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sol_old_exam3

# sol_old_exam3 - Georgia Institute of Technology School of...

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Unformatted text preview: Georgia Institute of Technology School of Electrical and Computer Engineering EE 3040A: Microelectronics Midterm Exam [[I 3 O l U +15 HS NOTE: Please show all your work to obtain partial credit. The exam is closed book and notes. You may use one handwritten sheet of formulas. NAME Social security number Grade: Problem 1 (20 points) A dimensioned energy band diagram for an ideal MOS—C operated at T = 300 K with VG ¢ 0 is shown below. Using this figure: (5) a. Determine (p1: and (pg. 4% “(E (b‘mé)_ “‘EF)/%/ 0-3eV _0. 3V 4%- b (bulk)- 21, (Sthce\ / 0°1+°3 — (3.3 f 1 ‘5 1 ‘t {4) b. Is the semiconductor n-type or p-type? Also, find the doping (donor or acceptor) concentration. A A» 0‘5 .oZSﬁ a-cboucwér—v ll’—+Y?cl;NA:M eff/Kr.» e A m3 (3) c. Determine V6. NA ' ‘ O7 7‘ ’0 M... M" _oﬁcv , ﬁVé. :: EFGnei’al) —— CF {Seat} (onAac‘lwhgpﬂeV 5V6 _ 1 i.) [VG ;o~m (3) d. Determine the oxide thickness x0. Use Ks = 11.8 and K0: 3. 9. .\1 Km W 21.16“ ammo of VG— (P544? ZQNﬁ¢S)x°.=70q..-.O>+ +17 x 7‘, 2-85’ “ammo; Xe: 03:13:; TS.— |03xlo Cm a) (5) e. Sketch the block charge diagram corresponding to the state pictured in the energy band diagram. What is the state of the MOS-C (accumulation, depletion, or inversion)? 0+ Tie Sow o C Semi (am my 1‘.) Mid: J'n’re/Chmﬁc} EA; | ovum N l n— “l:‘ F)lKT_lob°C .1214“? < As; [Aeeehonl (W <L".s<2""\=3 +C? (10) (5) Problem 2 (15 points) a. An n-channel MOSFET is used as a voltage-controlled resistor. In this range of operation, the voltage between drain and source is very2 small (vDs < 0.1V). The parameters of the device are Z/L= 100, C0 = 10‘8 F/cm2 ,VT- — 1 V, and 111.— — 180 cm2/(V. Sec) (assume that pm is independent of Va). Determine the gate voltage VG (or VGS) that results in a resistance RD: 2 K9. b. The source current in an n-channel MOSFET is IDS = 2 mA when V65 = 4 V. What is IDS when Vos = 6 V? Assume that the MOSFET is in saturation in both cases, and use VTn = 2 V. ‘ _\4 v ,v 7’ lbs__i“.(é5 T\ f) 15\$:- Kn (H-131): 5:) 1135:. “19,54 \4 n L4 ’13). 1‘35} 7.— T-DSL «1 :27 135';le Imcs’XlV‘A 2m Problem 3 (30 points) For the BJT ampliﬁer circuits shown below, find: (10) a. DC equivalent circuit and Q point (Ic and VCE). (6») b. AC equivalent circuit, amplifier conﬁguration (CE, CB or CC), and small signal model parameters. (4-) 0. Voltage gain (VG/V5). (3) (1. Input resistance R as speciﬁed in the circuit. (3) e. Output resistance Ro as speciﬁed in the circuit (4-) f. Current gain (Io/Ii) as specified in the circuit. For the transistor, assume that VEB = 0.7 in the active region, and BF = Bo = 100, and VA = 00 Other parameters are R3 = 100 K52, RE = 82 K52, RC = 39 K9, R3 = 1 K52, R3 = 470 K52, and VEE = VCC=12 V. a) :7 15' V“ 3 \1-0\ m ~|\3§¢lv MA Rﬁ't'ﬂé (ngl) (0‘3”?“7‘3‘82' ‘3 7 T \$0‘3émﬂ ‘7 1c: pigmlewl‘gﬁto ~o.\35mPl, TEAM-HM?) I‘VL In [email protected] Va 3 RE 136:): Vac + R911 “V355 Z’VFZC: VCWVEE ' KCICJEIE ﬁVEZcz 2H v82\$olgé 3‘VO‘351755V or VC€;_7 Sﬁ’v’ @PM ~. HZ; 35mA;v Court LMM Wuw,~m—4~ Extra work on Problem 3 comfnbﬂ BdSC ((13) "Q, ‘ 27 )1 V‘HA ' RE VS; jg...V\$:0.‘iXZ\/§ ,' R(+\:R\$HRE: “*3 -~:D‘CtXZk ﬁzz-His 33*“ RL: pan}, 3’qu‘70 :360kﬁ‘ M 301 4 H70 SmMLSu‘jMd moée‘ Pamm‘dc“? ...._...-——~v ’— 11 - 0""53 5.24%., h 9° W” swun- V VT . 0_c>254 T I C) ’ V V V ' " (”231753 Avi V: ,ﬁvx_\$b-o‘(32 Yn+(fﬁ°+\)?\+k \q 21— o 0 Hi: «:OJCMOI A) R YR ‘q‘2;0‘\ﬁk51~ 2L\$R\f\\\RE‘ “19*82 MA :— w,— .. 0+\ ‘0\ M :3qkﬁ ‘RC” KC 3) Kowkv; Y0((+jm R+k§z°05 no; 209*\ \$2 / ’ k P“ ‘16 ‘ 361 xltihxzdwﬁ H AW 33 5 197 “ ’7" ix; ; IZc-rV—35LES" “16% 34H7° ° b 1( 3,0 EL «L Problem 4 (25 points) For the MOSFET amplifier circuits shown below, ﬁnd: (.10) a. DC equivalent circuit and Q point (IDS and VDS). (5) b. AC equivalent circuit, ampliﬁer configuration (CS, CG or CD), and small signal model parameters. (4) 0. Voltage gain (Vo/Vs). (3) (1. Input resistance R, as specified in the circuit. (3) 6. Output resistance Ro as speciﬁed in the circuit For the MOSFET, assume that VTn = 1 V, and K, = 0.05 mA/VZ, and A = 0. Other parameters are R1=82KS2,R2=IOOKQ,R4=39KS2,RD=75KS2,R3=470K.Q,RS=IKSZandVDD=12V. a) 13¢ e1muulen+ (111qu 13a? 55533;” 393““‘§@ ' ., v50- ,ELVDD;££ “1:5 w" ' @3191 (004% WVS QéEV>VL 37W}; ~06l7°lVCr\$3 4‘50 ﬁVésrm‘r .lf)’ if? “Q N. ”Maw. am...» Z 3‘4“: ml 70%“? =77 "33-05" 57,”..“Vé “HA“ 0.01% M “WWW 35‘ 2:6! Extra work on Problem 4 U kvL {n INFO): V91): 12.9 1135+ VD5+ RH 1‘93 V -V :“éYV \/ QVDSLVL-C7§13ﬁ3¥000415Hle > as T Common Sowm -° (6) -- RG 143.05, ‘ a] \I 2%: 12m 27,: Rae ww‘ “‘l ”“ «em ”up: . I ' 7\$:‘+‘7°:‘ ~63” R+k=1Lsuzéz "MS"? =o-Q'I8 UM “6‘40”“? 73mg “ .444 {.05 sm“"'\$’j«u( Mod 9.“ PavameJ'e‘Ul jmsﬂ - .. — I 33 { l‘g:‘ ‘2 V~L .’ — .- ’ I I V5 V4» 0 5‘ A) Rina»,- 12.”: 12.3“ (16:26:49 SK :L Li) Rod’ti—ﬁao 1“ R9; Roo*\‘ KD:2.D:7> k D Problem 5 (10 points) The input and output voltages for a differential ampliﬁer (as shown below) in two sets of experiments are as follows: 1. When v1: vz = 1.5 mV, v0 :15 mV. 2. When v1 = 2 mV, and V2 = -1.6 mV, v0 = 360.2 mV. For this ampliﬁer, ﬁnd the common mode rejection ratio (CMRR). Vl !;A V‘ A \l +D'LFF- b \o c a A i 3 Case 1; \I\:\l1::) g WV :9 2. ...
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sol_old_exam3 - Georgia Institute of Technology School of...

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