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final_spr01_soln - Spring 2001 6.012 Microelectronic...

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Unformatted text preview: Spring 2001 6.012 Microelectronic Devices and Circuits Prof. J . A. del Alamo Name: Recitation: May 24, 2001 - Final Exam So L U T [ON 3 problem grade General guidelines (please read carefilliy before starting): 0 Make sure to write your name on the space designated above. Open book: you can use any material you wish. All answers should be given in the space provided. Please do not turn in any extra material. If you need more space, use the back page. You have 180 minutes to complete your exam. Make reasonable approximations and state them, i.e. quasi-neutrality. depletion approxima- tion. etc. Partial credit will be given for setting up problems without calculations. N0 credit will be given for answers without reasons. Use the symbols utilised in class for the various physical parameters, i.e. pm 19, E, etc. Every numerical answer must have the proper units next to it. Points will be subtracted for answers without units or with wrong units. Use 46 = 0 at no =.- p0 = n.- as potential reference. Use the following fundamental constants and physical parameters for silicon and silicon dioxide at room temperature: fl; = 1 X 1010 CHI-”3 kT/g = 0.025 V q = 1.60 x 10'19 C c, = 1.05 x 10‘12'F/cm em, = 3.45 x 10“13 F/cm 1. (15 points) The figure below shows the measured transconductanoe characteristics of the n- diannel MOSFET that you characterized in Device Characterization Project #2. Each of the lines represents a difi‘erent value of V33, starting with V93,men = 0.25 V, in steps of AVDS = 0.25 V. This device has L = 1.5 pm and W = 46.5 ,u-m. (1a) (10 points) In the space below, carefully sketch the gm vs. V93 characteristics predicted by the ideal MOSFET model presented in 6.012. Indicate the evolution of gm for several values of V35. Derive suitable equations for each of the branches that you identify. ' x ‘v L : F9“ we} ftJLd GLmd. +tfl'hktlu-l ’ “”14 {“91in \Il :V‘ 5"‘+"'r“t'rvv 1"" ‘J a], “A D -. E t V “V- ?‘v‘ .. [LP/“u- L- I. (- in! 1 ) val-($8.5 ll UwLfi-r") NMWMJ a.» VH maul leektpwvun‘l 4- VD! - \ ‘ ‘ (L nun-I _ A: V“ man no. r «1-H elk/la ¢cm+qu> Y3. o, lr-l *- 4"“ ”I’M" l \ I, PM}! “Spun-q: W 1'!“ 1 T- flnchi- V}! WWLM_ \u AgmLAr-Lga “MALI-J an V1” L-l‘ lflquMoUHJ ‘4' V5].- . . ~ ‘. L . TH tamer-v.3 Luluum +- um”, ma Jul-mat“ "km “ V” 3' ”m : V“ 'V7 er Vt: 1 Vin '* V: ‘ \ ‘ k on...»- u [41. (“um/x993 mug-Way, 'H“ 3"“ W‘fld‘mtw “L an MM!- MW“ ”3 +3 \0 oh UM '- (lb) (5 points) From the date. shown in the figure above, estimate VT and $5100: for the measured device. VT um 5* Mflmdm {rm 4N.- own} ”I a)“: V—I’; 0.2]. V W: n.» $41- mn Cnb’twi‘t {w 4-H VKLVI ‘1. fir“: $3 JVALVAHnd 4M liamILo-Aldrqc'l‘qnq 1‘". 4w um+ UM! 4N BIN. +Hh} L'rrhr‘ou It, #4 I v m: 0.?! v, FM 1‘“ 1 E)“ ("f VDA wh'uk 1] Henna-Alb. 2. (15 points) Consider the two-stage BiCMOS differential amplifier below. Vcc=+5 V + VI1 Veg—*5 V ,fl M1 and M2 are identical and are biased in the saturation regime. Q1 and Q2 are also identical and are biased in the forward active regime. Suitable parameters for these transistors are: nMOSFET: VT = 1 V and 3’1":an = 0.1 mA/V2 npn BJT: 5F = 250, V350... = 0.7 V and V05,“ = 0.2 V The current sources I; and I2 need at least 0.5 V across to operate properly. (2a) (5 points) Compute the power dissipation of this amplifier. T‘fi-L +t+~l tvrrlm‘l' “hurts 1M Vac. 1° Vi? 'I il+I1 3 Ema . Tb.- [I’Ww ourr-‘sr-oJLn-L 2; 4H... ‘_ W: (VtL*V—,) (II-Hi 1"]: ‘0 VA Gun-.31 _- 61: MW 5- (2b) (5 points) What constraint is imposed on RD so that the amplifier can properly handle 'a maximum comm-mode input of 3 V? (Express your answer as HQ > X or R9 < X. Give value of X). VHHA é. tomwxw—MOM :“IA'\., 4““ ILVPU‘M 8+ 0"] and M1 an 49“ka 4r} ‘ .‘ Fur Ml A-AJ M1 4% rpmninl "a JE'LWJ-HOVII VD 1“” 4" L? ALOVC 'H'“ ?}Q voHng I4. “Mr" *rn-nrkn‘lw.’ Ming +'n-L “Aral-NM qu'n-{V' TwL :1; Thu we ~44 -_ (2c) '( 5 points) If R9 = 5 k9, what. is the maximum possible voltage swing of node V02 with respect to ground? Give Voamgn and Vozmu- \ valid“ WE‘LL Ll #‘l‘ u.“ 4000 Why. FrrnL- :1 god). Le 3+ wLm,‘ .M JIM. t-nufl} 9‘- PM mead AJHMcuUfi-A NV} Jury? am i-n ELL Tm ¥h31 Lav-n. _. ‘3.<‘- H- g...L.l aha {fl yL¥ L3 Q1. 509% “I?“ led". Ju+~vc‘ix‘n\._ . gar nor-{kn f +N bmc a} Q?— ?s or} VB:- VLL" 2‘” gf—Slu‘h 0435A 1 2fv 7.. '1: Thu. L-‘d-m Q1. 9.“ ta}: Jfi+urswh VO‘LWG-n : V —\Ig‘”. -l- VGJ‘J. : 2.1‘ .. 0.1+ 0.2. -: L0 V {vi-mo. MAR; b lMé‘fi-W 4km Vouaj— (“lair—Rd «Jam, “4:1 L *L‘ m”; Vb2w3¢ tun-Le“ *H-r" ”UH-M 45M 4”- r-w/m", Jud-LA? (up? M Q! . 1‘42; thI' 7:”. Vow“. :IV. 2.; ~13; 5.th &' (Pa "1", J.J,r,.bfp.,‘ wt LAM-J .v '. It! ‘ Vt“ U -; 5-_2__'°;_ __ ,3: MA Kg, I k.- 3. (25 points) Consider the two-stage bipolar current amplifier shown below. At the input of this amplifier there is a signal source with an internal resistance R5 = 2.5 M2; at the output, there is a load characterized by R; = 2.5 M). . Voc=+5 V signal source Fls=2.5 K9 Veg—«5 V Both transistors in this amplifier are identical and are characterized by the following parameters: ,3]: = 100 and VA = 50 V. Treat all biasing current sources as ideal1 that is. with infinite internal resistance. (33.) (10 points) Draw a two-port low—frequency small-signal equivalent—circuit model of the first stage of this amplifier. Derive values for all elements of this two-port model. A Jari'lfilfill EE-lchhl “inf” wfirlbl far 9N dual Jim“ «‘1‘. \ ”an nun-u; Wind; AK} A?“ AH {manta-9h ,L 4,“ JUMAJJ QMHA reraWFL-M ol- +‘w. luv-Twin. irawiz‘l‘ar} NMLM an: a L? .9 015. ‘h “as '- 0-f— = L”— -_ 2.5 ML 3"“! ~10 mi [tfiis page left. blank intentionally} VJ! new unaw Lam «kw “JV- “Iv-3 1;»,- PM LOlefiu- in”: J4rtF: a“; ._ Sch. (H tic-Aug) ; 2.6 Mn- (3b) (10 points) Draw a two-port low-frequency small-signal equivalent-circuit model of the second stage of this amplifier. Derive values for all elements of this film-port model. TH MLDVLJ J+kv a¥ AJAX] AMA-Vuvar h a punch.» LNAHH 3H5}. HIM“, a J'v-JJ'ALKL 'LM -. “.f'k- cthr-l'lqb-tbu i) hug HM. LerLhJ 'I-lArDuJ'k +L~Q Jamal *(G-nrtr}nr L1 “1,9 JMA‘ ”4 J’Mnl/l-Pt‘rflfl/t pnquhJ ((u- a} haw: 'HN jaw! chJ+ w HA“! 4 Q1. Tm! \M +w “ha: (53c) (5 points) Calculate the loaded cun'ent gain A.- P‘JJWWv—k :54»! who“ *‘st man-u“ Fifi-Nu)». +w. 4-H J‘J‘LV.‘ ; M4, :3 {Cam \ \ VH1 “3' Ir“ “-4“- igf’» of this amplifier. Hand. M “M owl-p3- :L 4N PUJ‘V'U)‘ J40.“ I {3...}; “s: {er I H,“ ml 2 Em” v.3“ \ \ \ -. L H... VIN-.1 L‘ | h.-.‘°_:.—. _. ‘ ,4: mung“ \ I; i I: Vufl, ‘1‘. l5 — q!) M! ‘ 7.! It)!» :- “9° - 4. {15 points) The diagram below she-we an unloaded common-source amplifier with a. current ' s'ouroe supply. The adjoining table describes the relationship between device parameters and circuit parameters for this amplifier stage. Von JISUP signal source Circuit Parameters .— - — 31;; 1:0 . Parameters Imam-- I I I Vs I vour ._ _ _ _ _ I van I l Vss In this table, when changing one of the device parameters, adjustments are made to VGG. the gate _ bias, so that none of the other parameters are afiected. In this problem, you have to fill the fourth column of this table that contains the 3dB bandwidth of the amplifier. Use the same format as in the rest of the table and indicate in what direction war will _ -, change when the device parameters increase one at a time. Nothing else changes, except perhaps a for VGG as explained above. In the space below, provide an explanation for your entry. If there is no explanation, there are no points! (4a) (3 points) Iflsup T, how does wy change? Select:® —, 1. Why? TIM 344s LAD‘JWAIL‘ Ml A con/«wow. lawn-I. “why, iJ P‘Jttn Lb" 1| ””343 ’ r1“: C1; +C§A( I 4 Mandi] ~+ (fennel (.15 IL '31,»? 1‘, M 4N 4‘”; ‘ngt ,‘nruanl‘oq‘ IAvol L. M.) flai- : refill“: ll . «Ii-lbs 4‘wa RMPL‘D +kp,’ 04:?"87‘ . (4b) ‘(3 points) If W T. how does my change? Select: T. —, Q Why? T"? W? mw‘fl pd Luna-f} mmhd" M)! («pan—«46”; Inna-K La Fit and. mm + . 1m. Jul/k +V~1a Pwrvtic. }. +N {emu 44m) wye¢ (4c) (3 points) HpnCo, T, how does an; change? Select: T, —, @Why? :1 flu (by 1‘ WELL. “Wu”: 4“ para“;- [0-m+*—J' IMAM {-th 'IAVII’1‘ _ TN». “NHr . r'“ (4d) (3 points) If L T, how does war change? Select: T, —, OWhy? LP. Lr WW“ mm 4“ turbe-J' LOM+4¥A J C?) .IMLrnab-“J I “5‘ IA'” 1" (Lad; ‘- flnrlg 1‘ I 4““- ”E‘JGV'I' (4e) (3 points) If we now connect the output of the amplifier to a load resistance R1, (g roflroc, how does 2:}; change? Selectz® -, 1. Why? (*fl'hLL'ber! 5‘} 4-H 39‘1“; (2-,, “- (a H. r9; WV?) 'f’kql (1.“; $ Ifl‘r “(a ”(9L '2 {31, 4.4 (pf/{01, 1M“ 4M £‘M {9M tug: a.” - t;&}-\A M1 C.“ VJ. .lbhn— fink ii 4* H V‘Wwfi' aim. 1-, HAL 0,19% (“flux All “he“: inn-Law HM} 142.194 and 5. (30 points) Consider the following CMOS amplifier: Von signal source The devices are characterized by the following parameters: nMOS: Ln = 1 pm, W,1 = 4 pm, m0 = 50 pA/Vg, VTn =1 V,An = 0.1V‘1, Cg,“ = 6 fF, cm = 1 H", cm = 5 fF. pMOS: LP =1,m'1r1,I’Vp = 8 pm, ,upCm, = 25 pA/Vg, VT? = —1 V, AP = 0.1 V'1,09,P =12 fF. —-— ngp=2fF,Cdbp=10fF. Other values are: Vpp = 5 V, V33 = —5 V, R5 = 1 ME. and 1-2;, = 1 kn. (5a) { 5 points) Compute the value of Vac required to obtain a. quiescent output. voltage VOUT = 0 V. m- 1‘4...) l‘vlL’UML) 4h)“: U A tomnu‘l‘d) :WM'1H‘L Lava-H; .rkLF r h‘flu , .. -V \"H A WP . " Um - 1p, 1;!“ Lo, - I; ”M" x w k.“ , 8p. :m , m or-Mr J“. 99*0VI... A '4 w‘awl V5.1 : o V out. My! Vlrtr —_ o v . I W1 “NM ”MW“ 4%, ML... 1.) WEN} 4h...) .1 Vthrn 4m 4km. ‘1: n' Curve») +hro~JJh fiL— ”ML M A ”Ah”, I to“ = av" ' T 1 L .— ;wh _ } W ¢=-.. ,c av —v_V—— .__r v Jew J) “Jr Lu,” an 64 u I“) 2 Lrflp£°|££ W H I] A lovi—fiequency small-signal equivalent circuit model for this amplifier in an unloaded configuration sit the bias point specified in part (5a) is given below: ' with Gm = 1.6 m8 and Rm = 3.1 m. ' (5b) ( 5 points) Calculate the loaded voltage gain of the entire CMOS amplifier (don’t be alarmed if it comes out a bit. small, this is not a very good amplifier). TN \MV‘u‘Q imnbl- h‘rwJ Lfiuslvuuwj Quay“- may“! 1H 4-H eu‘lC-r: WWW”? ‘1; +Wg 4. . ”9.?- m. t: ' VM : _ cm. (autumn v3 Own-t V‘“ : V} 7H“! 4'” Dvbfluu with}! Wk .1le ”M“ M. ) ~ ’ Av“, __ ._._ __ GMO {fled} " ' — 1‘ _______._ 1 .— ll VJ if. 3.: I (5c) (15 points) Estimate the 3 dB bandwidth of the entire CMOS amplifier (that is, in its loaded donfiguration). To do this. calculate the time constant of each capacitor at a time (six capacitors at 2 points each). Then compute the 3 giB bandwidth in Hz (3 points). P- (D‘H‘thfih wt!“ " ‘fLfi gum.) JW- QWJ L‘— “:34 aha—.- Jr: m; M“:'-‘*‘ ‘34 I 4160 1+9»? :1: 3: :I 851-) 4’ CDMpv;t a“ 1km: LDVLJ+°‘“+‘ rl- (nLk thwafi+fv LtLG-WH. H" -11 [H M) : 2.2; an ,r“ [this 'page left intentionally blank] (lad- ' t ' CA5“ Tali.“ :. [4.5“ (2..” -.= C45“ flu. Hanna? 7. -- , -'l I 3x10 ”5 I '-, $33051 5 I I -—-+.—_ l". 3.1K CAL —- -17 __,_P_ we? :14di . Keno I NH ‘1: l-M '_ ’I I WK 7- '.‘— .————_.___...___w....___. - Z— T. 3’- éu'f'? + 3» ant-"1+ 3. Hunt” (5d) '(5 points) Calculate the voltage swing of the output node at this amplifier in its loaded confi- t'uration. "\‘N waivi I: 2&th L) 44“» ‘PMBJ sn‘uj ix}: Mu Lian, nsn‘mr_ TN cL°wnJ~v{»j wtli L4 as} L.) +H. ~Mo; “(SW 1,4, 4M UM“ ”Y9“: 53nd. {-Nrre ‘r; compu'l‘t H'mw-U‘rj , HNL urn—:73 gnJ. +H .{opmm‘k$ M; :muvl. \m out) NIL." 5—; amp-h an .L .LNM. \ Wt cam zmhflb wwflyh 4N J‘BwaIw-R-j M 4H [I'I‘Bufl‘l‘vx wimlu‘h»3 , TN. hM°J yo, tn}: +91 “MM-- ‘1 rpfiw‘. wk“ _..._..|, 1m.» = ‘Avl Mi“ _,.\ 3"; ll but“ 1‘ V6 5 = VT g.“ m yahwd 54-." [w 1..» m M} mm! M ”M ”H '9. WW ”k \ \ My. JV'NNY) hr t“) a"“nfll 4' .I VF:— 11V?“ vb ~. _ mm AV)... 1%“. ”5-D ._. (1+1Aul ) Av}, a V1- Au;_ _ V7 p ' _ msv H-U’wl H *1 Ll (5d) (5 points) Calculate the voltage swing of the output: node of this amplifier in its loaded confi- Euration. Pi“ ‘rhfi/lfii :J MM+IJ L“) wk PMbJ 3.n\"‘j in 0 4'“ 1/3ch ftlIA‘I-‘v‘hi ' \ “m cLowmwivj NEH L: M a.) “1 mm; W'fi ml. 4L4 DIM“. :tY‘w-t. Spud. +N-TG. \r) (nymph-+1, HEMMJ‘nJ I HM?» Brad-1:3 Hno-l- 4—qu pLDpunJm'fi Art :ma LO‘k _ W! Dual) KIA—r" L: Lmh.+t by!“ t L .LM Wt cam EKHLB KQquu“? '“M plkngvrtj :“ 4H lfilsuf‘" INA? ' TN I‘M") Y“ L33 4*“ “Mk-— cl 018M www— ? _.__¢ rm. .1» = \M M». ‘ Van bvl‘fi 1‘ I: WT (AM ha 5vam+ 54‘.“ [w gem m :MJ “,1 m MM 2. of Wm w. JV“:V“Y) '-r| L) n‘fl‘m’ 44"“— , V9. : 11an vb ’:. - IAV’ AV?“ TM \JY‘_b -_- (1+1Aul ) QV." -_- VT ‘i . maul"? —_ UV. VIVA“: ' L TWL tnrl o'l'N-I an)” 45 lawr.+l 4% LI ...
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