Microelectronic Circuits (The Oxford Series in Electrical and Computer Engineering) 7th edition
Microelectronic Circuits (The Oxford Series in Electrical and Computer Engineering) 7th edition
7th Edition
ISBN: 9780199339136
Author: Adel S. Sedra, Kenneth C. Smith
Publisher: Oxford University Press
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Chapter 3, Problem 3.1P
To determine

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

Expert Solution & Answer
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Answer to Problem 3.1P

The value of intrinsic carrier ni at T=55C is 26.84×105carriers/cm3 , T=0C is 1.523×109carriers/cm3 , T=20C is 8.595×109carriers/cm3 , T=75C is 3.7001×1011carriers/cm3 and T=125C is 4.7227×1012carriers/cm3 and the fraction of atom ionized at T=55C is 5.368×1017 , T=0C is 3.04×1014 , T=20C is 1.719×1013 , T=75C is 7.4002×1012 and T=125C is 9.4455×1010 .

Explanation of Solution

Calculation:

The intrinsic carrier density is given by

  ni=BT32eEg2kT …… (1)

Here, ni is the intrinsic carrier density, T is the given temperature, B is the material dependent parameter, B=7.3×1015cm3K32 , Eg is the bandgap energy, Eg=1.12eV and k is the Boltzmann’s constant, k=8.62×105eV/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 55C into K is given by,

  55C=55C+273K=218K

Substitute 7.3×1015cm3K32 for B , 1.12eV for Eg , 8.62×105eV/K for k and 218K 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)e29.8=(23496.74× 10 15)(1.1423× 10 13)

Solve it further as

  ni=26.84×105carriers/cm3

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

  F=26.84× 1055× 10 22=5.368×1017

For 0C into K is given by

  0C=0C+273K=273K

Substitute 7.3×1015cm3K32 for B , 1.12eV for Eg , 8.62×105eV/K for k and 273K 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)e23.79677=(23496.74× 10 15)(4.625× 10 11)

Solve it further as

  ni=1.523×109carriers/cm3

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

  F=1.523× 1095× 10 22=3.04×1014

For 20C into K is given by

  20C=20C+273K=293K

Substitute 7.3×1015cm3K32 for B , 1.12eV for Eg , 8.62×105eV/K for k and 293K 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)e22.1724=(36612.071× 10 15)(2.347× 10 10)

Solve it further as

  ni=8.595×109carriers/cm3

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

  F=8.595× 1095× 10 22=1.719×1013

For 75C into K is given by,

  75C=75C+273K=348K

Substitute 7.3×1015cm3K32 for B , 1.12eV for Eg , 8.62×105eV/K for k and 348K 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)e18.66=(47390.547× 10 15)(7.807× 10 9)

Solve it further as

  ni=3.7001×1011carriers/cm3

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

  F=3.7001× 10 115× 10 22=7.4002×1012

For 125C into K is given by,

  125C=125C+273K=398K

Substitute 7.3×1015cm3K32 for B , 1.12eV for Eg , 8.62×105eV/K for k and 398K 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)e16.32291=(57962.54796× 10 15)(8.1479× 10 8)

Solve it further as

  ni=4.7227×1012carriers/cm3

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

  F=4.72276× 10 125× 10 22=9.4455×1010

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