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Unformatted text preview: ECE 3040 Microelectronic Circuits Exam 1
February 5, 2004 Dr. W. Alan Doolittle Print your name clearly and largely: g7 /./ ‘i ((7 h 5
————————.—_________________
Instructiousi Read all the problems carefully and thoroughly before you begin working. You are allowed
to use I new sheet of notes (1 page front and back) as well as a calculator. There are 100 total points. 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 cannot read it, it will be considered 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: pegHob g/ #1
Number of Tests= 29 Point Value of problem= 2
Individual Problem Average= 90.9
Exam Average= 56.93 Exam Standard Deviatiou= 13.58
Exam Max= 86 Exam Miu= 30 Class Totals #2 #3 #4 #5 #6 #7 #8 #9 #10 #1] #12 #13a #13b #14
2 2 2 2 3 3 3 3 3 10 15 7 18 25
100.0 97.0 100.0 72.7 72.7 57.6 97.0 69.7 61.6 88.8 56.8 65.4 51.3 43.9 Number of exams 0 1o 20 so 40 so so 70 so 90 100
Test Score Class Totals
Number of Tests= Point Value of problem=
Individual Problem Average= Exam Average= 59.97 Exam Standard Deviation= Exam Max=
Exam Min= 2
90.9 100.0 1 1 .67
82
37 #3 #4 #5 #6 #7 #8 #9 #10 #11 #12 #13a #13b #14
2 2 2 3 3 3 3 3 10 15 T 18 25
97.0 97.0 60.6 72.7 60.6 97.0 75.8 68.7 96.1 55.2 68.8 58.8 40.7 Number of exams 20 30 40 so
Test Score 60 70 100 First 25% Multiple Choice and Truchalse (Circle the letter of the most correct answer)
1.) (2points.r False: An intrinsic material has equal concentrations of electrons and holes.
2.) (2points) rue o m' A semiconductor material has a wider bandgap than an insulator. 3.) 2 oints rue o .' An intrinsic semiconductor is also a de enerate semiconductor.
P g 4.) (2pointl ur a se: Direct bandgap semiconductors are used as light emitters (LEDS and
LASERS) r use of their efﬁcient conversion of electrons to photons. S.) (2points) True 0 w A zincblende crystal structure has more atoms in the unit cell than does
the diamond ctysta s ructure. Select the best answer or answerg for 6l 0:
6.) (ii—points) A “new” semiconductor consists of ﬁctitious group 11 elements Dr, Do, Is ‘and group V1 elements Go and Cd in equal composition. To within 1%, what is the correct reduced
semiconductor notation for this remarkable compound? a. DrDolsGoOd
DTMJ3000.333150333G0050d05 0 Dl’ncoDOozoISozoGon.200510.20
d.) Cannot be determined from the information given.
e.) You cannot have more than 4 elements making up a semiconductor. i'.) (3points) Which of the following mathematical expressions describes the free electron concentration? Hint: Gc is the density of states function, and f(E) is the fermi distribution
function a.) n = GCfUE')
b.) n = G(_. df(E%E (9 n = E G(_.f(E)dE d.) n = G? I: f(E)dE e.) Looks like I will be Working as a Walmart clerk! 8.) (3points) The following energy band diagram indicates the material is:
a.) Nondegenerate ntype b.) Degenerate ntype E“
c.) intrinsic EIi """"""""""""""""
d. Nondegenerate ptype Ev @Degenerate ptype Ef  9.) points) For the electron and hole shown in the following band diagram circle ELI that are true.
The electron experiences an electric ﬁeld that will accelerate it.
a. The hole experiences an electric field that will accelerate it.
mThe hole will tend to stay where it currently is.
. The device is in a transient state Ec ' """  e.) The device has to be in equilibrium 10.) (Cipoints) Effective mass...
of an electron is always smaller than for a hole. takes into account many different forces acting on the carrier that would otherwise not be
present in free space ..is always less than the mass of an electron in vacuum
.detennines (in part) how easy it is to move a carrier through the material
e.) .is a fundamental constant that does not change from material to material Second 25% Short Answer (“Plug and Chug”):
For the following problems (1 112) use the following material parameters: ni=‘1e14 cm"3 NDfSeIT cm"3 donors NA=6e12 cm'3 acceptors.
m1, =0.2mo mp =I.2mo
Electron mobility, is = 900 cmstec Hole mobility , 1.11, = 10 cmstec Temgeramre= 150 degrees C 11.)(10points) Assuming total ionization, what is the electron and hole concentrations and is the '  9
matenalporntype. ND ?7 NA N0 ?7h_‘ 12.) (1 5points) What is the bandgap of the semiconductor material? a 3/
MF a “fa/’17) 3/1 M; ; a[hn;m{AT)] J “[3 lNcVVE lap 12‘ _ 55/; H Slmlquly l 0 )(‘lrll 63—31,) (g_gl?e5)(J6314)(l§0+373)]
WP
“h : 11/3,”. 3.7% 89% (his; h:
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: —l(‘2.6ne5’) (I501 373’) [a lEj: 6:579Vl Third 25% Problems (3rd 25%)
13.) (ZSpoints total)
A 500 pm thick semiconductor at room temperature (27 degrees C) has the following parameters: Electron mobility, un=500 cmszec Substrate intrinsic concentration, ni=1e10 cm'3
The bandgap, Eg=l.l eV
The substrate doping is linearly graded such that Nd=100ni+lemx, where x is in cm. 5 #119ch
No )7 ,1. a.) (7 points)D;§wtf the band diagram in equilibrium and CLEARLY label Ec, Ev, Bi, and Ef l n: e 6!
EF E1? ATA (£134;— \ KL? x
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b.) (18 points) Determine an expression for the electric ﬁeld verses position in the material. 32:0 r/{IJ/AHE +/¢;/an”‘ Va : ATM“
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t When the dinosaurs wallged_t_l_i_e__plan§_t __a__1ight was turned onto ; 10 pm thick InP semiconductor held at room temperature (27 degrees C). 1111’ has a problem that the surfaces
are really efﬁcient recombination centers. Thus, the excess electron concentration at the
surfaces will always remain 0 cm'3. The light uniformly generates 1018 additional holes per
cm3 per second. Determine the excess electron concentration in the InP for all positions
:2! '9 ;=ue.w:ee=vm‘r;u;:= . Assume a minority carrier diffusion length of 0.1
pm and the mobility at room temperature is 3000 cmZNsec. General Solution is: An (x)=Ae_/L" +Be+A" L General Solution is: Ann (x): Ae‘A" +Be+A" + GL7"
__ _—‘—  '— ' h“—~——_..____,.,..__ General Solution is: Anp (x) = A + Bx General Solution is: Anp (x) = sz + Bx + C dzAn
Given: 0=Dfl dxzp +wa(x) General Solution is: Anp(x)=[%L£Hf(x)dx:’+8x+C
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 Doolittle

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