principles of semiconductor devices-l5

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  • 8/8/2019 Principles of Semiconductor Devices-L5

    1/21

    www.nanohub.org

    NCN

    Lecture5:EnergyBands

    MuhammadAshraful [email protected]

    Alam ECE606S09 1

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    Outline

    1) SchrodingerequationinperiodicU(x)

    2) Blochtheorem

    4) Propertiesof

    electronic

    bands

    5) Conclusions

    Alam ECE606S09 2

    . , .

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    GettingBacktoCrystals

    Original

    Problem

    PeriodicStructure

    Electronsinperiodic

    wewanttosolve

    Alam ECE606S09 3

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    Finallyan(almost)RealProblem

    =

    E

    n n n+ + +

    x x

    n n nC e D e = +

    U(x)

    a b

    But N atoms have two 2N unknown constants to find .

    p=a+

    Alam ECE606S09 4

    For large N, isnt there a better way ?

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    Outline

    1) Schrodinger

    equation

    in

    periodic

    U(x)

    2) Blochtheorem

    3) Bandstructure

    4) Propertiesofelectronicbands

    5) Conclusions

    Alam ECE606S09 5

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    FourStepsofFindingEnergyLevelsin Crystals

    d2 2 =1 Det coefficient matix =0

    ( ) 0x = =

    dx2

    2) Set2N2 equationsfor

    B B x x x x

    += =

    =3)

    B B x x x xdx dx += =

    =

    Nisverylargeforcrystal,butchanging steps2and3

    alittlebitwecanstillsolvethe robleminafewminutes!

    Alam ECE606S09 6

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    PeriodicU(x)andBlochsTheorem

    notourold(k)

    2 2( x ) ( x p ) = + ( ) ( ) eikp x p x + =

    E U(x)=U(x+a+b)=U(x+p)

    U(x)

    a b

    Alam ECE606S09 7

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    PhasefactorforNcells

    ( )x ( )x p +

    E U(x)

    N

    ikp

    = 2

    ikpx x e+=+

    2( )

    ikpx e

    =

    8] ( )[

    ikpN

    Np xx e =+

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    Step2:PeriodicBoundaryCondition

    E U(x)

    N-1

    [ ] ( ) ikpN x Np x e + =

    1 2 3

    21

    ikpN i ne e

    = 1

    23

    N-1 2 ..... 1,0,1,....2 2

    n N Nk nNp

    = =

    Alam ECE606S09 9

    max min,k k

    p p

    = =

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    Step3: BoundaryConditions

    2 2

    02 2 ( )mE i m U E

    0 0x x += ==

    sina a

    A x =sinb bA x =

    0 0x x

    d ddx dx

    += ==

    acosbB x+

    a b

    a b

    B B

    A A

    =

    =

    ik p=sin cos

    a a A a B a + =

    ax a x b

    ik pa b

    x a x b

    d de

    dx dx

    = =

    = =

    =

    ( )[ sin cos ]

    sin cos

    ik a b b b

    a a

    e A b B b

    A a B a

    + +

    =

    Alam ECE606S09 10

    ( )[ sin cos ]

    ik a b

    b be A b B b + +

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    Step4: Det(matrix)=0forEnergylevels

    sina aA x =sinb bA x =a b

    B B=

    acosbB x+

    a b

    A A =

    ( )

    sin cos

    [ sin cos ]

    a a

    ik a b

    b b

    A a B a

    e A b B b

    +

    + =

    + 0 1 0 01 aA 4)

    ( )

    sin cos

    [ sin cos ]

    a a

    ik a b

    A a B a

    e A b B b

    +

    =

    +* *

    *

    0

    0

    a

    bA

    B

    =

    021 2 mUE

    Alam ECE606S09 11

    0 2

    0

    ............2 1 U

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    Outline

    1) Solution

    of

    Schrodinger

    Equation

    in

    Periodic

    U(x)

    2) BlochTheorem

    3) Bandstructure

    ropert eso e ectron c an s

    5) Conclusions

    Alam ECE606S09 12

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    GraphicalsolutiontoEnergyLevels

    1 2 2 n N N............

    2 1

    ..... , , ,....2 2

    nNp

    = =

    k=0

    k=2/Np

    =E/U

    Alam ECE606S09 13

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    EnergyBandDiagram

    k=0

    = p

    =E/U

    0 k

    Alam ECE606S09 14

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    Brillouin ZoneandNumberofStates

    E

    2..... 1,0,1,....

    n N Nk n

    = =

    2m mx ina

    2

    tates pN

    band k Np

    = = =

    k

    Alam ECE606S09 15

    p

    p2

    k

    Np

    =

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    Outline

    1) SolutionofSchrodingerEquationinPeriodicU(x)

    2) BlochTheorem

    3) Bandstructure

    roper eso e ec ron c an s

    5) Conclusions

    Alam ECE606S09 16

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    WavePacketandGroupVelocity

    ( )E E

    i k k x iEikx i t t

    Ae Aex t

    + +

    = +

    ( )1

    EE i k x i t i tikx

    Ae e

    = +

    t1

    t

    Alam ECE606S09 17

    k

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    GroupVelocityforaGivenBand

    ( )x,t

    ( )

    1

    Ei k x iikx t t

    Ei

    Ae e

    = +

    1

    t21ikx i t

    i const . Ae e

    = +

    x E

    t k

    E

    = =

    cons an .x =

    1 1 d kd E d E F

    Alam ECE606S09 18

    2a

    t dt k dk k dt m*= = = =

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    EffectiveMassforaGivenBand

    1/m*vE

    kk

    1 E

    k

    =

    2

    2 2

    1 1 d E

    m* dk =

    Alam ECE606S09 19

    mass or eac an

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    EffectiveMassisnotEssential

    1/m*vE ??

    kk

    1 E

    k

    =

    2

    2 2

    1 1 d E

    *=

    kF

    t

    =

    0

    t

    x x dt = + 00

    t Fk k dt = +

    Alam ECE606S09 20

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    Conclusion

    1 SolutionofSchrodin ere uationisrelativel eas fors stems

    with

    well

    defined

    periodicity.

    .

    andbandgapssummarizeinformationaboutpossibleelectronic

    states.

    3) Effectivemassisnotafundamentalconcept.Therearesystems

    forwhicheffectivemasscannotbedefined.

    4) KPmodelisanalyticallysolvable.Realbandstructuresaresolvedoncomputer.Suchsolutionsarerelativelyeasy wewilldoHW

    Alam ECE606S09 21

    . .