1. Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/
EE5342 – Semiconductor Device Modeling and Characterization Lecture 03-Spring 2010
Professor Ronald L. Carter

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© L01 Aug 25

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L4 January 27

4. Semiconductor Electronics - concepts thus far
Conduction and Valence states due to symmetry of lattice
“Free-elec.” dynamics near band edge
Band Gap
direct or indirect
effective mass in curvature
Thermal carrier generation
Chemical carrier gen (donors/accept)
L03 January 25

5. Counting carriers - quantum density of states function
1 dim electron wave #s range for n+1 “atoms” is 2p/L < k < 2p/a where a is “interatomic” distance and L = na is the length of the assembly (k = 2p/l)
Shorter ls, would “oversample”
if n increases by 1, dp is h/L
Extn 3D: E = p2/2m = h2k2/2m so a vol of p-space of 4pp2dp has h3/LxLyLz
L03 January 25

6. QM density of states (cont.)
So density of states, gc(E) is (Vol in p-sp)/(Vol per state*V) = 4pp2dp/[(h3/LxLyLz)*V]
Noting that p2 = 2mE, this becomes gc(E) = {4p(2mn*)3/2/h3}(E-Ec)1/2 and E - Ec = h2k2/2mn*
Similar for the hole states where Ev - E = h2k2/2mp*
L03 January 25

7. Fermi-Dirac distribution fctn
The probability of an electron having an energy, E, is given by the F-D distr fF(E) = {1+exp[(E-EF)/kT]}-1
Note: fF (EF) = 1/2
EF is the equilibrium energy of the system
The sum of the hole probability and the electron probability is 1
L03 January 25

8. Fermi-Dirac DF (continued)
So the probability of a hole having energy E is 1 - fF(E)
At T = 0 K, fF (E) becomes a step function and 0 probability of E > EF
At T >> 0 K, there is a finite probability of E >> EF
L03 January 25

9. Maxwell-Boltzman Approximation
fF(E) = {1+exp[(E-EF)/kT]}-1
For E - EF > 3 kT, the exp > 20, so within a 5% error, fF(E) ~ exp[-(E-EF)/kT]
This is the MB distribution function
MB used when E-EF>75 meV (T=300K)
For electrons when Ec - EF > 75 meV and for holes when EF - Ev > 75 meV
L03 January 25

10. Electron Conc. in the MB approx.
Assuming the MB approx., the equilibrium electron concentration is
L03 January 25

11. Electron and Hole Conc in MB approx
Similarly, the equilibrium hole concentration is po = Nv exp[-(EF-Ev)/kT]
So that nopo = NcNv exp[-Eg/kT]
ni2 = nopo, Nc,v = 2{2pm*n,pkT/h2}3/2
Nc = 2.8E19/cm3, Nv = 1.04E19/cm3 and ni = 1E10/cm3
L03 January 25

12. Calculating the equilibrium no
The ideal is to calculate the equilibrium electron concentration no for the FD distribution, where
fF(E) = {1+exp[(E-EF)/kT]}-1
gc(E) = [4p(2mn*)3/2(E-Ec)1/2]/h3
L03 January 25

13. Equilibrium con- centration for no
Earlier quoted the MB approximation no = Nc exp[-(Ec - EF)/kT],(=Nc exp hF)
The exact solution is no = 2NcF1/2(hF)/p1/2
Where F1/2(hF) is the Fermi integral of order 1/2, and hF = (EF - Ec)/kT
Error in no, e, is smaller than for the DF: e = 31%, 12%, 5% for -hF = 0, 1, 2
L03 January 25

14. Equilibrium con- centration for po
Earlier quoted the MB approximation po = Nv exp[-(EF - Ev)/kT],(=Nv exp h’F)
The exact solution is po = 2NvF1/2(h’F)/p1/2
Note: F1/2(0) = 0.678, (p1/2/2) = 0.886
Where F1/2(h’F) is the Fermi integral of order 1/2, and h’F = (Ev - EF)/kT
Errors are the same as for po
L03 January 25

15. Degenerate and nondegenerate cases
Bohr-like doping model assumes no interaction between dopant sites
If adjacent dopant atoms are within 2 Bohr radii, then orbits overlap
This happens when Nd ~ Nc (EF ~ Ec), or when Na ~ Nv (EF ~ Ev)
The degenerate semiconductor is defined by EF ~/> Ec or EF ~/< Ev
L03 January 25

16. Donor ionization
The density of elec trapped at donors is nd = Nd/{1+[exp((Ed-EF)/kT)/2]}
Similar to FD DF except for factor of 2 due to degeneracy (4 for holes)
Furthermore nd = Nd - Nd+, also
For a shallow donor, can have Ed-EF >> kT AND Ec-EF >> kT: Typically EF-Ed ~ 2kT
L03 January 25

17. Donor ionization (continued)
Further, if Ed - EF > 2kT, then nd ~ 2Nd exp[-(Ed-EF)/kT], e < 5%
If the above is true, Ec - EF > 4kT, so no ~ Nc exp[-(Ec-EF)/kT], e < 2%
Consequently the fraction of un-ionized donors is nd/no = 2Nd exp[(Ec-Ed)/kT]/Nc = 0.4% for Nd(P) = 1e16/cm3
L03 January 25

18. Classes of semiconductors
Intrinsic: no = po = ni, since Na&Nd << ni =[NcNvexp(Eg/kT)]1/2,(not easy to get)
n-type: no > po, since Nd > Na
p-type: no < po, since Nd < Na
Compensated: no=po=ni, w/ Na- = Nd+ > 0
Note: n-type and p-type are usually partially compensated since there are usually some opposite- type dopants
L4 January 27

19. Equilibrium concentrations
Charge neutrality requires q(po + Nd+) + (-q)(no + Na-) = 0
Assuming complete ionization, so Nd+ = Nd and Na- = Na
Gives two equations to be solved simultaneously
1. Mass action, no po = ni2, and
2. Neutrality po + Nd = no + Na
L4 January 27

20. Equilibrium conc n-type
For Nd > Na
Let N = Nd-Na, and (taking the + root) no = (N)/2 + {[N/2]2+ni2}1/2
For Nd+= Nd >> ni >> Na we have
no = Nd, and
po = ni2/Nd
L4 January 27

21. Equilibrium conc p-type
For Na > Nd
Let N = Nd-Na, and (taking the + root) po = (-N)/2 + {[-N/2]2+ni2}1/2
For Na-= Na >> ni >> Nd we have
po = Na, and
no = ni2/Na
L4 January 27

22. Position of the Fermi Level
Efi is the Fermi level when no = po
Ef shown is a Fermi level for no > po
Ef < Efi when no < po
Efi < (Ec + Ev)/2, which is the mid-band
L4 January 27

23. EF relative to Ec and Ev
Inverting no = Nc exp[-(Ec-EF)/kT] gives Ec - EF = kT ln(Nc/no) For n-type material: Ec - EF =kTln(Nc/Nd)=kTln[(NcPo)/ni2]
Inverting po = Nv exp[-(EF-Ev)/kT] gives EF - Ev = kT ln(Nv/po) For p-type material: EF - Ev = kT ln(Nv/Na)
L4 January 27

24. EF relative to Efi
Letting ni = no gives  Ef = Efi ni = Nc exp[-(Ec-Efi)/kT], so Ec - Efi = kT ln(Nc/ni). Thus EF - Efi = kT ln(no/ni) and for n-type EF - Efi = kT ln(Nd/ni)
Likewise Efi - EF = kT ln(po/ni) and for p-type Efi - EF = kT ln(Na/ni)
L4 January 27

25. Locating Efi in the bandgap
Since Ec - Efi = kT ln(Nc/ni), and Efi - Ev = kT ln(Nv/ni)
The sum of the two equations gives Efi = (Ec + Ev)/2 - (kT/2) ln(Nc/Nv)
Since Nc = 2.8E19cm-3 > 1.04E19cm-3 = Nv, the intrinsic Fermi level lies below the middle of the band gap
L4 January 27

26. Sample calculations
Efi = (Ec + Ev)/2 - (kT/2) ln(Nc/Nv), so at 300K, kT = meV and Nc/Nv = 2.8/1.04, Efi is 12.8 meV or 1.1% below mid-band
For Nd = 3E17cm-3, given that Ec - EF = kT ln(Nc/Nd), we have Ec - EF = meV ln(280/3), Ec - EF = eV =117meV ~3x(Ec - ED) what Nd gives Ec-EF =Ec/3
L4 January 27

27. Equilibrium electron conc. and energies
L4 January 27

28. Equilibrium hole conc. and energies
L4 January 27

29. vx = axt = (qEx/m*)t, and the displ
Carrier Mobility
In an electric field, Ex, the velocity (since ax = Fx/m* = qEx/m*) is
vx = axt = (qEx/m*)t, and the displ
x = (qEx/m*)t2/2
If every tcoll, a collision occurs which “resets” the velocity to <vx(tcoll)> = 0, then <vx> = qExtcoll/m* = mEx
L4 January 27

30. Carrier mobility (cont.)
The response function m is the mobility.
The mean time between collisions, tcoll, may has several important causal events: Thermal vibrations, donor- or acceptor-like traps and lattice imperfections to name a few.
Hence mthermal = qtthermal/m*, etc.
L4 January 27

31. Carrier mobility (cont.)
If the rate of a single contribution to the scattering is 1/ti, then the total scattering rate, 1/tcoll is
L4 January 27

32. Drift Current
The drift current density (amp/cm2) is given by the point form of Ohm Law
J = (nqmn+pqmp)(Exi+ Eyj+ Ezk), so
J = (sn + sp)E = sE, where
s = nqmn+pqmp defines the conductivity
The net current is
L4 January 27

33. Drift current resistance
Given: a semiconductor resistor with length, l, and cross-section, A. What is the resistance?
As stated previously, the conductivity, s = nqmn + pqmp
So the resistivity, r = 1/s = 1/(nqmn + pqmp)
L4 January 27

34. Drift current resistance (cont.)
Consequently, since
R = rl/A
R = (nqmn + pqmp)-1(l/A)
For n >> p, (an n-type extrinsic s/c)
R = l/(nqmnA)
For p >> n, (a p-type extrinsic s/c) R = l/(pqmpA)
L4 January 27

35. Drift current resistance (cont.)
Note: for an extrinsic semiconductor and multiple scattering mechanisms, since
R = l/(nqmnA) or l/(pqmpA), and
(mn or p total)-1 = S mi-1, then
Rtotal = S Ri (series Rs)
The individual scattering mechanisms are: Lattice, ionized impurity, etc.
L4 January 27

36. Exp. mobility model function for Si1
Parameter As P B
mmin
mmax
Nref e e e17 a
L4 January 27

37. Exp. mobility model for P, As and B in Si
L4 January 27

38. Carrier mobility functions (cont.)
The parameter mmax models 1/tlattice the thermal collision rate
The parameters mmin, Nref and a model 1/timpur the impurity collision rate
The function is approximately of the ideal theoretical form: 1/mtotal = 1/mthermal + 1/mimpurity
L4 January 27

39. Carrier mobility functions (ex.)
Let Nd = 1.78E17/cm3 of phosphorous, so mmin = 68.5, mmax = 1414, Nref = 9.20e16 and a = Thus mn = 586 cm2/V-s
Let Na = 5.62E17/cm3 of boron, so mmin = 44.9, mmax = 470.5, Nref = 9.68e16 and a = Thus mn = 189 cm2/V-s
L4 January 27

40. Lattice mobility
The mlattice is the lattice scattering mobility due to thermal vibrations
Simple theory gives mlattice ~ T-3/2
Experimentally mn,lattice ~ T-n where n = 2.42 for electrons and 2.2 for holes
Consequently, the model equation is mlattice(T) = mlattice(300)(T/300)-n
L4 January 27

41. Ionized impurity mobility function
The mimpur is the scattering mobility due to ionized impurities
Simple theory gives mimpur ~ T3/2/Nimpur
Consequently, the model equation is mimpur(T) = mimpur(300)(T/300)3/2
L4 January 27

42. Mobility Summary
The concept of mobility introduced as a response function to the electric field in establishing a drift current
Resistivity and conductivity defined
Model equation def for m(Nd,Na,T)
Resistivity models developed for extrinsic and compensated materials
L4 January 27

43. Net silicon (ex- trinsic) resistivity
Since r = s-1 = (nqmn + pqmp)-1
The net conductivity can be obtained by using the model equation for the mobilities as functions of doping concentrations.
The model function gives agreement with the measured s(Nimpur)
L4 January 27

44. Net silicon extr resistivity (cont.)
L4 January 27

45. Net silicon extr resistivity (cont.)
Since
r = (nqmn + pqmp)-1, and
mn > mp, (m = qt/m*) we have
rp > rn
Note that since
1.6(high conc.) < rp/rn < 3(low conc.), so
1.6(high conc.) < mn/mp < 3(low conc.)
L4 January 27

46. Net silicon (com- pensated) res.
For an n-type (n >> p) compensated semiconductor, r = (nqmn)-1
But now n = N = Nd - Na, and the mobility must be considered to be determined by the total ionized impurity scattering Nd + Na = NI
Consequently, a good estimate is r = (nqmn)-1 = [Nqmn(NI)]-1
L4 January 27

47. Equipartition theorem
The thermodynamic energy per degree of freedom is kT/2
Consequently,
L4 January 27

48. Carrier velocity saturation1
The mobility relationship v = mE is limited to “low” fields
v < vth = (3kT/m*)1/2 defines “low”
v = moE[1+(E/Ec)b]-1/b, mo = v1/Ec for Si
parameter electrons holes
v1 (cm/s) E9 T E8 T-0.52
Ec (V/cm) T T1.68
b E-2 T T0.17
L4 January 27

49. vdrift [cm/s] vs. E [V/cm] (Sze2, fig. 29a)
L4 January 27

50. Carrier velocity saturation (cont.)
At 300K, for electrons, mo = v1/Ec = 1.53E9(300)-0.87/1.01(300) = 1504 cm2/V-s, the low-field mobility
The maximum velocity (300K) is vsat = moEc = v1 = 1.53E9 (300) = 1.07E7 cm/s
L4 January 27

51. Diffusion of carriers
In a gradient of electrons or holes, p and n are not zero
Diffusion current,`J =`Jp +`Jn (note Dp and Dn are diffusion coefficients)
L4 January 27

52. Diffusion of carriers (cont.)
Note (p)x has the magnitude of dp/dx and points in the direction of increasing p (uphill)
The diffusion current points in the direction of decreasing p or n (downhill) and hence the - sign in the definition of`Jp and the + sign in the definition of`Jn
L4 January 27

53. Diffusion of Carriers (cont.)
L4 January 27

54. Current density components
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55. Total current density
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56. Doping gradient induced E-field
If N = Nd-Na = N(x), then so is Ef-Efi
Define f = (Ef-Efi)/q = (kT/q)ln(no/ni)
For equilibrium, Efi = constant, but
for dN/dx not equal to zero,
Ex = -df/dx =- [d(Ef-Efi)/dx](kT/q) = -(kT/q) d[ln(no/ni)]/dx = -(kT/q) (1/no)[dno/dx] = -(kT/q) (1/N)[dN/dx], N > 0
L4 January 27

57. Induced E-field (continued)
Let Vt = kT/q, then since
nopo = ni2 gives no/ni = ni/po
Ex = - Vt d[ln(no/ni)]/dx = - Vt d[ln(ni/po)]/dx = - Vt d[ln(ni/|N|)]/dx, N = -Na < 0
Ex = - Vt (-1/po)dpo/dx = Vt(1/po)dpo/dx = Vt(1/Na)dNa/dx
L4 January 27

58. The Einstein relationship
For Ex = - Vt (1/no)dno/dx, and
Jn,x = nqmnEx + qDn(dn/dx) = 0
This requires that nqmn[Vt (1/n)dn/dx] = qDn(dn/dx)
Which is satisfied if
L4 January 27

59. References
*Fundamentals of Semiconductor Theory and Device Physics, by Shyh Wang, Prentice Hall, 1989.
**Semiconductor Physics & Devices, by Donald A. Neamen, 2nd ed., Irwin, Chicago.
M&K = Device Electronics for Integrated Circuits, 3rd ed., by Richard S. Muller, Theodore I. Kamins, and Mansun Chan, John Wiley and Sons, New York, 2003.
1Device Electronics for Integrated Circuits, 2 ed., by Muller and Kamins, Wiley, New York, 1986.
2Physics of Semiconductor Devices, by S. M. Sze, Wiley, New York, 1981.
L3 January 25