Exam 1. It is important that you clearly show your work and mark the final answer clearly, closed book, closed notes, no calculator.

Transcription

1 Exam N a m : _ S O L U T I O N P U I D : I n s t r u c t i o n s : It is important that you clarly show your work and mark th final answr clarly, closd book, closd nots, no calculator. T i m : h o u r 5 m i n u t s P r o b l m ( T o t a l 3 p o i n t s ) P a r t a ) 5 p o i n t s _ P a r t b ) 5 p o i n t s _ P r o b l m ( T o t a l 3 p o i n t s ) P a r t a ) 5 p o i n t s _ P a r t b ) 5 p o i n t s _ P r o b l m 3 ( T o t a l 4 p o i n t s ) P a r t a ) 3 p o i n t s _ P a r t b ) p o i n t s _ T o t a l : p o i n t s _

2 W 3 Exam ECE 559 (Fall 9), Purdu Univrsity P r o b l m : For th circuits and conditions givn blow, dtrmin th nrgy dissipatd. Clarly spcify your assumptions, if any. 3 p o i n t s ] P a r t a ) Inputs A and B ar switching simultanously from D D t o f o l l o w d b y t o D D. Assum out initially. 5 p o i n t s ] D D A B W / L / L K p K n o u t t p p C o x µ n C o x µ -. 3, 3-6 A / 6-6 A / t n. 3 D D A W / L C L f F B W / L A n s w r Whn th inputs A and B simultanously switch from to,th PMOS ntwork gts ON and currnt is drawn from supply that chargs th output nod capacitanc C L to. Enrgy drawn from supply ( ). L out L E i t dt C d C Enrgy stord in th capacitor C L, E i t dt C d C stord, C ( ). L out L out out L Pag of 3

3 A A B B B B Exam ECE 559 (Fall 9), Purdu Univrsity D D i ( t ) d D u r i n g C h a r g i n g out i( t) CL. dt D D A i ( t ) o u t C L f F So, E E E C L dissipatd, stord, C L. D D d D u r i n g d i s c h a r g i n g out i( t) CL. dt D D A i ( t ) o u t C L f F i ( t ) Pag 3 of 3

4 A B W Exam ECE 559 (Fall 9), Purdu Univrsity Whn th inputs A and B simultanously switch from to,th NMOS ntwork gts ON and th output nod capacitanc C L gts dischargd to. E i t dt C d C dissipatd, ( ) out L out out L So total nrgy dissipatd E E + E C + C C total dissipatd, dissipatd, L L L Putting C L f F and, E dissipatd,total f Joul P a r t b ) Th voltag at th supply trminal is switching from t o D D. Assum A and B. 5 p o i n t s ] D D A B W / L / L K p K n o u t t p p C o x µ n C o x µ -. 3, 3-6 A / 6-6 A / t n. 3 A W / L C L f F B W / L 3 Pag 4 of 3

5 Exam ECE 559 (Fall 9), Purdu Univrsity Whn th inputs A and B, th PMOS ntwork gts ON and currnt is drawn from supply that chargs th output nod capacitanc C L to. Enrgy drawn from sourc ( ). L out L E i t dt C d C Enrgy stord in th capacitor C L, E i t dt C d C So, stord, C ( ). L out L out out L E E E C L dissipatd, stord, C L Putting C L f F and, E dissipatd f Joul. P r o b l m : 3 p o i n t s ] P a r t a ) For th circuit shown blow, find th m i n i m u m v a l u o f so that OL.. OL rprsnts th output low voltag. Clarly stat all of your assumptions. What will b th OH? OH rprsnts th output high voltag. Explain your answr. 5 p o i n t s ] Pag 5 of 3

6 B B W W L Exam ECE 559 (Fall 9), Purdu Univrsity D D K n µ n C o x 4-6 A / o u t t n. 3 A W / L B W / L C W / L A n s w r D D I K n µ n C o x 4-6 A / I A O. t n. 3 A I A I C x / L I B C W / L / L I OL..8. Pag 6 of 3

7 Exam ECE 559 (Fall 9), Purdu Univrsity I has to b qual to th currnt flowing through th NMOS ntwork as thy ar srially connctd, i.. I I. NMOS I AB + IC To gt th m i n i m u m v a l u o f, I has to b maximum, i.., I NMOS has to b maximum. Th output low voltag is achivd whn A B or C and of cours whn A B C. Accordingly, if all th NMOS transistors ar ON, i.., A B C (as OFF currnt is assumd to b zro), th minimum valu of is achivd. To calculat I AB w hav to dtrmin th voltag at th intrmdiat nod, X. Both th NMOSs with inputs A and B will b apparntly in linar rgion. This is bcaus of th rason that thr is only. to b droppd across both th transistors and for both of thm th linar rgion condition would b satisfid, i.., ds,a < GS,A - tn and ds,b < GS,B - tn. Anyway, w can doubl-chck latr on aftr finding th voltag X. Sinc th NMOSs with inputs A and B ar srially connctd I I I AB A B (. vx) vx (..3 vx) *(. vx) (..3 )* vx (. vx) vx (.7 vx) *(. vx).7* vx. Sinc, X has to b lss than. and th transistor with input A is ON, w can nglct X with rspct to.7. Also, body ffct is not considrd hr. Accordingly, ( ). vx vx.7* (. vx).7* vx. vx vx vx. W can vrify now that both of th NMOSs ar indd in linar rgion. Pag 7 of 3

8 Exam ECE 559 (Fall 9), Purdu Univrsity I I I + I I + I NMOS AB C B C * 4*.7*. + *.7* * ( 4*.65 + *.).8 6 4* * * *.4 4 5* Ω So, minimum 5 KΩ In th prvious calculation w hav assumd diffrnt OL for diffrnt input vctor combinations and calculatd th minimum. If OL is assumd to b th output low voltag for all th possibl input combinations, thn w hav to considr only th highst-rsistanc path from output to ground whn th output is low bcaus that will caus th highst OL at th output. W nd choos th minimum for which th highst OL will occur bcaus if is mor than a minimum valu, OL will just go lowr. Accordingly, w nd to considr only I C bcaus that corrsponds to th lowst ON currnt through th NMOS ntwork whn th NMOS ntwork is ON and th output is low. So, * **.7* * ** * Ω. Accordingly, minimum 67 KΩ Pag 8 of 3

9 L H Exam ECE 559 (Fall 9), Purdu Univrsity N o t : Full crdit will b givn to any of th two approachs. Whn NMOS ntwork is off (for crtain combinations of th inputs A, B, and C), th currnt flowing through th rsistor will vntually charg th output nod capacitanc. Th charging tim will vary dpnding on th valu of but vntually th output is going to charg up to. So, OH P a r t b ) For th part a), xplain qualitativly th r s u l t o f b o d y f f c t on OL and OH. 5 p o i n t s ] s u l t o f b o d y f f c t o n O Sinc th substrat and sourc nods for th NMOS with input A ar not at th sam potntial, th NMOS will xprinc an incras in thrshold voltag. So vntually th total currnt flowing through th NMOS ntwork, I NMOS would dcras. But, I I. NMOS So would incras. Hnc, O out I L w i l l i n c r a s. s u l t o f b o d y f f c t o n O Sinc during th dtrmination of OH w considr th NMOS ntwork to b off and w hav only a rsistor conncting to out, b o d y f f c t w i l l n o t a f f c t O H. Pag 9 of 3

10 o E Exam P r o b l m 3 : ECE 559 (Fall 9), Purdu Univrsity 4 p o i n t s ] g s E ε ε f r g s G S D µ ε i n c r a s i n g g s µ d s f ( E ) f ( E ) P a r t a ) For th NMOS shown abov, driv an p r s s i o n o f l c t r o n c u r r n t x flowing from sourc (S) to Drain (D). µ, µ ar lctro-chmical potntial nrgis at th sourc and drain trminals, rspctivly. Th Frmi distribution function f(e) spcifis, undr quilibrium conditions, th probability that an availabl stat at an nrgy E will b occupid by an lctron. f ( E) + ( EE F ) whr E F is th Frmi lvl. Clarly show your stps, spcify your assumptions, and nam th paramtrs you ar using. A n s w r 3 p o i n t s ] Th lctron currnt that is flowing from sourc to drain (I SD ) and drain to sourc (I DS ) can b writtn as I I sd ds qan v + + qan v Pag of 3

11 Exam ECE 559 (Fall 9), Purdu Univrsity whr q singl lctron charg (C), A cross-sctional ara (cm ) of sourc/drain, n + numbr of lctrons/cm 3 travrsing from sourc to drain, n - numbr of lctrons/cm 3 travrsing from drain to sourc, v + avrag vlocity (cm/sc) of th lctrons travrsing from sourc to drain, v - avrag vlocity (cm/sc) of th lctrons travrsing from drain to sourc. Not that, th unit of I sd and I ds ar in Ampr. In gnral w can assum, v + v - v. W know that th numbr of lctrons, n dnsity of stats*frmi function. Sinc, w hav only on nrgy lvl, w will hav dnsity of stats as. Accordingly, n f ( ε ) + ( ε µ ) + ( ε µ ) for (ε - µ ) >> n f ( ε ) ( ε µ ) + ( ε µ ) for (ε - µ ) >> Sinc, according to th figur providd, w hav put som positiv voltag at nod D with rspct to sourc nod S, µ > µ n > n + Accordingly, thr will b a nt lctron currnt flowing from sourc to drain, i.., Pag of 3

12 Exam ECE 559 (Fall 9), Purdu Univrsity I I I qav( n n ) sd ds I qav + ( ε µ ) ( ε µ ) I qav ( ε µ ) ( µ µ ) I qav ( ε µ ) ( µ µ ) I qav ( ε µ ) ( ε ε ) ( µ µ ) Now, w can writ µ µ q ε ε q m GS DS whr, m is th body-bias cofficint. It ariss from th fact that whn a voltag GS (gat-tosourc voltag) is applid, thr is som voltag drop that happns across th gat oxid as wll. Thus only a part of th voltag applid is rsponsibl to shifting th gat nrgy lvl downward. Thus, m > in gnral. So, whr, ( ε µ ) GS m I qav GS m I I I ( ε µ ) qav. q DS q DS Pag of 3

13 E E Exam ECE 559 (Fall 9), Purdu Univrsity P a r t b ) Explain qualitativly th f f c t o f D r a i n I n d u c d B a r r i r L o w r i n g ( D I B L ) that you hav drivd in part a). on th currnt p o i n t s ] Whn a positiv voltag at drain is applid (with rspct to sourc), it lowrs th ffctiv barrir hight as shown in th blow figur. Idally gat only should hav control ovr dtrmining th barrir hight. But, drain gts also som control as it is lctrostatically coupld to th gat. D r a i n I n d u c d G a t B a r r i r L o w r i n g S o u r c D r a i n f ( E ) f ( E ) So du to DIBL, th lctrons going from sourc-to-drain and drain-to-sourc both ar (xponntially) incrasd. But as th Frmi lvl at drain is at lowr potntial than that of th sourc, th incras in th numbr of lctrons travrsing from sourc-to-drain is mor than that of for drain-to-sourc. Accordingly, t h c u r r n t t h a t i s c a l c u l a t d i n p a r t a ) w o u l d g t i n c r a s d. Pag 3 of 3