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Problems 1. (Pierret 2.2) Using the energy band model for a

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Problems 1. (Pierret 2.2) Using the energy band model for a Powered By Docstoc
					                              The University of Western Australia
                    School of Electrical, Electronic and Computer Engineering
J.H 2006                                        1                            TS 4 Solutions
______________________________________________________________________________________

Problems

1.     (Pierret 2.2) Using the energy band model for a semiconductor indicate how
one visualises a) an electron,b) a hole, c) donor sites, d) acceptor sites, e) an intrinsic
semiconductor, f) an n-type semiconductor, g) a p-type semiconductor

Solution
Donor Level eg Phosphorus doped             Acceptor Level eg Boron doped


          - Electron
                               E
                                   D



                                                                    E
                                                Hole                    A

                                                 +

            n - type                                   p-type
                  Intrinsic
 CB



EG



 VB




2.    Explain in general terms how we might produce an n-type and a p-type
semiconductor from an ultra-pure Gr IV semiconductor.

Solution:
     Si            Si                      Si          Si
                              Substitute         -
             Si                                 P

     Si            Si                      Si          Si
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                    School of Electrical, Electronic and Computer Engineering
J.H 2006                                        2                            TS 4 Solutions
______________________________________________________________________________________




3.       (Callister Eg 19.2)     Phosphorus is added to high purity silicon to give a
                     23    3
concentration of 10 /m charge carriers at room temperature.
a) is this material n or p type.
b) calculate the room temperature conductivity of this material, assuming that electron
and hole mobilities are the same as for the intrinsic material.

Solution
a) P Group V                 Si  Group IV           n-type material

b) Conductivity = neµn (µn is the same as for an intrinsic material)
µn = 0.15 m2/V-s = 1500 cm2/V-s
σ = 1023. 1.6 x 10-19. 0.15 = 2400 (Ω-m)-1


4.     (Callister 19.27) Will each of the following elements act as a donor or an
acceptor when added to the indicated semiconducting material?     Assume that the
impurity elements are substitutional.

(Note for example, GaAs is a II-V semiconductor and Cd is a Group IIb. Cd will
substitute for Ga, and being an electron short will give rise to p-type conduction)
Solution
        Impurity              Semiconductor
        Al                    Si              Acceptor
        P                     Ge              Donor
        Cd                    GaAs            Acceptor (Cd replaces Ga - 1 el short)
        S                     AlP             Donor (S subs for P - 1 el extra)
        Sb                    ZnSe            Acceptor (Sb subs for Se 1 el short)
        In                    CdTe            Donor (In subs for Cd, 1 extra el)


5.     (Callister 19.32)              The following electrical characteristics have been
determined for both intrinsic and p-type gallium antimonide (GaSb) at room temperature:
                              The University of Western Australia
                    School of Electrical, Electronic and Computer Engineering
J.H 2006                                        3                            TS 4 Solutions
______________________________________________________________________________________

                                  σ (Ω-m)-1       n (m-3)       p (m-3)

             Intrinsic          8.9 x 104        8.7 x 1023   8.7 x 1023

             Extrinsic          2.3 x 105        7.6 x 1022   1.0 x 1025

Calculate the electron and hole mobilities.
Intrinsic:                   σ = n|e|µe + pqµh
                         4
To start: 8.9 x 10 = 8.7 x 1023 . 1.6 x 10-19 µe + 8.7 x 1023 . 1.6 x 10-19 µp
So we need µe and µh
Finding µh:      Extrinsic         σ = pqµh
2.3x 105 = 1 x 1025.1.6x10-19µh                   µh=0.1438 m2/V-s
Therefore µe=0.50 m2/V-s


6.      Callister Eg19.1    For intrinsic silicon the room temperature conductivity is 4
     -4      -1
x 10 (Ω-m) ; the electron and hole mobilities are respectively 0.14 and 0.048 m2/V-s.
Compute the electron and hole concentration at room temperature.
Solution
a) σ = 4 x 10-4 (Ω-m)-1
µe = 0.14 m2/V-s
µh = 0.048 m2/V-s
σ = neµe + pqµh
4 x 10-4 = n . 1.6 x 10-19 (0.14 + 0.048)
 n=p = 1.33 x 1016 /m3

7.      (Pierret pge 127 (5)) Which diagram shown below best shows the position of
electrons at 0 K in an n-type material?




Solution
Only (b) and (c) show a donor level . AT 0oK there will be no ionisation of donors  all
electrons will be in the donor level so the answer is (b)
                              The University of Western Australia
                    School of Electrical, Electronic and Computer Engineering
J.H 2006                                        4                            TS 4 Solutions
______________________________________________________________________________________

8.      Using energy band diagrams, explain the mechanisms of conduction in an
intrinsic semiconductor, an extrinsic p-type material and an extrinsic n-type
semiconductor. Which mechanism is dominant in each case?
Solution
Extrinsic p-type
        Major mechanism          Minor mechanism
           Many holes            Small number of electrons

         CB                        CB
                                                !




                             EA

         VB         ++++++         VB           +




Intrinsic Mechanism
  Applied field E


         -




             +

The two mechanisms that occur in the same crystal and at the same time, with the
extrinsic generation producing the majority hole carriers and the intrinsic producing the
minority electrons.

Extrinsic n-type
         Major mechanism                Minor mechanism
          Many electrons                Small number of holes
                              The University of Western Australia
                    School of Electrical, Electronic and Computer Engineering
J.H 2006                                        5                            TS 4 Solutions
______________________________________________________________________________________

9.     Pierret pge 127        The illustration below is likely to depict the situation inside
an n-type semiconductor at approximately what temperature?
a) T<100K       b) T about room temperature         c) T> 400K d) T about 0 K




Solution
0K no ionisation takes place
Room Temp More ionisation occurs
>400K fully ionised
So the answer is (a)

                                                                              1
10.    The Fermi-Dirac distribution function is given by : f(E) =
                                                                               E ! Ef $
                                                                     1+ exp"
                                                                             # kT %
Show that at 0oK all states below Ef are filled and all states above Ef are empty.
Solution
                1
f(E) =          "E ! E f %
        1+ exp $
                # kT '   &
                        1           1
At T=0K f(E) =             ! + $ ' 1+ ( ' 0
                   1+ exp # &
                           "0 %
ie at 0K all the states above the Fermi level are empty

                          1         1
Say E<Ef    f(E) =          "! % ( 1 ( 1
                     1+ exp $ '
                            #0 &
ie at 0K all the states below the Fermi level are full

11.    Calculate the number of holes in the valence band of pure Ge at 300oK, given
Eg=0.7eV and mp*=m.
Solution
                                                !E
                                   N p = N v exp f
                                                kT
                              3
               " 2 ! m * kT % 2
        N v = 2$         p
with                   2
                            '
               #     h      &
                              The University of Western Australia
                    School of Electrical, Electronic and Computer Engineering
J.H 2006                                        6                            TS 4 Solutions
______________________________________________________________________________________

                    3                                   3
       " 2 ! m* kT % 2  "            -31       -23 %2
N v = 2$         p
                   ' = 2$ 2 ! 9.1x10 .1.38x10 300 '
                        $                          '
       $     h 2   '                     (34 2
                                 (6.62x10 )
       #           &    #                          &
    = 2.51 x 1025 states/m3

            !E f                 !0.35x1.6x10!19
N p = N v exp    = 2.51x1025 exp
            kT                    1.38x10-23.300
    = 3.35x1019 /m3


12.     In questions 3 and 4, a simplified distribution function is used to obtain carrier
densities i.e Maxwell-Boltzmann. In what region of space would this be invalid?
Solution
This approximation is valid far from the Fermi level as the “1” becomes insignificant in
comparison with the exponential term. This approximation would be invalid close to the
Fermi level.


13.    (Pierret 2.4) At T>0 K what is the probability of an electron state being
occupied if it is located at the Fermi level?

Solution
By definition, te Fermi level is the “half filled” level ie there is a 50% chance of finding
an electron at the Fermi level.
                1                1           1      1
f(E) =                     =              =      =
                "E ! E f %         " 0 % 1 +1 2
       1+ exp $          ' 1 + exp $ '
                # kT &             # kT &
                              The University of Western Australia
                    School of Electrical, Electronic and Computer Engineering
J.H 2006                                        7                            TS 4 Solutions
______________________________________________________________________________________

14.     A silicon chip is doped with one phosphorus atom per 108 host atoms. Assume
that the device operates at a temperature where complete impurity ionisation occurs, and
calculate the resistivity and conductivity.   Silicon has an atomic density of 5 1022
          3
atoms/cm and an electron mobility of 1500 cm2/volt sec.

Solution
1/108 = 10-8 P atoms per silicon.
Complete Ionisation
Si has 5 x 1022 atoms per cm3
µ = 15 cm2/V-s
No of P atoms = 10-8 x 5x1022 /cm3 = 5 x1014 atoms/cm3
Complete ionisation: so we have n = 5 x1014 /cm3
σ = neµ = 5 x1014x1.6x10-19 = 1500=0.12 (Ω-cm)-1
ρ=8.33 Ω-cm

15.     If a field of 500 mV/mm is applied across the device of Q21, calculate the current
density J flowing through the device.

E=500V/mm = 0.5 V/mm = 5 V/cm
J=σE = 0.12.5 = 0.6A/cm2


16.    Explain how minority carriers are thermally generated in doped semiconductors
and the precise meaning of the terms n>>p and also p>>n.

Solution
These terms refer to minority and majority carriers in an extrinsic semiconductor.

Let us take a p-type material. There will be many holes left behind in the VB from the
promotion of electrons from the VB to the acceptor level - this is the main mechanisn of
carrier promotion. There is however a finite probability that some electrons will have
enough energy to bridge the large gap from the VB to the CB giving us some conducting
electrons - minor mechanism. As a result there are many holes and a few electrons
p>>n.

Let us take a n-type material.      There will be many electrons in the CB from the
promotion of electrons from the donor level to the CB - this is the main mechanisn of
carrier promotion. There is however a finite probability that some electrons will have
enough energy to bridge the large gap from the VB to the CB giving us some conducting
holes - minor mechanism. As a result there are many electrons and a few holes n>>p

17.   Show that a donor impurity level in an n-type semiconductor is approximately
95% ionised when the fermi level lies 3kT below ED.
                              The University of Western Australia
                    School of Electrical, Electronic and Computer Engineering
J.H 2006                                        8                            TS 4 Solutions
______________________________________________________________________________________

Solution
Ef = 3kT below ED
                                    1
Ef =ED - 3kT f(E D ) =        " E D ! E D ! 3kT % = 0.048
                       1+ exp $                 '
                              #       kT        &
There is a 5% chance of finding an electron in the donor level ie a 95% chance of NOT
finding an electron.
This means that the donor level is 95% ionised



18. (Neaman 3.1)          Calculate the intrinsic carrier concentration ni at T=200, 400 and
600K for Si and Ge.
               !(E G - E f )       ! Ef
                   kT               kT                             3
np = Nc Nv e                   e                                                                  3
                                                      " 2 ! m* kT % 2                 " 2 ! m*kT % 2
                                                               p
               !E G                            N v = 2$
                                                      $       2
                                                                  '
                                                                  '            N c = 2$       e  '
                                                                                      $      2   '
  = Nc Nv e     kT   = n2
                        i                             #     h     &                   #    h     &
There are large discrepancies for values of effective mass (mainly because they are
experimentally measured) in these results, we have used: mp = 0.16m and me = 0.98m.
As long as you have the right order of magnitude that is ok.


T(K)      Eg (Si)     Nv                  Nc            Eg/kT      ni (m-3)      ni (cm-3)
200       1.12eV      8.74x 1023          1.325x1025 64.9          2.74 x 1010 2.74 x 104
400       1.12eV      2.47x1024           3.75x1025     32.46      8.6x1017      8.6x1011
600       1.12eV      4.54x1024           6.89x1025     21.6       3.53x1020     3.53x1014


Another (easier) way to calculate is to group the terms together:
               !(E G - E f )       ! Ef
np = Nc Nv e       kT          e    kT

                      3                  3
      # 2 " m* kT & 2      # 2 " m* kT & 2 !E G
                p
   = 2%
      %       2
                  (
                  (       2%
                           %       2
                                     e ( e kT
                                       (
      $     h     '        $     h     '
                     G    !E
           # T & 3 kT
   = N VN C%     ( e
           $ 300 '
NC300K = 1.04x1019 /cm3                        NV300K = 2.8x1019 /cm3
                              The University of Western Australia
                    School of Electrical, Electronic and Computer Engineering
J.H 2006                                        9                            TS 4 Solutions
______________________________________________________________________________________


T(K)      Eg (Si)      Eg/kT        ni (cm-3)
200       1.12eV       64.9         7.5 x 104
400       1.12eV       32.46        2.35x1012
600       1.12eV       21.6         9.84x1014


NC300K = 6.0x1018 /cm3                       NV300K = 1.04x1019 /cm3


T(K)      Eg (Ge) Eg/kT             ni (cm-3)
200       0.66eV       38.26        2.1 x 1010
400       0.66eV       19.13        8.5x1014
600       0.66eV       12.75        3.8x1016

19.    (Neaman 3.6)                The carrier effective masses in a semiconductor are

m * = 0.62m 0 and m * = 1.4m 0 . Determine the position of the intrinsic Fermi level with
  e                 p

respect to the centre of the bandgap at T=300K.

     EG   3      m*
                  p
Ef =    +   kT ln *
      2   4      me

        EG   3      m*
                     p        (1.38x10!23)         " 1.4 %
Ef !       =   kT ln * = 0.75              (300)ln $        = 0.0158eV
         2   4      me          1.6x10!19          # 0.62 '
                                                          &

So it is 0.0158e above midgap and therefore an n-type material.

20.    (Neaman 3.13) The electron concentration in Si at T=300K is.n 0 = 5 ! 104 cm"3
a) Determine p. Is this an n or p-type material? b) Determine the position of the Fermi
level with respect to the intrinsic Fermi level.


ni = 1.4 x 1010 /cm3
                                             2
  i                            (
n 2 = n 0p 0 = 5 ! 104 p0 = 1.4x1010     )
p0 = 3.92x1015 / cm3

Therefore this is a p type material.
                              The University of Western Australia
                    School of Electrical, Electronic and Computer Engineering
J.H 2006                                        10                           TS 4 Solutions
______________________________________________________________________________________


                 kT   n      1.38x10 !23       " 5x10 4 %
E f ! E fi =        ln = 0.5        !19  300ln $           ' = !0.32eV
                  2   p      1.6x10            $ 3.92x1015 '
                                               #           &
As expected the Fermi level is below the intrinsic Fermi level (midgap).

Note that if you used ni = 1.5 x 1010 /cm3 then the difference in energy levels would be:
0.3266eV



21.    (Neaman 3.15) a) If Ec - EF =0.25eV in GaAs at T=400K, calculate the values
of n and p b) Assuming that n doesn’t change from part (a) determine Ec - EF at 300K


               !(E G ! E f )
N e = Nc exp
                   kT


From Neaman:                       GaAs: Egap =1.42eV; mh = 0.48m, mn = 0.067m


So calculating:
1)
                        3                        3
        " 2 ! m*kT % 2      "                  %
N c = 2$        e   ' = 2$ 2 ! 0.067mk400 ' 2 = 6.7x1023
        $           '
        #    h2     &       #       h2         &
                    !(0.25)
N e = 6.7x10 23 exp          = 6.7x10 23 • 7.13x10 !4 = 4.81x1020 / m 3 = 4.8x1014 / cm3
                      kT
2)                                                      3
                                                ! T $2
                                                #     &
                                                " 300 %
Or use: NC300 = 4.7x1017/cm3: NC400 = 4.7x1017            = 7.24x1017 /cm3
So
                      !(0.25)
N e = 6.7x10 23 exp           = 7.24x1017 • 7.13x10 !4 = 5.16x1014 / cm3
                        kT

22.    (Neaman 3.22)           The Fermi level in n-type Si at T=300K is 245meV below the
conduction band and 200meV below the donor level.             Determine the probability of
finding an electron a) in the donor level and b) in a state in the conduction band kT above
the conduction band edge.
                              The University of Western Australia
                    School of Electrical, Electronic and Computer Engineering
J.H 2006                                        11                           TS 4 Solutions
______________________________________________________________________________________




EC - Efn = 0.245eV                      ED - EEfn = 0.2eV so therefore EC - ED = 0.045eV


                                  Ec

                                  E
0.245eV                               D
                        0.2eV
                                E
                                    F

We can use FD or a MB approximation since we are quite far from the Fermi level:
f(E) =        1         =     1       = 4.42x10 !4
             E - Ef            0.2
       (1 + e       kT ) (1 + e kT )



b) E-EC = kT=0.0259eV, E-EF = 0.245+kT
f(E) =         1         =          1            = 2.87x10 !5
              E - Ef            0.245+.0259
        (1 + e       kT ) (1 + e            kT )
23.    (Neaman 3.28) A GaAs device is doped with a donor concentration of 3 x1015
cm-3. For the device to operate properly the intrinsic carrier concentration must remain
at less than5% of the total electron concentration. What is the maximum temperature at
which the device may operate?

ND = 3x1015 /cm3 so ni = 0.05x3 1015 /m3= 1.5x1014 /m3
                 !E G
np = Nc Nv e      kT    = n2
                           i
                                           G        !E
                                 " T % 3 kT
n2            28
 i   = 2.25x10     = NV300 NC300 $     ' e
                                 # 300 &
                                                 !1.42
                              "          T %3
2.25x1028   = 7x1018 4.7x1017         $     ' e kT
                                      # 300 &
                          !1.42
             " T % 3 kT
6.8x10 !9   =$     ' e
             # 300 &
using trial and error:
T=500K:          2.3x10-14 < 6.8x 10-9
T=1000K:         2.62x10-6 > 6.8x 10-9
T=800K:          2.19x10-8> 6.8x 10-9
T=700K           7.75x10-10< 6.8x 10-9
                              The University of Western Australia
                    School of Electrical, Electronic and Computer Engineering
J.H 2006                                        12                           TS 4 Solutions
______________________________________________________________________________________


T=750K          4.57x10-9< 6.8x 10-9
etc
Final answer is about 762K

24. (Neaman 3.31) Silicon at T=300K is doped with acceptor atoms at a concentration
of NA = 7 x 1015 cm-3 a) Determine EF - EV b) Calculate the concentration of additional
acceptor atoms that must be added to move the Fermi level a distance kT closer to the
valence band edge.

a)
              !E f
N p = N v exp
              kT
At 300K, Si has NV 1.04 x 1019 / cm3: therefore
                            "E f
7 ! 1015 = 1.04 ! 1019 exp
                           0.0259
E f = 0.189eV
b) The full form of this is
               !(E f ! E v )
N p = Nv exp
                   kT
So the energy gap would be Ef-Ev = 0.189-0.0.259=0.1633eV
                       "0.1633
N p = 1.04 ! 1019 exp
                        0.0259
N p = 1.90 ! 1016

#Np = 1.90 ! 1016 " 7 ! 1015 = 1.2 ! 1016 / cm3

				
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