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Unformatted text preview: ECE 3040 Microelectronic Circuits
Exam 3 April 18, 2002 Dr. W. Alan Doolittle Print your name clearly and largely: /M 4 i Q 3;! 3 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), your note sheets from the previous exams 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 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: ' Ex I observed an ethical violation during this exam: First 25% Multiple Choice, True/False (Select the mos
\'\l.). r(3'points) The gate current in the MOSF ET is a.) Highest under large positive gate bias b.) Highest under large negative gate bias oi Is determined by the change in fermi level position at the semiconductoroxide interface t correct answer) Is always zero
e. Is zero except in inversion 2.) (3 points) A depletion mode PMOS transistor has a...
a. ...ptype substrate.
...ntype substrate c. ntype channel when biased into cutoff (either depletion or accumulation)
(1.) huge gate current. 3.) (4points each, 12points total) For the three transistor crosssections I, II and III shown below,
answer each of the three questions ac: S D
I g . .
am“. a___;
III. a.) For VGS>VDS>0 which crosssectional View (I, II or III) is correct. :1
b.) For 0<VGS<VGgVt<VDs which crosssectional View (I, II or III) is correct. I.
c.) For VGS<0 and VDS>0 which crosssectional View (I, II or III) is correct. . m: 4.) (4—points) Which energy band diagram corres
a.) b.)
/
l ponds to a PMOS capacitor biased into accumulation. c.) d.)
f) 5.) (3 points) For a Transresistance ampliﬁer designed for maximum gain (circle all that apply):
a. You should have a high input impedance? You should have a low input impedance?
0. You should have a high output impedance? You should have a low output impedance?
' The output is a voltage f. 17/6, 5:! §$frare
Second 10% Short ; wer' //,/>
6.) (10 points) ‘ W ansistor at room temperature has an oxide thickness of 1000 angstroms (1 angstrom =lO'8 cm), determine the doping required to make the transistor begin conducting at 0.75
V. Use the following: Substrate relative Dielectric Constant, sr_semiconducto,=Ks=1 1.7
Oxide relative Dielectric Constant, sundae =Kox=3.9
Dielectric Constant of free space, so =8.854e14 F/cm
Substrate intrinsic concentration, ni=1e10 cm'3 Also recall that the total dielectric constant of the semiconductor is: as =Srserniconductoreo the total dielectric constant for the oxide is: 80X :Sroxidego
r NA
(M: 64‘ 3.7/{.me44) é; T
XM ’ \F/C: s [V4
momcm (9.035? A law)
‘ 3""1' 8? 13/0": 0.7;: M: + nitriteWW“ 2M 9) g:
3.1—r5’e6’ F/WI (NJXIJ'Jﬂe—IQJF/m
0.75’= MI: 4 3e5' (5473.2an N4 Problems (3rd 20%)
7.) You may assume that the OpAmps are ideal. a.) (15 points) Determine an equation relatin
input voltages, Vinl, Vin2, Vin3.
b.) (5 g the output voltage, Vout in terms of the three points) Determine the input impedance seen by the source Vin3. Extra work can be done here, but clearly indicate with problem you are solving. é “For? I: L94 ’0": 27/7} =0 Man J». Jeri/t5
, 6R1 C L
5 A t Le+ A/‘y _ 4}" = ﬂ . 5 /’ ‘2 MUMX (L', R I R
ffkce ‘er v.'rmq19mm”j a§SUf€9 ’V‘=OJ m3: 3 a,» New! a: Arm + 449.: m3) —: 4»: k3 Pulling all the concepts together for a useful purpose: (4th portion) 8.) 14520171“) Given the following circuit, what is the AC voltage gain, Vout/VinAc? You may
assume all capacitors have inﬁnite capacitance and are thus, AC shorts. Additionally consider
the circuit to be operated at low frequencies where you can neglect all small signal capacitances.
Grading will be based as such: 20 points for DC solution (VGS, VDS along with IDS for both transistors), 5 points for the conversion to the small signal model and 20 points for the small
signal analysis. Hint: Assume and then verify that both transistors are saturated. If M2 is saturated, identify
it’s functionality (i.e. what is M2 being used for) in the DC circuit. Do not panic! Performing this identiﬁcation will make the AC solution easier but is not
necessary to get the correct answer. VinAC R1
1meg Use the following parameters:
For NMOS Depletion Transistors: [Vi R’9Kn’=20 uA/v2 VT=1.0V 7t=o.o V'1 Length (L)=1 um Width (W)=100 um
M1 For NMOS Enhancement Transistors:
 7Kn’=20 uA/v2 VT=1.0V x=o.o V'1 Length (L)=1 um Width (W)=100 um
For PMOS Depletion Transistors:
K,’=40 uA/Vz VT=+3.ov 7t=0.1 V'1 Length (L)=10 um Width (W)=10 um
For PMOS Enhancement Transistors:
K,’=40 uA/VZ vT=—3.0v h=0.1 V" Length (L)=10 urn Width (W)=‘10 urn Extra work can be done here, but clearly indicate with problem you are solving. éolwid'x.’ ‘
06 Mi :7 NM“; E""‘WCM@M A=0,Kh=&0e6,Vw‘l ’Vi "3 MIME (0801;? /€7‘2(0*\ >) :0 z k“ :_"we'62 WW?"
/7 :ﬁway‘i W59»: 591%»{1’47‘60/ (Now V95> V‘s“ VT) >04 101/ > 0 — ~l) , WV > ﬂggumﬂ#.0kw wam/ fw‘V Extra work can be done here, but clearly indicate with problem you are solving. AC éa/WVWWI «as. JIM? (/Curren+ $3ch ano/ 4,1ij
V is an AC ﬂaﬂen chew}
éa‘i‘wftl‘ley/ ’*‘ V‘VTI‘ V65360A6‘T4‘un'b
I05 = cansran+ ’V'in Bonus: (IOpoints  Absolutely no partial credit) Determine an expression form
He transfer voltage gain, VON [5y éué—‘ﬂW’JWS’E‘aM/ .. R 13:. . R s.
NW1?" +21? + R‘)Mn '— WM“ Mﬂwx/
Va: M59 anviﬂﬂf' an + fhfu+ vﬁuﬁ' : _\
4N“ 1+ RC5 3 3 RC5 — (RCsH)
\+R65 W 45:.” 2 RCs—l V/\/
4’73“ RC§+\ ...
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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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