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Unformatted text preview: ECE 271
ExamII (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 ;  (ﬁgﬂ: 39(8554'kliyAEUH)
5“: : “Him“ :lui ‘5‘~ =11): *9 =: fﬁﬁG i  1 ’5' ' A  '
‘ ' IE Tm. V — sec 211415."! “371(miksxs‘mz)
, a F _ 1' ‘ , Jigi
Kﬁ=3l’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 depletionmode 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 operating 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 {52 ,. , ‘33., a _ FQF'preEn: 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" @
 .ﬁiAF .f’w E’W , W ..... 2001K}
. , 1.4 9.0338
Checlmsg: D = Sat—i z: —1— : : poo28 = 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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 Spring '05
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