(Oxford series in electrical and computer engineering) Adel S Sedra Kenneth Carless Smith - Microelectronic circuits-Oxford University Press (2010 ), Exams of Electronics

Download (Oxford series in electrical and computer engineering) Adel S Sedra Kenneth Carless Smith - Microelectronic circuits-Oxford University Press (2010 ) and more Exams Electronics in PDF only on Docsity! —_— IN i sctrical Engineering P an ( , Dept. of Electrical Engineering Electric | i ain at i \ Faculty of Engineering Course , { ea } Assiut University Hours hk » \ 5 Mid Term Exam — November 22, 2018 fime:2 FQue +9 Marks: 30 points Instructor Dr. Wagih G, Girgis No of Pages: No of Questions: 9 Solve the folloy problems: : ERT circul nin fig. 1, What is the logic function performed by the following MOSFET circuits show! iB (4 points). Y1=(AB) ae vy fond Y Y=YI=AB a fm tH (Vy AND ( )NAND ( )OR ( )NOR (| )XOR ( )XNOR bt, The function is : he =D+(A+ B.C) The function is: ( ) AND ( )NAND ( JOR ( )NOR (| )XOR ( )XNOR ( Vv ) NON Of the above Page 1 of 7 Scanned by CamScanner  6 ns, , Vbi=0.7V, Cyo = 0.4 pF, and R= 5 K Q. [2 Points] . . we . e stored char} 2.a. Estimate the storage time ts, that is, the time it takes to remove the si in the circuit shown in Fig., when, forward biasing Vr = 5 V, rev! ge in the diode connected erse biasing Vr = -5 V, carrier lifetime T: = ! Diode current Vp-0.7% ran R K—> Diode voltage 407. > pK time tefl 4 we ee Ve < > ty bry ty Forward Voltage = 5 Reverse Voltage = -5 Resistance K ohm = 5 Depletion capacitance Pico Farad =0.4 Life time nano Sec = 6 If=(V£-0.7)/R =(5-0.7)/5 = 0.8600 Ir=(Vr-0.7)/R = (-5-0.7)/5 = -1.1400 Ts = Tau * log ((If-Ir) /(-Ir) )= 3.3727 Tr = (10E~-3) *R*Cj*1log(10)= 0.0461 Trr= Ts+Tr = 3.4188 Page 2 of 7 Scanned by CamScanner  . Calculate the overall resistance between points A and B if Rsquare = 2KQ, [2 Points} co] ee ee Number of corners = 3. Number of squares = 6 Total resistance = (6+3x0.6) x Rsquare = 7.8X2= 15.6 KQ 3.c. For the MOSFET layout seen in fig, , write the SPICE statement including the areas and number of squares in the source/drain regions with or without scale factor of 50 nm. [3 Points] ) Ss G L N-select Estimates for the areas of the drain/source are Ap = 40 and As = 45 The perimeters of the drain/source are estimated as Pp =28 and Ps = 36 The length of the MOSFET is | and the width is 10, or, L=1 and W= 10 The number of squares of n+ in series with the drain/source is NRD = + +0 and NRS=41 x4 The SPICE statement for the MOSFET is M1 0GSB NMOS L=1 W=10 AD=40 AS=45 PD=28 PS=36 NRD=0 NRS=4 Page $ of 7 Scanned by CamScanner  rd. Estimate the delay through a 400 KQ resistor made using an n-well with width of 15 and length 740. Assume that the capacitance of a 15 by 15 square of n-well to subsrate is 6 fF. |2 Points} Number of squares = 450/15 = 30 square Total capacitance = Number of squares X square capacitance = 30 X 6 = 180 {F Delay time = 0.35 R C = 0.35 x 400x10° x 180x107'S = 25.2x107? = 25.2 ps 4. State if each sentence of the following is correct (\) or incorrect(x). [5 points] ( ) The n-well resistor value decreases with the increase of its width. ( ) Infra-red rays incidence on a photo-resistive liquid converts it to solid isolator. ( ) A photo-resistive material is representing the isolator above the p-substrate passive area. ( ) In spice programing the subcircuit call begins by the word GOTO. ( Vv ) The n-substrate should always be connected to Vau. ( ( ( ( ee X_ ) The n-well resistor value decreases with the increase of its thickness. X ) The p-substrate should always be connected to Vaa X_) The n-well can form a parasitic diode and or transistor with an N-substrate 4X) The n-well can be used as a resistor, The voltage on either side of the resistor must be large enough to keep the substrate/well diode from forward biasing. ( X ) The growth rate of oxide increases with temperature. The main benefit of the wet oxide is fast growing time. The main drawback of the wet oxide is the hydrogen byproduct. Page 6 of 7 Scanned by CamScanner  ind Kr = 25 pa/V?. Determine: - The threshold voltage (switching point) Vm - The output and the input voltages at the transition point of the MOSFET (Vion , Vuin)- - The output voltage at an input voltage of 4 V - The average static power dissipation. 5 points). (5 p ) A Yaa Driver NMOSFET Constant KN uA/V2 : 100 8 Load NMOSFET Constant KP uA/V2 : 25 Gly p-channel Driver Threshold Voltage VTN: 2 Load Threshold Voltage VTP unsigned : 1 p Power Source Voltage VDD : 5 Vin | Ys Nort e——— -—— Results | D| k 2_k 2 ) 2) (Vj, —Vron) =F (Vin -Vpp-Vro.p) LL meta _ ‘s { 7) f \k ik, | Vin 1 i |= Vega tle “(Yop + V0,» | wing teeny oor . ji Vron i (Von +Vr0.0) Vv. = R = n= ( [i \1+.j— ! \ Vk } Vm = Vin = Vsp = 2.6667 Vit = Vsp = 2.6667 Votn = Vit — Vrx = 2.6667 -2 = 0.66667 a Votp = Vit + Vir = 3.6667 \ k 2 fa [2-(Vs —Vyo.n)* Vout ~Vou® |= “(Vi -Vop~ Vr0,») Input Voltage Vin > Vit : 4 Vo=0 ‘Consider a CMOS Inverter shown in fig. with Von = 5 V, Vix =2,[Vie]=1.V,5 Ky = 100 pa/v? Page 7 of 7 Scanned by CamScanner