4.3 MOSFET Circuits at DC

Transcription

1 10//004 4_3 MOSFETs Circuis a C empy.doc 1/1 4.3 MOSFET Circuis a C Reading Assignmen: pp K i Q: A: HO: Seps for C Analysis of MOSFET Circuis K = 04. ma 1K = Example: NMOS Circui Analysis Example: PMOS Circui Analysis Example: Anoher PMOS Circui Analysis

2 10//004 Seps for C Analysis of MOSFET Circuis.doc 1/7 Seps for.c Analysis of MOSFET Circuis To analyze MOSFET circui wih.c. sources, we mus follow hese five seps: 1. ASSUME an operaing mode. ENFORCE he equaliy condiions of ha mode. 3. ANALYZE he circui wih he enforced condiions. 4. CHECK he inequaliy condiions of he mode for consisency wih original assumpion. If consisen, he analysis is complee; if inconsisen, go o sep MOIFY your original assumpion and repea all seps. Le s specifically look a each sep in deail. 1. ASSUME Here we have hree choices cuoff, riode, or sauraion. You can make an educaed guess here, bu remember, unil you CHECK, i s jus a guess!

3 10//004 Seps for C Analysis of MOSFET Circuis.doc /7. ENFORCE For all hree operaing regions, we mus ENFORCE jus one equaliy. Cuoff Since no channel is induced, we ENFORCE he equaliy: I = 0 Triode Since he conducing channel is induced bu no in pinch-off, we ENFORCE he equaliy: Sauraion ( ) I K = S S Since he conducing channel is induced and is in pinch-off, we ENFORCE he equaliy: ( ) I = K

4 10//004 Seps for C Analysis of MOSFET Circuis.doc 3/7 Noe for all cases he consan K is: 1 W K k L and is he MOSFET hreshold volage. 3. ANALYZE The ask in.c. analysis of a MOSFET circui is o find one curren and wo volages! a) Since he gae curren I is zero ( I = 0) for all G G MOSFETS in all modes, we need only o find he drain curren I --his curren value mus be posiive (or zero). b) We also need o find wo of he hree volages associaed wih he MOSFET. Typically, hese wo volages are and S, bu given any wo volages, we can find he hird using KL: = + S G Some hins for MOSFET C analysis: 1) Gae curren I = 0 always!!! G ) Equaions someimes have wo soluions! Choose soluion ha is consisen wih he original ASSUMPTION.

5 10//004 Seps for C Analysis of MOSFET Circuis.doc 4/7 4. CHECK You do no know if your.c. analysis is correc unless you CHECK o see if i is consisen wih your original assumpion! WARNING!-Failure o CHECK he original assumpion will resul in a SIGNIFICANT REUCTION in credi on exams, regardless of he accuracy of he analysis!!! Q: Wha exacly do we CHECK? A: We ENFORCE he mode equaliies, we CHECK he mode inequaliies. We mus CHECK wo separae inequaliies afer analyzing a MOSFET circui. Essenially, we check if we have/have no induced a conducing channel, and hen we check if we have/have no pinched-off he channel (if i is conducing). Cuoff We mus only CHECK o see if he MOSFET has a conducing channel. If no, he MOSFET is indeed in cuoff. We herefore CHECK o see if: < (NMOS) > (PMOS)

6 10//004 Seps for C Analysis of MOSFET Circuis.doc 5/7 Triode Here we mus firs CHECK o see if a channel has been induced, i.e.: > (NMOS) < (PMOS) Likewise, we mus CHECK o see if he channel has reached pinchoff. If no, he MOSFET is indeed in he riode region. We herefore CHECK o see if: < S (NMOS) > S (PMOS)

7 10//004 Seps for C Analysis of MOSFET Circuis.doc 6/7 Sauraion Here we mus firs CHECK o see if a channel has been induced, i.e.: > (NMOS) < (PMOS) Likewise, we mus CHECK o see if he channel has reached pinchoff. If i has, he MOSFET is indeed in he sauraion region. We herefore CHECK o see if: > S (NMOS) < S (PMOS)

8 10//004 Seps for C Analysis of MOSFET Circuis.doc 7/7 If he resuls of our analysis are consisen wih each of hese inequaliies, hen we have made he correc assumpion! The numeric resuls of our analysis are hen likewise correc. We can sop working! However, if even one of he resuls of our analysis is inconsisen wih our ASSUMPTION, hen we have made he wrong assumpion! Time o move o sep MOIFY If one or more of he circui MOSFETSs are no in heir ASSUME mode, we mus change our assumpions and sar compleely over! In general, all of he resuls of our previous analysis are incorrec, and hus mus be compleely scraped!

9 10//004 Example NMOS Circui Analysis.doc 1/4 Example: NMOS Circui Analysis Consider his C MOSFET circui: 5.0 1K i K = 04. ma = 0. 1K -5.0 Le s ASSUME he NMOS device is in sauraion.

10 10//004 Example NMOS Circui Analysis.doc /4 Thus, we mus ENFORCE he condiion ha: ( ) I = K Now we mus ANALYZE he circui. Q: Wha now? How do we proceed wih his analysis? A: I s cerainly no clear. Le s wrie he circui equaions and see wha happens. From he Gae-Source loop KL: (1) I = 5.0 Therefore, rearranging: I = 50. 1K I And from he rain-source loop KL: 5.0 (1) I (1) I = 5.0 S 1K Therefore, rearranging: S = I -5.0

11 10//004 Example NMOS Circui Analysis.doc 3/4 Look! We can equae he NMOS device equaion and G-S equaion o find. ( ) I = K = 50. ( ) (. ) = K + K + K A quadraic equaion! The soluions o his equaion are: = 376. or = 6. Q: Yikes! Two soluions! Which one is correc? A: Noe we assumed sauraion. If he MOSFET is in sauraion, we know ha: > = 0. Only one soluion of he quadraic saisfies his conidion, i.e.: = 376>. Thus, we use = he soluion ha is consisen wih our original assumpion.

12 10//004 Example NMOS Circui Analysis.doc 4/4 Insering his volage ino he Gae-Source KL equaion, we find ha he drain curren is: I = 50. = = 4. ma And using he rain-source KL, we find he remaining volage: S = I = (. 4) = 55. Even hough we have answers (one curren and wo volages), we sill are no finished, as we now mus CHECK our soluion o see if i is consisen wih he sauraion mode inequaliies = > = = > = 176. S Boh answers are consisen! Our soluions are correc!

13 10//004 Example PMOS Circui Analysis.doc 1/8 Example: PMOS Circui Analysis Consider his PMOS circui: 5 K = 0. ma/ = GG 10 K I =4.0 4K For his problem, we know ha he drain volage = 4.0 (wih respec o ground), bu we do no know he value of he volage source GG. Le s aemp o find his value GG! Firs, le s ASSUME ha he PMOS is in sauraion mode. Therefore, we ENFORCE he sauraion drain curren equaion I K ( ) =.

14 10//004 Example PMOS Circui Analysis.doc /8 Now we mus ANALYZE his circui! 5 Q: Yikes! Where do we sar? I G A: The bes way o sar is by picking he low-hanging frui. In oher words, deermine he obvious and easy values. on ask, Wha is GG?, bu insead ask, Wha do I know?! + - GG 10 K I =4.0 4K There are los of hings ha we can quickly deermine abou his circui! I G = 0.0 S = 5.0 ma I = = = 1 ma 4 4 ( ) = 10I = 10 0 = G GG G GG GG Therefore, we can likewise deermine: = = = 10. S S = = 50. G S GG

15 10//004 Example PMOS Circui Analysis.doc 3/8 Noe wha we have quickly deermined he numeric value of drain curren (I =1.0 ma) and he volage drain-o-source ( S =-1.0) Moreover, we have deermined he value in erms of unknown volage GG ( = 50. ). GG We ve deermined all he imporan suff (i.e.,, S, I )! We can now relae hese values using our PMOS drain curren equaion. Recall ha we ASSUME sauraion, so if his assumpion is correc: ( ) I = K Insering ino his equaion our knowledge from above, along wih our PMOS values K=0. ma/ and =-.0, we ge: ( ) ( ) ( ) I = K ( ) 10. = = + 0. Be careful here! Noe in he above equaion ha hreshold volage is negaive (since PMOS) and ha I and K are boh wrien in erms of milliamps (ma). Now, we solve his equaion o find he value of!

16 10//004 Example PMOS Circui Analysis.doc 4/8 ( ) 50. = + 0. ± 5 = + 0. ± 5 0. = Q: So is boh 5 0. = 0 4. and 5 0. = 4 3.? How can his be possible? A: I s no possible! The soluion is eiher =0.4 or = Q: Bu how can we ell which soluion is correc? A: We mus choose a soluion ha is consisen wih our original ASSUMPTION. Noe ha neiher of he soluions mus be consisen wih he sauraion ASSUMPTION, an even meaning ha our ASSUMPTION was wrong. However, one (bu only one!) of he wo soluions may be consisen wih our sauraion ASSUMPTION his is he value ha we choose for! For his example, where we have ASSUME ha he PMOS device is in sauraion, he volage gae-o-source mus be less (remember, i s a PMOS device!) han he hreshold volage: < < 0.

17 10//004 Example PMOS Circui Analysis.doc 5/8 Clearly, one of our soluions does saisfy his equaion ( = 43. < 0. ), and herefore we choose he soluion = 43.. Q: oes his mean our sauraion ASSUMPTION is correc? A: NO! I merely means ha our sauraion ASSUMPTION migh be correc! We need o CHECK he oher inequaliies o know for sure. Now, reurning o our circui analysis, we can quickly deermine he unknown value of GG. Recall ha we earlier deermined ha: = 50. GG And now, since we know ha he =-4.3, we can deermine ha: = GG = = 077. This soluion ( GG =0.77 ) is of course rue only if our original ASSUMPTION was correc. Thus, we mus CHECK o see if our inequaliies are valid: We of course already know ha he firs inequaliy is rue a p-ype channel is induced: = 43. < 0. =

18 10//004 Example PMOS Circui Analysis.doc 6/8 And, since he excess gae volage is = 3., he second inequaliy: = 10. > 3. = S shows us ha our ASSUMPTION was incorrec! Time o make a new ASSUMPTION and sar over! So, le s now ASSUME he PMOS device is in riode region. Therefore ENFORCE he drain curren equaion: ( ) i K = S S Now le s ANALYZE our circui! Noe ha mos of our original analysis was independen of our PMOS mode ASSUMPTION. Thus, we again conclude ha: I G = 0.0 ma S = 5.0 I = = = 1 ma 4 4 ( ) = 10I = 10 0 = G GG G GG GG

19 10//004 Example PMOS Circui Analysis.doc 7/8 Therefore, = = = 10. S S = = 50. G S GG Now, insering hese values in he riode drain curren equaion: i = K ( ) S S 10. = 0. ( ( ) )( 1) ( 1) 50. = ( + ) 1 Look! One equaion and one unknown! Solving for we find: ( ) ( ) 50. = = = = Thus, we find ha = -5.0, so ha we can find he value of volage source GG : = GG = GG = GG The volage source GG is equal o zero provided ha our riode ASSUMPTION was correc.

20 10//004 Example PMOS Circui Analysis.doc 8/8 To find ou if he ASSUMPTION is correc, we mus CHECK our riode inequaliies. Firs, we CHECK o see if a channel has indeed been induced: = 50. < 0. = Nex, we CHECK o make sure ha our channel is no in pinchoff. Noing ha he excess gae volage is = 50. ( 0). = 30., we find ha: = 10. > 30. = S Our riode ASSUMPTION is correc! Thus, he volage source GG = 0.0.

21 10//004 Example Anoher PMOS Circui Analysis.doc 1/6 Example: Anoher PMOS Circui Analysis Consider he PMOS circui below, where we know (somehow) ha = -4.0, bu don know (for some reason) he value of resisor R. R R =1 K 15 R 1 =1 K I R 3 =1 K K = 0.75 ma/ = -.0 = -4.0 Le s see if we can deermine he value of resisor R. Firs, le s ASSUME ha he MOSFET is in sauraion, and herefore ENFORCE he drain curren equaion: ( ) I = K Now we ANALYZE he circui:

22 10//004 Example Anoher PMOS Circui Analysis.doc /6 R =1 K S R G I I 1 + R 1 =1 K - = I G =0 I - S + R 3 =1 K I 15 Since we know ha =-4.0, and we ASSUME ha he PMOS device was in sauraion, we can direcly deermine he drain curren I : I = K ( ) ( ( )) ( ) = = = 3mA and hus he drain volage is: = I R 3 ( ) = = 30.

23 10//004 Example Anoher PMOS Circui Analysis.doc 3/6 Q: OK, his firs par was easy, bu wha do we do now? How can we deermine he value of resisor R? A: The key o unlocking his circui analysis is recognizing ha he poenial difference across resisor R is simply he volage and we know he value of ( =-4.0)! S R =1 K R G I I 1 + R 1 =1 K - = I G =0 I = 3mA - S + 1 K I = Thus, we can immediaely deermine ha curren I is: 40. I = = = 40. ma R 1 Likewise, from KCL, we find: I + I = I 1 G

24 10//004 Example Anoher PMOS Circui Analysis.doc 4/6 Bu since gae curren I G = 0, we conclude: I1 = I = 40. ma Now we can deermine much abou his circui! R =1 K S R -4.0mA G -4.0mA + R 1 =1 K - = I G =0 I = 3mA - S + 1 K I = For example, from KL, we find he gae volage: = 00. IR G 1 1 (. ) = 40 1 = 40. And likewise he source volage: = I R S G ( ) = = 80.

25 10//004 Example Anoher PMOS Circui Analysis.doc 5/6 R =1 K S =8.0 R G = mA -4.0mA + R 1 =1 K - = I G =0 I = 3mA - S + 1 K I = Likewise, from KCL, we can deermine he curren hrough resisor R: I = I I ( ) = = 70. ma And hus from Ohm s Law we can find he value of R: R = S I = 70. = 1 K Bu wai! We re sill no done! We mus CHECK o see if our original ASSUMPTION was correc.

26 10//004 Example Anoher PMOS Circui Analysis.doc 6/6 Firs, we CHECK o see if he channel is induced: = 40. < 0. = Nex, we CHECK o see if he channel is pinched off. Here, we noe ha = = = 50., and excess gae S volage is ( ) S = = 0.. Therefore: = 50. < 0. = S Hence, our ASSUMPTION is correc, and R =1K.