final_sample_soln

final_sample_soln - 3. [30] A source-degenerated PMOS...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 3. [30] A source-degenerated PMOS differential amplifier is shown below. Assume Id 2 ék’[%J(Vp — V,h)2, k' = 40,uA/V2 , and = 5 for M1 and Al; in your calculation. You may ignore capacitance in your calculation for low frequencies. Answer the following questions. 1) Assuming sine—wave input drive, what input amplitude would produce 1% M3 at the output? [20] 2) If two identical stages of the amplifier are cascaded, repeat the calculation in 1). Take the output from the second stage. [10] " K 9‘ _ «A. " 7 ‘ 7 , X ’t \Ijfiigti a lac L/\° "Lib / '— 5" r \ ‘ iwv r \ Ir e/J \ " ‘w kgk _' + I .l _. _ _:' x \n r ‘ a w, __ _‘ w x’) ,' T L, S ,> L _ k. g | \ L L _I l, _.\p K. i u - I f \ \ i ., L 7 "\ , l I 5,: {£121 Q l ~ _ \ N ‘ ' t n l \. v' x ‘: r" ~ iCK' 4" \‘L 5 "v 1 'v’ \ ~‘ I 1’ k? T.‘ ’1 a “r 'E H "a .i... ‘ ‘ L» A QLT \Ii _€. L,_ \ -_— u- I ’ M. I! i P ’ ‘ ‘I'L "L! if\L.§T'-. v 'V Lt i “inr' ’ I ‘. -‘ ' . a -\_ .., r . r —— 1. “ i.’ — i 7‘ *: k . a ‘i‘ " “"7 “ I ' : J. \. [/\! i i [\l f r g\ \ l l ik \r‘:/ J — , 1 i i m- - . a 'i" W ~- —r- c c v t 1’ 4 ’ * c s 4. [2m Fnr Lhu uuLivc cuauude anuilifici' shown b'clow, aHSIImu that lliu EUXiiifiFf amplifier is. ideal with .1 guin ni' aid. WiLh |J1I_' 11:11pm V. attached II) an H's": gmund. li'll: Efffli‘ti‘u'fl traitisumuiucmncc [Fm uE'Iiic umpiiiiur imiufincd as i, {Illl L —I V. Answer I'll: i'ulluwing qua-minus using 1.11": i'clurii-ralin HER} Inuilmci. Afiuuruu {11:11 1* r unl'l" in and 'r un: known. [LI-min: Llic 1mm! effect. 35..., r. r.- . | :I Fur Lliu feedback loop [bi-med by A" and M1. what is llic RR? [5] RR I Am. a . 6W1trfllifrug2 i'i‘ fimzflr—flu‘ri rm) and; “(‘91 Va: : Ari n MLTGL 2] C'fllfluime {£qu when RH --:rJ. 9W (flu—mtm Hit-1W RQ2<3<3 Tim mdLGD 1,15 and imfl mva fiwhmai Qwfiq : 3PM ll 3} ['alculate Gm... when HR =U [S] (vi/‘7 A“: r: WM M {iii/{:0 ‘W fit/E. Wiuf‘kfln” fiUfiJr‘WWJ—Ch {-‘0 :51 Elfimrflfi (migade ill—ELQEF WWK HLEQaui—e M2 gwmwmfi a L _._. film I re J 53mm, LTTIS‘E_"1"|I. ) ”—‘ Tweet} "— I : @=.fi_Ffffiz+fi>r mam rm+mI1-m 4] USE {he asytrlmeic guin i'nrmuhl In derive [Ln uxprussinn I'm [EL-L- G,” “1'11: fining: _-J_ [5] ER { _,_____, 1': [fimpq + @WIUHQR n fiw‘vw Cfiwfl Fm fifi‘ ' fiwfllmlrfll 4r n11.1— rfiz + fimflh 31.1“ hit TV" 2. {—1- i :1— m-l T—a I. T0 L @mzrmT—cl‘l-‘I’flm TVS: (flaw-rakrflz + rfli+rfl2 Am fimrolflmraz + fimlmmmmmfl) A131 @Heurflllrfll + aHQVmEL+ “(GIL—‘3 T91 <5?)me L@‘fifl)flmlroz+‘3 <5— I‘SW (A‘CH) fi'MLroarez 4?“ rm ‘1— rag 47' Ra H NQLL; “PM VLL‘LMELIFWV‘ r333 AV '2 aflinmwfimflw— “1.233 W EWM’E R0 03L "AAA g'LWjfl ' I _ Av ‘ {FEW J RD _ E“ ' 3‘ . nu1 require umnmun—nmdc fhcrlbuuk as explaincd in Chflfi. ‘r’nu may assume that flli lranxislnrs EI'E biased in Ibrwnrdrncliuc rug-gum. "r’nu can [game .I'“ but include r_T and 1'” In " [35] A dnublc-balancud l‘ully—dif‘t‘crenriu] ampiificr is Shuwn buimaa The uirru'u Linea your derivnrinn. H DOES 1hr: half—circuil INCIch axis-.1 Far :3 purEIy dil'l'urunlia! input? If 50, dmw [his mudfl']; il'nur, explain why. [5} 7‘] Derive [11c sumll—siglml diffel'cuiiul—Inudu guilt AIM = [[5] 3] Dues [he half-circuit mndel aim for u cumnmn—nmdc lupul'? Il'sn. draw this. mudul; IF Hut. explain 1.v|1y.[5j r 4} Darifi 1hr: snmII-signu] common-1de guin Am = l [[3] V.+V_ f"_+ll_ l1.=fh—ll_,l’“=_-_j ‘1”:[H_I ‘[ = j fi fifinffl Q42 [Q9 {ELIE ctman* cammfiwf fig Uo+ ZUUH "for L-mnmmL—made, fitpmi’, 3. [25] In the fully-differential amplifier shown below, all transistors are biased in the forward active region. Assume that the circuit is balanced and the gm of Q1 and Q; are known. Ignore the ro’s of Q1 and Q2. Consider only the capacitors shown in the schematic. :bO 1) What is the low-frequency small—signal voltage gain? [5] 2) What is the small-signal voltage gain at very high frequency? [5] 3) Sketch the frequency response (both magnitude and phase) of the voltage gain of this amplifier. Calculate the frequencies of the poles and zeros if any. You may assume gm -REE >> 1 to simplify your final results. [10] .4) A fully—differential amplifier typically requires output common-mode control. Explain ‘ what problem this circuit may have if four simple BIT current mirrors are used to implement the 13’s. Suggest a fix for the problem. [5] l) MVM drown" = 3M T“ [ -‘- fiMEE/Z 12» v0 1 Ro= 35 AA) 2 —G[VV\ E0 :3 1; “BE :_ ametlz 2‘ iii fimlEE/Z. 2. @LF 7.) @HF, 6155 9xme REE . VMM WWI “mm W- Wm M WWW, (M: 63m . r20 = 3% => An) = — QMRL/z @ HF Mmmx/ elf «Mug m1 «2 M as; glnor‘m m ero MH-org 34; (9‘ 52/ L97, ,CW m #11: pour w/ a bi! mama“ Somme oj— LL; 6; my my m. 9 WM m (m m m 7,2419 _ WM fink/L AV T: H [Shaka]: Lamb—ll) (‘TSWWEECEQ - 4M3”) * Q REE moo H‘ “‘1 + ‘w KEECEE ‘HW _ 1 _ W2 — REE (BE J WP :_ (H. Quin) W2} ,2: 5m ’LCEE 4) Nam/am plvm Mgeergrlower La’; [emu ’rv W-mw WW uwwn—MQwU-nfle,_ Col _‘ mpg _ —_— 401m 1 __ Mm QCwmer . Vac g vow) 10 ...
View Full Document

Page1 / 9

final_sample_soln - 3. [30] A source-degenerated PMOS...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online