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# midterm1solution - E13456 Midterm 1 March 3rd 09 Name Exam...

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Unformatted text preview: E13456 Midterm 1 March 3rd, 09 Name: Exam time: 75 minutes Answer all 3 problems. You will loose points if your answers are not legible. Problem 1 In this problem we are using a standard 0.25pm CMOS process. A minimum size inverter in this technology has an NMOS transistor size of WI1 / Ln:0.5 um / 0.25 pm (W: transistor Width, L transistor length), a PMOS transistor size of WI) / 1150.8 um / 0.25 pm, a gate capacitance of 5.2fF and an intrinsic delay of 20pSec. The ratio of intrinsic capacitance at the output of the inverter to its input gate capacitance is 1.2. A digital data signal generated by a microprocessor designed in this technology is buffered through a chain of inverters in order to drive an overall pad and wire capacitance of 333tF. Assume that the ﬁrst inverter in this chain has minimum size. 1. Calculate the number of inverters in the chain to achieve minimum overall propagation delay. 2. Draw the entire circuit and clearly indicate NMOS and PMOS transistors sizes (both with and length). 3. Calculate the overall prepagation delay of this chain. a on reﬁt—{Vs W “I’M £354— (Mater u“- Chou'n is W'\:nimum size. .2, r1 - , N 10. '£:4=£/:: 64 "Z" N’ “(Ex-02:5 EE456 Midterm 1 - March 3rd, 09 Name: Exam time: 75 minutes begow? WA [5 Shown wt ha. Dnqu {is gm» {= ‘EP; 37.1!” 4.33:— 160 PSLc l E13456 Midterm 1 March 3rd: 09 Name: Exam time: 75 minutes Problem 2 In this problem we are again using a standard 0.25 pm CMOS process. A minimum size inverter in this technology has an NMOS transistor size of WD / Ln=0.5 um / 0.25pm and a PMOS transistor size of WI) /Lp=0.8 um / 0.25pm. Implement the logic function F : AEE + 135+ EEC + ABC with minimum number of transistors XOR in complementary CMOS logic. Draw the entire circuit and Specify only the transistor Width (in um) next to every transistor. Make sure all inputs and the output are clearly marked. Assume that inverted input signals are provided. This is a. 3-1}! X0}? 9M Cam 67‘va aimptieg it 0.- little I‘m—whknj 'gg-Unvh‘m M: We Heirs? Wplmt FXOR Pull up Ltd'wOJ’K : “DR: 4 we did ml’ Mplﬁ‘lﬁ—VM %n¢'m~, ode war/Jo! have W0 Thar-e {tramsn‘sl-dv‘s we. E13456 Midterm 1 March 3'1109 Name: Exam time: 75 minutes PM — down NUme A304}. “0* Aim-‘9“?an w “Prvnchm MAJ W!— M 2 -€x hra ‘Wamsfsl-ws. add am Cnvakb ad- in Now wt pud' wagﬁw‘nj toJc'lIw, Md WW? “OM Ea cowshud— “him Ema {m mmgsm is wow M W perm Ina»: W ﬂaw wish/neg FMR Tlruz 52¢ (w) 0{_ 5w min SIZE ink/(Ala. 05> A ’ 3.771%. 3.74:: 3‘ We mu — ’- . ’ a. A 3W9 yuc _ Wm V A 39;“8 32,5: N) 621*: Elana/u FXO‘R A Wu A “5)! A Ina/1 E- LIE/u §* 16/“ B “37“ 3 “3):. §— 135/4 2 “51* c":- l'ﬁ'yt 9.2 trmnSCSWQ EE456 Midterm 1 March 3rd, 09 Name: Exam time: 75 minutes Problem 3 In this problem we are again using a standard 0.25um CMOS process. A minimum size inverter in this technology has an NMOS transistor size of WD 1’ Ln = 0.5 pm / 0.25u1n and a PMOS transistor size of WD / LP = 0.8 pm / 0.2511111. Consider the following circuit: M2 Wr WPIL P = 0.8pm1025pm WPIL P = 2.4pml0.25pm Input WniL n = wnlL n = 0.5}JmID.25pm 1.5pml0.25pm MI Assume that all four transistors have 0 identical drain contact area of l pm2, 0 Keq : 0.5 0 identical gate capacitance density of 3 fF / umz, - identical Gate—Drain and Gate—Source overlap capacitance density of 0.1 ﬂ3 / um, - identical bottom junction capacitance density of 2 fF / urnz. Ignore sidewall capacitance and wire capacitance and assume an average equivalent resistance m: 2.5“; for the ﬁrst inverter. Calculate the propagation delay from input terminal to output 2 terminal (Note Where the Output terminal is marked). Fn'rsi' we 11in. Ismail-’5 D-e M3 Mot ’ This humid W 0m wife/Jaye CaPaCIi-vwiq Wt- woaiol be aha/53,64 Quad deal/1N2ng (Bimini?) L—rH Glmci 'H-E L tramsH‘fans- E13456 Midterm 1 ' March 3rd, 09 Name: Exam time: 75 minutes CL s 2“" [ CMMoS»jak-Salr "r CPMoS—ﬁak-ﬁyi '1' CMMDS—qu- DH- 31— CPMas—ﬁaIe—de s it ‘zg'cox W“; L 1—2 CowM3)+(C¢KwP4L‘\'2C-awp4) .1. (Coxwug L4 ZQWNg) / - / W5 l-‘JHWI 1' Cox quL—rz Co - i _. -2- [tau 3 x Luau} 2x34 x I-§)+(gx2.1ho,1s+zm.\x 2.2;)+ (3“.qmgjf1. 2" 0" 3 “37+ (%K3x2-4x0-2§ + 2*0"‘2-4j II %(l.gq+1.28+ l-415'+l.6'8) : 3.117%"??? N u.) we Cox/Qde UM di'wuﬁim cwaCIWm 0"— ﬂu 'ELI‘rS“ 34R' 0 Cd'ref- :— Keel 3 Drain Area w (1,01% = O-GMMKZf-F +0.5; Hz: 2191: : Cw 0:01.601; C(M‘i‘ C:L = 2.43.2113: 52:76 *3: .— . I .2 R Chord“- 0.695 ...
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## This note was uploaded on 09/24/2009 for the course ECE 456 taught by Professor Mohammadi during the Spring '09 term at Purdue.

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midterm1solution - E13456 Midterm 1 March 3rd 09 Name Exam...

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