Rail-to-Rail Op Amps

115 Pages·2001·0.51 MB·English

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