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chapter2-device01

Course: MR 310, Spring 2010
School: Shanghai Jiao Tong...
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VLSI CMOS Design 2.CMOS Transistor Theory Fu yuzhuo School of microelectronics,SJTU Introduction Digital IC 2: Device omar fadhil,Baghdad outline PN junction principle CMOS transistor introduction Ideal I-V characteristics under static conditions Dynamic Characteristics Non-ideal I-V effects Digital IC 2: Device 2/74 Diffusion&Drift activity PN junction diffusion P-type hole...

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VLSI CMOS Design 2.CMOS Transistor Theory Fu yuzhuo School of microelectronics,SJTU Introduction Digital IC 2: Device omar fadhil,Baghdad outline PN junction principle CMOS transistor introduction Ideal I-V characteristics under static conditions Dynamic Characteristics Non-ideal I-V effects Digital IC 2: Device 2/74 Diffusion&Drift activity PN junction diffusion P-type hole concentration>>N-type hole concentration N-type electron concentration>>P-type electron concentration Diffusion result Build up space-charge region(depletion) Stronger diffusion current, wider space-charge region Drift result Opposite to the diffusion current Resulting in a zero net flow Digital IC 2: Device 3/74 Depletion Region hole diffusion electron diffusion p B p hole drift electron drift n r + x Distance - (b) Charge density. Cross-section of pn-junction in an IC process A Electrical Field SiO 2 (a) Current flow. n Charge Density A Al x x (c) Electric field. Al p A n B V Potential -W 1 y0 x W2 Digital IC (d) Electrostatic potential. 2: Device One-dimensional representation B diode symbol 4/74 PN junction energy band Example an abrupt junction has doping densities of NA=1015 atoms/cm3, and ND=1016 atoms/cm3, calculate the built-in potential at 300K, ni is the intrinsic carrier concentration in a pure sample of semi and equals approximately 1.5X1010 atoms/cm3 0 = Digital IC kT N D N A NN ln( ) = 0.026 ln( D 2 A ) 600 mV q n i2 ni 2: Device 5/74 Forward-bias Mode Applied potential lowers the potential barrier Diffusion current dominates the drift component Digital IC 2: Device 6 Reverse-bias mode Potential barrier is raised Drift current becomes dominant The number of minority carriers in the neutral regions is very small, so drift current component can almost be ignored Digital IC 2: Device 7/74 Pn-juntion outline PN junction analysis PN junction current PN junction capacitance Digital IC 2: Device 8/74 Diode static behavior Minority carrier concentration in the neutral regions near the pn-junction under forward-bias Minority carrier concentration conditions is supported by p-zone majority carrier concentration Digital IC 2: Device 9 How to calculate diffusion current? N-region diffusion computingp/nare mobility of hole/electron Diffusion coefficiency Dn 34cm 2 / s, Dp 13cm 2 / s dp n dp = qA D D p n dx dx p n (W2 ) - p n 0 p (W ) W - p W p n (x) = x+ n 2 n n 0 2 Wn - W2 Wn - W2 I D,p T pn0 N D n p 0 N A ni2 pn ( W2 ) pn0eVD / T N A Digital IC 2: Device 10 How to calculate diffusion current? I D p qAD D p pn0 Wn W2 (e VD /T 1) I Dn qAD Dn n p0 W p W1 (e VD /T 1) V / n p0 pn0 I D I D p I Dn qAD ( Dn Dp ) (e D T 1) W p W1 Wn W2 VD /T I S (e 1) Digital IC 2: Device 11/74 Diode static behavior Minority carrier concentration in the neutral regions near the pn-junction under reverse-bias conditions Digital IC 2: Device 12/74 Diode Current T = kT/q = 26mV at 300K IS is the saturation current of the diode Digital IC 2: Device 13/74 Models for Manual Analysis + ID = IS(eV D/T 1) ID + VD + VD VDon (a) Ideal diode model (b) First-order diode model Example 3.2 Digital IC 2: Device 14/74 PN junction outline PN junction analysis PN junction current PN junction capacitance Digital IC 2: Device 15/74 Dynamic behavior of pn-junction Junction capacitance/depletion capacitance -Cj Diffusion capacitance/dope capacitance -Cd dQ C dV Digital IC 2: Device 16/74 Junction capacitance P +N + The boundary of PN-junction could accumulate charge when bias voltage was changed, which show capacitance characteristic When forward voltage improvedmore charges pass,just like these charges are saved When forward voltage decreased, less charges pass, just like these charges are leaved 2011/3/15 Digital IC 2: Device 17 Junction capacitance Depletion-region width 2 si N A N D W j W2 W1 q N N 0 VD AD Maximum electric field 2q N A N D Ej N N 0 VD D si A Depletion-region charge Cj dQ j dVD AD Q j AD N AND 2 si q 0 VD N A ND C j0 C j0 si q N A N D 0 VD 1 0.5 2 N N 1 VD 0 1 VD 0 A D Zero-bias conditions C j0 AD Digital IC si q N A N D 20 N A N D 2: Device 18/74 Junction Capacitance Digital IC 2: Device 19/74 Diffusion capacitance Diffusion capacitance is dominant when PN-juntion works under forwardbias mode, because its current is supported by minority carrier Charge of diffusion region p n (x) = Q p qAD Wn p ( x ) p dx n n0 Wn W2 pn 0 e W2 qAD Wn W2 2 I 2Dp VD T 2 Dp p n (W2 ) - p n 0 p (W ) W - p W x+ n 2 n n 0 2 Wn - W2 Wn - W2 1 Qn Wp 2 2 Dn I Dn 2 Wn I Dp 2Dp Digital IC I D p qAD D p 2: Device pn0 Wn W2 (e VD /T 1) 20/74 Diffusion capacitance cont. Transmit time Tn Wn2 Tp 2Dp Diffusion current Qn QD ID Tp Tn T ID IS ( e Digital IC 2: Device 2 Dn Qn dQD dI I Cd T D T D dVD dVD T Qp W p2 VD T Wp 2 2 Dn 1 ) 21 I Dn Junction and diffusion capacitance Forward-bias mode Diffusion cap. Is dominant small RC effect for ignoring Junction cap. Reverse-bias mode Minority carrier concentration is very small, so its diffusion cap. Can be ignored Junction cap. Is dominant large RC effect Digital IC 2: Device 22/74 PN junction switch model Digital IC 2: Device 23 PN junction switching model Rsrc VD V1 ID Vsrc V2 t=0 t=T VD Excess charge Space charge OFF ON Digital IC Time 2: Device ON 24 Diode Model Cd dQD dI I T D T D dVD dVD T RS + VD ID CD - Cdiode T ID T Digital IC C j0 1 VD 2: Device 0 m 25/74 outline PN junction principle CMOS transistor introduction Ideal I-V characteristics under static conditions Dynamic Characteristics Nonideal I-V effects Digital IC 2: Device 26/74 What is a Transistor? A Switch! An MOS Transistor VGS V T |V GS| Ron S D Digital IC 2: Device 27/74 Why MOS transistor? MOSMetal Oxide Semiconductor type Field Effective Transistor-MOSTFET) Merits Simple manufacturing process yield Low power, high density High input resistor Digital IC 2: Device 28/74 Terminal Voltages Mode of operation depends on Vgs, Vgd, Vds Vds = Vd Vs = Vgs - Vgd Source and drain are symmetric diffusion terminals source is terminal at lower voltage Hence Vds 0 nMOS body is grounded. First assume source is 0 too. Three regions of operation Vg Cutoff + Vgd - + Vgs - Linear Saturation Vs Digital IC 2: Device - Vds + Vd 29/74 nMOS Cutoff No channel Ids = 0 Vgs = 0 + - g + - s d n+ Vgd n+ p-type body b Digital IC 2: Device 30/74 nMOS threshold voltage Vt>Vgs>0 depletion region(inversion) is formed below the gate Vgs=Vt A strong inversion is built up, the potential at the silicon surface reaches a critical value Further increases the gate voltage produce no further changes in the depletion layer width Digital IC 2: Device 31/74 Threshold Voltage + S VGS - D G n+ n+ n-channel Depletion Region p-substrate B VT is a function of several components Digital IC 2: Device 32 outline PN junction principle CMOS transistor introduction Ideal I-V characteristics under static conditions Velocity Saturation Dynamic Characteristics Nonideal I-V effects Digital IC 2: Device 33/74 nMOS Linear Channel forms V Current flows from d to s e- from s to d Ids increases with Vds Similar to linear resistor V gs > Vt + - g + - s d n+ Vgd = Vgs n+ Vds = 0 p-type body b gs > Vt + - g s + d n+ n+ Vgs > Vgd > Vt Ids 0 < Vds < Vgs-Vt p-type body b Digital IC 2: Device 34/74 nMOS in linear area n = 3800 cm 2 /v . s, p = 1800 cm 2 /v . s Charge per unit area: Qi ( x) Co x [Vgs V ( x) VT ] dV vn ( x) n E ( x) n I D vn ( x)Qi ( x)W dx dV I D n Co x [Vgs V ( x ) VT ]W dx L I 0 D dx VDS C n ox [Vgs V ( x ) VT ]WdV 0 2 2 VDS VDS W W ' I D=nCo x [(V gs-VT )VDS ]=kn [(V gs-VT )VDS ] L 2 L 2 Digital IC 2: Device 35/74 The Threshold Voltage where VT = VT0 + (|-2F + VSB| - |-2F|) VT0 is the threshold voltage at VSB = 0 and is mostly a function of the manufacturing process Difference in work-function between gate and substrate material, oxide thickness, Fermi voltage, charge of impurities trapped at the surface, dosage of implanted ions, etc. VSB is the source-bulk voltage F = -Tln(NA/ni) is the Fermi potential (T = kT/q = 26mV at 300K is the thermal voltage; NA is the acceptor ion concentration; ni 1.5x1010 cm-3 at 300K is the intrinsic carrier concentration in pure silicon) (2qsiNA)/Cox = is the body-effect coefficient (impact of changes in VSB) (si=1.053x10-10F/m is the permittivity of silicon; Cox = ox/tox is the gate oxide capacitance with ox=3.5x10-11F/m) Digital IC 2: Device 36/74 I-V character in resistive or linear region Page 92 'W I D kn L 2 2 VDS VDS (VGS VT )VDS k n (VGS VT )VDS 2 2 ' k n nCox n kn = k' n ox tox W W W = nC ox = n ox L L tox L Linear dependence between Vds and ID Digital IC 2: Device 37/74 nMOS Saturation Channel pinches off Ids independent of Vds We say current saturates Similar to current source Vgs > Vt g + - + - Vgd < Vt d Ids s n+ n+ Vds > Vgs-Vt p-type body b Digital IC 2: Device 38/74 I-V relation under Saturation condition VdsVgs-VT 2 2 VDS (VGS VT ) I D kn (VGS VT )VDS Vds Vgs VT kn 2 2 ox k = unCox = un to x ' n Digital IC 2: Device 39 I-V characteristic of saturation Charge per unit area: Qi ( x) Co x [Vgs V ( x) VT ] dV vn ( x ) n E ( x ) n I D vn ( x)Qi ( x)W dx dV I D n Co x [Vgs V ( x ) VT ]W dx L I D dx VGS VT C n 0 ox [Vgs V ( x ) VT ]WdV 0 I Dsat k 'n W 2 VGS - VT = 2L Digital IC 2: Device 40/74 Current-Voltage Relations Long-Channel Device 6 x 10 -4 VGS= 2.5 V 5 Resistive Saturation 4 ID (A) VGS= 2.0 V 3 2 VGS= 1.5 V 1 0 Quadratic Relationship VDS = VGS - VT VGS= 1.0 V 0 0.5 1 1.5 2 2.5 VDS (V) Digital IC 2: Device 41/74 Another method for giving I-V Characteristics In Linear region, Ids depends on How much charge is in the channel? How fast is the charge moving? Digital IC 2: Device 42/74 Another method for giving I-V Characteristics MOS structure looks like parallel plate capacitor while operating in inversion Gate oxide channel Qchannel = gate Vg polysilicon gate W tox n+ L n+ SiO2 gate oxide (good insulator, ox = 3.9) p-type body Digital IC 2: Device + + Cg Vgd drain source Vgs Vs Vd channel + n+ n+ Vds p-type body 43 Channel Charge MOS structure looks like parallel plate capacitor while operating in inversion Gate oxide channel Qchannel = CV C= gate Vg polysilicon gate W tox n+ L n+ SiO2 gate oxide (good insulator, ox = 3.9) p-type body Digital IC 2: Device + + Cg Vgd drain source Vgs Vs Vd channel + n+ n+ Vds p-type body 44 Channel Charge MOS structure looks like parallel plate capacitor while operating in inversion Gate oxide channel Qchannel = CV Cox = ox / tox C = Cg = oxWL/tox = CoxWL V= gate Vg polysilicon gate W tox n+ L n+ SiO2 gate oxide (good insulator, ox = 3.9) p-type body Digital IC 2: Device + + Cg Vgd drain source Vgs Vs Vd channel + n+ n+ Vds p-type body 45 Channel Charge MOS structure looks like parallel plate capacitor while operating in inversion Gate oxide channel Qchannel = CV C = Cg = eoxWL/tox = CoxWL Cox = ox / tox V = Vgc Vt = (Vgs Vds/2) Vt gate Vg polysilicon gate W tox n+ L n+ SiO2 gate oxide (good insulator, ox = 3.9) p-type body Digital IC 2: Device + + Cg Vgd drain source Vgs Vs Vd channel + n+ n+ Vds p-type body 46 Carrier velocity Charge is carried by eCarrier velocity v proportional to lateral E-field between source and drain v= Digital IC 2: Device 47/74 Carrier velocity Charge is carried by eCarrier velocity v proportional to lateral E-field between source and drain v = E called mobility E= Digital IC 2: Device 48/74 Carrier velocity Charge is carried by eCarrier velocity v proportional to lateral E-field between source and drain v = E called mobility E = Vds/L Time for carrier to cross channel: t= Digital IC 2: Device 49/74 Carrier velocity Charge is carried by eCarrier velocity v proportional to lateral E-field between source and drain v = E called mobility E = Vds/L Time for carrier to cross channel: t=L/v Digital IC 2: Device 50/74 nMOS Linear I-V Now we know How much charge Qchannel is in the channel How much time t each carrier takes to cross I ds Digital IC 2: Device 51/74 nMOS Linear I-V Now we know How much charge Qchannel is in the channel How much time t each carrier takes to cross Qchannel I ds t Digital IC 2: Device 52/74 nMOS Linear I-V Now we know How much charge Qchannel is in the channel How much time t each carrier takes to cross Qc VDS W I ds = = Cox (VGS - VT )VDS t L 2 VDS = k(VGS - VT )VDS 2 ox w w w kn = k = unCox = un L L tox L ' n Digital IC 2: Device 53/74 nMOS Saturation I-V If Vgd < Vt, channel pinches off near drain When Vds > Vdsat = Vgs Vt Now drain voltage no longer increases current Ids= Digital IC 2: Device 54/74 nMOS Saturation I-V If Vgd < Vt, channel pinches off near drain When Vds > Vdsat = Vgs Vt Now drain voltage no longer increases current Ids=k(Vgs-Vt-Vdsat/2)Vdsat Digital IC 2: Device 55/74 nMOS Saturation I-V If Vgd < Vt, channel pinches off near drain When Vds > Vdsat = Vgs Vt Now drain voltage no longer increases current Ids=k(Vgs-Vt-Vdsat/2)Vdsat=k(Vgs-Vt)2/2 Digital IC 2: Device 56/74 summary Shockley 1st order transistor models 0 Ids= Vgs<Vt cutoff k(Vgs-Vt-Vdsat/2)Vdsat Vds<Vdsat linear k(Vgs-Vt)2/2 saturation Digital IC Vds>Vdsat 2: Device 57/74 outline PN junction principle CMOS transistor introduction Ideal I-V characteristics under static conditions Velocity Saturation Dynamic Characteristics Nonideal I-V effects Digital IC 2: Device 58/74 Current-Voltage Relations The Deep-Submicron Era 2.5 x 10 -4 VGS= 2.5 V Early Saturation 2 VGS= 2.0 V ID (A) 1.5 1 0.5 0 Linear Relationship VGS= 1.5 V VGS= 1.0 V 0 0.5 1 1.5 2 2.5 VDS (V) Digital IC 2: Device 59/74 Attention : velocity position n = 3800 cm 2 /v . s, p = 1800 cm 2 /v . s Qi ( x) Co x [Vgs V ( x) VT ] Charge per unit area: dV I D vn ( x)Qi ( x)W vn ( x) n E ( x) n dx dV I D n Co x [Vgs V ( x ) VT ]W dx L I 0 I D satCoxW (VGT D dx VDS C n ox [Vgs V ( x ) VT ]WdV 0 W VDSAT ) nCox L Digital IC 2 VDSAT (VGS VT )VDSAT (VDSAT ) 2 2: Device 60/74 un ( m /s) Velocity Saturation u = 10 5 sat nz v for z z c 1 z zc vsat for z z c Constant velocity Constant mobility (slope = ) x = 1.5 x (V/m) c The critical field depends upon the doping levels and the vertical electrical field applied(1-5V/um) Digital IC 2: Device 61/74 Velocity Saturation ID nCox V 1 ( DS W nCox L Lx c W )L 2 VDS (VGS VT )VDS 2 2 VDS (VGS VT )VDS (VDS ) 2 1 (V) = 1+ V L c Digital IC 2: Device 62/74 Velocity Saturation I D satCoxW (VGT W VDSAT ) nCox L 2 VDSAT (VGS VT )VDSAT (VDSAT ) 2 VDSAT (VGT )VGT Digital IC 2: Device 63/74 Perspective ID Long-channel device VGS = VDD Short-channel device V DSAT VGS - V T Digital IC 2: Device VDS 64/74 ID versus VGS -4 6 x 10 -4 x 10 2.5 5 2 4 linear quadratic ID (A) ID (A) 1.5 3 1 2 0.5 1 0 0 quadratic 0.5 1 1.5 2 2.5 0 0 0.5 1 1.5 2 2.5 VGS(V) VGS(V) Long Channel Short Channel Digital IC 2: Device 65/74 Another two assumptions nz v for z z c 1 z zc vsat for z z c v nz for z z c vsat for z z c VDSAT (VGT )VGT VDSAT VGT VGT 1 xc L Digital IC 2: Device VGT xc L LvSAT n 66/74 Modify the velocity formula to be coherent with the familiar long-channel equations v nz for z z c vsat for z z c VDSAT x c L W I D nCox L ID VDS VDSAT n 2 VDSAT (VGS VT )VDSAT 2 W nCox L vsatCoxW (VGT LvSAT 2 VDSAT (VGS VT )VDSAT 2 VDSAT ) 2 Digital IC 2: Device 67/74 ID versus VDS 6 x 104 2.5 VGS= 2.5 V x 10-4 VGS= 2.5 V 5 3 VDS = VGS - VT 2 VGS= 1.5 V 1 VGS= 1.0 V 0 0 0.5 1 VDSV) ( 1.5 2 2.5 VGS= 2.0 V Linear dependence ID(A) VGS= 2.0 V 1.5 ID (A) 4 2 Saturation Quadratic dependence Resistive VGS= 1.5 V 1 0.5 0 0 VGS= 1.0 V 0.5 1 1.5 2 2.5 VDS(V) Long Channel Short Channel Digital IC 2: Device 68/74 A unified model for manual analysis G S D B Digital IC 2: Device 69 Simple Model versus SPICE 2.5 x 10 -4 VDS=VDSAT 2 Velocity Saturated I (A) 1.5 D Linear 1 VDSAT=VGT 0.5 VDS=VGT Saturated 0 0 0.5 1 1.5 2 2.5 V DS (V) Digital IC 2: Device 70/74 A pMOS Transistor -4 0 x 10 -0.2 VGS = -1.5V Assume all variables negative! ID (A) -0.4 VGS = -1.0V -0.6 VGS = -2.0V -0.8 -1 -2.5 VGS = -2.5V -2 -1.5 -1 -0.5 0 VDS (V) Digital IC 2: Device 71/74 Summary Strong Inversion VGS > VT Linear (Resistive) VDS < VDSAT Saturated (Constant Current) VDS VDSAT Weak Inversion (Sub-Threshold) VGS VT Exponential in VGS with linear VDS dependence Digital IC 2: Device 72/74
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C ompSci 101 FCTHE UNIVERSITY OF AUCKLANDFIRST SEMESTER, 2001City CampusCOMPUTER SCIENCEPrinciples of Programming(Time allowed: TWO h ours)Surname:Forenames:Student ID number:INSTRUCTIONS:Attempt ALL questions - write your answers in the space
University of Auckland - COMPSCI - 101
C ompSci 101 FC 2001Terms Test AnswersQuestion 1 (10 marks)Part A (5 marks)The contents of a simple HTML file is given below. The page contains an applet called Test, which hasbeen compiled, and the class file for this applet is in the same directory
University of Auckland - COMPSCI - 101
COMPSCI 101 SC/STTHE UNIVERSITY OF AUCKLANDSECOND SEMESTER, 2001Campus: City/TamakiCOMPUTER SCIENCEPrinciples of Programming(Time allowed: TWO hours)NOTE:Attempt ALL questions.Write your answers in the space provided.There is space at the back f
University of Auckland - COMPSCI - 101
COMPSCI 101 SC/STTHE UNIVERSITY OF AUCKLANDSECOND SEMESTER, 2001Campus: City/TamakiCOMPUTER SCIENCEPrinciples of Programming(Time allowed: TWO hours)NOTE:Attempt ALL questions.Write your answers in the space provided.There is space at the back f
University of Auckland - COMPSCI - 101
COMPSCI 101 TESTTHE UNIVERSITY OF AUCKLANDSECOND SEMESTER, 2001Campus: City/TamakiCOMPUTER SCIENCETEST SOLUTIONSPrinciples of Programming(Time allowed: ONE hour and 15 minutes)NOTE:Attempt ALL questions.Write your answers in the space provided.
University of Auckland - COMPSCI - 101
COMPSCI 101 TESTTHE UNIVERSITY OF AUCKLANDSECOND SEMESTER, 2001Campus: City/TamakiCOMPUTER SCIENCETESTPrinciples of Programming(Time allowed: ONE hour and 15 minutes)NOTE:Attempt ALL questions.Write your answers in the space provided.There is s
University of Auckland - COMPSCI - 101
Question/Answer SheetSURNAME: .- Page 25 -CompSci 101 ACFORENAMES: .CompSci 101 AC 2002 Final Exam Model AnswersQUESTION 1a)b)c)d)e)-115.105d is: 12.0i is : 1232QUESTION 2private static String randomString(String[] words) cfw_int randIn
University of Auckland - COMPSCI - 101
C ompSci 101 ACTHE UNIVERSITY OF AUCKLANDSummer School, 2002City CampusCOMPUTER SCIENCEPrinciples of Programming(Time allowed: TWO h ours)Surname:Forenames:Student ID number:INSTRUCTIONS:Attempt ALL questions - write your answers in the space p
University of Auckland - COMPSCI - 101
C ompSci 101 AC 2002 Terms TestMODEL ANSWERSQUESTION 1a)b)c)d)e)f)g)h)i)2ter3.34THE CODE WILL COMPILE123456System.out.println(&quot;\n\n\n&quot;);i is: 80d is: 4.5-1655QUESTION 24 syntax errors:1) String allStars = ' *' ;single quotes, sh
University of Auckland - COMPSCI - 101
C ompSci 101 ACTHE UNIVERSITY OF AUCKLANDSUMMER SCHOOL, 2002COMPUTER SCIENCEPrinciples of ProgrammingTERMS TEST(Time allowed: 75 MINUTES)Surname:Forenames:Student ID number:Circle ONE of the following:Lab time:T hursday 11-1Friday 11-1Thursd
University of Auckland - COMPSCI - 101
- 13 -COMPSCI 101FCAnswer bookSURNAME:FORENAMES:DEGREE (BSc, COP, Etc):STUDENT IDENTIFICATION NUMBER:SIGNATURE:Examiner to complete:QuestionMarkQuestion1627384Mark95TOTALCONTINUEDANSWER BOOK- 14 -Surname:Forenames:COMPSCI 101FC
University of Auckland - COMPSCI - 101
COMPSCI 101FCTHE UNIVERSITY OF AUCKLAND_EXAMINATION FOR BSc ETC 2002_COMPUTER SCIENCEPrinciples of Programming(Time allowed: TWO hours)NOTE:Attempt ALL questions.Write your answers in the answer book provided at the end of the exam paper. Youma
University of Auckland - COMPSCI - 101
COMPSCI 101FC 2002 TEST AND ANSWER BOOKPAGE 1STUDENT UPI:Question 1 (11 marks)a) What would be printed out by the following code segment?int i = 10;double d = i / 3;System.out.println(d);3.0(1 mark)b) What is printed by the following?System.out
University of Auckland - COMPSCI - 101
C OMPSCI 101FC Principles ofP rogrammingTest and Answer BookT est - Thursday 18th April 6:30pm-8:00pmFamily name or surnameGiven namesLab Day and Time (e.g. Monday 9)I nstructionsThis test constitutes 15% of your final grade for the course.No one
University of Auckland - COMPSCI - 101
COMPSCI 101THE UNIVERSITY OF AUCKLANDSECOND SEMESTER, 2002Campus: City and TamakiCOMPUTER SCIENCEPrinciples of Programming(Time allowed: TWO hours)NOTE: Attempt ALL questions.Write your answers in the space provided.There is space at the back for
University of Auckland - COMPSCI - 101
COMPSCI 101THE UNIVERSITY OF AUCKLANDSECOND SEMESTER, 2002Campus: City and TamakiCOMPUTER SCIENCEPrinciples of Programming(Time allowed: TWO hours)NOTE: Attempt ALL questions.Write your answers in the space provided.There is space at the back for
University of Auckland - COMPSCI - 101
COMPSCI 101THE UNIVERSITY OF AUCKLANDSECOND SEMESTER, 2002Campus: City and TamakiCOMPUTER SCIENCETESTPrinciples of Programming(Time allowed: 75 minutes)NOTE:Attempt ALL questions.Write your answers in the space provided.There is space at the ba
University of Auckland - COMPSCI - 101
COMPSCI 101THE UNIVERSITY OF AUCKLANDSECOND SEMESTER, 2002Campus: City and TamakiCOMPUTER SCIENCETESTPrinciples of Programming(Time allowed: 75 minutes)NOTE: Attempt ALL questions.Write your answers in the space provided.There is space at the ba
University of Auckland - COMPSCI - 101
CompSci 101 SS CTHE UNIVERSITY OF AUCKLANDSummer School, 2003City CampusCOMPUTER SCIENCEPrinciples of Programming(Time allowed: TWO HOURS)Surname:Forenames:Student ID number:Login name (UPI):INSTRUCTIONS:Attempt ALL questions - write your answ
University of Auckland - COMPSCI - 101
CompSci 101 SS CTHE UNIVERSITY OF AUCKLANDSummer School, 2003City CampusCOMPUTER SCIENCEPrinciples of Programming(Time allowed: TWO HOURS)Surname:Forenames:Student ID number:Login name (UPI):INSTRUCTIONS:Attempt ALL questions - write your answ
University of Auckland - COMPSCI - 101
CompSci 101 SS C Terms Test 2003Answers to question 1, 2, 10 and 11QUESTION 1a)b)c)d)e)f)g)h)5.0Total = 51.5Total = &quot; + 5 + 1.5n\\n10097int rand = (int)(Math.random() * 50) * 2) + 1;System.out.println(rand);QUESTION 24 syntax errors:
University of Auckland - COMPSCI - 101
C ompSci 1 01 S S CTHE UNIVERSITY OF AUCKLANDSUMMER SCHOOL, 2003COMPUTER SCIENCEPrinciples of ProgrammingTERMS TEST(Time allowed: 75 MINUTES)Surname:Forenames:Student ID number:Login name (UPI):INSTRUCTIONS:Attempt ALL questions - write your a
University of Auckland - COMPSCI - 101
CompSci 101 S1 CTHE UNIVERSITY OF AUCKLANDFirst Semester, 2003City CampusCOMPUTER SCIENCEPrinciples of Programming(Time allowed: TWO HOURS)Surname:Forenames:Student ID number:Login name (UPI):INSTRUCTIONS:Attempt ALL questions - write your ans
University of Auckland - COMPSCI - 101
CompSci 101 S1 CTHE UNIVERSITY OF AUCKLANDFirst Semester, 2003City CampusCOMPUTER SCIENCEPrinciples of Programming(Time allowed: TWO HOURS)Surname:Forenames:Student ID number:Login name (UPI):INSTRUCTIONS:Attempt ALL questions - write your ans
University of Auckland - COMPSCI - 101
CompSci 101 S1 CTHE UNIVERSITY OF AUCKLANDFIRST SEMESTER, 2003COMPUTER SCIENCEPrinciples of ProgrammingTERMS TEST(Time allowed: 60 MINUTES)Surname:SOLUTIONSForenames:Student ID number:Login name (UPI):INSTRUCTIONS:Attempt ALL questions - writ
University of Auckland - COMPSCI - 101
CompSci 101 S1 CTHE UNIVERSITY OF AUCKLANDFIRST SEMESTER, 2003COMPUTER SCIENCEPrinciples of ProgrammingTERMS TEST(Time allowed: 60 MINUTES)Surname:Forenames:Student ID number:Login name (UPI):INSTRUCTIONS:Attempt ALL questions - write your ans
University of Auckland - COMPSCI - 101
COMPSCI 101 S2 C/TTHE UNIVERSITY OF AUCKLANDSecond Semester, 2003City/Tamaki CampusCOMPUTER SCIENCEPrinciples of Programming(Time allowed: TWO HOURS)Surname:Forenames:Student ID number:Login name (UPI):INSTRUCTIONS:Attempt ALL questions - writ
University of Auckland - COMPSCI - 101
COMPSCI 101 S2 C/TTHE UNIVERSITY OF AUCKLANDSecond Semester, 2003City/Tamaki CampusCOMPUTER SCIENCEPrinciples of Programming(Time allowed: TWO HOURS)Surname:Forenames:Student ID number:Login name (UPI):INSTRUCTIONS:Attempt ALL questions - writ