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Unformatted text preview: Name Section: (check one). Cooper Schwartz
» (T, Th) (MWF) EE 305 Exam 1 Fall 2003 DO NOT OPEN THIS EXAM UNTIL INSTRUCTED TO DO SO 1) Closed book
2) Allowed: a calculator and one 8.5x11 crib sheet 3) Be sure you have 13 test pages including the cover and score sheet. 4) Write only on the question sheets. Show all your work. If you need more
room for a particular problem, use the reverse side of the same sheet. 5) Write neatly, if your writing is illegible then print. 6) This exam is worth 100 points. I 7) There are useful equations and constants at the back of the. exam. TEST SCORE ABET CRITERIA SATISFIED (1) ____
(2) __
(3) ___. (13 pts) Prob. 1. Match the following with the best deﬁnition. Energy band edge Freeze out Degenerate semiconductor
Direct gap semiconductor The sign of the charge on an ionized acceptor Energy band gap Fermi level
Effective mass Mobility Minority carrier lifetime Einstein relationship
' Low injection
Quasi—Fermi level a. b.
c.
d. we‘ prpar‘ The highest energy state in the valence band or the lowest energy state in the
conduction band The energy level at which the probability of occupation is 1/2. Relates the diffusion coefﬁcient to the mobility The excess minority carrier concentration is much smaller than the majority
carrier concentration A condition that occurs when the temperature of a semiconductor is so low
that there is not enough thermal energy to ionize the donors and acceptors
A constant that relates the acceleration of a hole or electron to force applied
by a macroscopic electric ﬁeld. ' A constant that expresses the relationship between the average velocity of a
hole or electron and applied macroscopic electric ﬁeld. The average distance an excess minority carrier will diffuse before it
recombines. An energy level that is used to describe the density of holes and electrons
when the semiconductor is not in equilibrium. A semiconductor with its Fermi level within 3kT of either band edge. A semiconductor in which the bottom of the conduction band is directly above
the top of the valence band in a Ek diagram The minimum energy required breaking a covalent semiconductor bond. . Positive. Negative The energy between the bottom and the top of the conductionband.
The energy at which recombination levels are found. None of the definitions are appropriate (7 pts) Prob. 2. a. (3 pts) For the face centered cubic crystal lattice shown below, what is the distance from the atom at the origin to the nearest atom in the [111] direction in
units of a? Answer b. I (4 pts) If the lattice constant, a, is 3 x 10‘7 cm, what is the density of the atoms in
the (001 ) plane in atoms/cmz? Answer ' (16 pts) Prob. 3 The density of states functions for the conduction and valence bands of a ﬁctitious
semiconductor are shown below. The density of states in the conduction band is a
constant, N c, for energies above EC. The density of states in the valence band is a
constant, NV, for energies below EV. The semiconductor is non~degenerate. a. (8 pts) Derive expressions for n and p in terms E1: and the constants NC and
NV , EC, and kT. \
Energy + NC Density of States b. (8 ts) What is the band gap, Eg, of this semiconductor, in units of kT,
if NC =2 x 10 9 cm'3 eV’1 and NV = 1x 1020 cm‘3 eV'1 , kT = .026 eV, and the intrinsic
carrier concentration, ni, is 1 X 108 cm‘3. Answer (16 pts) Prob. 4 Uniformly doped silicon, at T=3OO K, has a donor density, ND, of 5 x 1014 cm"3 and an
acceptor density of 1.5 X 1015 cm'3. The band gap of silicon is 1.12eV.
a. (2 pts) Is the semiconductor ntype, ptype or intrinsic?
( Check one) b. (6 pts) What is the minority carrier concentration? Answer 0. (2 pts) Is the semiConductor degenerate? yes no
(1. (6 pts) Show Why your answer in ( c ) is correct. (16 pts) Prob. 5. A semiconductor at T=300 K is uniformly dOped with an acceptor density, NA; of 1 x
1014 01113. The resistivity is measured to be 45 ohmcm. If a concentration gradient of 1 X 1014 holes cm'4 is present as a result of an external
inﬂuence, what is the resultant hole current density, J P/DIFF, due to diffusion ? Answer (16 pts) Proh. 6. A slab of p—type silicon with NA = 1 x 10 17 cm'3 is exactly one minority carrier diffusion
length thick. The excess minority carrier concentration at x = Ln is An(x= Ln) = 1 x 1011 cm'3. At X = 0, An(x= 0 ) = 0. The semiconductor is in steady state
and no carriers are generated by light. Use the minority carrier diffusion equation to derive the expression for An(x ) in terms of
the concentrations at x = 0 and X = Ln. An (16 pts) Prob. 7. The silicon pn junction band diagram has the quasiFermi levels as shown in the figure. a. (2 pts) Is the semiconductor in equilibrium? yes
b. (2 pts) Explain your answer in (a) n0 c. (12 pts) Sketch the hole current density, Jp, between X3 and X4. Jp 10 Catalog of General Solutions for the
Minority Carrier Diffusion Equation '5'Anp 072Anp _ Anp +G
at (2C2 2'” L Case I: “Uniform” Conditions ' ' am An
Simpliﬁed MCDE: a} p = — p + GL
 T In General Solution: Anp(t) = Ae‘W" + tnGL '
Case II: “Steady State” Conditions
(9%,, _ Anp ’SimpliﬁedMCDE: 0:1)" 0V T +_G,; General Solution: Anp(x)'.= Ae”"/L" +Be+x/L" + TnGL ‘ ,wher‘e Ln = Dntn Case 111: “Uniform” and “Steady State” .  An
Simpliﬁed MCDE: 0 = — 'p + GL
2' 7: General Solution: An = 7 G p n L Full MCDE: =Dn Note that GL may be zero (“Dark” Conditions) in any of the situations above. In this case,
set GL = 0 in the'General Solution. 11 Table 2.4 Carrier Modeling Equation Summary. . Density of States and Fermi Function m§\/2m_§ (E‘ — in) gC(E) = ’ ‘c ‘\ if 1
m*\/2m*(EV — E)" ﬂE) _ 1 + e(E‘EF)/kT
ME) = .—"’——"—————— E s E \ #2ﬁ3 ’ Carrier Concentration Relationships
’1' = Nce(EF_Ec)/kT 3/2 .
2 m*kT . _
'n = No V; 171/2010) NC = 2[27‘;h2] p — NveUiv Emlkr . 3/2
' 2 ' m*kT _ I
1": NV V77 F1/2(77v) NV = 2[27:h2] nieaip E,)/kT niem—EFVH ni, npProduct, and Charge Neutrality
ni=\/N—C1TVe‘EG’2kT np=ni2 p—n+ND—NA;—0 n, p, and Fermi Level Computational Relationships 2 1/2
ND—NA ND—NA E+E 3 mi‘,‘
=..___ + 2 ._ C V+_ V__
n. 2 +[< 2‘ n1 2 4len * mn EF — E = kT1n(n/ni) =' — len(p/ni) EF — E = len(ND/ni) ND>NA,ND> ni 12 Table. carrier'Actibn Eduétibn Surﬁrﬁéry. Eguéztibns bf’State ‘ an 1 ' an art ' 6An 62An An
_:_V‘JN+—thermal+;—Vqther 4': at q _ at'RjG 6t processes at 6x, Tn_
8p 1 > 3}; 3p BAP _ 62A}? Ap
_:_"V'JP+T“therma1+'_"other "*DPﬂ 2n_ “+01.
at  q . 6t R_G at memes at 3x 779 Currgnt and R~G Relationships JN = JNldrift + JNIdiff = (11%"? qDNVh I . 1% = __ ‘ ' ' at 2m 1  11 m It diffusion mi? Tn
JP = JPldrift JPldiff = ‘I/MJPfg ’" qDPVp 93 = _~A_}_’
' ' at i—thennal _ 77
J = JN fi JP R—G. '1’
Key Parametric Reldiionships
' "& '_' kl. T = 1
LN E V .H‘n n LPEVDP'ﬁ» &~§I 7:.1
. _ ‘ [up q» ,p  CPNT
Resistivity‘and Eléctréktqtic Relationshigk ’ L
,1 ‘
p =. V  . . , n—ty ,e semiconductor
p = 1 anND p
p — . . . ptype sennconductor
1 dE 1 dE 1 dE~ 1 ' 
%=—'°=—:V=~—f =—(Ec—E,ef)
4 dx 4 4x q ‘dx ' q ' .
' Qudsi—Fermi Level Relatibnships
FN a E! .+ kT ln<$> JN = gnnvEN l: ‘é‘
I e MPRVFP E.
."i’ r .—len( '1 13 Semiconductor Constants Fundamental constants: m0=9.11 x10'3‘ kg h=6.63x10‘34Js ’ﬁzh/2n
q=1.6x1o19C 80:8.85x10'14F/cm
k = 8.62 x 10'5 eV/°K kT = 0.026 eV (at room temperature) Constants for silicon at room temperature:
ni=101ocm‘3 EG=1.12ev KS=11.8 Useful conversion factors: 1ym=10'4 cm=1o6m ' 1 eV = 1.6x1o19 Joules Useful integrals = gum/a“ ...
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This note was uploaded on 03/25/2011 for the course ECE 305 taught by Professor Melloch during the Spring '08 term at Purdue.
 Spring '08
 MELLOCH

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