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Midterm1AS06

Midterm1AS06 - (1 point(1 point(3 points(3 points(3...

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Unformatted text preview: (1 point) (1 point) (3 points) (3 points) (3 points) (3 points) EE 307A 806 MIDTERM EXAM #1 CLOSED BOOK + 1 Cheat Sheet State assumptions and show work. Part 1 12 Points Part 2 14 Points Print Your Name: <5 17 L M 77 0%) Part 3 l4 Pomts Total 40 Points No unauthorized help given or received. I will not discuss this exam with another student until after 4:00 today. 1. Inverter VTC Analysis. Consider an inverter with the VTC shown below. Signature: A) Determine the value of the gate switching threshold voltage to within 10 mV. 4.00 V VIN <1 V VOUT = 4.80V —0.8 VIN VIN 21V M > VIN Vour E; 6 VM _: V *0/3MI/ > 4.50 V W: T; aZé7i/ B) Compared with the abilities of CMOS inverters we’ve discussed in class to regenerate logic levels, the above inverter has (circle the best answer): a. a better ability b. the same ability . orse b'l' Explain. . 4/5417 / 72¢ f/gﬂé’ mi ﬂ; V76 I; foe? j/a/ﬂmﬁ, 217;; 4 j 5” 7%? [Que/5 few/ﬁwzr/V , (/7675r zékﬁrﬁéi’s kiwi/76 [m V: fie gam/ L/T(5 7%«2‘ )435ZL01/8 /0//c é/e/. C) If one of the above inverters drives another, producing a ﬁnal output voltage Voyp, plot the resulting VTC consisting of Vow; vs. VIN for the series combination on the same set of axes above. VOUT Var) 747" VOW; —‘ 44/ MM 265,157/ VIN I; é 1.9V VOUTZ Va). >4? VMD 31/,él/ #2; D) If one of the above inverters drives another, producing a ﬁnal output voltage Vow), determine the value of the gate switching threshold voltage for the series combination to within 10 mV. 519/475 #5 ﬁg” ﬂ “Ewiv EE 307A 806 VTN= — VTP = 0.5 V, ksz = 40 uA/Vz, kp’/2 = 20 pA/vz, ,1 = 0 V'1,-2¢F= 0.6 v, and y = 0.0 v1”. Assume that body connections of all NMOS transistors are grounded and PMOS bodies go to +VDD. 2. NMOS inverter with several loads VDD: 4 V Please use the transistor parameters shown above and on the schematic. W/L ‘ 2 ' 1.. ﬁRL _ , L=1 7,. VGG"5V “‘7 mom (5 points) A) Determine the power dissipation in transistor M1 M2 J: M3 __ when Vm = 1.057 V and Vow = 2V. ' VOUT i.— V135): W147 -' 2V > V51“ V7 : 29;;V VIN —‘l:f”—1W/L=50 I, 3 £1, £1), (Véﬂ ‘l/WIJL ; [40%)50) [ﬂ Burl/)2: 520/44 ﬂ 3 V951 '3 (5 points) B) Determine the total power dissipation in the entire gate when VIN: V and VOUT= am : 1,04,; : (WWW/Md”? W (4 points) C) Compared with VIN: 1.057 V and VOUT = 2V, determine the inﬂuence on the value of VOUT of Check 3 r0 riate boxes) Increase Decrease remains Insufﬁcient n..- mesameEXPLAIN V 311 ‘15 how/e Pf 171’ 0/4 1/7Z I, , Increase W/L \/ fir pW/c’h pa//etw// az‘ “MI ~>> v 14” Increase W/L \/ Sfra/ygr lag/e £90 561‘ ()sz “'9 7‘ 167’ Increase value 3 i l l 1, Weakn- ,M/ﬂtpa at '9 ¢ My EE 307A 806 3. CMOS inverter driving CL = 4 pF. Please use the following transistor parameters: Vm = — Vm = 0.5 V, kN = 503 uA/VZ, kP = 200 uA/VZ, A: 0 V", -2¢F = 0.6 V, and y = 0.4 V“2 V. Assume that body connections of all NMOS transistors are grounded and PMOS bodies go to +VDD. VDD = 2.5 V (3 points) A) Determine the value of VOH M/n-ri /%V 9/35) /% 4/”, Léur = Me»: M» p 339/” Vm Vow N CL (3 points) B) Determine the value of VOL mm x?» we Ma ma W = e «'W e (6 points) C) Estimate the value of tpHL. gr 2.: p)‘ M” 3' l/pHI2ISV: [4‘47 1% 5447/ ma 0;; l A: V 2 IL”)? ; 14/ Z KUéVi—Slv ‘ V7”) :— S’ﬂﬁ/If/(f/ZEV~J,5V) : 9239/2 M/7 47 :l‘ V/I/JMr’32'5V Z" F52} 1/2””;1 W’Héb :A‘ZFV 2. % é/ﬂz/Za/W ‘ Z I ) ” .Z—xz/ Z MZ’EKI/ésy‘ 74914;” y;;;/_/ L 41¢ -’ .2 573% 2(391/va5'W/125‘V-K/t V : E” [12 ﬂ”? +L/w4] : jgmzm/r Wayﬂr/ﬂ : 487W»? 4/ (l’oy: j ifﬂ; Mig/ W/. n may : /, ﬁfwi (2 points) D) Circle and cross out the necessary words to make the following sentence true: Halving the value of capacitor CL would inereese / Jeane—unchanged / (decrease) the value of tpHL. Explain why. W [5 fora/mﬁgw/ 96 /m/ 61/4'5/ rfer/m: (A ...
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