This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: lECE 3040 Microelectronic Circuits Exam 1 February 6, 2007 Print your name clearly and largely: 50 [Dirk i O VL Instructions:
1 . Closed book, closed notes. You are allowed to use 1 sheet of notes (1 page front
and bazk, letter size) as well as a calculator. Show all work in order to receive full credit for your answers. Incorrect answers
may be: awarded partial credit when appropriate. CIRCLE YOUR FINAL ANSWER WITH THE PROPER UNITS INDICATED.
Write legibly. If I cannot read it, it will be considered a wrong answer. [)0 all work on the paper provided. Clearly indicate which problem you are
solving; on each page. Turn in all scratch paper, even if it did not lead to an
answer. . There are 9 pages to this exam, including the cover sheet and 3 blank pages. There are 100 total points in this exam. Observe the point value of each problem
and all Jcate your time accordingly. Good luck! Part 1:( 30 points) Multiple Choice and True/False Circle the lette f the most correct answer: 1.) (2—points) True r False: Drift current results from movement of electrons and holes in
response to an applied electric ﬁeld. 2.) (2—points) True 0 ® An intrinsic semiconductor is also a degenerate semiconductor. 3.) (2points) (True )or a se: A semiconductor has an energy bandgap greater than a metal and
less than an insulator. 4.) (2—points)r False: If Magnesium (Mg is group 2 element) is used to dope GaN (Ga is
group 3, N is group S , a ptype semiconductor will result if the Mg replaces a Ga atom. S.) (2points) True [email protected] An n—type Silicon semiconductor has more electrons than holes
AND has more holes than an intrinsic silicon semiconductor. 6.) (2points) True or. The electrostatic potential is same everywhere in a pn junction
diode under equilibrium condition. Select the M answer:
7.) (3—points) 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 b Genscoj
@ Sioszeosscono .) Sl3G€3C4 8.) 3 oints) The valence band...
@HJS mostly ﬁlled with electrons.
.) ...is mostly empty.
c.) ...is higher in energy than the conduction band.
(1.) .. . is above the fermi energy in a degenerately doped semiconductor. 9.) 3—points) The following energy band diagram indicates the material is:
a. ptype En ) ntype E. ______________________________
c.) intrinsic Ef
d.) Silicon E, 10.) (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  .) There is no diffusion current in this material. Bf ....... l 1.) (3points)
true. The ho .) The ho
e.) Therei f.) The device has to be in equilibrium The electron will move to the right The electron will move to the left For the electron and hole shown in the following band diagram circle all that are a :1 e will move to the right 6 will move to the left
; no current ﬂow in this device 12.) (3points) Circle all that are true for a pn diode.
a.) The depletion width increases under forward biasing. 8T1“: ele d.) The ele :tric ﬁeld decreases inside the depletion region under forward biasing. The depletion width increases under reverse biasing. :tric ﬁeld decreases inside the depletion region under reverse biasing. Part 2: (25 points) Short Answer (“Plug and Chug”)
For the following problems (13 15) use the following material parameters. n,= 8e15 em'3
Effective densit
Effective densit Electron mobility, u,“ = 800 cmZ/Vsec
Temperature=2’ 13.)(10points) ND=l e16 cm 3donors NA=2616 cm «acceptors
y of states in the conduction band is Nc=lel9 cm‘3
5/ of states in the valence band is NV=2e 19 cm"3 Hole mobility , 1.11, =‘200 cmstec
7 degrees C Assuming total ionization, what is the electron and hole concentrations and is the material p or ntype? NA 7" l’ \— Np NA _l\lb )liVliZ 14.)(10points) What is E‘sEv (where Er is the fermi energy and Ev is the top of the valence band)? 
‘ LEV‘E‘FJ/k’l—
l7: NV 8 Evajl: : JaTbxl‘Ev)
2 0.02.9? LAP—”WWW ) 18w .
:: —0.)g7 Erntv :: 0.!“ 15.) (Spoints) What length of material is needed to make a resistor with resistance 1000 ohms
using a cylinder with crosssectional area 0.0001 cmz. E’ejistwij p: fU/LMKWLEDID) W 1.5 KW ( gthrrg €13 + 200 x‘ AWN} _ 1 0. ‘77 .S’LCM Part 3: (20 points) 16.) (EDpoints total) A semiconductor has the following parameters: Hole Mobility, rp=4oo cmZ/VSec Substrate relative Dielectric Constant, ar_semimducm=l(g=l 1.7
Dielectric Cons :ant of free space, so =8.854e14 F/cm Substrate intrinsic concentration, ni=le10 cm‘3
The hole concentration in NONEQUILIBRIUM in the material is maintained at p(x)=1e15e("“°°”m) cm3 a.) (15 points) Calculate the hole current density if an electric ﬁeld of 10 V/cm is applied across the material.
b.) (5 points) Can we determine the electron concentration as a ﬁmction of position? Why? “3 j: ill/(PFE ‘jDer:
:. TEMPE * j(JSI)MF W l? 7" a
WHO )(‘WOX ”)0 wel A “ﬂwp I l * Mm x0 02?? MW x) €(%F¢ll (0.651— 045976 )elxgll
0479 9 Kg” A/Culz X00] ll ll M
,5 tomcegm‘M’hm, 14/9 ff ”it Part 4: (25 points) Pulling all the concepts together for a useful purpose 17.) (ZS—points) Light is absorbed in a silicon wafer of thickness 500 um (the wafer is similar to that passed
around in class). The wafer is p—type and is uniformly doped with 1017 cm'3 acceptors. The light
has been on for a very long time and can be approximated as being absorbed uniformly
throughout the material. If the excess HOLE generation rate is 101'1 cm'3/sec and the minority
carrier lifetime is 1 milliseconds (Ie—3 seconds); a.) (10 points) What is the excess electron concentration for all positions in the wafer? b.) (15 points) What would be the electron concentration as a function of time for all
times: after the light is turned off? cllAn Ar: _x +x
Given: 0 = D" — dxz 7’ — ’7 General Solution is: Anp (x): A6 A” +Be A"
T"
clen An a x x
Given: 0 = D" — 2 p  ’7 +G, General Solution is: An (x): A3 A" + Beef/5" +Gfr,1
dx ’1'" : p 4
clenp _ _
Given: 0 = D" — dxz General Solution 15: Amp (x) = A + Bx
alen 2
Given: 0 = 0,, ~ dxz ‘0 + GL General Solution is: Anp (x): Ax + Bx + C
6123!: An _r
Given: ‘0 = — p General Solution is: An (I): An (I = 0)e A"
dt T" p p
. A“), . .
Given: 0 = ——— + G1. General Solutton is: Anp = GLrn
7: Extra work can be done here but clearly indicate with problem you are solving. 5) MW W 5W W ”M if . 7V
djjcl/L __ DH ~45}? +/6/
/ Ax: Gleamt 91"” H)“:
”H“: ﬁhpwzo) 8 *‘L/‘Cn {7/08 {fa/{”96 Jul/1+ W LMPHwJ eﬁmb +9
10a [WA W6 hmwi ”76L,
1/ ~(fre—3) All/LFH); [0 Q 6”th ...
View
Full Document
 Spring '07
 HAMBLEN

Click to edit the document details