ES 330 Electronics II Homework # 9 (Fall 2017 Due Monday, December 4, 2017)

Transcription

1 Pag1 Na OLUTON E 330 Elctronics Howork # 9 (Fall 017 Du Monday, Dcbr 4, 017) Probl 1 (14 points) Dsign a MO diffrntial aplifir illsuratd in th schatic blow to oprat at O = 0.5 olt with a transconductanc g = 1 A/. Find th (W/L) ratio and th bias currnt to t ths rquirnts. Th procss tchnology usd to construct th NMO dics ha paratrs alus of t = 0.5 olt and nox = 0.4 A/. Assu th two dics ar ntical and ignor th drain-to-sourc rsistanc r0. g D / A A / / 0.5 and thn O O O 1 W L D n OX O 1 1 A W W 0.5 A L L = _0.5_ A (W/L) = _10_

2 Pag Probl (1 points) Dsign a MO diffrntial aplifir to oprat fro powr supply oltags of DD = 1 and = -1, and dissipat at ost 1 W of powr P at its quiscnt stat (i.., no applid input signal). Find th alu of O so that a alu for th diffrntial input oltag, naly, = 0.5 olt, strs all currnt to on dic (i.., no currnt to th othr dic so that on dic is on and othr is off). Th agnitud of th diffrntial oltag gain Adiff is to b 10 /. Assu dic paratrs of nox = 0.4 A/ and nglct th Early ffct (i.., ignor r0). Find currnt, th drain load rsistor RD and th (W/L) ratio. Th quiscnt powr dissipation is P ( ) ; DD olts, and th axiu allowabl currnt is 1 W 0.5 A Th alu of O can b found fro th rlationship, 0.5 O 0.5 O 0.18 ( /) 0.5 A Thn g is gin by g Th diffrntial gain is gin by A g R ; A diff D D A 10 so 10 =.8 R R 3.6 k Finally, w dtrin th ( W / L) ratio fro 1 W L D n OX O Thus, ; W 38.6 L O DD diff D 1 A W 0.5 A ) L A (0.4 (0.18) = _0.5_ A RD = _3.6_ k (W/L) = _38.6_

3 Pag3 Probl 3 (1 points) (a) Draw th A schatic circuit of th diffrntial half-circuit for th diffrntial aplifir cll shown idiatly blow. [This aks us of sytry in a DA.] To do this probl w bgin by drawing th schatic quialnt circuit: (b) Dri an xprssion for th diffrntial gain Ad (dfind as od /) as a function of g, RD and brging rsistor R. Nglct th Early ffct (i.., ignor r0). Using sytry, a irtual ground appars at th point of rsistor R (giing two rsistors of alu R / ach). Hnc, od RD grd Ad 1 R gr 1 g (c) Ealuat th diffrntial oltag gain with R = 0?

4 Pag4 For rsistor R 0. od RD RD Ad grd ; as would b xpctd. 1 R 1 g g (d) What is th alu of R (xprss it in trs of 1/g) that rducs th gain calulatd in part (b) abo to on-half of that alu? To rduc th gain to on-half ( i.., A /), w rquir th condition R 1 ; thrfor, R g g d Probl 4 (1 points) onsr th BJT diffrntial aplifir shown blow. nitially assu is ry larg. (a) What is th largst input coon-od signal that can b applid whil th BJTs rain confortably within th acti rgion of opration with B = 0 olt? Writ an xprssion in trs of, currnt and rsistanc R. M ax R (b) f th aailabl powr supply is.0 olts, what is th alu of R that should b chosn to allow a coon-od input signal of 1 olt?

5 Pag5 For olts, M ax ( R ); thrfor W ha R olts (c) Using th R alu you found in part (b), slct alus for and R. Now w assu that th currnt gain = 100. Us th largst possibl alu for currnt subjct to th constraint that th bas currnt of ach transistor (gin that dis qually) should not xcd A (0.00 A). W can writ a rlationship for bas currnt B ( /) 1 μa Gin: = μa uppos w slct = 0.4 A, thn R = A B Probl 5 (3 points) Dsign a BJT diffrntial aplifir to aplify a diffrntial input signal of 0.1 olt and pro a diffrntial output signal of olts. To nsur adquat linarlity it is rquird to liit th signal aplitud across ach bas-ittr junction to a axiu of 5 illiolts. Yt anothr dsign rquirnt is that th diffrntial input rsistanc b at last 100 k. Th BJTs that ar usd in this diffrntial aplifir ha a = 100 (how connint!). lct a circuit configuration and spcify th coponnt alus (.g., powr supply oltag and rsistor alus) for all coponnts. You ay us an al currnt sourc for stting th tail currnt (EE) in th diffrntial aplifir. [Hint: You will probably want to us ittr dgnration rsistors.] nput oltag 100 appars across (r R ). Thus, th signal across ( r R ) is 50. Bcaus th signal across rsistor r is 5 for sall-signal condition, it follows that th signal across rsistor R is (50-5) = 45. Thrfor, R = 9 r

6 Pag6 Th input rsistanc R is R = ( +1)( r + R ) R = 101 ( r + R ) 0 ( r 9 r ) 0 (10 r ) Now to obtain R = 100 k, r TH W gt r 50 ; Fro r E 0.5 A o =, giing = 1 A ; E Th Gain is gin by bcaus alpha is ( 1) 1 W rquir a oltag gain of 0 R 10 k Hnc, to choos f th D oltag drop across R E od R R Gain ( r + R ) r + R and R = 9 r = 450 ; thus 0 R 500, w nd inforation for th M rang. is 5 olts and th collctor oltag swing is liitd to 1 olt, = 10 olts will b adquat.