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Unformatted text preview: ECE 3040B Microelectronic Circuits
Exam 1 September 20, 2001 Dr. W. Alan Doolittle /M Home NVV Print your name clearly and largely: 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 in this exam plus 5 point bonus. Observe the point value of each
problem and allocate your time accordingly. SHOW ALL WORK AND CIRCLE YOUR
FINAL ANSWER WITH THE PRQPER UNITS INDICATED. Write legibly. IfI
cannot read it, it will be comdered a wrong answer. Do all work on the paper provided
Turn m all scratch paper, even ifit 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: 5.3 0 .3
mu
2* 0,: 12131311111 1 ﬂoomummh c
r N N n n w a. w m 0 3’ ... x a m a 1 1111.11111n1111111ms11 h S
133% a First 25% Multiple Choice and True/False (Circle the letter of the most correct answer) 1.) (2points,r False: Drift current results from movement of electrons and holes in
response to :u . applied electric ﬁeld. 2.) (2points) @r False: The unit of kT is energy and can be expressed in electron volt (eV)
or joules ( . 3.) (2points) r False: If Magnesium (Mg is group 2 element) is used to dope GaN (Ga is
group 3, N is group a  type semiconductor will result if the Mg replaces a Ga atom. 4.) (2points) True 0 I‘m A ntype Silicon semiconductor has more electrons than holes
AND has more holes ii . u an intrinsic silicon semiconductor. 5.) (2points) True or @ ilicon is a compound semiconductor. Select the My; answer for 610: 6.) (3points) Given Si, C and Ge are all from group 4, which of the following semiconductors
is a valid semiconductor representation of a ternary compound semiconductor?
a.) Si 13 G'eoscus
loszeossCMo
. Si3Ge3C4 7.)  ints) The valence band...
.is mostly ﬁlled with electrons.
. . . .is mostly empty.
c.) ...is higher in energy than the conduction band. d.) .. .is above the fermi energy in a degenerame doped semiconductor. 8.)  . ints) The following energy band diagram indicates the material is:
a"... E. ———————————
'  n'type 1 ...............................
c.) intrinsic E;
d.) Silicon Ev 9.) (3points) For to the following band diagram, what is known from the information given: a.) The device is bent.
b There is no electric ﬁeld in this material There is no current ﬂow in this device EC MW WMNNR
.) There is no diffusion current in this material. &H
E 10.) (3points) In equilibrium: a The drift current is equal in magnitude and in the same direction as the diffusion current.
The driﬁ current is equal in magnitude and opposite in direction as the diffusion current. c.) There is always no driﬁ current. d.) There is always no diffusion current. e.) I hear WalMart'“ is hiring sales clerks — that could be my new career. EV Second 25% Short Answer (“Plug and Chug”): For the following problems (1114) use the following material parameters: n.=lelé cm'3 ND=l.1el6 cm 3donors NA=2e16 cm‘3 acceptors.
Effective density of states in the conduction band is Nc=le19 cm'3
Effective density of states in the valence band is Nv=2el9 cm"3 Electron mobility, p1,: 800 cm N sec Hole mobility, up: 200 cm2/Vsec Temperature=27 degrees C 
11. )(Spoints) Assuming total ionization, what is the electron and hole concentrations and is the
. 9
material p or 11 type. NR 7 NO
P : NA ' ND ”6 N 3 3
 + a h,
'3. a "‘wf‘_‘imyr*“m .j.;/.v;::: ..,.., z . . _ Ié— In:  J
 39 ' __________€_. W(sel§ Mel ‘46“! —.sn._ﬂ e.e—m —' .
 ﬁﬁewj: I '
=>lh= 6.96 e :5 cm) 12. )(5points)WhatisEgEv(whereEtisthe fermienergyandEvisthetopofthe valence band)? P? NYC (EvE Ea/AT Ev Er: AT!” (“ﬂ”) : 0.0”? A(———— ‘3‘??? x
I
l
s
s 13. )(5points) What length of material is needed to make a resistor with resistance 1000 ohms
using a cylinder with crosssectional area 0.0001 cmz. F:_.L______ «a if
?(Jl:n+#ff) I000: ’2076JZCM)L ' ‘ :“Vkmv‘t‘e'g WW 0 000! a... ;
L: 0: '39. CM 2
'0‘“ ’1’“ s 14. )(Spoints) If 1M olts is placed across the resistor in part 13, what are the electron current
density, hole current density and the electron driﬁ velocity? (If you found no answer in part
13, then use a resistor length of 1 cm in your calculation). (In: (We l‘l) (300N752 757%. 466215)
I.. (a 61 Mo» 3,941.42 IQ)Q00)(75.757)( ”8“)
3p: '37. 7) lar/cm2 15.) (5points) A semiconductor is doped with 5el9 cm'3 very deep acceptors (large binding
energy) which are only partially ionized at room temperature. Ifthe fermi energy is 0. leV
above the valence band, and the acceptor energy (EA) is 0.16 eV above the valence hand,
what is the number of ionized acceptors in this material? You can assume the degeneracy
factor, gA=4. Hint: do not make this problem harder than it is. Na NA ‘ ‘ +9“? e(Fa Er)/4r £351....—
= (0.16— 0 l) Micam l+l+e ' ’E Noet LR 6'3 (“—3 ' a Third 25% Problems (3.“I 25%) 16.) (ZSpoints total) A semiconductor has the follo parameters: Hole Diffusion coefficient,2 Dp— cmz/Sec , ,4 T 0/)
Hole Mobility, 11,5200 cm lVSec \> 5.19 v" ; ‘
Substrate relative Dielectric Constant, ammmFKﬁl 1.7 7’ 4“” Dielectric Constant of free space, so =8. 854e14 F/cm
Substrate intrinsic concentration, n,=1e10 cm'3 The hole concentration in NONEQUILIBRIUM 1n the material 1s maintained at
p(x)=1e15e°‘“°°um)cm {or x= o +0 x: IOOuM a.) (20 points) Plot and label (label maximum and minimum values) the hole current
; density if an electric ﬁeld of 1 Wm is applied across the material.
a b. ) (5 points) Explain why we cannot determine the electron concentration as a function of
I position. ’ a) 3p=15€PE‘$DPVo° m
=<1.6eI‘l)@00)(\0)5[Iel‘5€x '0'"?
(I 6:24?) (6‘16) '5—g—fe Mm] = [033:1  aomﬂe WW“ = 0.237eMy'm‘“ ft/m1 Pulling all the concepts together for a useful purpose. (4th 25%)
17. ) (ZSpoints)
Light is absorbed in a silicon wafer of thickness 500 um (the wafer is similar to that passed around in class). The wafer is ptype and is uniformly dog with 1017 cm'3 acce tors. The 1i t
$359,053 ' umoﬁl has been on for a very lon time and can be approximated as bein
Tﬁfouﬁout Hie maEﬁal. 1% the excess HOLE generation rate is 10%7 cm 5/sec and the minority \ 2t _ )0 ')erlifetime is 1 milliseconds (le3 seconds); ’“n' _, 1000 Chi/Km; “ti/Q I‘M
affla I"? a.) (4 points) Should the absorption coefﬁcient be large or small for such a condition to a how . 0A
“a“? {or "x b. ) (5 points) What is the excess electron concentration for all positions in the wafer. "7 “m c. ) (16 points) What would be the electron concentration as a flmction of tune for all A ’\
times after the light is turned off? Bonus: (5 pointsno partial credit on this bonus) Determine what electron current density ﬂows in the material. Support your answer thoroughly for credit. E
l
l
!
(1213:: An  + l
Given: 0=D,, 0&2" — 7" General Solution is: Anp(x)= Ae 7/" +Be 77‘ I
dzAn An _ +
Given: 0=Dn 2p — 1’ +GL ' General SolutioniszAnp(x)=Ae 7‘"+Be ﬂ” +GL1', i
dzAnp . .
dxz General Solution is: Anp(x)= A + Bx
dzAn .
Given: 0: D P+GL General Solutionis: Anp(x)=Ax2+Bx+C
dAn An _¢
Given: dtp =— 1’ General Solution is: Anp(t)=Anp(t=0)e /"
An
Given: 0= p +GL General Solutioisz? An], =GL1',‘ . l ,
AlmoﬁI IJQ... i +0 +k¢ P0,; 9M one .' n 6,155 Extra work can be done here, but clearly indicate with problem you are solving.0 C ) U 3AM Quad”
' x1 an L Genera‘ 30/14th "
/_________ AW“ We ' 69“” +0 Va’ut Ju5+ Befdre +Le 13 A1 was wmeO‘ﬂ,
=7 Armin): 10'”e4‘/'“’) m3 Bums: MM.\{6(‘M‘/ , agree! “:3 N0 Eleanzﬁ‘e/Jzi 7'”: 1’ D Vn lanipvrm 0’40???)
4/ ‘4“ (arm n Hum Ku"""/6emm9:ao\
:? 0n Mil) 21> 7h = 10,10) ...
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 Spring '11
 Doolittle

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