1. EE 5340 Semiconductor Device Theory Lecture 5 - Fall 2010
Professor Ronald L. Carter

2. Second Assignment
Please print and bring to class a signed copy of the document appearing at
L05 08Sep10

3. Intrinsic carrier conc. (MB limit)
ni2 = no po = Nc Nv e-Eg/kT
Nc = 2{2pm*nkT/h2}3/2
Nv = 2{2pm*pkT/h2}3/2
Eg = 1.17 eV - aT2/(T+b)
a = 4.73E-4 eV/K
b = 636K
L05 08Sep10

4. Classes of semiconductors
Intrinsic: no = po = ni, since Na&Nd << ni, ni2 = NcNve-Eg/kT, ~1E-13 dopant level !
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
L05 08Sep10

5. n-type equilibrium concentrations
N ≡ Nd - Na , n type  N > 0
For all N, no = N/2 + {[N/2]2+ni2}1/2
In most cases, N >> ni, so
no = N, and
po = ni2/no = ni2/N, (Law of Mass Action is al- ways true in equilibrium)
L05 08Sep10

6. p-type equilibrium concentrations
N ≡ Nd - Na , p type  N < 0
For all N, po = |N|/2 + {[|N|/2]2+ni2}1/2
In most cases, |N| >> ni, so
po = |N|, and
no = ni2/po = ni2/|N|, (Law of Mass Action is al- ways true in equilibrium)
L05 08Sep10

7. 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
L05 08Sep10

8. 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)
L05 08Sep10

9. 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)
L05 08Sep10

10. Locating Efi in the bandgap
Since Ec - Efi = kT ln(Nc/ni), and Efi - Ev = kT ln(Nv/ni)
The 1st equation minus the 2nd 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
L05 08Sep10

11. 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
L05 08Sep10

12. Equilibrium electron conc. and energies
L05 08Sep10

13. Equilibrium hole conc. and energies
L05 08Sep10

14. 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
L05 08Sep10

15. 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.
L05 08Sep10

16. Carrier mobility (cont.)
If the rate of a single contribution to the scattering is 1/ti, then the total scattering rate, 1/tcoll is
L05 08Sep10

17. Figure (p. 31 M&K) Electron and hole mobilities in silicon at 300 K as functions of the total dopant concentration. The values plotted are the results of curve fitting measurements from several sources. The mobility curves can be generated using Equation with the following values of the parameters [3] (see table on next slide).
L05 08Sep10

18. Parameter Arsenic Phosphorus Boron
μmin
52.2
68.5
44.9
μmax
1417
1414
470.5
Nref
9.68 X 1016
9.20 X 1016
2.23 X 1017 α
0.680
0.711
0.719
Figure (cont. M&K)
L05 08Sep10

19. 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
L05 08Sep10

20. 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)
L05 08Sep10

21. 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)
L05 08Sep10

22. 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.
L05 08Sep10

23. Net intrinsic mobility
Considering only lattice scattering
L05 08Sep10

24. 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
L05 08Sep10

25. Net extrinsic mobility
Considering only lattice and impurity scattering
L05 08Sep10

26. 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.)
L05 08Sep10

27. 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
L05 08Sep10

28. Figure (p. 32 in M&K1) Low-field mobility in silicon as a function of temperature for electrons (a), and for holes (b). The solid lines represent the theoretical predictions for pure lattice scattering [5].
L05 08Sep10

29. Exp. m(T=300K) model for P, As and B in Si1
L05 08Sep10

30. Exp. mobility model function for Si1
Parameter As P B
mmin
mmax
Nref e e e17 a
L05 08Sep10

31. 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
L05 08Sep10

32. 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 mp = 189 cm2/V-s
L05 08Sep10

33. 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)
L05 08Sep10

34. Figure (p. 29) M&K Dopant density versus resistivity at 23°C (296 K) for silicon doped with phosphorus and with boron. The curves can be used with little error to represent conditions at 300 K. [W. R. Thurber, R. L. Mattis, and Y. M. Liu, National Bureau of Standards Special Publication 400–64, 42 (May 1981).]
L05 08Sep10

35. Net silicon extr resistivity (cont.)
Since
r = (nqmn + pqmp)-1, and
mn > mp, (m = qt/m*) we have
rp > rn, for the same NI
Note that since
1.6(high conc.) < rp/rn < 3(low conc.), so
1.6(high conc.) < mn/mp < 3(low conc.)
L05 08Sep10

36. 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
L05 08Sep10

37. Figure (p. 31 M&K) Electron and hole mobilities in silicon at 300 K as functions of the total dopant concentration. The values plotted are the results of curve fitting measurements from several sources. The mobility curves can be generated using Equation with the following values of the parameters [3] (see table on next slide).
L05 08Sep10

38. 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
L05 08Sep10

39. Equipartition theorem
The thermodynamic energy per degree of freedom is kT/2
Consequently,
L05 08Sep10

40. 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
L05 08Sep10

41. Carrier velocity2 carrier velocity vs E for Si, Ge, and GaAs (after
Sze2)
L05 08Sep10

42. 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
L05 08Sep10

43. 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)
L05 08Sep10

44. 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
L05 08Sep10

45. Diffusion of Carriers (cont.)
L05 08Sep10

46. Current density components
L05 08Sep10

47. Total current density
L05 08Sep10

48. 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
L05 08Sep10

49. 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
L05 08Sep10

50. 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
L05 08Sep10

51. 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.
L05 08Sep10

52. References
M&K and 1Device Electronics for Integrated Circuits, 2 ed., by Muller and Kamins, Wiley, New York, 1986.
See Semiconductor Device Fundamen-tals, by Pierret, Addison-Wesley, 1996, for another treatment of the m model.
2Physics of Semiconductor Devices, by S. M. Sze, Wiley, New York, 1981.
L05 08Sep10