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Unformatted text preview: ECE 3040 Dr. Alan Doolittle
Microelectronic Circuits Exam #1 ECE 3040B Microelectronic Circuits amp/in: Cm—ern‘a Exam] t .
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mewmzu {err(.97: C 3 m ad m.nw+ej [Va Clue. = 61002,) Dr. W.Alan Doolittle Print your name clearly and largely: k6 ‘1 Instructions: a Read all the problems carefully and thoroughly before you begin working. You are allowed to
use 2 sheets of notes (1 page front and back) as well as a calculator. There are 100 total points in
this exam. Observe the point value of each problem and allocate your time accordingly. SHOW
ALL WORK AND CIRCLE YOUR FINAL ANSWER WITH THE PROPER UNITS
INDICATED. Write legibly. If I can not read it, it will be considered to be a wrong answer. Do all work on the paper provided. Turn in all scratch paper, even if it did not lead to an answer.
Report any and all ethics violations to the instructor. Good luck! Sign your name on ONE of the two following cases: I did not observe any ethical violations during this exam: —————————————__—__—__—__—_ I observed an ethical violation during this exam: N6 2:20 59505... E mem mowommom moumuﬁm Em ESQ SGJOQS 1391 JO JaqwnN mm @038 8 $530.8 5% 5% s8 Sm Q8 8 E08 BF 8* 9.: “m2 3 2 NF : S N w mm
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Haggai E .53. 91003 1391 ECE 3040 Dr. Alan Doolittle
Microelectronic Circuits Exam #1 First 25%, T rue/F alse and Multiple choice: Circle the correct answer (only one) 1. (2 points) True cw An insulator has a smaller energy bandgap than does a
semiconductor. 2. (2 points) True 0 A ptype semiconductor has more electrons than holes. 3. (2 points) True 0 An intrinsic semiconductor has a higher number of donors than
acceptors. 4. (2 points) @or False: Antimony (symbol Sb from group 5 of the periodic table) would act
as a donor in 51 Icon (symbol Si from group 4 of the periodic table). 5. (2 points) True 0@ Silicon Carbide, SiC, is an example of an elemental semiconductor.
(Both Si and C are from group 4 of the periodic table). Select the best answer for 610. 6. (3 points) For a plane intersecting the coordinate axes at x=a, y=2a and z=3a, where a is the
lattice constant, the Miller indexes are: a.) (123) b. 321
t. enough information to answer this problem. 7. (3 points) When a formerly intrinsic material is converted to an extrinsic material by doping
with acceptors, a.) There are more electrons than holes
b. The fermi level is above the intrinsic energy
@There are fewer electrons than when the material was intrinsic.
. None of the above 8. (3 points) The electrons in an intrinsic material come from: a.) Donors b.) Acceptors c Photogenerated by breaking the bonds that hold the atoms in the crystal together.
Thermally generated by breaking the bonds that hold the atoms in the crystal together. e.) The electron factory. f.) None of the above 9. (3 points) The effective mass...
a.) of an electron is always smaller than the mass of an electron in a vacuum.
b.) of an electron is always larger than the mass of an electron in a vacuum. is different from the mass of an electron in a vacuum because it is physically smaller. .. is different from the mass of an electron in a vacuum because it sees different electrostatic
potentials. f.) is the same for electrons as it is for holes. ECE 3040 Dr. Alan Doolittle
Microelectronic Circuits Exam #1 10. (3 points) A semiconductor doped with donors has it’s fermi energy, E, above it’s
conduction band energy, E. This semiconductor
a.) is nondegenerate.
b. is a compound semiconductor.
$5.. is a degenerate semiconductor.
d.) obeys the minority carrier diffusion equations.
e.) is really messed up! Second 25%, Short answer
11. (5 points) A semiconductor has 12.5% of the atomic concentration Indium (In) with the rest
of the material made up of Aluminum (Al) and Arsenic (As). What is the properly reduced
semiconductor formula for this material? (By reduced, I mean put the formula in the format
discussed in class, not the standard chemical equation/i) I"n.5"/a Alum. 55W. \ﬂ, ’Calucea/
I»; ’4 Ions A 5 12. (10 points) The ﬁgure below and to the
right is the zincblende unit cell of the
semiconductor in question (1 1). If the
smaller atoms are arsenic, label the larger
atoms numbered 1, 2, 3, and 4 as Al or In
such that the structure is consistent with
your answer in question 11. Note, your
answer may or may not be unique and
atoms 1, 2, 3, and 4 are all completely
contained inside the unit cell. Atoml: I,‘ Atomz; or any , 8 a+om5/una'* ‘9'”
Cambihaﬁtﬁ’“ #416045 Muff Atom 3: ,‘kvalvt'nj ZINCBLENDE ya a Txese Atom4' i In + .5 are a” on cuée 3' Li Larje I'h‘l‘em‘df a +0
13. (5 points) In equilibrium, what restriction is placed on the product of the electron,” m (w + be 3 “M, 3 concentration, 11, and hole concentration, p regardless of doping concentrations? “47: "fr; I ECE 3040 Dr. Alan Doolittle
Microelectronic Circuits Exam #1 14. (5 points) If a material is doped with donors, draw the energy band diagram, labeling the
conduction band energy E, the valence band energy E, the intrinsic energy E, the fermi energy
Ef and the donor energy Ed at (a)... room temperature assuming total ionization (b) ...0 degrees Kelvin . c_,a_____ ,~___.,__~Ea mg,
war: 0 E _ p” _’_ __ _. _ 4‘ h “*9 A~~ E4, Third 25%, Problem Solutions
15. You are given the following information about a semiconductor sample at 27 degrees Celsius: N3 = 1.6615 cm‘3 Nd: 2615 cm‘3 m*n=0.01m0 m*p=5mo Eg=0.5 eV
Nc=2.51x1019 (m"‘n / m0)3/2 Nv=2.51x1019 (m"‘p / m0)3/2
un=2000 cmz/V second pp=500 cmz/V second Cross sectional area 0.01 cm2 by 1 cm long a.(8 points) Find the intrinsic, electron and hole concentration (assume total ionization) — _. 0540.03513 3)
14:: New e E’AAL; n; : Erma”(a.m)”°][2.s'lxw'YrWﬂ e m.“ : 2.6 5613 6 a 5—)17L7ew):L
: 96114 + @el‘t)a #— 4.704): ECE 3040 Dr. Alan Doolittle
Microelectronic Circuits Exam #1 b.(5 points) Find the intrinsic energy. m f
,1.
Ex. 52 + gaérw a) e.(4 points) Find the fermi energy position relative to the intrinsic energy. (5,2 15;) =4TA(%~) %éwﬂ[% : 0,029? A “'5 3’“ 566444.53 Np "’ "a' d.(4 points) The resistance of a piece of material. .3 I I ,
F ?(9~6’00 (4.681%) 3‘ 505(613813D féchh + ,afﬂ) __ 6L. 6,570) ‘
R” A 2 m9: e.(4 points) The hole drift velocity of the sample when 1000 volts are applied along the sample’s length. _3
E: VA“ : twav
lam A /V}['—' Mao E
= 5790 “709 A111: 5'xl05 Lin/{co ECE 3040 Dr. Alan Doolittle
Microelectronic Circuits Exam #1 Fourth 25%, Problem Solutions
16. (25 points) A single “pixel” from semiconductor camera (a CCD camera) has a 10 pm long
region of silicon that is at 27 degrees C, is in nonequilibrium and is uniformly doped. The material is p—type, is doped with 1016 cm'3 acceptors, and has a minority carrier lifetime that is 4
long enoungh to ne lect recombination when com ared with the deminsions of the material. ;’> A
C 0 The electron mobility is
500 cmz/V Second and
the intrinsic
concentration is 1610
cm'3. The material is continu Minority carrier
extraction Minority carrier
extraction 0 illuminated with light
AA for a very long time.
The light is uniformly
I“ absorbed throughout the volume of the
semiconductor. The
light generates 1017 cm‘3
per second extra holes. Minority carriers are extracted from the sample at x=0 and x=10 pm
maintaining the minority carrier concentration at a constant value equal to the equilibrium
minority carrier concentration at both edges of the material. Write an numerical expression (no
unknowns) for the minority carrier concentration as a ﬁmction of position. ( The distance
variable, “x” should be the only alphabetic variable in your answer. I.E. no unsolved constants. ) AA 0 )lAhp _ A” a
22;: Dn ’Sx—a +61. X=0 X=10um 4A ._ (lei7
IV?! —7,7;e '9 ECE 3040 Dr. Alan Doolittle
Microelectronic Circuits Exam #1 Bonus: (5 points) In problem 16, could you solve the problem if the equilibrium carrier
concentration was given as, p (X): 1 018 e(x/(2 um)) cm3 Support your answer. ...
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 Spring '11
 Doolittle

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