Solve both 2.10 Find the transistor schematic for the CMOS logic circuit realized by the layout...

the equanity,0-4 um Congidev 2-10) bransistore Ps game and Pa The lengts, L of all tfe given as follows L 2 2 (o-4m) O 8um the lengt, E, is Hence,

following are the wielts of the tronsisto1s The follow ing Ps #e widt of upper trans stoNs The - 8 Co 4 um) 3 2 m The following s te widtt, of lowev tians fetos 2.4 um The following Po the transietov Schemrahc fr te cmos Logic cvcuit Yealizecl by, fke layout shown in fguve 1 3.2U 3-2um O-8 um 0 8 um 4) 3-2 Hm Ou 2 ) 2 Am 2-4 Atm o.8Hm O 8 um OC b o 2 4 uM Figurre 2

The followng tble the are and perimeter Of each Jun Table 1 perimeter (um) Junction ode Avea Cum)2 1 3.2x2-4C32+2-4)-8 = 768 Vpр J2 (3 2x2 4C2 4+2-4)>48| J3 3.2x24) C3-2 +2-444 8 2. 15 C2.4x 2.4 5.70 C3.2 +3.2 +o 8) -2 2. GND J7 C2.4x8) 92 Co-8t08)A 3 |(2.4x2.4) = 5.76 (3-2+3.2+0.8 3.2 J8 2

1-33) Consrcdev the bistable multivibrutov ccuit as figuv 1 shown in R2 Vout 2 WAN art wW Fisure The output Vout f te a sjuare wrave wit peak ampletudles V and V Heve, R2 Rz Fr and -( V+ R¢TR2 R1 +R2 Since R, R2 = R3 - Ry =R5 =R =R.

R andl V R tR RtR

+15 V 1 ane v 2. V+ - +7.5 V omol v- 5 -7.5 V Thu he pezik Volge of t5e Square wave fa 5 V Thus square wave output Vout wfth peak mplitucle t-5 v f input to #e integvatov. Hence, the oufput Veutz of the fntegyatov fe a biangulav -wih perik amplitude t 5 V wave The Yesisttnces, R, = R2 = R3= Ru= Rs - Re=R. Detevmine n zenev Voltoge V VzVo V 6.8V Theis, e Zeney Volttge v, is the period T by Sebstituhng 10x 03 Hz-fov f Detemine T- Ox103 Thus, tte tme peviod of the output tieingutav waveform ia oo us The pulse widt5 of the squave re fa Jf a square uare fa appliecl to an mtegratur, e autpet of tke ntesiatov is a tríangulav wave

The aprctev chavges and olischavgeA to te uppey Valuea yespecively. Hephold Lowev The Voltage acrobs te capzucitov s Vet Cto) Vc Ct RC The Capacifto changeB from tte Sauave wave uhoae pea amplitudes ave 75 V .5 7.5 t 2.

RC 2 7-5 7.5 + 15)-to) 2 RCo-5x16 75 2 R Co-5X16 1 R Co-5X16 1 0 Sx (5-ט 200 ka Thevefore, e yesiplance Values qye, RIR2 Rs =Ri, R R = 200 K Apply kichheff's Curret law to e culput node ef t5e bistable martivibyatov ccuft Vpeak min . Vpenk -VE Vpeak R1tR2 7.5 -535 /3-.5 200t200 200 5.5 0485 +0.05625 R7 5.5 I05

R, 1kn Therefore, e vesistan ce e, fa 5-1 kn Aebfstable muHvfbratov cřcuit as shoun in favea Consfcle 2oukA 200kn Vouh 200kA 5.1 kn 20ok MM -ww wY Figre 2 Draw the ntegyator outpLt Yoltage wAvefom as shown in fguve -5 Appuet 7.5 -+7.5 OUTPUT T = 100 LIs Thus, te fntegratov output Psa bríangutzar wave uf a pevia) of lou s aye ke peak anplitude (Vpeake) of 5 v