ni is the intrinsic carrier density, T is the given temperature, B is the material dependent parameter, B=7.3×1015 cm−3 K−32 , Eg is the bandgap energy, Eg=1.12 eV and k is the Boltzmann’s constant, k=8.62×10−5 eV/K .

The fraction of atom ionized is calculated as,

F=ni5×1022 …… (2)

Here, ni is the concentration, F is the fraction of ionized atom and 5×1022 is the atoms in a silicon crystal.

For −55∘C into K is given by,

−55∘C=−55∘ C+273 K=218 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 218 K for T in equation (1).

ni=(7.3× 10 15)(218)32e( −1.12 2( 8.62× 10 −5 )( 218 ) )=(7.3× 10 15)(3218.73)e( −1.12 3758.32× 10 −5 )=(7.3× 10 15)(3218.73)e−29.8=(23496.74× 10 15)(1.1423× 10 −13)

Solve it further as

ni=26.84×105 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=26.84× 1055× 10 22=5.368×10−17

For 0∘C into K is given by

0∘C=0∘ C+273 K=273 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 273 K for T in equation (1).

ni=(7.3× 10 15)(273)32e( −1.12 2( 8.62× 10 −5 )( 273 ) )=(7.3× 10 15)(4510.7)e( −1.12 4706.2× 10 −5 )=(32928.112)e−23.79677=(23496.74× 10 15)(4.625× 10 −11)

Solve it further as

ni=1.523×109 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=1.523× 1095× 10 22=3.04×10−14

For 20∘C into K is given by

20∘C=20∘ C+273 K=293 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 293 K for T in equation (1).

ni=(7.3× 10 15)(293)32e( −1.12 2( 8.62× 10 −5 )( 293 ) )=(7.3× 10 15)(5015.35)e( −1.12 5051.32× 10 −5 )=(7.3× 10 15)(5015.35)e−22.1724=(36612.071× 10 15)(2.347× 10 −10)

Solve it further as

ni=8.595×109 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=8.595× 1095× 10 22=1.719×10−13

For 75∘C into K is given by,

75∘C=75∘ C+273 K=348 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 348 K for T in equation (1).

ni=(7.3× 10 15)(348)32e( −1.12 2( 8.62× 10 −5 )( 348 ) )=(7.3× 10 15)(6491.85)e( −1.12 5999.52× 10 −5 )=(47390.547× 10 15)e−18.66=(47390.547× 10 15)(7.807× 10 −9)

Solve it further as

ni=3.7001×1011 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=3.7001× 10 115× 10 22=7.4002×10−12

For 125∘C into K is given by,

125∘C=125∘ C+273 K=398 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 398 K for T in equation (1).

ni=(7.3× 10 15)(398)32e( −1.12 2( 8.62× 10 −5 )( 398 ) )=(7.3× 10 15)(7940.07)e( −1.12 6861.52× 10 −5 )=(57962.54796× 10 15)e−16.32291=(57962.54796× 10 15)(8.1479× 10 −8)

Solve it further as

ni=4.7227×1012 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=4.72276× 10 125× 10 22=9.4455×10−10

Answer 6

Chapter 3, Problem 3.1P

To determine

The intrinsic carrier ni for silicon and the fraction value of atom ionized at T=−55∘ C , T=0∘ C , T=20∘ C , T=75∘ C and T=125∘ C .

Expert Solution & Answer

Answer to Problem 3.1P

The value of intrinsic carrier ni at T=−55∘ C is 26.84×105 carriers/cm3 , T=0∘ C is 1.523×109 carriers/cm3 , T=20∘ C is 8.595×109 carriers/cm3 , T=75∘ C is 3.7001×1011 carriers/cm3 and T=125∘ C is 4.7227×1012 carriers/cm3 and the fraction of atom ionized at T=−55∘ C is 5.368×10−17 , T=0∘ C is 3.04×10−14 , T=20∘ C is 1.719×10−13 , T=75∘ C is 7.4002×10−12 and T=125∘ C is 9.4455×10−10 .

Explanation of Solution

Calculation:

The intrinsic carrier density is given by

ni=BT32e−Eg2kT …… (1)

Here, ni is the intrinsic carrier density, T is the given temperature, B is the material dependent parameter, B=7.3×1015 cm−3 K−32 , Eg is the bandgap energy, Eg=1.12 eV and k is the Boltzmann’s constant, k=8.62×10−5 eV/K .

The fraction of atom ionized is calculated as,

F=ni5×1022 …… (2)

Here, ni is the concentration, F is the fraction of ionized atom and 5×1022 is the atoms in a silicon crystal.

For −55∘C into K is given by,

−55∘C=−55∘ C+273 K=218 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 218 K for T in equation (1).

ni=(7.3× 10 15)(218)32e( −1.12 2( 8.62× 10 −5 )( 218 ) )=(7.3× 10 15)(3218.73)e( −1.12 3758.32× 10 −5 )=(7.3× 10 15)(3218.73)e−29.8=(23496.74× 10 15)(1.1423× 10 −13)

Solve it further as

ni=26.84×105 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=26.84× 1055× 10 22=5.368×10−17

For 0∘C into K is given by

0∘C=0∘ C+273 K=273 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 273 K for T in equation (1).

ni=(7.3× 10 15)(273)32e( −1.12 2( 8.62× 10 −5 )( 273 ) )=(7.3× 10 15)(4510.7)e( −1.12 4706.2× 10 −5 )=(32928.112)e−23.79677=(23496.74× 10 15)(4.625× 10 −11)

Solve it further as

ni=1.523×109 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=1.523× 1095× 10 22=3.04×10−14

For 20∘C into K is given by

20∘C=20∘ C+273 K=293 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 293 K for T in equation (1).

ni=(7.3× 10 15)(293)32e( −1.12 2( 8.62× 10 −5 )( 293 ) )=(7.3× 10 15)(5015.35)e( −1.12 5051.32× 10 −5 )=(7.3× 10 15)(5015.35)e−22.1724=(36612.071× 10 15)(2.347× 10 −10)

Solve it further as

ni=8.595×109 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=8.595× 1095× 10 22=1.719×10−13

For 75∘C into K is given by,

75∘C=75∘ C+273 K=348 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 348 K for T in equation (1).

ni=(7.3× 10 15)(348)32e( −1.12 2( 8.62× 10 −5 )( 348 ) )=(7.3× 10 15)(6491.85)e( −1.12 5999.52× 10 −5 )=(47390.547× 10 15)e−18.66=(47390.547× 10 15)(7.807× 10 −9)

Solve it further as

ni=3.7001×1011 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=3.7001× 10 115× 10 22=7.4002×10−12

For 125∘C into K is given by,

125∘C=125∘ C+273 K=398 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 398 K for T in equation (1).

ni=(7.3× 10 15)(398)32e( −1.12 2( 8.62× 10 −5 )( 398 ) )=(7.3× 10 15)(7940.07)e( −1.12 6861.52× 10 −5 )=(57962.54796× 10 15)e−16.32291=(57962.54796× 10 15)(8.1479× 10 −8)

Solve it further as

ni=4.7227×1012 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=4.72276× 10 125× 10 22=9.4455×10−10

Answer 7

Chapter 3, Problem 3.1P

To determine

The intrinsic carrier ni for silicon and the fraction value of atom ionized at T=−55∘ C , T=0∘ C , T=20∘ C , T=75∘ C and T=125∘ C .

Answer 8

Expert Solution & Answer

Answer to Problem 3.1P

The value of intrinsic carrier ni at T=−55∘ C is 26.84×105 carriers/cm3 , T=0∘ C is 1.523×109 carriers/cm3 , T=20∘ C is 8.595×109 carriers/cm3 , T=75∘ C is 3.7001×1011 carriers/cm3 and T=125∘ C is 4.7227×1012 carriers/cm3 and the fraction of atom ionized at T=−55∘ C is 5.368×10−17 , T=0∘ C is 3.04×10−14 , T=20∘ C is 1.719×10−13 , T=75∘ C is 7.4002×10−12 and T=125∘ C is 9.4455×10−10 .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Calculation:

The intrinsic carrier density is given by

ni=BT32e−Eg2kT …… (1)

Here, ni is the intrinsic carrier density, T is the given temperature, B is the material dependent parameter, B=7.3×1015 cm−3 K−32 , Eg is the bandgap energy, Eg=1.12 eV and k is the Boltzmann’s constant, k=8.62×10−5 eV/K .

The fraction of atom ionized is calculated as,

F=ni5×1022 …… (2)

Here, ni is the concentration, F is the fraction of ionized atom and 5×1022 is the atoms in a silicon crystal.

For −55∘C into K is given by,

−55∘C=−55∘ C+273 K=218 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 218 K for T in equation (1).

ni=(7.3× 10 15)(218)32e( −1.12 2( 8.62× 10 −5 )( 218 ) )=(7.3× 10 15)(3218.73)e( −1.12 3758.32× 10 −5 )=(7.3× 10 15)(3218.73)e−29.8=(23496.74× 10 15)(1.1423× 10 −13)

Solve it further as

ni=26.84×105 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=26.84× 1055× 10 22=5.368×10−17

For 0∘C into K is given by

0∘C=0∘ C+273 K=273 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 273 K for T in equation (1).

ni=(7.3× 10 15)(273)32e( −1.12 2( 8.62× 10 −5 )( 273 ) )=(7.3× 10 15)(4510.7)e( −1.12 4706.2× 10 −5 )=(32928.112)e−23.79677=(23496.74× 10 15)(4.625× 10 −11)

Solve it further as

ni=1.523×109 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=1.523× 1095× 10 22=3.04×10−14

For 20∘C into K is given by

20∘C=20∘ C+273 K=293 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 293 K for T in equation (1).

ni=(7.3× 10 15)(293)32e( −1.12 2( 8.62× 10 −5 )( 293 ) )=(7.3× 10 15)(5015.35)e( −1.12 5051.32× 10 −5 )=(7.3× 10 15)(5015.35)e−22.1724=(36612.071× 10 15)(2.347× 10 −10)

Solve it further as

ni=8.595×109 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=8.595× 1095× 10 22=1.719×10−13

For 75∘C into K is given by,

75∘C=75∘ C+273 K=348 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 348 K for T in equation (1).

ni=(7.3× 10 15)(348)32e( −1.12 2( 8.62× 10 −5 )( 348 ) )=(7.3× 10 15)(6491.85)e( −1.12 5999.52× 10 −5 )=(47390.547× 10 15)e−18.66=(47390.547× 10 15)(7.807× 10 −9)

Solve it further as

ni=3.7001×1011 carriers/cm3

Substitute 26.84×105 carriers/cm3 in equation (2).

F=3.7001× 10 115× 10 22=7.4002×10−12

For 125∘C into K is given by,

125∘C=125∘ C+273 K=398 K

Substitute 7.3×1015 cm−3 K−32 for B , 1.12 eV for Eg , 8.62×10−5 eV/K for k and 398 K for T in equation (1).

ni=(7.3× 10 15)(398)32e( −1.12 2( 8.62× 10 −5 )( 398 ) )=(7.3× 10 15)(7940.07)e( −1.12 6861.52× 10 −5 )=(57962.54796× 10 15)e−16.32291=(57962.54796× 10 15)(8.1479× 10 −8)