Category Archives: Principles of Electronic Materials and Devices
Constants and Useful Information
Physical Constants 

Atomic mass unit 
Amu 
1.66054 x 10~27 kg 
Avogadro’s number 
Na 
6.02214 x 1023 mol1 
Bohr magneton 
Я 
9.2740 x 1024JT"’ 
Boltzmann constant 
K 
1.3807 x 10“23 J K"1 = 8.6174 x 105 eV K 
Electron mass in free space 
Me 
9.10939 x 1031 kg 
Electron charge 
E 
1.60218 x 10~19C 
Gas constant 
R 
8.3145 J K1 mol“1 or m3 Pa K~’ mol“1 
Gravitational constant 
G 
6.6742 x 10“" N m2 kg~2 
Permeability of vacuum or 
Яo 
An x 107 H m_1 (or Wb A1 m1) 
Absolute permeability 

Permittivity of vacuum or 
E0 
8.8542 x 1012 F m_1 
Absolute permittivity 

Planck’s constant 
H 
6.626 x 1034 J s = 4.136 x 1015 eV s 
Planck’s constant/2n 
Ft 
1.055 x 1034 J s = 6.582 x 1016 eV s 
Proton rest mass 
Mp 
1.67262 x 10“27 kg 
Rydberg constant 
Roc 
1.0974 x 107 m1 
Speed of light 
C 
2.9979 x 108 m s_1 
Stefan’s constant 
Os 
5.6704 x 10“8 W m“2 K4 
Useful Information 

Acceleration due to gravity at 
8 
9.81 m s2 
45° latitude 

KT at 7 = 293 K (20 °C) 
KT 
0.02525 eV 
KT at r = 300 K (27 °C) 
KT 
0.02585 eV 
Bohr radius 
Qo 
0.0529 nm 
1 angstrom 
В 
10‘°m 
1 micron 
xm 
106m 
EV= 1.6022 x 10"19J
KJ mol1 = 0.010364 eV atom
Atmosphere (pressure)
= 1.013 x 105 Pa
The table gives the wavelength ranges and colors as usually specified for LEDs.
Color 
Emerald Blue green 
Green 
Yellow 
Amber 
Orange 
Red Orange 
Red 
Deep red 
Infrared 
А (nm) 
А < 500 530564 
565579 
580587 
588594 
595606 
607615 
616632 
633700 
K > 700 
The table gives the typical wavelength ranges and color perception by an average person.
Color Violet Blue Green Yellow Orange Red
K (nm) 390455 455492 492577 577597 597622 622780
J = (i)*’2 j* = – i exp(jO) = cos 6 + j sin 0
Z = a + jb = reje r = (a2 + b2)l/2 tan 0 = –
A
(:) 
Z* = a — jb = re~je Re(Z) = a lm(Z) = b
Magnitude2 == IZ 2 = Z Z* = a2 + b2 Argument = 0 = arctan
Cos 0 = (eje + e~je) sin 0 = — e~jB)
X 1 1 2 1 3
Ex = 1 + * + —X + — л: + .. •
_ n(n — 1) 7 n(n — 1 )(n — 2) .
(1 + *)" = 1 + nx + У V + – Ј V + ■■■
Small x: (1 + x)n « 1 + nx sinx&x tan jc ~ jc cosjc^I
Small Ax in x = x0 + A*: f(x) « /(*<,) + Д* ( J
dx / x0
"We have a habit in writing articles published in scientific journals to make the work as finished as possible, to cover up all the tracks, to not worry about the blind alleys or describe how you had the wrong idea first, and so on. So there isn’t any place to publish, in a dignified manner, what you actually did in order to get to do the work."
Richard P. Feynman Nobel Lecture, 1966
I 5 Remember that the change dE in the PE is the work done against the force, dE = Fn dr.
I 7 The mathematics and a more rigorous description may be found in the textbook’s CD.
SPEED OF SOUND IN AIR Calculate the root mean square (rms) velocity of nitrogen molecules in atmospheric air at 27 °C. Also calculate the root mean square velocity in one direction (i>rms,*) Compare the speed of propagation of sound waves in air, 350 ms1, with Urms. jc and explain the difference.
SOLUTION
From the kinetic theory
VOLUME EXPANSION COEFFICIENT Suppose that the volume of a solid body at temperature
T0 is V0. The volume expansion coefficient av of a solid body characterizes the change in its
Volume from VQ to V due to a temperature change from TQ to T by
V = V0[l+av(T – T0)] [1.20]
Show that av is given by
Cev = 3X [1.21]
Aluminum has a density of 2.70 g cm3 at 25 °C. Its thermal expansion coefficient is 24 x 10"6 °C_1. Calculate the density of A1 at 350 °C.
SOLUTION
Consider the solid body in the form of a rectangular parallelepiped with sides x0, y0, and z0• Then at T0,
Vo = X0y0Zo
And at T, V = [x0(l + X AT)][y0(l + X AT)][za( + X AT)]
= x0y0Zo( 1 + a AT)3
That is V = x0y0z0[l + 3A AT + 3X2(AT)2 + X3(AT)3]
We can now substitute for V from Equation 1.20, use V0 = x0y0z0, and neglect the
X2(AT)2 and X3(AT)3 terms compared with the X AT term (X 1) to obtain,
V = v0[l + 3X(T – To)] = Voll + av(T – To)]
Since density p is mass/volume, volume expansion leads to a density reduction. Thus,
P = 1 + rr r. * S’*1 ~a^T~ To>]
1 +av{T – T0)
For Al, the density at 350 °C is
P = 2.70[1 – 3(24 x 106)(350 – 25)] = 2.637 g cm ‘3
MEAN AND RMS SPEEDS OF MOLECULES Given the MaxwellBoltzmann distribution law for the velocities of molecules in a gas, derive expressions for the mean speed (vav)> most probable speed (u*), and rms velocity (v^) of the molecules and calculate the corresponding values for a gas of noninteracting electrons.
SOLUTION
The number of molecules with speeds in the range v to (ь + dv) is
3/2 / 2
/ m N /
DN = nvdv = 4tt NI
2tt kT
By definition, then, the mean speed is given by
/ vdN f vnv dv 18kT
• = TZjat = "7 7“ = V Mean sPeed
J dN J nv dv V Km
Where the integration is over all speeds (v = 0 to oo). The mean square velocity is given by
— f v1 dN f v2nv dv 3kT
I 8 A lowpass filter allows all signal frequencies up to the cutoff frequency 8 to pass. 8 is ]/[2nRC).
12 It is possible to define a unit cell on the surface of a carbon nanotube and apply various crystalline concepts, as some scientists have done. To date, however, there seems to be no single crystal of carbon nanotubes in the same way that there is a fullerene crystal in which the C60 molecules are bonded to form an FCC structure.
The mean KE has half the magnitude of the mean PE if the only forces acting on the particles are such that they
Follow an inverse square law dependence on the particleparticle separation (as in Coulombic and gravitational
Forces).
DRIFT VELOCITY IN A FIELD: A CLOSER LOOK There is another way to explain the observed dependence of the drift velocity on the field, and Equation 2.3. Consider the path of a conduction electron in an applied field Ј as shown in Figure 2.4. Suppose that at time t = 0 the electron has just been scattered from a lattice vibration. Let uxX be the initial velocity in the x direction just after this initial collision (to which we assign a collision number of zero). We will assume that immediately after a collision, the velocity of the electron is in a random direction. Suppose that the first collision occurs at time t. Since is the acceleration, the dis
Tance s covered in the x direction during the free time t will be
RESISTIVITY DUE TO IMPURITIES The mean speed of conduction electrons in copper is about 1.5 x 106 m s_1. Its room temperature resistivity is 17 nЈ2 m, and the atomic concentration Nat in the crystal is 8.5 x 1022 cm3. Suppose that we add 1 at.% Au to form a solid solution. What is the resistivity of the alloy, the effective mean free path, and the mean free path due to collisions with Au atoms only?
SOLUTION
According to Table 2.3, the Nordheim coefficient C of Au in Cu is 5500 nQ m. With X = 0.01 (1 at.%), the overall resistivity from Equation 2.22 is
P = Pmatrix + CX(1 – X) = 17 nQ m + (5500 nft m)(0.01)(l – 0.01)
= 17 nЈ2 m + 54.45 nЈl m = 71.45 nЈ2 m
Suppose that i is the overall or effective mean free path and r is the effective mean free time between scattering events (includes both scattering from lattice vibrations and impurities). Since
I = mt, and the effective drift mobility = ex/me, the expression for the conductivity becomes
E2nr e2ni
A = en/id = =
Me meu
We can now calculate the effective mean free path i in the alloy given that copper has a valency of I and the electron concentration n = Nat,
1 (1.6 x 10_19C)2(8.5 x 1028 m"3K
71.5 x 10~9 Q m ” (9.1 x 10~31kg)(1.5 x 106 m s"1)
Which gives i = 8.8 nm. We can repeat the calculation for pure copper using o = 1/Anatrix = 1/(17 x 109 Ј2 m) to find ЈCu = 37 nm. The mean free path is reduced approximately by 4 times by adding only 1 at.% Au. The mean free path ii due to scattering from im –
7 Recall that J=a‘E which is equivalent to Equation 2.35.
I 9 The Q value refers to the quality factor of an inductor, which is defined by Q = a>0L/R, where oj0 is the resonant
I 13 The mixture rules mentioned in this chapter turn up again in a different but recognizable form for predicting the
I overall relative permittivity of porous dielectrics.
Electrical and thermal conductivity of In Electron drift mobility in indium has been measured to be 6 cm2 V1 s1. The room temperature (27° C) resistivity of In is 8.37 x 10“8 Q m, and its atomic mass and density are 114.82 amu or g mol“1 and 7.31 g cm3, respectively.
Based on the resistivity value, determine how many free electrons are donated by each In atom in the crystal. How does this compare with the position of In in the Periodic Table (Group IIIB)?
If the mean speed of conduction electrons in In is 1.74 x 108 cm s"1, what is the mean free path?
Calculate the thermal conductivity of In. How does this compare with the experimental value of 81.6 Wm"1 K"1?
Electrical and thermal conductivity of Ag The electron drift mobility in silver has been measured to
Be 56 cm2 V1 s1 at 27 °C. The atomic mass and density of Ag are given as 107.87 amu or g mol1 and 10.50 g cm3, respectively.
Assuming that each Ag atom contributes one conduction electron, calculate the resistivity of Ag at 27 °C. Compare this value with the measured value of 1.6 x 108 Ј2m at the same temperature and suggest reasons for the difference.
Calculate the thermal conductivity of silver at 27 °C and at 0 °C.
NOTE: Compositions are in wt.%. Ag10% Ni means 90% Ag10% Ni.
Vickers hardness number (VHN) is a measure of the hardness or strength of the alloy, and d is density.
AgW alloys (contact materials) and the mixture rule Silvertungsten alloys are frequently used in heavyduty switching applications (e. g., currentcarrying contacts and oil circuit breakers) and in arcing tips. AgW is a twophase alloy, a mixture of Agrich and Wrich phases. The measured resistivity and density for various AgW compositions are summarized in Table 2.12.
Plot the resistivity and density of the AgW alloy against the W content (wt.%)
Show that the density of the mixture, d, is given by
1 = wad~l + wpdj1
Where wa is the weight fraction of phase a, wp is the weight fraction of phase f$, da is the density of phase a, and dp is the density of phase
Thermal conduction Consider brass alloys with an X atomic fraction of Zn. Since Zn addition increases the number of conduction electrons, we have to scale the final alloy resistivity calculated from the simple MatthiessenNordheim rule in Equation 2.22 down by a factor (1 + X) (see Question 2.8) so that the resistivity of the alloy is p « [p0 + CX(l — X)]/(l + X) in which C = 300 nЈ2 m and p0 = PCu = 17 nQ m.
An 80 at.% Cu20 at.% Zn brass disk of 40 mm diameter and 5 mm thickness is used to conduct heat from a heat source to a heat sink.
NOTE: Film annealed at 150 °C.
SOURCE: Data extracted from J. W. Lim etal., Appl. Surf. Sci. 217, 95, 2003.
Interconnects Consider a hightransistordensity CMOS chip in which the interconnects are copper with a pitch P of 500 nm, interconnect thickness T of 400 nm, aspect ratio 1.4, and H = X. The dielectric is FSG with er = 3.6. Consider two cases, L = 1mm and L = 10 mm, and calculate the overall effective interconnect capacitance Ceff and the RC delay time. Suppose that Al, which is normally Al with about 4 wt.% Cu in the microelectronics industry with a resistivity 31 nЈ2 m, is used as the interconnect. What is the corresponding /?Cdelay time?
ENERGY OF A BLUE PHOTON What is the energy of a blue photon that has a wavelength of
450 nm?
SOLUTION
The energy of the photon is given by
„ , he (6.6 x 1034Js)(3 x 108ms‘) , ,
Eph = hv = — = ———— = 4.4 x 10 J
K 450 x 10 9m
Generally, with such small energy values, we prefer electronvolts (eV), so the energy of the photon is
4.4 x 1019 J „
= 2.75 eV
X 1019 J/eV
Net rate of radiative power emission per unit surface = a ST4 — osT*
Further, not all surfaces are perfect black bodies. Black body emission is the maximum possible emission from a surface at a given temperature. A real surface emits less than a black body. Emissivity e of a surface measures the efficiency of a surface in terms of a black body emitter; it is the ratio of the emitted radiation from a real surface to that emitted from a black body at a given temperature and over the same wavelength range. The total net rate of radiative power emission becomes
S = тг(63.5 X 1(T6 m)(0.579 m) = 1.155 x 104 m2
The emissivity e of tungsten is about 0.35. Assuming that under steadystate operation all the electric power that is input into the bulb’s filament is radiated away,
100 W = Pradiation = Ssos(t; – Г4)
= (1.155 x 10~4 m2)(0.35)(5.67 x 10~8 W m2 K_4)(7’/4 – 3004)
Solving we find,
Tf = 2570 К or 2297 °C
Which is well below the melting temperature of tungsten which is 3422 °C. The second term that has Го4 has very little effect on the calculation as radiation absorption from the environment is practically nil compared with the emitted radiation at Tf.
The shift in the spectral intensity emitted from a black body with temperature is of particular interest to many photoinstrumention engineers. The peak spectral intensity in Figure 3.11 occurs at a wavelength Amax, which, by virtue of Equation 3.9, depends on the temperature of
Was for the scanning tunneling microscope by Gerd Binning and Heinrich Rohrer. The second was awarded to Georg
Bednorz and Alex Mьller for the discovery of hightemperature superconductors which we will examine in Chapter 8.
1 is
3 3s 3p 3 d
4 4s 4p Ad 4/
5 5s 5p 5 d 5/ 5 g
I 10 This qualitative explanation is discussed in Chapter 8.
Ef = 5.5 eV, so from Equation 4.10, the density of states at EF is g(EF) = 1.60 x 1028 m’3
EV1. The velocity of Fermi electrons, vF = (2EF/me)l/2 = 1.39 x 106 m s1. The conductivity a of Ag at room temperature is 62.5 x 106 ft1 m“1. Substituting for cr, g(EF), and vF in Equation 4.27,
, 1 , , 1 1Q * * , / 1.60 x 1028
A = 62.5 x 106 = – e2v2Fzg{EP) = (1.6 x 1(T19)2(1.39 x 106)2r — j
Lattice waves and heat capacity
Consider an aluminum sample. The nearest separation 2R (2 x atomic radius) between the AlAl atoms in the crystal is 0.286 nm. Taking a to be 2R, and given the sound velocity in A1 as 5100 m s“1, calculate the force constant p in Equation 4.66. Use the group velocity vg from the actual dispersion relation, Equation 4.55, to calculate the “sound velocity” at wavelengths of A = 1mm, 1 fxm, and 1 nm. What is your conclusion?
Aluminum has a Debye temperature of 394 K. Calculate its specific heat at 30 °C (Darwin, Australia) and at — 30 °C (January, Resolute Nunavut, Canada).
Calculate the specific heat capacity of a germanium crystal at 25 °C and compare it with the experimental value in Table 2.5.
Specific heat capacity of GaAs and InSb
The Debye temperature To of GaAs is 344 K. Calculate its specific heat capacity at 300 K and at
30 °C.
For InSb, Td = 203 K. Calculate the room temperature specific heat capacity of InSb and compare it with the value expected from the DulongPetit rule (T > To).
Thermal conductivity
Given that silicon has a Young’s modulus of about 110 GPa and a density of 2.3 g cm3, calculate the mean free path of phonons in Si at room temperature.
Diamond has the same crystal structure as Si but has a very large thermal conductivity, about
1000 W m1 K1 at room temperature. Given that diamond has a specific heat capacity cs of
0.50 J K1 g_1, Young’s modulus Y of 830 GPa, and density p of 0.35 g cm3, calculate the mean free path of phonons in diamond.
GaAs has a thermal conductivity of 200 W m1 K1 at 100 K and 80 W m1 K1 at 200 K. Calculate its thermal conductivity at 25 °C and compare with the experimental value of 44 W m1
K1. (Hint: Take k <x T~n in the temperature region of interest; see Figure 4.48.)
RESISTIVITY OF INTRINSIC AND DOPED Si Find the resistance of a 1 cm3 pure silicon crystal. What is the resistance when the crystal is doped with arsenic if the doping is 1 in 109, that is,
1 part per billion (ppb) (note that this doping corresponds to one foreigner living in China)? Given data: Atomic concentration in silicon is 5 x 1022 cm"3, nt — 1.0 x 1010 cm“3, tie = 1350 cm2 V"1 s1, and ixh — 450 cm2 V“1 s“1.
COMPENSATION DOPING An ntype Si semiconductor containing 1016 phosphorus (donor) atoms cm3 has been doped with 1017 boron (acceptor) atoms cm3. Calculate the electron and hole concentrations in this semiconductor.
SOLUTION
This semiconductor has been compensation doped with excess acceptors over donors, so
A! aNa = 1017 – IQ16 = 9 x 1016 cm3
2 x 103
In the dark: np = nf*2 (b) In light: np * n?
S (excess Sb) that will result in GaSb having a conductivity of 100 Ј21 cm1. Will this be an n
Doping.
Tions. Assume that the minority carrier recombination time zt, is constant (independent of injection—
= [5.87]
Dt xh
Nside, that is, Na Nd. Since the amount of charge Q on both sides of the metallurgical junc
Tion must be the same (so that the junction is overall neutral)
It is clear that the depletion region essentially extends into the nside. According to Equation 6.7,
When Nd < Na, the width is
[2eV0V/2
What is the depletion width for a/w junction Si diode that has been doped with 1018 acceptor
Atoms cm3 on the pside and 1016 donor atoms cm3 on the wside?
SOLUTION
To apply the above equation for WQ, we need the builtin potential, which is
K,(") JViЈ) , ,0.0259 V, „[112^11 = 0.835 V
For W0
2eV0~l/2 I”2(11.9)(8.85 x 10~12)(0.835)T1/2
~ [ (1.6 x 1019)(1022) J
= 3.32 x 107m or 0.33 im
Nearly all of this region (99 percent of it) is on the nside.
AVALANCHE BREAKDOWN Consider a uniformly doped abrupt p+n junction (Na » Nd) reverse biased by V = — Vr.
What is the relationship between the depletion width W and the potential difference (VQ + Vr) across W1
If avalanche breakdown occurs when the maximum field in the depletion region reaches the breakdown field Јbr> show that the breakdown voltage Vbr (» VQ) is then given by
ST2
Vbr =
2eNd
An abrupt Si p+n junction has boron doping of 1019 cm“3 on the /?side and phosphorus doping of 1016 cm3 on the «side. The dependence of the avalanche breakdown field on the impurity concentration is shown in Figure 6.19.
What is the reverse breakdown voltage of this Si diode?
Calculate the reverse breakdown voltage when the phosphorus doping is increased to
1 rv17
I 7 The actual concentration profile can be calculated by solving the steadystate continuity equation, which can be
A pnp TRANSISTOR Consider a pnp Si BJT that has the following properties. The emitter region mean acceptor doping is 2 x 1018 cm3, the base region mean donor doping is
1 x 1016 cm3, and the collector region mean acceptor doping is 1 x 1016 cm3. The hole drift mobility in the base is 400 cm2 V1 s“1, and the electron drift mobility in the emitter is 200 cm2 V1 s“1. The transistor emitter and base neutral region widths are about 2 fom each when the transistor is under normal operating conditions, that is, when the EB junction is forwardbiased and the BC junction is reversebiased. The effective crosssectional area of the device is 0.02 mm2. The hole lifetime in the base is approximately 400 ns. Assume that the emitter has 100 percent injection efficiency, y — 1. Calculate the CB current transfer ratio a and the current gain p. What is the emitterbase voltage if the emitter current is 1 mA?
SOLUTION
The hole drift mobility fih = 400 cm2 V~1 s1 (minority carriers in the base). From the Einstein relationship we can easily find the diffusion coefficient of holes,
Dh = nh = (0.0259 V)(400 cm2 V1 s“1) = 10.36
The minority carrier transit time xt across the base is
^ Total emitter current
Is given by
1
=
Na We /X/, (base)
How would you modify the CB current gain a to include the emitter injection efficiency?
Calculate the emitter injection efficiency for the pnp transistor in Example 6.8, which has an acceptor doping of 2 x 1018 cm3 in the emitter, donor doping of 1 x 1016 cm3 in the
I 10 Surface concentration of atoms (atoms per unit area) can be found from nsurf % (nbuik)^3
VV, C = 8 V
05 – Vth = 4 V
THE ENHANCEMENT NMOSFET A particular enhancement NMOS transistor has a gate with a width (Z) of 50 /im, length (L) of 10 /Ltm, and Si02 thickness of 450 A. The relative permittivity of Si02 is 3.9. The ptype bulk is doped with 1016 acceptors cm3. Its threshold voltage is
I 1 It may be noticed that even if Fr is a complicated function of x, it can still be expanded in a series in terms of
I 3 Electronic polarization at optical frequencies controls the optical properties such as the refractive index, as will be
I covered in Chapter 9.
ELECTRONIC POLARIZABILITY OF A VAN DER WAALS SOLID The electronic polarizability of the Ar atom is 1.7 x 1040 F m2. What is the static dielectric constant of solid Ar (below 84 K) if its density is 1.8 g cm3?
SOLUTION
To calculate er we need the number of Ar atoms per unit volume N from the density d. If Mat = 39.95 is the relative atomic mass of Ar and N& is Avogadro’s number, then
N = ^ = (6.02 x 1023 moP’Xl. Sgcm"3) = ^ ^ ^ ^
A/« (39.95 g mol"1)
With N = 2.71 x 1028 m“3 and a, = 1.7 x 1040 F m2, we have
, , Nae, , (2.71 x 1028)(1.7 x KT40) ,
Sr = l + ~7T = l+ lh. s~x~o~) = 152
If we use the ClausiusMossotti equation, we get
2Note
Јr = 7T^r = 1,63
3 eQ
The two values are different by about 7 percent. The simple relationship in Equation 7.14 underestimates the relative permittivity.
I 5 The oppositely directed forces also slightly stretch the Cl – H+ bond, but we neglect this effect.
IONIC AND ELECTRONIC POLARIZABILITY Consider the CsCl crystal which has one Cs+Cl" pair per unit cell and a lattice parameter a of 0.412 nm. The electronic polarizability of Cs+ and Cl“ ions is 3.35 x 1040 F m2 and 3.40 x 10“40 F m2, respectively, and the mean ionic polarizability per ion pair is 6 x 1040 F m2. What is the dielectric constant at low frequencies and that at optical frequencies?
SOLUTION
The CsCl structure has one cation (Cs+) and one anion (Cl) in the unit cell. Given the lattice parameter a = 0.412 x 109 m, the number of ion pairs Ni per unit volume is 1/a3 = 1 /(0.412 x 10~9m)3 = 1.43 x 1028 m3. Ni is also the concentration of cations and anions individually. From the ClausiusMossotti equation,
I 6 The polarization P lags behind •Ј by some angle <f>, that is determined by Equation 7.26 as shown in Figure 7.13.
DIELECTRIC LOSS PER UNIT CAPACITANCE AND THE LOSS ANGLE 8 Obtain the dielectric loss per unit capacitance in a capacitor in terms of the loss tangent. Obtain the phase difference between the current through the capacitor and that through Rp. What is the significance of 81
SOLUTION
We consider the equivalent circuit in Figure 7.14. The power loss in the capacitor is due to Rp. If V is the rms value of the voltage across the capacitor, then the power dissipated per unit capacitance Wcap is
V2 1 – j 0)6o6 Ad ry u/o
Wcap = — x – = V ——— x = V2—r
P RP C d e0efrA e’
Cos’!
Or
Int/nltiac /~lr> aI A/’+ri’Mt knnrinn /il imninnl fmm n Mn3+ fn Mn4+ inn – wp Hn nnt neeH the nhvsiml details
THE QUARTZ CRYSTAL AND ITS EQUIVALENT CIRCUIT From the following equivalent definition of the coupling coefficient,
2 Mechanical energy stored Total energy stored
Show that
F2
J a
Given that typically for an Xcut quartz crystal, k = 0.1, what is fa for fs = 1 MHz? What is your conclusion?
SOLUTION
C represents the mechanical mass where the mechanical energy is stored, whereas CQ is where the electrical energy is stored. If V is the applied voltage, then
.2 Mechanical energy stored C ^ f2
Total energy stored ±CV2 + C0V2 C + C0 f2
Rearranging this equation, we find
Fs 1 MHz aac x/ru
Fa = ■ = y — = 1.005 MHz
Vl – k2 y/ – (O. lj2
Thus, fa — fs is only 5 kHz. The two frequencies fs and fa in Figure 7.43d are very close. An oscillator designed to oscillate at fs, that is, at 1 MHz, therefore, cannot drift far (for example, a few kHz) because that would change the reactance enormously, which would upset the oscillation conditions.
I 13 In analogy with the ferromagnetic crystals that already possess magnetization.
Medium theories (or approximations). The theory of finding an effective dielectric constant of a mixture has intrigued
Many famous scientists in the past. Over the years, many quite complicated mixture rules have been developed, and there is no shortage of formulas in this field. Many engineers however still tend to use simple empirical rules to model a composite dielectric. The primary reason is that many theoretical mixture rules depend on the exact knowledge of
*7.21 Dielectric breakdown in a coaxial cable Consider a coaxial underwater highvoltage cable as in Figure 7.64a. The current flowing through the inner conductor generates heat, which has to flow through the dielectric insulation to the outer conductor where it will be carried away by conduction and convection. We will assume that steady state has been reached and the inner conductor is carrying a dc current /. Heat generated per unit second Qf = dQ/dt by Joule heating of the inner conductor is
Q! = RI2 = ^ [7.97]
Nal
Where p is the resistivity, a the radius of the conductor, and L the cable length.
This heat flows radially out from the inner conductor through the dielectric insulator to the outer conductor, then to the ambient. This heat flow is by thermal conduction through the dielectric. The rate of heat flow Q’ depends on the temperature difference 7} — Ta between the inner and outer conductors;
Permeabilities will be denoted by fiQ and /zr.
AN INDUCTOR WITH A FERRITE CORE Consider a toroidal coil with a ferrite core. Suppose that the coil has 200 turns and is used in HF work with small signals. The mean diameter of the toroid is 2.5 cm and the core diameter is 0.5 cm. If the core is a MnZn ferrite, what is the approximate inductance of the coil?
SOLUTION
The inductance L of a toroidal coil is given by
I _ ^ri^oN2A
*8.16 Permanent magnet with yoke and air gap Consider a permanent magnet bar that has Lshaped ferromagnetic (high permeability) pieces attached to its ends to direct the magnetic field to an air gap as depicted in Figure 8.72. The Lshaped high /xr pieces for directing the magnetic field are called yokes. Suppose that Am, Ay, and Ag are the crosssectional areas of the magnet, yoke, and gap as indicated in the figure. The lengths of the magnet, yoke, and air gap are lm, y, and g, respectively. The magnet, the two yokes, and the gap can be considered to be all connected endtoend or in series. Applying Ampere’s circuital law for H we can write,
Hmim + 2 Hyly + Hglg = 0
Since all four components, magnet, yokes, and gap, are in series, we can assume that the magnetic flux <I> through each of them is the same,
^ = Am = By Ay — Bg Ag
Show that
(softening temperature), f$T = 7 x 10“11 m2 N“1 (at high temperatures), n = 1.4446 at 1.5 xm?
SOLUTION
We simply calculate the Rayleigh scattering attenuation using
8 7T3 9 9
“* * 3F( “ V fckTt
So
8jt*
A* ** TTTTi <L44462 “ 1)2(7 x 10n)(1.38 X 10“®)(1730 + 273)
3(1.55 x 10”6)4
= 3.27 x 10“5 m"1 or 3.27 x 102 km“1
Attenuation in dB per km is then
«dB = 4.34a« = (4.34)(3.27 x 10“2 km“1) = 0.142 dB km“1
This represents the lowest possible attenuation for a silica glass fiber at 1.55 jxm.
[1] By the way the second law of thermodynamics simply says that you cannot extract heat from a system in thermal equilibrium and do work [i. e., charge x voltage).
[2] A change in any type of PE can, in principle, be used to do work, that is, A (PE) = work done. Chemical PE is the potential to do nonmechanical work [e. g., electrical work) by virtue of physical or chemical processes. The chemical PE per electron is Ef and AEf = electrical work per electron.
[3] The conduction electrons around the Fermi energy have a mean speed that has only a small temperature dependence. This small change in the mean speed with temperature is, nonetheless, intuitively significant in appreciating the thermoelectric effect. The actual effect, however, depends on the mean free path as discussed later.
[4] Thomas Seebeck observed the thermoelectric effect in 1821 using two different metals as in the thermocouple, which is the only way to observe the phenomenon. It was William Thomson (Lord Kelvin) who explained the observed effect.
[5] An electron at a distance x from the surface of a conductor experiences a force as if there were a positive charge of +e at a distance 2x from it. The force is e2/[4^e0(2x)2] or e*/[6ne0)?. The result is called the image charge theorem. Integrating the force gives the potential energy in Equation 4.39.
[6] In Chapter 3 we showed that the transmission probability T—TQ exp(2aa) where a2 = 2m (Vc – E)/h2 and a is the barrier width. The preexponential constant TQ can be taken to be. Clearly Va – E = <J>eff since electrons with E — Ef are tunneling and a = xp.
[7] Carbon nanotubes can be singlewalled or multiwalled (when the graphite sheets are wrapped more than once) and can have quite complicated structures. There is no doubt that they possess some remarkable properties, so it is likely that CNTs will eventually be used in various engineering applications. See, for example, M. Baxendale,
J. Mater. Sci.: Mater Electron, 14# 657, 2003.
[8] Constant volume in the definition means that the heat added to the system increases the internal energy without doing mechanical work by changing the volume.
[9] Sometimes it is stated that the Debye temperature is a characteristics temperature for each material at which all the atoms are able to possess vibrational kinetic energies in accordance with the Maxwell equipartition of energy principle; that is, the average vibrational kinetic energy will beikT per atom and average potential energy will also be 2 kJ. This means that the average energy per atom is 3kT, and hence the heat capacity is 3kNA or 3R per mole which is the DulongPetit rule.
[10] Wellknown exceptions are glasses, noncrystalline solids, whose heat capacity is proportional to a)T + 027"3, where ai and 02 are constants.
[11] We use Miller indices in two dimensions by dropping the third digit but keeping the same interpretation. The direction along x is [10] and the plane perpendicular to x is (10).
[12] K is a measure of the elastic change in the volume of a body in response to an applied pressure; large K means a small change in volume for a given pressure. Y is a measure of the elastic change in the length of the body in response to an applied stress; large Y means a small change in length. Both involve stretching or compressing bonds.
[13] 1 The formation of energy bands in the silicon crystal was described in detail in Chapter 4.
[14] The correct value appears to be 1.0 x 1010 cm"3 as discussed by M. A. Green (J. Appf. Phys., 67, 2944, 1990)
And A. B. Sproul and M. A. Green (J. Appl. Phys., 70, 846, 1991).
[16] 3 The proof can be found in advanced solidstate physics texts.
[17] The reader may have observed that the currents in Table 5.3 do not add exactly to zero. The analysis here is only approximate and, further, it was based on neglecting the hole drift current and taking the field as nearly zero to use Equation 5.47 in deriving the carrier concentration profiles. Note that hole drift current is much smaller than the other current components.
[18] Mechanical stress is defined as the applied force per unit area, am = F/A, and the resulting strain em is the fractional change in the length of a sample caused by <7m; sm = SL/L, where L is the sample length. The two are related through the elastic modulus Y; crm = Yem. Subscript m is used to distinguish the stress am and strain em from the conductivity a and permittivity s.
[19] The actual momentum of the electron, however, is not fik because
Fexternal ^internal
At
Where Fexternal + internal are all forces acting on the electron. The true momentum pe satisfies
Dpe _ r, r
, — ”external i “internal
At
However, as we are interested in interactions with external forces such as an applied field, we treat fik as if it were the momentum of the electron in the crystal and use the name crystal momentum.
*5.20 Compound semiconductor devices Silicon and germanium crystalline semiconductors are what are called elemental Group IV semiconductors. It is possible to have compound semiconductors from atoms in Groups III and V. For example, GaAs is a compound semiconductor that has Ga from Group III and As from Group V, so in the crystalline structure we have an “effective” or “mean” valency of IV per atom and the solid behaves like a semiconductor. Similarly GaSb (gallium anti monide) would be a IIIV type semiconductor. Provided we have a stoichiometric compound, the semiconductor will be ideally intrinsic. If, however, there is an excess of Sb atoms in the solid GaSb, then we will have nonstoichiometry and the semiconductor will be extrinsic. In this case, excess Sb atoms will act as donors in the GaSb structure. There are many useful compound semiconductors, the most important of which is GaAs. Some can be doped both n – and ptype, but many are one type only. For example, ZnO is a IIVI compound semiconductor with a direct bandgap of
EV, but unfortunately, due to the presence of excess Zn, it is naturally «type and cannot be doped to ptype.
GaSb (gallium antimonide) is an interesting direct bandgap semiconductor with an energy bandgap Eg = 0.67 eV, almost equal to that of germanium. It can be used as an light emitting diode (LED) or laser diode material. What would be the wavelength of emission from a GaSb LED? Will this be visible?
Calculate the intrinsic conductivity of GaSb at 300 K taking Nc = 2.3 x 1019cm3, Nv =
X 1019 cm3, ixe — 5000 cm2 V1 s"1, and fih — 1000 cm2 V“1 s”1. Compare with the intrinsic conductivity of Ge.
Excess Sb atoms will make gallium antimonide nonstoichiometric, that is, GaSbi+s, which will result in an extrinsic semiconductor. Given that the density of GaSb is 5.4 gem3, calculate
Or ptype semiconductor? You may assume that the drift mobilities are relatively unaffected by the
Excess minority carrier concentration Consider an «type semiconductor and weak injection condi
Hence the weak injection assumption). The rate of change of the instantaneous hole concentration
Dpn/dt due to recombination is given by
[26]5.29 Seebeck coefficient of semiconductors and thermal drift in semiconductor devices Consider an «type
Semiconductor that has a temperature gradient across it. The right end is hot and the left end is cold, as depicted in Figure 5.55. There are more energetic electrons in the hot region than in the cold region. Consequently, electron diffusion occurs from hot to cold regions, which immediately exposes negatively charged
Donors in the hot region and therefore builds up an internal field and a builtin voltage, as shown in Figure
5.55. Eventually an equilibrium is reached when the diffusion of electrons is balanced by their drift driven by the builtin field. The net current must be zero. The Seebeck coefficient (or thermoelectric power) S measures
[27] This is called Gauss’s law in point form and comes from Gauss’s law in electrostatics. Gauss’s law is discussed in Section 7.5.
[28] We use Boltzmann statistics, that is, n[E) a exp(E/kT), because the concentration of electrons in the conduction band, whether on the nside or pside, is never so large that the Pauli exclusion principle becomes important. As long as the carrier concentration in the conduction band is much smaller than NC/ we can use Boltzmann statistics.
THE p+n JUNCTION Consider a p+n junction, which has a heavily doped pside relative to the
Then with Nd = 1016 cm3, that is, 1022 m3, V0 = 0.835 V, and er = 11.9 in the equation
[34] This is simply the solution of the continuity equation in the absence of an electric field, which is discussed in Chapter 5. Equation 6.11 is identical to Equation 5.48.
I 4 This is generally proved in advanced texts.
[36] 5 The derivation is similar to that for the Schottky diode, but there were more assumptions here.
[37] 6 CB should not be confused with the conduction band abbreviation.
[38] found in more advanced texts.
[39] Various saturation effects are ignored in this approximate discussion.
[40] V. Estimate the drain current when VGS = 8 V and Vds = 20 V, given X = 0.01. Due to the strong scattering of electrons near the crystal surface assume that the electron drift mobility iie in the channel is half the drift mobility in the bulk.
SOLUTION
Since VDS > Vth, we can assume that the drain current is saturated and we can use the Ids versus Vgs relationship in Equation 6.64,
Ids = K(VGS – ^(1 + kVDS)
ZfleЈ
K =
1 11 The intensity is not necessarily maximum when both the electron and hole concentrations are maximum, but it will I be close.
[42] Much of the pioneering work for highefficiency PERL solar cells was done by Martin Green and coworkers at the University of New South Wales.
[43] Partial reflections of waves from the corrugations in the DBR can interfere constructively and constitute a reflected wave only for certain wavelengths, called Bragg wavelengths, that are related to the periodicity of the corrugations. A DBR acts like a reflection grating in optics.
[44] powers of x, that is, x, x2, x3, and so on, and for small x only the x term is significant, Fr = – fix.
[45] The term natural frequency refers to a system’s characteristic frequency of oscillation when it is excited. A mass attached to a spring and then let go will execute simple harmonic motion with a certain natural frequency co0. If we then decide to oscillate this mass with an applied force, the maximum energy transfer will occur when the applied force has the same frequency as co0; the system will be put in resonance. co0 is also a resonant frequency. Strictly, a) = 2nf is the angular frequency and / is the frequency. It is quite common to simply refer to co as a frequency because the literature is dominated by co; the meaning should be obvious within context.
Figure 7,4 Electronic polarizability and its resonance frequency versus the number of electrons in the atom (Z).
The dashed line is the bestfit line.
[47] This field is called the Lorentz field and the proof, though not difficult, is not necessary for the present introductory treatment of dielectrics. This local field expression does not apply to dipolar dielectrics discussed in Section 7.3.2.
[48] + ir
[49] 7 These currents are phasors, each with a rms magnitude and phase angle.
Debye Equations, ColeCole Plots, and Equivalent Series Circuit
Consider a dipolar dielectric in which there are both orientational and electronic polarizations, a a and ae, respectively, contributing to the overall polarizability. Electronic polarization ae will be independent of frequency over the typical frequency range of
[50] This simple relationship is used because the Lorentz local field equation does not apply in dipolar dielectrics and the local field problem is particularly complicated in these dielectrics.
[51] The departure is simply due to the fact that a simple relaxation process with a single relaxation time cannot describe the dielectric behavior accurately. (A good overview of nonDebye relaxations is given by Andrew Jonscher in J. PhysD, 32, R57, 1999.)
[52] Z. C. Xia et al, J. Phys. Cond. Matter, 13, 4359, 2001. The origin of the dipolar activity in this ceramic is quite
[53] The emission of electrons by tunneling from an electrode in the presence of a large field was treated in Chapter 4 as FowlerNordheim field emission.
[54] The equivalence of the coefficients in Equations 7.54 and 7.55 can be shown by using thermodynamics and is not considered in this textbook. For rigorous piezoelectric definitions see IEEE Standard 1761987 (IEEE Trans, on Ultrasonics, Ferroelectrics and Frequency Control, September 1996).
[55] A finite Q on the plates of a capacitor when V= 0 implies an infinite capacitance, C = oo. However, C = c/Q/dV definition avoids this infinity.
[56] More rigorous theories of ionic polarization would consider the interactions of a propagating electromagnetic wave with various phonon modes within the crystal, which is beyond the scope of this book.
The geometrical shapes, sizes, and distributions of the mixed phases.
[58] The theories that try to represent a heterogeneous medium in terms of effective quantities are called effective
[59]7.28 Pyroelectric detectors Consider a typical pyroelectric radiation detector circuit as shown in Figure 7.66. The FET circuit acts as a voltage follower (source follower). The resistance R represents the input resistance of the FET in parallel with a bias resistance that is usually inserted between the gate and source. Ci is the overall input capacitance of the FET including any stray capacitance but excluding the capacitance of the pyroelectric detector. Suppose that the incident radiation intensity is constant and equal to I. Emissivity r] of a surface characterizes what fraction of the incident radiation that is absorbed? r)I is the energy absorbed per unit area per unit time. Some of the absorbed energy will increase
[60] The symbol x for the magnetic dipole moment should not be confused with the permeability. Absolute and relative
[61] According to H. P. Myers, Introductory Solid State Physics 2nd ed.; London: Taylor and Francis Ltd., 1997, p. 362, there have been no theoretical calculations of the exchange integral Je for any real magnetic substance.
[62] See, for example, D. Jiles, Introduction to Magnetism and Magnetic Materials, London, England: Chapman and
I 6 This is the power engineers Steinmetz equation for commercial magnetic steels. It has been applied not only to I silicon irons (Fe + few percent Si) but also to a wide range of magnetic materials.
[65] 7 1 , /0.005 V
(2 x 10 )(4jt x 10"7 Hm )(200) ;r I—— ml
L = — = 0.025 H or 25 mH
(jr0.025 m)
I 7 SQUID is a superconducting quantum interference device that can detect very small magnetic fluxes.
[67] There is a third law to thermodynamics that is not as emphasized as the first two laws, which dominate all branches of engineering. That is, one can never reach the absolute zero of temperature.
[68] Designing a superconducting solenoid is by no means trivial, and the enthusiastic student is referred to a very readable description given by James D. Doss, Engineer’s Guide to High Temperature Superconductivity, New York: John Wiley & Sons, 1989, ch. 4. Photographs and descriptions of catastrophic failure in high field solenoids can be found in an article by G. Broebinger, A. Passner, and J. Bevk, "Building WorldRecord Magnets" in Scientific American, June 1995, pp. 5966.
[69] GMR was discovered in the late 1980s by Peter Grunberg Pulich, Germany), and Albert Fert (University of ParisSud) and their coworkers. Magnetoresistance itself, however, has been well known, and dates back to Lord Kelvin’s experiments in 1857.
[70] The physics of the coupling process between the two magnetic layers is an indirect exchange interaction, the details of which are not needed to understand the basics of the GMR phenomenon.
[71] One highly recommended book on magnetic recording is R. L. Comstock, Introduction to Magnetism and Magnetic Recording, New York: Wiley, 1999. See also R. L. Comstock, "Modern Magnetic Materials in Data Storage," J. Mater. Sci: Mater. Elecron. 12, 509, 2002.
[72]8.13 A permanent magnet with an air gap The magnetic field energy in the gap of a permanent magnet is available to do work. Suppose that Bm and Bg are the magnetic field in the magnet and the gap, Hm and Hg are the field intensities in the magnet and the gap, and Vm and Vg are the volumes of the magnet
[73] This chapter uses E for the electric field which was reserved for energy in previous chapters. There should be no confusion with Eg that represents the energy bandgap. In addition, n is used to represent the refractive index rather than the electron concentration.
[74] Maxwell’s equations formulate electromagnetic phenomena and provide relationships between the electric and magnetic fields and their space and time derivatives. We only need to use a few selected results from Maxwell’s equations without delving into their derivations. The magnetic field B is also called the magnetic induction or magnetic flux density.
[75] Willebrord van Roijen Snell (15811626), a Dutch physicist and mathematician, was born in Leiden and eventually became a professor at Leiden University. He obtained his refraction law in 1621 which was published by Rene Descartes in France in 1637; it is not known whether Descartes knew of Snell’s law or formulated it independently.
OPTICAL FIBERS IN COMMUNICATIONS Figure 9.10 shows a simplified view of a modem optical communications system. Information is converted into a digital signal (e. g., current pulses) which drives a light emitter such as a semiconductor laser. The light pulses from the emitter are coupled into an optical fiber, which acts as a light guide. The optical fiber is a very thin glass fiber [made of silica (Si02)], almost as thin as your hair, that is able to optically guide the light pulses to their destination. The photodetector at the destination converts the light pulses into an electric signal, which is then decoded into the original information.
The core of the optical fiber has a higher refractive index than the surrounding region, which is called the cladding as shown in Figure 9.10. Optical fibers for shortdistance applications (e. g., communications in local area networks within a large building) usually have a core region that has a diameter of about 100 xm, and the whole fiber would be about 150200 jxm in diameter. The core and cladding refractive indices, nx and n2, respectively, are normally only 13 percent different. The light propagates along the fiber core because light rays experience total internal reflections at the corecladding interface as shown in Figure 9.10. Only those light rays that can exercise TIR travel along the fiber length and can reach the destination. Consider a fiber with Ai(core) = 1.455, and n2(cladding) = 1.440. The critical angle for a ray traveling
[76] The light propagation in an optical fiber is much more complicated than the simple zigzagging of light rays with TIRs at the corecladding interface. The waves in the core have to satisfy not only TIR but also have to avoid destructive interference so that they are not destroyed as they travel along the guide; see for example, S. O. Kasap, Optoelectronics and Photonics: Principles and Practices, Upper Saddle River: Prentice Hall, 2001, chap. 2.
[77] The definitions of the field components follow those of S. G. Lipson et al., Optical Physics, 3rd ed., Cambridge, MA, Cambridge University Press, 1995, and Grant Fowles, Introduction to Modern Optics, 2nd ed., New York, Dover Publications, Inc., 1975, whose clear treatments of this subject are highly recommended. The majority of the authors use a different convention which leads to different signs later in the equations; Fresnel’s equations are related to the specific electric field directions from which they are derived.
[78] These equations are readily available in any electromagnetism textbook. Their derivation from the two boundary conditions involves extensive algebraic manipulation whicn we will not carry out here. The electric and magnetic field components on both sides of the boundary are resolved tangentially to the boundary surface and the boundary conditions are then applied. We then use such relations as cos 9t = (1 – sin 0f)1/2 and sin 9t as determined by Snell’s law, etc.
REFLECTION AT NORMAL INCIDENCE. INTERNAL AND EXTERNAL REFLECTION Consider the reflection of light at normal incidence on a boundary between a glass medium of refractive index 1.5 and air of refractive index 1.
If light is traveling from air to glass, what is the reflection coefficient and the intensity of the reflected light with respect to that of the incident light?
If light is traveling from glass to air, what is the reflection coefficient and the intensity of the reflected light with respect to that of the incident light?
What is the polarization angle in the external reflection in part (a)l How would you make a polaroid device that polarizes light based on the polarization angle?
SOLUTION
The light travels in air and becomes partially reflected at the surface of the glass which corresponds to external reflection. Thus n — 1 and n2 = 1.5. Then,
Tti — n2 1 — 1.5
R. = r± = — = = 0.2
" 1 m+n2 1 + 1.5
This is negative which means that there is a 180° phase shift. The reflectance (R),
Which gives the fractional reflected power, is
R = r = 0.04 or 4%
The light travels in glass and becomes partially reflected at the glassair interface which corresponds to internal reflection. Thus n = 1.5 and n2 = 1. Then,
N — n2 1.5—1 ^ ^
Tii = fj_ = = — 0.2
Ti n2 1.5 J – 1
There is no phase shift. The reflectance is again 0.04 or 4 percent. In both cases (a) and
, the amount of reflected light is the same.
Light is traveling in air and is incident on the glass surface at the polarization angle. Here n — 1, n2 = 1.5, and tan 9P = (n2/n) = 1.5, so 9P = 56.3°.
If we were to reflect light from a glass plate keeping the angle of incidence at 56.3°, then the reflected light will be polarized with an electric field component perpendicular to the plane of incidence. The transmitted light will have the field greater in the plane of incidence; that is,
[80] Both electronic and ionic polarizabilities have similar expressions. The ionic polarizability in an oscillating field was derived in Chapter 7, and looks almost exactly like Equation 9.66.
[81] As much as an electromagnetic radiation is quantized in terms of photons, lattice vibrations in the crystal are quantized in terms of phonons. A phonon is a quantum of lattice vibration. If K is the wavevector of a vibrational wave in a crystal lattice and co is its angular frequency, then the momentum of the wave is fiK and its energy is fico.
RAYLEIGH SCATTERING LIMIT What is the attenuation due to Rayleigh scattering at around the k = 1.55 xm window given that pure silica (Si02) has the following properties: 7}= 1730 °C
[83] There is a difference in this definition in optics and engineering. The definition here follows that in optics which is more prevalent in optoelectronics.
Elements to Uranium Atomic
Element 
Symbol 
Z 
Mass (g mol“1) 
Hydrogen 
H 
1 
1.008 
Helium 
He 
2 
4.003 
Lithium 
Li 
3 
6.941 
Beryllium 
Be 
4 
9.012 
Boron 
B 
5 
10.81 
Carbon 
C 
6 
12.01 
Nitrogen 
N 
7 
14.007 
Oxygen 
O 
8 
16.00 
Fluorine 
F 
9 
18.99 
Neon 
Ne 
10 
20.18 
Sodium 
Na 
11 
22.99 
Magnesium 
Mg 
12 
24.31 
Aluminum 
Al 
13 
26.98 
Silicon 
Si 
14 
28.09 
Phosphorus 
P 
15 
30.97 
Sulfur 
S 
16 
32.06 
Chlorine 
CI 
17 
35.45 
Argon 
Ar 
18 
39.95 
Potassium 
K 
19 
39.09 
Calcium 
Ca 
20 
40.08 
Scandium 
Sc 
21 
44.96 
Titanium 
Ti 
22 
47.87 
Vanadium 
V 
23 
50.94 
Chromium 
Cr 
24 
52.00 
Manganese 
Mn 
25 
54.95 
Iron 
Fe 
26 
55.85 
Cobalt 
Co 
27 
58.93 
Nickel 
Ni 
28 
58.69 
Copper 
Cu 
29 
63.55 
Zinc 
Zn 
30 
65.39 
Gallium 
Ga 
31 
69.72 
Germanium 
Ge 
32 
72.61 
Electronic 
Density (g cm 3) 
Crystal in 
Structure 
(*at 0 °C, 1 atm) 
Solid State 
I*1 
0.00009* 
HCP 
Is2 
0.00018* 
FCP 
[He]2^1 
0.54 
BCC 
[He]2s2 
1.85 
HCP 
[He]2 s2pl 
2.5 
Rhombohedral 
[He]2 s2p2 
2.3 
Hexagonal 
[He]2 s2p3 
0.00125* 
HCP 
[He]2 s2p4 
0.00143* 
Monoclinic 
[He]2 s2p5 
0.00170* 
Monoclinic 
[He]2 s2p6 
0.00090* 
FCC 
[N e]3sl 
0.97 
BCC 
[Ne] 3s2 
1.74 
HCP 
[Ne] 3s2/?1 
2.70 
FCC 
[Ne]3 s2p2 
2.33 
Diamond 
[Ne]3 s2p3 
1.82 
Triclinic 
[Ne]3s2/?4 
2.0 
Orthorhombic 
[Ne]3 s2p5 
0.0032* 
Orthorhombic 
[Ne]3 s2p6 
0.0018* 
FCC 
[Ar]4 sl 
0.86 
BCC 
[Ar]4s2 
1.55 
FCC 
[Ar]3dl4s2 
3.0 
HCP 
[Ar]3d24s2 
4.5 
HCP 
[Ar]3цf34^2 
5.8 
BCC 
[Ar]3d54sl 
7.19 
BCC 
[Ar]3цf54^2 
7.43 
BCC 
[Ar]3цf64^2 
7.86 
BCC 
[Ar]3rf74s2 
8.90 
HCP 
[Ar]3d84,s2 
8.90 
FCC 
[Ar]3dl04sl 
8.96 
FCC 
[Ar]3dl04s2 
7.14 
HCP 
[Ar]3dl04s2pl 
5.91 
Orthorhombic 
[Ai]3dl04s2p2 
5.32 
Diamond 
Atomic
Mass
Element 
Symbol 
Z 
(g mol ‘) 
Arsenic 
As 
33 
74.92 
Selenium 
Se 
34 
78.96 
Bromine 
Br 
35 
79.90 
Krypton 
Kr 
36 
83.80 
Rubidium 
Rb 
37 
85.47 
Strontium 
Sr 
38 
87.62 
Yttrium 
Y 
39 
88.90 
Zirconium 
Zr 
40 
91.22 
Niobium 
Nb 
41 
92.91 
Molybdenum 
Mo 
42 
95.94 
Technetium 
Tc 
43 
(97.91) 
Ruthenium 
Ru 
44 
101.07 
Rhodium 
Rh 
45 
102.91 
Palladium 
Pd 
46 
106.42 
Silver 
Ag 
47 
107.87 
Cadmium 
Cd 
48 
112.41 
Indium 
In 
49 
114.82 
Tin 
Sn 
50 
118.71 
Antimony 
Sb 
51 
121.75 
Tellurium 
Te 
52 
127.60 
Iodine 
I 
53 
126.91 
Xenon 
Xe 
54 
131.29 
Cesium 
Cs 
55 
132.90 
Barium 
Ba 
56 
137.33 
Lanthanum 
La 
57 
138.91 
Cerium 
Ce 
58 
140.12 
Praseodymium 
Pr 
59 
140.91 
Neodymium 
Nd 
60 
144.24 
Promethium 
Pm 
61 
(145) 
Samarium 
Sm 
62 
150.4 
Europium 
Eu 
63 
151.97 
Gadolinium 
Gd 
64 
157.25 
Terbium 
Tb 
65 
158.92 
Dysprosium 
Dy 
66 
162.50 
Holmium 
Ho 
67 
164.93 
Erbium 
Er 
68 
167.26 
Thulium 
Tm 
69 
168.93 
Ytterbium 
Yb 
70 
173.04 
Lutetium 
Lu 
71 
174.97 
Hafnium 
Hf 
72 
178.49 
Tantalum 
Ta 
73 
180.95 
Tungsten N i : 
W N _ 
74 Nc 
183.84 1 n/T O 1 
Electronic 
Density (g cm 3) 
Crystal in 
Structure 
(*at 0 °C, 1 atm) 
Solid State 
[Ar]3rf’°4s2p3 
5.72 
Rhombohedral 
[Ar]3d’°4s2p4 
4.80 
Hexagonal 
[Ai]3d,04s2p5 
3.12 
Orthorhombic 
[Ai]3dl04s2p6 
3.74 
FCC 
[Ki]5s‘ 
1.53 
BCC 
[Ki]5s2 
2.6 
FCC 
[Kr]4d’5s2 
4.5 
HCP 
[Kr]4d25s2 
6.50 
HCP 
[Kr]4d45s’ 
8.55 
BCC 
[Kr]4d55s’ 
10.2 
BCC 
[Kr]4d55j2 
11.5 
HCP 
[Kr]4rf75i> 
12.2 
HCP 
[Kr]4d85s’ 
12.4 
FCC 
[Kr]4rf10 
12.0 
FCC 
[Kr]4d105*> 
10.5 
FCC 
[Kr]4rf105*2 
8.65 
HCP 
[Kr]4rf1055y 
7.31 
FCT 
[Kr]4rf105*y 
7.30 
BCT 
[KT]4dl05s2p3 
6.68 
Rhombohedral 
[Kr]4rf105*2/j4 
6.24 
Hexagonal 
[Kr]4rf105fV 
4.92 
Orthorhombic 
[Kr]4d’°5sy 
0.0059* 
FCC 
[Xe]6 s’ 
1.87 
BCC 
[Xe]6s2 
3.62 
BCC 
[Xe]5</’6i2 
6.15 
HCP 
[Xt}4fx5dx6s2 
6.77 
FCC 
[Xe]4/36*2 
6.77 
HCP 
[Xe]4/46s2 
7.00 
HCP 
[Xe]4/S6i2 
7.26 
Hexagonal 
[Xe]4/66s2 
7.5 
Rhombohedral 
[Xe]4/76.s2 
5.24 
BCC 
[Xe]4/75d>6s2 
7.90 
HCP 
[Xe]4/96s2 
8.22 
HCP 
[Xe]4/106s2 
8.55 
HCP 
[Xe]4/"6s2 
8.80 
HCP 
[Xe]4/I26s2 
9.06 
HCP 
[Xe]4/136i2 
9.32 
HCP 
[Xe]4/I46i2 
6.90 
FCC 
[Xe4fu5d’6s2 
9.84 
HCP 
[Xe4fH5d26s2 
13.3 
HCP 
[Xe]4fH5d36s2 
16.4 
BCC 
[Xe]4/I45<f6s2 Aj’) Ac iSr 7 
19.3 <11 A 
BCC TT/^n 
Atomic
Mass
Element 
Symbol 
Z 
(g mol x) 
Osmium 
Os 
76 
190.2 
Iridium 
Ir 
77 
192.22 
Platinum 
Pt 
78 
195.08 
Gold 
Au 
79 
196.97 
Mercury 
Hg 
80 
200.59 
Thallium 
TI 
81 
204.38 
Lead 
Pb 
82 
207.2 
Bismuth 
Bi 
83 
208.98 
Polonium 
Po 
84 
(209) 
Astatine 
At 
85 
(210) 
Radon 
Rn 
86 
(222) 
Francium 
Fr 
87 
(223) 
Radium 
Ra 
88 
226.02 
Actinium 
Ac 
89 
227.02 
Thorium 
Th 
90 
232.04 
Protactinium 
Pa 
91 
(231.03) 
Uranium 
U 
92 
(238.05) 
Electronic 
Density (g cm 3) 
Crystal in 
Structure 
(*at 0 °C, 1 atm) 
Solid State 
[Xe]4/I45d66s2 
22.6 
HCP 
[Xe]4/145d76s2 
22.5 
FCC 
[Xe]4/I45^6i‘ 
21.4 
FCC 
[Xe]4fI45</106s‘ 
19.3 
FCC 
[Xe]4/145d106i2 
13.55 
Rhombohedral 
[Xe]4/145d106*y 
11.8 
HCP 
[Xe]4/145d106s2p2 
11.34 
FCC 
[Xe]4/I45d106s2p3 
9.8 
Rhombohedral 
[Xй]4fH5dl06sY 
9.2 
SC 
[Xe]4fu5d’%s2p5 
— 

[Xe]4fu5dl%s2p6 
0.0099* 
Rhombohedral 
[Rn]7i‘ 
— 
— 
[Rn]7i2 
5 
BCC 
[Rn]6d’7i2 
10.0 
FCC 
[Rn]6d27s2 
11.7 
FCC 
[Rn]5/2W7s2 
15.4 
BCT 
[Rn]5/36rf‘7s2 
19.07 
Orthorhombic 
"I don’t really start until I get my proofs back from the printers. Then I can begin serious writing."
John Maynard Keynes (18831946)
Curie temperature critical temperature (K) mechanical stress along direction j (Pa) thermocouple
Temperature coefficient of capacitance (K1) temperature coefficient of resistivity (K1)
Total internal energy –
Mean speed (of electron) (ms1)
Voltage; volume; PE function of the electron, PE(x)
Breakdown voltage
Builtin voltage
Pinchoff voltage
Reverse bias voltage
Velocity (m s1); instantaneous voltage (V)
Mean square velocity; mean square voltage
Drift velocity in the x direction
Effective velocity or rms velocity of the electron
Fermi speed
Group velocity
Thermal velocity
Valence band
Width; width of depletion layer with applied voltage; dielectric loss width of depletion region with no applied voltage
Width of depletion region on the «side and on the pside with no applied voltage
Atomic fraction
Admittance (Њ1); Young’s modulus (Pa)
Impedance (Њ); atomic number, number of electrons in the atom
Polarizability; temperature coefficient of resistivity (K1); absorption coefficient (m1); gain or current transfer ratio from emitter to collector of a BJT current gain Ic/h of a BJT; Bohr magneton (9.2740 x 10“24 J T_1); spring constant (Ch. 4)
Schottky coefficient
Emitter injection efficiency (Ch. 6); gyromagnetic ratio (Ch. 8); Griineisen parameter (Ch. 4); loss coefficient in the Lorentz oscillator model flux (m2 s1), photon flux (photons m2 s_1)
Small change; skin depth (Ch. 2); loss angle (Ch. 7); domain wall thickness (Ch. 8); penetration depth (Ch. 9) change, excess {e. g., An = excess electron concentration)
D2/dx2 + d2/dy2 + d2/dz2
E0sr, permittivity of a medium (C V“1 m1 or F m“1); elastic strain permittivity of free space or absolute permittivity (8.8542 x 10“12 C V“1 m“1 or Fm"1)
Sr relative permittivity or dielectric constant
R) efficiency; quantum efficiency; ideality factor
6 angle; an angular spherical coordinate; thermal resistance; angle between a
Light ray and normal to a surface (Ch. 9) k thermal conductivity (W m1 K"1); dielectric constant
X wavelength (m); thermal coefficient of linear expansion (K_l); electron
Mean free path in the bulk crystal (Ch. 2); characteristic length (Ch. 8) fi, fji magnetic dipole moment (A m2) (Ch. 3)
Fi HoVr, magnetic permeability (H m“1); chemical potential (Ch. 5)
FjL0 absolute permeability (4tt x 10“7 H m_l)
Fir relative permeability
Lim, magnetic dipole moment (A m2) (Ch. 8)
Ixd drift mobility (m2 V“1 s1)
Fxh, fjie hole drift mobility, electron drift mobility (m2 V“1 s“1)
V frequency (Hz); Poisson’s ratio; volume fraction (Ch. 7)
It pi, 3.14159… ; piezoresistive coefficient (Pa1)
7tL,7tT longitudinal and transverse piezoresistive coefficients (Pa~!)
N Peltier coefficient (V)
P resistivity (Q m); density (kg m“3); charge density (C m3)
PE energy density (J m3)
Pnet net space charge density (C m3)
PJ2 Joule heating per unit volume (W m3)
A electrical conductivity (Q~l m1); surface concentration of charge (C m2)
(Ch. 7)
Op polarization charge density (C m2)
A0 free surface charge density (C m2)
0’s Stefan’s constant (5.670 x 10“8 W m~2 K4)
R time constant; mean electron scattering time; relaxation time; torque (N m)
Rg mean time to generate an electronhole pair
Angle; an angular spherical coordinate
$ work function (J or eV), magnetic flux (Wb)
4>e radiant flux (W)
Metal work function (J or eV)
<Pn energy required to remove an electron from an «type semiconductor (J or eV)
<PV luminous flux (lumens)
X volume fraction; electron affinity; susceptibility (Xe is electrical; Xm is
Magnetic)
(jc, f) total wavefunction
Ir (x ) spatial dependence of the electron wavefunction under steadystate conditions
Яk(x) Bloch wavefunction, electron wavefunction in a crystal
T^hyb hybrid orbital
Co angular frequency (2nv); oscillation frequency (rad s"1)
Coi ionic polarization resonance frequency (angular)
Co0 resonance or natural frequency (angular) of an oscillating system.
Major Symbols and Abbreviations
A area; crosssectional area; amplification
A lattice parameter; acceleration; amplitude of vibrations; halfchannel thick
Ness in a JFET (Ch. 6) a (subscript) acceptor, e. g., Na = acceptor concentration (m3)
Ac alternating current
A0 Bohr radius (0.0529 nm)
Ay, Ap voltage amplification, power amplification
APF atomic packing factor
B, B magnetic field vector (T), magnetic field
B frequency bandwidth
Bc critical magnetic field
Bm maximum magnetic field
B0y Be RichardsonDushman constant, effective RichardsonDushman constant
BC base collector
BCC bodycentered cubic
BE base emitter
BJT bipolar junction transistor
C capacitance; composition; the Nordheim coefficient (Ј2 m)
C speed of light (3 x 108 m s“1); specific heat capacity (J K1 kg1)
Cdep depletion layer capacitance
Cm molar heat capacity (J K1 mol1)
Cdifr diffusion (storage) capacitance of a forwardbiased pn junction
Cs specific heat capacity (J K“1 kg~l)
Cv heat capacity per unit volume (J K“1 m“3) .
CB conduction band; common base
CE common emitter
CMOS complementary MOS
CN coordination number
CVD chemical vapor deposition
D diffusion coefficient (m2 s_1); thickness; electric displacement (C m~2)
D density (kg m3); distance; separation of the atomic planes in a crystal,
Separation of capacitor plates; piezoelectric coefficient; mean grain size (Ch. 2)
Donor, e. g., Nd = donor concentration (m~3) direct current piezoelectric coefficients
D (subscript) dc dij E Eat Ed EC, EV Јex Ef, Epo Eg ^mag Ј ^br Јloc E E (subscript) eff (subscript) EHP EM EMF (emf) F F № FCC FET G GPh GP 9(E) 9 9d 9m H, H h n H (subscript) HFE, hfe HCP HF / I /, i (subscript) /br Ib, I & h 
Energy; electric field (V m1) (Ch. 9) acceptor and donor energy levels conduction band edge, valence band edge exchange interaction energy
Fermi energy, Fermi energy at О К – k
Bandgap energy magnetic energy electric field (V m1)
Dielectric strength or breakdown field (V m1) local electric field
Electronic charge (1.602 x 10"19 C)
Electron, e. g., xe — electron drift mobility; electronic
Effective, e. g., /xeff = effective drift mobility *
Electronhole pair
Electromagnetic
Electromagnetic force (V)
Force (N); function frequency; function FermiDirac function facecentered cubic field effect transistor
Rate of generation rate of photogeneration parallel conductance (Ј21) density of states
Conductance; transconductance (А/V); piezoelectric voltage coefficient (Ch. 7) incremental or dynamic conductance (А/V) mutual transconductance (A/V)
Magnetic field intensity (strength), magnetizing field (A m1)
Planck’s constant (6.6261 x 1034 J s)
Planck’s constant divided by 2n(fi = 1.0546 x 10“34 J s) hole, e. g., xh = hole drift mobility
Dc current gain, smallsignal (ac) current gain in the common emitter configuration hexagonal closepacked high frequency
Electric current (A); moment of inertia (kg m2) (Ch. 1) light intensity (W m2) quantity related to ionic polarization breakdown current
Base, collector, and emitter currents in a BJT
I instantaneous current (A); smallsignal (ac) current, i = 81
I (subscript) intrinsic, e. g., rii = intrinsic concentration
Ib, ic, ie small signal base, collector, and emitter currents (8IB, 8IC, SIE) in a BJT
IC integrated circuit
J current density (A m2)
J total angular momentum vector
Imaginary constant: V—I Jc critical current density (A m“2)
Jp pyroelectric current density
JFET junction FET
K spring constant (Ch. 1); phonon wavevector (m1); bulk modulus (Pa);
Dielectric constant (Ch. 7)
Boltzmann constant (k = R/NA — 1.3807 x 10~23 J K1); wavenumber (k = 2n/k), wavevector (m_1); electromechanical coupling factor (Ch. 7) KE kinetic energy
L total orbital angular momentum
L length; inductance
Length; mean free path; orbital angular momentum quantum number Lch channel length in a FET
Le, Lh electron and hole diffusion lengths
In, ip lengths of the n – and /7regions outside depletion region in a pn junction
In (jc) natural logarithm of x
LCAO linear combination of atomic orbitals
M, M magnetization vector (A m1), magnetization (A m1)
M multiplication in avalanche effect
Mat relative atomic mass; atomic mass; “atomic weight” (g mol1)
Mr remanent or residual magnetization (A m1); reduced mass of two bodies A
And B, Mr = MAMB/(MA + MB)
Msat saturation magnetization (Am1)
M mass (kg)
Me mass of the electron in free space (9.10939 x 10“31 kg)
M* effective mass of the electron in a crystal
M*h effective mass of a hole in a crystal
Mt magnetic quantum number
Ms spin magnetic quantum number
MOS (MOST) metaloxidesemiconductor (transistor)
MOSFET metaloxidesemiconductor FET
N number of atoms or molecules; number of atoms per unit volume (m~3)
(Chs. 7 and 9); number of turns on a coil (Ch. 8)
Na Avogadro’s number (6.022 x 1023 mol1)
N electron concentration (number per unit volume); atomic concentration;
Principal quantum number; integer number; refractive index (Ch. 9) n r heavily doped nregion
Nat number of atoms per unit volume
NC, NV 
Effective density of states at the conduction and valence band edges (m3) 
No, N+ 
Donor and ionized donor concentrations (m3) 
Ne, n0 
Refractive index for extraordinary and ordinary waves in a birefringent crystal 
Hi 
Intrinsic concentration (m3) 
Ft/lO) Ppo 
Equilibrium majority carrier concentrations (m~3) 
Ftpoi pno 
Equilibrium minority carrier concentrations (m~3) 
Ns 
Concentration of electron scattering centers 
Nv 
Velocity density function; vacancy concentration (m3) 
P 
Probability; pressure (Pa); power (W) or power loss (W) 
P, P 
Electric dipole moment (C m) 
P 
Hole concentration (m~3); momentum (kg m s“1); pyroelectric coefficient (C m’2K"1) (Ch. 7) 
P+ 
Heavily doped pregion 
Pav 
Average dipole moment per molecule 
Pe 
Electron momentum (kg ms1) 
PE 
Potential energy 
/^induced 
Induced dipole moment (C m) 
Po 
Permanent dipole moment (C m) 
PET 
Polyester, polyethylene terephthalate 
PZT 
Lead zirconate titanate 
Q 
Charge (C); heat (J); quality factor 
Q’ 
Rate of heat flow (W) 
<1 
Charge (C); an integer number used in lattice vibrations (Ch. 4) 
R 
Gas constant (NAk = 8.3145 J mol1 K“1); resistance; radius; reflection coefficient (Ch. 3); rate of recombination (Ch. 5) 
R 
Reflectance (Ch. 9) 
Pyroelectric current and voltage responsivities 

R 
Position vector 
R 
Radial distance; radius; interatomic separation; resistance per unit length 
R 
Reflection coefficient (Ch. 9) 
Rh 
Hall coefficient (m3 C1) 
RQ 
Bond length, equilibrium separation 
Rm:> 
Root mean square 
S 
Total spin momentum, intrinsic angular momentum; Poynting vector (Ch. 9) 
S 
Crosssectional area of a scattering center; Seebeck coefficient, thermoelectric power (V m1); strain (Ch. 7) 
^band 
Number of states per unit volume in the band 
Sj 
Strain along direction j 
SCL 
Space charge layer 
T 
Temperature in Kelvin; transmission coefficient 
T 
Transmittance 
T 
Time (s); thickness (m) 
T 
Transmission coefficient 
Tan 8 
Loss tangent 
Flux, Luminous Flux, and the Brightness of Radiation
Many optoelectronic light emitting devices are compared by their luminous efficiencies, which requires a knowledge of photometry. Radiometry is the science of radiation measurement, for example, the measurement of emitted, absorbed, reflected, transmitted radiation energy; radiation is understood to mean electromagnetic energy in the optical frequency range (ultraviolet, visible, and infrared). Photometry, on the other hand, is a subset of radiometry in which radiation is measured with respect to the spectral responsivity of the eye, that is, over the visible spectrum and by taking into account the spectral visual sensitivity of the eye under normal light adapted conditions, i. e.,photopic conditions.
Flux (<J>) in radiometry has three related definitions, radiant, luminous and photon flux, which correspond to the rate of flow of radiation energy, perceptible visual energy, and photons, respectively. (Notice that, in radiometry, these fluxes are not defined in terms of flow per unit area.) For example, radiant flux is the energy flow per unit time in the units of Watts. Radiometric quantities, such as radiant flux <&e, radiant energy flow per unit time, usually have a subscript e and invariably involve energy or power. Radiometric spectral quantities, such as spectral radiant flux <J>*, refer to the radiometric quantity per unit wavelength; i. e., = d$>e/dk is the
Radiant flux per unit wavelength.
Luminous flux in lumens 
Luminous flux or photometric flux 4>„, is the visual “brightness” of a source as observed by an average daylight adapted eye and is proportional to the radiant flux (radiation power emitted) of the source and the efficiency of the eye to detect the spectrum of the emitted radiation. While the eye can see a red color source, it cannot see an infrared source, and the luminous flux of the infrared source would be zero. Similarly, the eye is less efficient in the violet than in the green region, and less radiant flux is needed to have a green source at the same luminous flux as the blue source. Luminous flux <!>„ is measured in lumens (lm), and at a particular wavelength it is given by ‘
<!>„ = <J>e X K X TJeye
Where <t>e is the radiant flux (in Watts), K is a conversion constant (standardized to be 633 lm/W), ye (also denoted as V) is the luminous efficiency (luminous efficacy) of the daylight adapted eye, which is unity at 555 nm; r)cyc depends on the wavelength. By definition, a 1 W light source emitting at 555 nm (green, where rjeyc = 1) emits a luminous flux of 633 lm. The same 1 W light source at 650 nm (red), where rjtye =0.11, emits only 70 lm. When we buy a light bulb, we are essentially paying for lumens because it is luminous flux that the eye perceives. A typical 60 W incandescent lamp provides roughly 900 lm. Fluorescent tubes provide more luminous flux
output than incandescent lamps for the same electric power input as they have more spectral emission in the visible region and make better use of the eye’s spectral sensitivity. Some examples are 100 W incandescent lamps, 13001800 lm, depending on the filament operating temperature (hence bulb design), and 25 W compact fluorescent lamps, 15001750 lm.
Luminous efficacy of a light source (such as a lamp) in the lighting industry is the efficiency with which an electric light source converts the input electric power (W) into an emitted luminous flux (lm). A 100 W light bulb producing 1700 lm has an efficacy of 17 lm/W. While at present the LED efficacies are below those of fluorescent tubes, rapid advances in LED technologies are bringing the expected efficacies to around 50 lm/W or higher. LEDs as solidstate lamps have much longer lifetimes and much higher reliability, and hence are expected to be more economical than incandescent and fluorescent lamps.
From left to right: Michael Faraday, Thomas Henry Huxley, Charles Wheatstone, David Brewster and John Tyndall. Professor Tyndall has been attributed with the first demonstration (1854) of light being guided along a water jet, which is based in total internal reflection.
I SOURCE: Courtesy of AIP Emilio Segre Visual Archives, Zeleny Collection.