1 4/5/011 secion 4_5 Biasing in MOS Amplifier Circuis 1/ 4.5 Biasing in BJT Amplifier Circuis eading Assignmen: 8086 Now le s examine how we C bias MOSFETs amplifiers! f we don bias properly, disorion can resul! EXAMPLE: MOSFET AMPLFE STOTON There is a classic bias circui for MOSFET amplifiers; le s see wha i is! HO: MOSFET BASNG USNG A SNGLE POWE SUPPLY Le s do an example C bias design. EXAMPLE: BASNG OF SCETE MOSFET AMPLFES We can also use a C curren source o bias he Amplifier. Q: Yes, bu jus how do we consruc a curren source? A: HO:THE MOSFET CUENT MO HO: CUENT STEENG CCUTS Jim Siles The Univ. of Kansas ep. of EECS
2 4/5/011 secion 4_5 Biasing in MOS Amplifier Circuis / HO: MOSFET BASNG USNG CUENT MOS EXAMPLE: MOSFET BASNG USNG CUENT MOS Jim Siles The Univ. of Kansas ep. of EECS
3 4/5/011 Example MOSFET Amplifier isorion 1/9 Example: MOSFET Amplifier isorion ecall his circui from a previous handou: i ( ) = i ( ) d 15.0 = 5K v ( ) = v ( ) O o vi( ) K = 0.5 ma/ = We found ha he smallsignal volage gain is: A vo vo( ) = = 5.0 v ( ) i
4 4/5/011 Example MOSFET Amplifier isorion /9 Say he inpu volage o his amplifier is: v ( ) = cosω i Q: Wha is he larges value ha i can ake wihou producing a disored oupu? i A: Well, we know ha he smallsignal oupu is: v ( ) = A v ( ) o vo i = 5.0 cosω BUT, his is no he oupu volage! i The oal oupu volage is he sum of he smallsignal oupu volage and he C oupu volage! Noe for his example, he C oupu volage is he C drain volage, and we recall we deermined in an earlier handou ha is value is: O = = 10 Thus, he oal oupu volage is : v ( ) = v ( ) O o = cosω i
5 4/5/011 Example MOSFET Amplifier isorion 3/9 is very imporan ha you realize here is a limi on boh how high and how low he oal oupu volage v ( ) can go! O Tha s righ! f he oal oupu volage v ( ) ries o exceed O hese limis even for a momen hemosfet will leave sauraion mode. And leaving sauraion mode resuls in signal disorion!
6 4/5/011 Example MOSFET Amplifier isorion 4/9 Le s break he problem down ino wo separae problems: 1) f oal oupu volage v ( ) becomes oo small, he MOSFET will ener he riode mode. ) f oal oupu volage v ( ) becomes oo large, he MOSFET will ener cuoff. O O We ll firs consider problem 1. For a MOSFET o remain in sauraion, v ( ) mus remain greaer han he excess gae volage ( ) v ( ) > S S for all ime. Since he source erminal of he MOSFET in his circui is conneced o ground, we know ha = 0. Theore: vs ( ) = v( ) = vo ( ) and = G And so he MOSFET will remain in sauraion only if he oal oupu volage remains larger han = G! S v ( ) > O G
7 4/5/011 Example MOSFET Amplifier isorion 5/9 Thus, we conclude for his amplifier ha he oupu floor is: L = G And since = 4.0 and =.0, we find: G L = = 4 =.0 G Thus, o remain in sauraion, he oal oupu volage mus remain larger han he floor volage L for all ime : Since his oal volage is: v ( ) > L =.0 O v ( ) = cosω O we can deermine he maximum value of smallsignal inpu magniude i : i cosω > > 5.0 cosω cosω < 1.6 i i i
8 4/5/011 Example MOSFET Amplifier isorion 6/9 Since cosω can be as large as 1.0, we find ha he magniude of he inpu volage can be no larger han 1.6, i.e., i < 1.6 f he inpu magniude exceeds his value, he MOSFET will (momenarily) leave he sauraion region and ener he dreaded riode mode! Now le s consider problem For he MOSFET o remain in sauraion, he drain curren mus be greaer han zero (i.e., i > 0 ). Oherwise, he MOSFET will ener cuoff mode. Applying Ohm s Law o he drain resisor, we find he drain curren is: vo 15 vo i = = 5 C i is eviden ha drain curren is posiive only if v < 15. n oher words, he upper limi (i.e., he ceiling ) on he oal oupu volage is: L = = 15.0 O Since: v ( ) = cosω O i
9 4/5/011 Example MOSFET Amplifier isorion 7/9 we can conclude ha in order for he MOSFET o remain in sauraion mode: cosω > 15.0 Theore, we find: i 5.0 s cosω > = Since cosω 1, he above equaion means ha he inpu signal magniude i can be no larger han: i < 1.0 f he inpu magniude exceeds 1.0, he MOSFET will (momenarily) leave he sauraion and ener he cuoff region! n summary: 1) f i > 1.6, he MOSFET will a imes ener riode, and disorion will occur! ) f 1.0 i >, he MOSFET will a imes ener cuoff, and even more disorion will occur!
10 4/5/011 Example MOSFET Amplifier isorion 8/9 To demonsrae his, le s consider hree examples: 1. i < 1.0 The oupu signal in his case remains beween = 15.0 and =.0 for all ime. Theore, he oupu signal is no G disored. L = = 15 v ( ) O = O 10 L = = G. 1.6 > i > 1.0 The oupu signal in his case remains greaer han L = G = for all ime. However, he smallsignal oupu is now large enough so ha he oal oupu volage a imes ries o exceed L = = 15. For hese imes, he MOSFET will ener cuoff, and he oupu signal will be disored.
11 4/5/011 Example MOSFET Amplifier isorion 9/9 L = = 15 v ( ) O = O 10 L = = G 3. > 1.6 i n his case, he smallsignal inpu signal is sufficienly large so ha he oal oupu will aemp o exceed boh limis (i.e., = 15.0 and G =.0 ). Theore, here are periods of ime when he MOSFE will be in cuoff, and periods when he MOSFET will be in sauraion. L = = 15 v ( ) O = O 10 L = = G
12 4/5/011 MOSFET Biasing using a Single Power Supply 1/9 MOSFET Biasing using a Single Power Supply The general form of a singlesupply MOSFET amplifier biasing circui is: 1 S S Jus like BJT biasing, we ypically aemp o saisfy hree main bias design goals: 1) Maximize Gain Typically, he smallsignal volage gain of a MOSFET amplifier will be proporional o ransconducance g m :
13 4/5/011 MOSFET Biasing using a Single Power Supply /9 A vo g m Thus, o maximize he amplifier volage gain, we mus maximize he MOSFET ransconducance. Q: Wha does his have o do wih.c. biasing? A: ecall ha he ransconducance depends on he C excess gae volage: ( ) g = K m Anoher way o consider ransconducance is o express i in erms of C drain curren. ecall his C curren is relaed o he C excess gae volage (in saureaion!) as: ( ) ( ) = = K And so ransconducance can be alernaively expressed as: K g = K ( ) = K = K K m Theore, he amplifier volage gain is ypically proporional o he squareroo of he C drain curren: A vo
14 4/5/011 MOSFET Biasing using a Single Power Supply 3/9 To maximize A vo, maximize ) Maximize olage Swing ecall ha if he C drain volage is biased oo close o, hen even a small smallsignal drain volage vd ( ) can resul in a oal drain volage ha is oo large, i.e.: v ( ) = v ( ) d n oher words, he MOSFET eners cuoff, and he resul is a disored signal! To avoid his (o allow vd ( ) o be as large as possible wihou MOSFET enering cuoff), we need o bias our MOSFET such ha he C drain volage is as small as possible. Noe ha he drain volage is: = Theore is minimized by designing he bias circui such ha he C drain curren is as large as possible.
15 4/5/011 MOSFET Biasing using a Single Power Supply 4/9 However, we mus also consider he signal disorion ha occurs when he MOSFET eners riod. This of course is avoided if he oal volage drainosource remains greaer han he excess gae volage, i.e.: ( ) vs( ) = S vds( ) > Thus, o avoid he MOSFET riode mode and he resuling signal disorion we need o bias our MOSFET such ha he C volage S is as large as possible. To minimize signal disorion, maximize S 3) Minimize Sensiiviy o changes in K, We find ha MOSFETs are sensiive o emperaure specifically, he value of K is a funcion of emperaure. Likewise, he values of K and hreshold volage are no paricularly consan wih regard o he manufacuring process. Boh of hese facs lead o he requiremen ha our bias design be insensiive o he values of K and. Specifically, we wan o design he bias nework such ha he C bias curren does no change values when K and/or does.
16 4/5/011 MOSFET Biasing using a Single Power Supply 5/9 Mahemaically, we can express his requiremen as minimizing he value: d d and dk d Similar o he BJT, we find ha hese derivaives are minimized by maximizing he value of source resisor S. To minimize d dk, maximize S So, le s recap wha we have learned abou designing our bias nework: 1. Make as large as possible.. Make S as large as possible. 3. Make S as large as possible. Again analogous o BJT biasing, we find ha hese hree goals are conflicing, as hey are consrained by he KL equaion of he bias circui:
17 4/5/011 MOSFET Biasing using a Single Power Supply 6/9 1 S S = 0 S S or S S = Maximize A vo by maximizing his erm. Minimize disorion by maximizing his erm. Minimize sensiiviy by maximizing his erm. Bu he oal of he hree erms mus equal his!
18 4/5/011 MOSFET Biasing using a Single Power Supply 7/9 esolving his conflic is a subjec choice of he amplifier designer. However, here is a ruleofhumb procedure. However, verify ha hese resuls saisfy your design requiremens (or he requiremens assigned o you by your boss and/or professor!). 1. Given he desired value of, make source volage S = 4, i.e. se he source resisor S o: S = s = 4 (1) This value reduces he sensiiviy d dk!. Now deermine he required value of. Since = K( ), we find ha should be: = K 3. Se he required value of gae volage G. Noe ha: = G S Thus, we can add he resuls of he previous wo seps o find he required value of he gae volage G.
19 4/5/011 MOSFET Biasing using a Single Power Supply 8/9 To se he gae volage o his value, we mus selec he proper values of resisors 1 and. Since he gae curren is zero ( i = 0), we find from volage G division ha: G = = () Noe his equaion deermines he raio of resisors 1 and, bu no he resisors hemselves. We need a second equaion o explicily deermine he resisors values he sum of he wo resisances, for example. We find ha making he resisances 1 and as large as possible is very desirable! This will ypically maximize he amplifier inpu resisance, as well as resul in minimum power dissipaion. As a resul, we make he resisors as large a pracicable. For example: = 50 K (3) 1 4. Se he required value of C drain volage. ecall ha: a) we require = L o avoid cuoff mode.
20 4/5/011 MOSFET Biasing using a Single Power Supply 9/9 b) and, we require ha G = L o avoid riode mode. Soluion: se he drain volage o a value halfway beween and! G n oher words, se he C drain volage o be: ( ) G = To achieve his, we mus selec he drain resisor so ha: ( ) G = = (4) Thus, use equaions (1), (), (3), and (4) o deermine he sandard C bias design (i.e., 1,, S, and ) for MOSFET amplifiers. f were you, d make sure undersood his maerial well enough ha could also bias a nonsandard MOSFET amplifier problem. s no enough o simply know how, you mus also know why!
21 4/5/011 Example Biasing of iscree MOSFET Amplifiers 1/4 Example: Biasing of iscree MOSFET Amplifiers S S f he MOSFET has device values K = 10. ma/ and = 10., deermine he resisor values o bias his MOSFET wih a C drain curren of: = 4 ma 1. Given he desired value of, make source volage s = 4 = 4. 0, i.e. se he source resisor S o: S s 40. = = = 1K Ω 40.
22 4/5/011 Example Biasing of iscree MOSFET Amplifiers /4. Now deermine he required value of. Since = K( ), we find ha should be: = K 40. = = Se he required value of gae volage G. = G S = = 70. Since he gae curren is zero ( i = 0), we find from volage G division ha: = G 1 1 Theore: 1 = 1 G = = 7
23 4/5/011 Example Biasing of iscree MOSFET Amplifiers 3/4 We need a second equaion o explicily deermine he resisors values he sum of he wo resisances, for example. We make he resisors as large a pracicable. For example: Theore: and hus: = 40 K 1 9 = = 40 7 = 135 K Ω and = 105 KΩ 1 4. Se he required value of C drain volage. Se he drain volage o a value halfway beween and! G n oher words, se he C drain volage o be: ( ) G = = = (.. )
24 4/5/011 Example Biasing of iscree MOSFET Amplifiers 4/4 To achieve his, we mus selec he drain resisor so ha: = = 40. = 15. KΩ K 15K. 105 K 1K S
25 4/5/011 The MOSFET Curren Mirror 1/6 The MOSFET Curren Mirror Consider he following MOSFET circui: Noe =, heore: G S = and hus: > S So, if >, hen he MOSFET is in sauraion! We know ha for a MOSFET in sauraion, he drain curren is equal o: = K ( ) Say we wan his curren o be a specific value call i. Since s = 0, we find ha from he above equaion, he drain volage mus be: = K
26 4/5/011 The MOSFET Curren Mirror /6 Likewise, from KL we find ha: = And hus he resisor value o achieve he desired drain curren is: = where: Q: Why are we doing his? = K A: Say we now add anoher componen o he circui, wih a second MOSFET ha is idenical o he firs : L L 1
27 4/5/011 The MOSFET Curren Mirror 3/6 Q: So wha is curren L? A: Noe ha he gae volage of each MOSFET is he same (i.e., = ), and if he MOSFETS are he same (i.e., 1 K = K = ), and if he second MOSFET is likewise in, 1 1 sauraion, is drain curren is: L ( ) ( ) = K L = K = Theore, he drain curren of he second MOSFET is equal o he curren of he firs! = L Q: Wai a minue! You mean o say ha he curren hrough he resisor L is independen of he value of resisor L? A: Absoluely! As long as he second MOSFET is in sauraion, he curren hrough L is equal o period. The curren hrough L is independen on he value of L (provided ha he MOSFET remains in sauraion). Think abou wha his means his device is a curren source!
28 4/5/011 The MOSFET Curren Mirror 4/6 L L Curren Source emember, he second MOSFET mus be in sauraion for he curren hrough L o be a consan value. As a resul, we find ha: > S or for his example, since s = 0 : > G Since =, we find ha in order for he MOSFET L o be in sauraion: > = L G G1 1 Or, saed anoher way, we find ha he load resisor L can be no larger han:
29 4/5/011 The MOSFET Curren Mirror 5/6 G1 1 < L Where we know ha: = G 1 and hus we can alernaively wrie he above equaion as: 1 < L f he load resisor becomes larger han, he 1 volage will drop below he excess gae volage, S and hus he second MOSFET will ener he riode region. As a resul, he drain curren will no equal he curren source will sop working! Alhough he circui presened here is someimes erred o as a curren sink, undersand ha he circui is clearly a way of designing a curren source.
30 4/5/011 The MOSFET Curren Mirror 6/6 We can also use PMOS devices o consruc a curren mirror! SS 1 SS This beer be in sauraion! L L L =, regardless of he value of L!!!
31 4/7/011 Curren Seering Circuis 1/3 Curren Seering Circuis A curren mirror may consis of many MOSFET curren sources! L1 = L = L1 L L3 L3 = Q Q 1 Q Q 3 This circui is paricularly useful in inegraed circui design, where one resisor is used o make muliple curren sources. Q: Wha if we wan o make he sources have differen curren values? o we need o make addiional curren mirrors? A: NO!! ecall ha he curren mirror simply ensures ha he gae o source volages of each ransisor is equal o he gae o source volage of he erence:
32 4/7/011 Curren Seering Circuis /3 = = = = 1 3 Theore, if each ransisor is idenical (i.e., K = K1 =, and = = = ) hen: 1 n ( ) ( n n ) = K = K = n oher words, if each ransisor Q n is idenical o Q, hen each curren n will equal erence curren. n Bu, consider wha happens if he MOSFETS are no idenical. Specifically, consider he case where Kn K (bu = ). emember, we know ha n = sill, even when Kn K. Thus, he drain curren n will now be: n n n ( ) = K n n ( ) = K = Kn K K n = K The drain curren is a scaled value of! n
33 4/7/011 Curren Seering Circuis 3/3 For example, if K 1 is wice ha of K (i.e., K 1 = K ), hen 1 will be wice as large as (i.e., 1 = ). From he sandpoin of inegraed circui design, we can change he value of K by modifying he MOSFET channel widholengh raio (W/L) for each ransisor. W ( ) W ( ) W ( ) W ( ) 1 K k L L = = K 1 k L L n n n L1 = K K 1 L1 L L3 L = K K L3 = K K 3 Q Q 1 Q Q 3
34 4/7/011 MOSFET Biasing using a Curren Mirror 1/5 MOSFET Biasing using a Curren Mirror Jus as wih BJT amplifiers, we can likewise bias a MOSFET amplifier using a curren source: 1 K, S is eviden ha he C drain curren, is equal o he curren source, regardless of he MOSFET values K or! Thus, his bias design maximizes drain curren sabiliy! We now know how o implemen his bias design wih MOSFETs we use he curren mirror o consruc he curren source!
35 4/7/011 MOSFET Biasing using a Curren Mirror /5 1 K, S Q Q 1 1 Since =, i is eviden ha mus be equal o: = K and since he C gae volage is: G = 1 is eviden ha he C source volage S is hus:
36 4/7/011 MOSFET Biasing using a Curren Mirror 3/5 = S G = K 1 Since we are biasing wih a curren source, we do no need o worry abou drain curren sabiliy he curren source will deermine he C drain curren for all condiions (i.e., = ). We migh conclude heore, ha we should make C source volage S as small as possible. Afer all, his would allow us o maximize he oupu volage swing (i.e., maximize and S ). Noe however, ha he source volage S of he MOSFET is numerically equal o he drain volage (and hus S ) of he second MOSFET of he curren mirror. Q: So wha?! A: The volage = S mus be greaer han: S = 1 1 ( ) = 1 in order for he second MOSFET o remain in sauraion. There is a minimum volage across he curren source in order for he curren source o properly operae!
37 4/7/011 MOSFET Biasing using a Curren Mirror 4/5 Thus, o maximize oupu swing, we migh wish o se: 1 S > 1 1 = S 1 1 (alhough o be pracical, we should make S slighly greaer han his o allow for some design margin). Q: How do we se he C source volage S?? A: By seing he C gae volage G!! ecall ha he C volage is deermined by he C curren source value : = K and he C gae volage is deermined by he wo resisors 1 and : = G 1
38 4/7/011 MOSFET Biasing using a Curren Mirror 5/5 Thus, we should selec hese resisors such ha: = G S = K ( ) 1 1 Q: So wha should he value of resisor be? A: ecall ha we should se he C drain volage : a) much less han o avoid cuoff. b) much greaer han o avoid riode. G Thus, we compromise by seing he C drain volage o a poin halfway in beween! ( ) G = To achieve his, we mus selec he drain resisor so ha: = = ( ) G
39 4/7/011 Example MOSFET Biasing using a Curren Mirror 1/3 Example: MOSFET Biasing using a Curren Mirror Le s deermine he proper resisor values o C bias his MOSFET. The curren source is 5.0 ma and has a minimum volage of.0 ols in order o operae properly K =0. ma/ =1.0 Since = = 5.0 ma, we know ha will be (if in sauraion): = K 5.0 ma min = = = 60. Assuming ha we wan he C source volage o be he minimum value of S =.0, we nee for he C gae volage o be: G = S = = 8.0
40 4/7/011 Example MOSFET Biasing using a Curren Mirror /3 Thus, we need o selec resisors 1 and so ha: G = 80. = 1 or in oher words, we wan: 80. = Since we can make 1 and large, le s assume ha we wan: 1 = 300K So ha 1 = 140 KΩ and = 160 KΩ. Finally, we wan he C drain volage o be: ( ) G = = = ( ) So ha he resisor is: = = 50. = 08K. Ω
41 4/7/011 Example MOSFET Biasing using a Curren Mirror 3/ = 140 K = 08K. K =0. ma/ =1.0 = 160 K 5.0 ma min =.0
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