Rail-to-Rail Op Amps PDF
Preview Rail-to-Rail Op Amps
AMSC/TAMU Rail-to-Rail Op Amps Mb3 Mb4 I Ib P d es ng aa Vi+ Vi- rrent ation ent st umu C q me M1 M2 s M3 M4 ub Su s I N Mb2 Mb1 Edgar Sánchez-Sinencio TAMU, AMSC 1 Thanks to Shouli Yan for his valuable input in helping in generating part of this material Rail-to-Rail Op Amps • There are 2 basic configurations for Op Amp applications: (a) inverting configuration, and, (b) non-inverting configuration. R2 Vout Vin Vout R1 Vin Vout Vin R2 R1 (a) Inverting Configuration (b) Non-Inverting (c) Voltage Follower Configuration ( a special case of non- inverting configuration ) 2 Analog and Mixed-Signal Center,TAMU Why Rail-to-Rail Differential Input Stage? • The input and output swings of inverting and non-inverting configurations Configuration Input common mode Output voltage swing voltage swing » Inverting 0 Rail-to-rail Non-inverting R1/(R1+R2) * Vsup Rail-to-rail Voltage follower Rail-to-rail Rail-to-rail • From the table, we see that for inverting configuration, rail-to-rail input common mode range is not needed. But for non-inverting configuration, some input common mode voltage swing is required, especially for a voltage follower which usually works as an output buffer, we need a rail- to-rail input common mode voltage range! To make an Op Amp work under any circumstance, a differential input with rail-to-rail common mode range is needed. 3 Analog and Mixed-Signal Center,TAMU How to Obtain a Rail-to-Rail Input Common Mode Range? • We know that usually the input stage of an op amp consists of a differential pair. There are two types of differential pairs. To the next stage Vi+ Vi- Vi+ Vi- I b1 To the next stage (a) P-type differential input (b) N-type differential input stage stage 4 How to Obtain a Rail-to-Rail Input Common Mode Range? ( cont’d ) • First, let us observe how a differential pair works with different input common mode voltage – P-type input differential pair V V V ( Common Mode Range ) dsat GS CMR Vdd I V tail dsat,Ib gm I b V GS,M1,2 Vi+ M1 M2 Vi- V CMR Input common mode voltage range To the next Input Common stage Mode Voltage -Vss -Vss Vdd Vicm 5 Where V =V +V GS dsat T How to Obtain a Rail-to-Rail Input Common Mode Range? ( cont’d ) – N-type differential input stage V V V ( Common Mode Range ) dsat GS CMR Vdd I To the next tail gm stage V CMR Vi- Input common Vi+ mode voltage range V I GS Input Common b V Mode Voltage dsat -Vss -Vss Vdd 6 How to Obtain a Rail-to-Rail Input Common Mode Range? ( cont’d ) • Why not connect these two pairs in parallel and try to get a full rail-to- rail range? Yes, this is one way! Mb3 V V V Mb4 dsat GS CMR Vdd There should be an I Ib P d es overlap between ng N aa CMR, VCMR,P and VCMR,N , Vi+ Vi- rrent ation ent st so the minimum umu V C q me power supply M1 M3 M4 M2 ubs P Su , s R voltage requirement M C is V I N ( 4V +V +V ) dsat TN TP Mb2 Mb1 -Vss P Pair N Pair Simple N-P complementary input stage Almost all of the rail-to-rail input stages are doing in this way by some variations! But how well V ‡‡ 4V +V +V SUP dsat TN TP does it work? 7 How to Obtain a Rail-to-Rail Input Common Mode Range? ( cont’d ) • Transconductance vs. Vicm 1 W 1 W = = • If K KP ( ) KP ( ) N N P P 2 L 2 L and gm I =I =I Gm, the sum of N P TAIL Region II gm and gm then gm =gm =gm= 2 K I . N P N P TAIL Region I Region III Region I. When Vicm is close to the negative rail, only P- channel pair operates.The N channel pair is off because its V is less than V . The total transconductance of the GS T differential pair is given by gm = gm =gm. T P gm gm P N Region II. When Vicm is in the middle range, both of the P and N pairs operate. The total transconductance is given by gm = gm +gm =2gm. -Vss T N P Vdd Common Mode Voltage Region III.When Vicm is close to the positive rail, only N- The total transconductance of the input channel pair operates. The total transconductance is stage varies from gm to 2gm, the given by gm = gm =gm. T N variation is 100% ! 8 Why is a Constant Gm needed ? • The total transconductance, g , of the input stage shown in the mT previous slide varies as much as twice for the common mode range! • For an operational amplifier, constant transconductance of the input stage is very important for the functionality of the amplifier. • As an example, we will analyze a simple 2-stage CMOS operational amplifier. The conceptual model of the amplifier is shown below. Cm Vi+ gm1 gm2 Vi- R L C L 9 Why is a Constant Gm needed ? ( cont’d ) • The transfer function of the amplifier is given by C - 1 g g (1 s m ) - 1 s m1 m2 g » = z A(s) m2 A s2C C + sC g + g g 0 1 1 L m m m2 o1 L s2 + s +1 p p p 1 2 1 g g = where A m 1 m 2 , which is the DC gain of the amplifier. 0 g g o1 L GBW g /C g g = = = = p m1 m , p m2 , and z m 2 , 1 2 A A C C 0 0 L m p and p are the dominant pole and non-dominant pole of the amplifier 1 2 respectively, and p << p . 1 2 z is the zero generated by the direct high frequency path through C . m 10
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