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Exam-2_students_Sol_F06

Exam-2_students_Sol_F06 - ECE 271 Exam-II Solutions Prof...

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Unformatted text preview: ECE 271 Exam-II (11/08/2006) Solutions Prof. Hanaﬁ 1) Calculate Kn” for an NMOS transistor with pm = 500 cmZ/Vs for an oxide thickness of 50 nm. What is the value of Kﬂ if W = 10 pm and L = 1 pm. (5 points) .- t. E , 3,93 ; - (ﬁg-ﬂ: 39(8554'kliy-AEUH) 5“: : “Him“ :lui ‘5-‘~ =11): *9 =: fﬁﬁG i - 1 -’5' ' A - '- ‘ ' IE Tm. V — sec 211415."! “371(miksxs‘m-z) , a F _ 1' ‘ , Jig-i Kﬁ=3-l’3‘61{3—5 —34< ~ 19“ 1:54.50? ' V —sec 11' ’2’ 2 v» 1"" ET“ A]? = A i; — v j: 34.5 x 10/ 1 = 345 LLA/V2 M 2) Calculate K; for a PMOS transistor with up = 200 cmZ/Vs for an oxide thickness of 50 nm. What is the value of KD if W = 25 pm and L : 1 pm. (5 points) '2 " E 33\$ , M €352: 3%39(38543'18'HF563;?l Rf; :‘Urf‘n‘ 2;!” w =‘{[.F 9 :3 213%} - 2;____:g——’————-—-— ’ - F - .th 3w~=33~5‘1‘-}£ =13.5‘ , V— sec y - MI ”W ,. = K ? _ Exp. .31- L 13.8 X 25/1: 345 [LA/V2 W 3) For the transistor in problem 1 if VTN = 1 V, ﬁnd the region of operation and drain cunent when: a) VGS : 2 V and V333 : 0.5 V (5 points) b) VGS : 1 V and VDS : 2 V (5 13011113) 0) VGS : 2 V and VDS = 2 V (5 points) __ V 2 I V0! F a) V65 T V f; 0 ‘ 5 V DS .. 1/ V P? VG S T > D S 7/77 5,) VcerTf ° WEE” Egg; ___ :2 [ Val/‘— C) L38 ”VT : 2 VDS : 2V \/ Salad/L m 5% U09 71/63" 7. ﬂ 4) For the transistor in problem 2 if VTP = - 1 V, ﬁnd the region of operation and drain current when: a) VGS 3 - 4 V and VDS = - 5 V (5 points) b) VGS 1” - 5 V and VDS = - 3 V (5 points) C) VGS : - 1 V and VDS = - 0.2 V (5 pOiI'ltS) 5) Estimate the transeonductance for the transistor in Figure 1 for VGS = 4 V and VDS = 4 V. (5 points) 80G - A 600 <2: i E. 400 5 E 200 0 Dminusource voltage (V) Figure 1 \$1} 760 — 140 L41" -. g}, =_,_ = —~———£[—— = 31C} “us ... ,1 ‘ I: _ w W in (ﬁg _; 3 V w—‘FW 6) Calculate the transconductance for the transistor in probiem 2 for VGS : — 2 V, with VDS = - 3.3 V if VTp = - 0.7 V. Check the saturation region assumption. (5 points) 7"sz 5 gm = kp |VGSHVTP| = 345 12 — 0.71:. _ .5 us lVGSHVTPl = 1.3 v < |VDS| = 3.3 V 7) Find the drain current for the NMOS depletion-mode transistor in Figure 2 if K,” = 25 ”AN2 and VTN : —2 V. (20 points) I EEXIQT'WO'L. "1.: <00 J. {a} If the transistor were saturated, than 3 _ a, {—1—;(._“} _ “ H“ but this would require a power supply of greater than 50 V, Thus the transistor must be ope-rating in the ‘ . ___,_...:h. MWFW tricde regton. 1M”... , mV— K; _—i .. . If.“a ——i%:250.~;10°;0—t—2}— A gym 10‘ Q \ - 3 10— V35 =12)5V33(4 3153.} and 7,5. = (1206517 n 3311 g the gnadmz‘ic egamrforz. 1‘ ' {1.2665 _ I... z 2503‘10‘53 2 — 2.9.2965 = 939 Aug _, :\ Z .3 ¢ a...» 8) Find VH, V1, and the power dissipation (for 00 = V1,) for the logic inverter with resistor load in Figure 3. The NMOS transistor (MS) has K; = 25 uA/VZ and VTN = 1 V (15 points) For Mg :33". I, = om; try: 53/. 5 - , 5—K ,..5 {5-2- ,. , ‘33., a _- FQF'pre-En: A :KVELZJ—Ifnr—AE F l/-;:|_§23—'Té:ab—T ‘ “' 2503.11 " 2 _.‘ ' ' 1 - .3" " r? "' , , .,. _ J, ,1343", , ,, , , z , 3—K; 2(2319') s51}, 33—1-:fﬁ _,-;.:~.Ig‘—~61VE.+3=G 2 ‘ L " 5' W'.\A\,,M\.~ :— 033: . ,. , . If; #108381” g I, =,_,_‘ G" 5 =24,ﬁ‘[email protected]=5V{24.613%):133yﬂ" @ - .ﬁiA-F .f’w E’W , W ....-. 2001K} . , 1.4 9.0338 Checlmsg: D = Sat—i z: —1— : : poo-28 = 34.6 3:3 9) Find VH, VL and the power dissipation (for oo : VH) for the logic inverter with resistor load in Figure 4. The PMOS transistor (MS) has WfL = 7.5, Kp’ = 10 uA/V2 and VTp = - I V (15 points) ...
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