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Unformatted text preview: ECE 3040 Microelectronic Circuits Exam 1
February 19, 2007 Dr. W. Alan Doolittle Print your name clearly and largely: /(l J. i. :J . L. «s. 1. in _ _ : . Instructions: Read all the problems carefully and thoroughly before you begin working. You are allowed
to use 1 new sheet of notes (1 page front and back) as well as a calculator. There are 100
total points. Obsewe 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. lfl 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. Repon 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: First 33% Multiple Choice and True/False
(Circle the letter of the most correct answer or answers) _,,.':‘\T . 1.) (3points) True ((EFa/lseii MBE is an epitaxy technique performed at very high pressures above atmospheric pressure?“
2.) (3points) True dngalse:_..’lfa semiconductor has a large interatomic spacing it will likely have a large bandgap. 3.) (3points) True fig) The density of states describes the probability that a particular state at energy, E, i cup' . 4.) (3pointsZJTryE or False: An indirect bandgap material (such as Germanium and Silicon) results
in a longer/Tﬁmgrity carrier lifetime than a direct bandgap material. 5.) (3points)’ Truebr False: For direct bandgap materials there is no difference in momentum
between eTe'ﬂr/ons andsholes. 6.) (3points) True or False: The probability of occupying a state located at the fermienergy is
always 1. 7.) (3points) True or False: ‘Auger recombination occurs when an electron is captured by a defect inside the bandgap'tlfen' drops to the valence band killing off a hole. Select the pe_st answer or answer§ for 610:
8.) L4points) The law of mass action... describes the balance between electrons and holes. indicates that an increase of electron concentration beyond the intrinsic concentration results in a decrease in the hole concentration ﬂay... indicates that at constant temperature the product of electrons and holes is constant
"" only applies to nondegenerate doping conditions. Who knowsllli 9.) (4points) The electrons in the conduction band ...?
a.) Are immobile
i1.) _,Are most often found exactly at the lowest energy, at the conduction band edge
___'e;_j_.~iCan be heavier than the free space (vacuum) electron mass
;j""d.)_ \Can be lighter than the free space (vacuum) electron mass
'73.) Are always lighter than the free space (vacuum) electron mass due to their interaction with
the atoms in the crystal lO.)(4points) The following energy band diagram indicates the material is: a.) Degenerate and n—type 13c _._______ In equilibriun] EC_ ................................................................................................... ..
c.) Nondegenerate n—type ____________________________ “FV _ d.) Degenerate and ptype 5'62) In low level injection
'f.) Nondegenerate ptype E:  
g.) In steady state m .' a in 6 2 1?;
Seconm Short Answer (“Plug and Chug”): For the fol owing problems (1 112) use the following material parameters:
l3 €5L‘. “xi? 7:” cg , §;.,.2'C.f'1'r.f}'h ll ni=le6 cm‘3 NA=1e18 cm'3 acceptors mp'=0.5m0 mn'=0.5mO
EG=1.45 eV Electron mobility, p.“ = 900 cmZ/Vsec Hole mobility , p1, = 10 cmZ/Vsec
Temperature= 27 degrees C G l 1.)(}§=points) Where is the intrinsic energy (relative to the valence band which is referenced to zero .'  I en_e_rgx)l_d.__ IT): V0“. I'f‘cn'Jjﬁmif‘EIf II; \ _f } 7:: ﬂipJ; I. f’ pﬂv;
A " — r _, . .
VHF  yang?" ) _ " .ﬂ. . . _ r /" r t.— \
r liq}?  I r I’ )l
P): Em ’ “T” 4 (275 (m k I w L €17
~:' 0."?3ﬁev \
i
l
l
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K G: 12.) Wpoints) What is the resistivity of the semiconductor? I —\
 I :96) I" ,
ﬂo— Mu _ l5"? J00 Third 25% j .
' r53 CI"\ . . _ low 
l3.)(25pomts total) Sometimes semiconductors are used for radiation detectors. A 100 um x 100 um x 100 um cube of intrinsic siliconcarbide (SiC) material with bandgap Eg=3.0 eV is to be
used as an xray detector. The semiconductor has an intrinsic concentration, ni=le9 cm'3, an
electron mobility of 500 ch/Vsec and a hole mobility of 50 cmz/V—sec. The cube is biased on
two opposing sides with 3,000 volts. Each xray photon has an energy of 30,000 eV meaning
that each xray photon has enough energy to produce many electron hole pairs. Ifa single x—ray
photon strikes the semiconductor... a) (3 points) What type (diffusion, drift, electron dominated, hole dominated, etc...) of current
results? in _.r'~ , _g_ r +
Q‘rqrm ._./_/_3’.,'~ r 1 _ T)". ilwl (I.
In“; ,_ L—L ﬁler v11" “9’ *" " ‘ "
II T I l ‘
, x I, ,. ,_,. b) (7 points) Draw the ludimensional energy band diagram in the direction of the electric field
indicating the actual slope of the bands (numeric answer). f5.) ..._ I f5 . ' ,a than I; I' II: a} .
' i F‘ ' I‘l ' . KP" II
_ a 645‘ ‘ f. E : Ag 514 (.3.. lax, ! {3’56} VA r131)
__'F ___ _ _ H" k “(If .. / f I. gKJIfhix 7 _ gaggocﬂc/ eU/; 2:: l .41”; r?“ L 5 5' —
6 Wyn?" " a “7
 ..."f ‘
.' d/ ’ l/
e) (15 — points) Assuming'all photogenerated electrons and holes are collected instantaneously as current, what is the current (not current density) that results? Hint: consider how much
energy it takes to generate an electronhole pair verses how much energy the xray photon has. ‘
n = m + an r J). 4. if,
KHH__‘§;¢::TP7 ’ ___7?
i3:‘*‘ Fr L1“ [I l
' "362.5%. 3" , \
 «I “l” 3 ‘9‘? + 1. .3” (1 .L
H: 0.114 I..r 3 ,0 fig [‘II;‘ VI LC... \
(1' ’ i " '6] ‘ , 'l: h H" 44 \I‘_,. . i f' l r‘ ‘ ﬂ; ‘ i/giw/
\, , 4.. . Pulling all the concepts together for a useful purpose: l4.) (30points)
Surfaces of a crystalline semiconductor represent regions where recombination is higher than
in the bulk (inside the semiconductor) and thus, lower the excess minority carrier concentrations near the crystal surfaces. GeorgLVYashirlthnjAmerica’s 715‘ president) lit a candle that shines on a 5?}1m thick ptype GaAs semiconductor held at roomtemperature
(27 degrees C). The lig t
throughout the semiconductor but the surface recombination is such that there are no excess
carriers at both the surfaces. Determine the minority carrier current density at all positions in
the semiconductor. Assume a minority carrier lifetime of 10 nanoseconds (le—8 seconds),
and minority carrier diffusion coefficient at room temperature of 4.0 cmZ/sec. d 2A}: An , x 7 3‘
Given: 0 2 D77 7 ’1 — P General Solution is: Anp (x): Ae + Be dx‘ 1,7
l dzAn 7 An _x +1: V '
Given: 0 = D” d 7’ — p + G,‘ General Solution iszAnp(x)= Ae +Be +GLT”
.._———#—— x———_....:u_..__.. __ ~ 77 ~
W if d 2An 7 _ .
Given: 0 2 D77 2] 7 ’ General Solution is: Anp (x) = A + Bx
(x
dZAn 7 . . 2
Given: 0 2 D77 d 7 ’ + GL General Solution is: Anp(x) = Ax + Bx + C
x:
d 2An 7 7 . G I (7
Given: 0 = D" I 7 ’ + GLOf(x) General Solution is: Anp (x): — D‘ Hf(x)dx + Bx + C
(x‘ N
dAn An — 1
Given: d p 2 — p General Solution is: Ann (1‘) 2 AMP (t = 0)e t I"
All . .
Given: 0 = — p + GL I General Solution is: Anp = GL1”
T”  L' I '
'H'f—FJ‘wm ILL.
7’5 . I
a.__,._.__ .\
6W.” " ~__ \d._ t 1 " “2m : 9“ ' m Kr” . :_ , ‘ —,._ . .7 j l N LJ{r I. 4‘ QC Mr; I " Z "  k E) Q + 1‘ 4— It: :
A —_ WV 6 X)
(I e H ~6} e ‘ ' + l} :_" _
— uniformly generates 1022 additional holes _p_ei_‘___c_m3_ per second : f.
I. Extra work can be done here, but clearly indicate which problem you are solving. 
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This note was uploaded on 08/23/2011 for the course ECE 3040 taught by Professor Doolittle during the Spring '11 term at University of Florida.
 Spring '11
 Doolittle

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