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IEEE TRANSACTIONS ON CIRCUITS AND SYSTEMS, VOL. 37, NO. 9, SFPTEMBER 1990
Power Supply Rejection Ratio in
Operational Transconductance Amplifiers
1077
Abstract The
amplifiers is analyzed. An analyzing technique based on cuts in suhcir
cuits is presented that allows hand calculation of PSRR of any circuit.
In this paper it is shown that the PSRR of the single stage operational
amplifier (OTA) has an order of magnitude better PSRR than the
commonly used Miller OTA. The analyses are compared with hand
calculations and SPICE level2 simulations on a realized improved
Miller OTA structure.
power supply rejection ratio (PSRR) of operational
I. INTRODUCTION
HE DESIGN of complex systems with analog, digital,
T
one chip suffers from large signal variations on the power
supply lines (up to 100 mVpeak [l]) that are introduced by
the digital and the switchedcapacitor networks.
Especially in those cases where lowlevel signals have to
be measured, the use and development of high perfor
mance amplifiers are necessary. In analog building blocks
and switchedcapacitor structures, the main building
blocks are operational transconductance amplifiers
(OTA's). For this reason the performance of such ampli
fiers must be studied and analyzed as function of the
power supply variations. The power supply rejection ratio
(PSRR) is very often only specified at dc or at very low
frequencies ( f < 1 Hz). In order to reduce the influences
of 50/60 Hz clockfrequencies and high frequency power
supply noise, the performance at frequencies up to the
bandwidth of the system, which is for amplifiers the gain
bandwidth (GBW), must be studied. This is especially
important in aliasing (sampled data) contexts where high
frequency power ~ ~ p p l y noise can be folded back into the
signal band. This effect can drastically decrease the PSRR
performance, as has been demonstrated in switched
capacitor filters [ll.
The performance of a system influenced by power
supply variations can be described by the PSRR. In the
next section the definition of the PSRR is described.
From this definition it can be concluded that for ampli
fiers, the PSRR at high frequencies (f do mini,n, < f < GBW)
can be improved by increasing the GBW of the amplifier.
Therefore, the PSRR can be best normalized to 2. T.
and switchedcapacitor building blocks integrated on
Manuscript received August 19. 1987; revised February 6, 1989. This
paper was recommended by Associate Editor C. A. T. Salama.
The authors are with the Department of Electrical Engineering,
Katholieke Universiteit Leuven, 3030 Heverlee. Belgium.
IEEE Log Number 9037126.
VIN.m$Frn
vPOWER
Fig. 1. A block diagram of a general electrical circuit.
GBW/s in order to be able to compare different ampli
fier structures. This normalization results in the parame
ter l/Ap(sl: i.e., the reciprocal of the power supply gain
(PSR, as distinct from PSRR). Secondly, a technique to
calculate the PSRR is discussed. Using this technique,
several opamps, such as an OTA and a Miller opamp,
are analyzed. Finally, an improved twostage amplifier is
studied and analyzed.
11. DEFINITION
OF THE PSRR
A general electrical circuit as presented in Fig. 1 has an
input, an output, and a power node. Hence it has voltage
transfer functions from any node to any other node. In
many cases, only the transfer function from the input to
the output and from the power node to the output node
are important. If the transfer function of the power node
to the output node is called the power supply gain (Ap),
and the transfer function of the input node to the output
node is called the openloop transfer function (A), the
PSRR is defined as (in the frequency domain s = j . 01
which is normally given in decibels ( = 20.log(A /Ap)).
By increasing the GBW of an amplifier, A b ) increases
( A h ) = 2..sr.GBW/s for f > fdominant), and as a result the
PSRR increases, too. Thus to compare different amplifier
structures, the PSRR can be best normalized to 2 . 7 ~
GBW/s. From the definition, this results in the parame
ter l/Ap(s), i.e., the reciprocal of the power supply gain.
In this text this parameter is called the PSR. If both
functions (A(s) and PSRR(s)) are assumed to be first
order, the PSR at high frequencies is a constant (see Fig.
2). The smaller A,(s) is (or the higher the PSR is), the
better the structure performance is.
The equivalent mathematical equation for the output
node as function of the input and of the power supply
node is described by the superposition of the power
supply gain and the openloop gain, or uOut = A;u, +
00984094/90/09001077$01.00 01990 IEEE
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IEEE TRANSACTIONS ON CIRCUITS AND SYSTEMS, VOL. 37, NO. 9, SEPTEMBER. 1990
......................................
.
.
:
.
i
.
.
:
.
.
.
/.. ............... : ...................

i
RI
P2
A 7
Fig. 2. Bode dia ram of the o en loop gain (Ah)), the power supply
rejection ratio &SRR(s)) an8 thk PSR: the reciprocal of the power
supply gain (AJs)).
Mw A
PSRR(s)
Fig. 3. A block diagram of a twostage amplifier.
Aav,. If the system has different power supplies (e.g., an
OTA has a positive ( V D D ) and a negative (Vss) power
supply), a power supply gain for each power node can be
defined separately. In this case A,,vdd (Ap, uss) is called
the transfer function from the V,,
output node, whereby the Vss (V,,)
the basis of the knowledge of these transfer functions, the
PSRR of each power supply can be defined as
(Vss) node to the
is acgrounded. On
A A
PSRR, Udd =
~
and PSRR, U,, = 
. (2)
7 Udd
A, , U,,
111. PSRR ON CIRCUIT
In order to be able to calculate the PSRR of the total
system, the system can be divided into subcircuits or into
a block diagram, using controlsystem theory. In Fig. 3, an
example of a multistage amplifier with a resistance feed
back network is shown. The subcircuits in the figure must
not contain feedback loops and the characteristics of the
different subcircuits must not be influenced by each other.
In some cases it is necessary to use the input impedance
of the next stage in the subcircuit, e.g., the load capaci
tance of an OTA. Once the system is split, the different
subcircuits can be separately calculated and evaluated. As
an example, the circuit in Fig. 3 can be split into two
subcircuits without any feedback loop. The PSRR of
these subcircuits ( A , /Apl and A , /Ap2) being known, it
is possible to calculate the PSRR of the total system. The
output ( U , ) of the system shown is given by
LEVEL
U , =
42 + A 2 4 1
/( R, + R2) '"
+ 1 + A2*A,.R2 /( R, + R,) *'in
1 + A2*A1R2
Al.A2
or
U , =  . [ (  +
RI + R2
R2
1
1
PSRRl PSRR2Al
or if A1 x= 1, the PSRR of this system is approximately
equal to the PSRR of the first stage. At high frequencies
the gain of amplifiers usually decreases. So if PSRR2 >
PSRR1, the PSRR of the system can be dominated by the
second stage. A twostage Miller amplifier, which is dis
cussed later, is one example of this effect.
In this paper, a method to calculate the PSRR of
subcircuits is studied. Since in complex analog systems the
subcircuits are amplifiers, different amplifier structures
up to a frequency range equal to their GBW are analyzed
and compared.
IV. THE PSRR OF A SUBCIRCUIT
In a MOS transistor circuit, the voltage transfer func
tions are always realized by a transconductance (g, of a
transistor) and an admittance. For example, the gain of
an OTA is given by
gm
A( S) = go + s * c 1
*
(4)
In fact, a MOS transistor converts this input gatesource
voltage into a current. At the gain node (and there is only
one gain node, otherwise it can be divided in two subcir
cuits as is presented in Fig. 3), this current is converted
into a voltage by the admittance at that node. The power
supply variation also introduces in that circuit a current
or there exists a power supply transconductance: GM,.
This introduced current is converted into a voltage at that
same gain node as for the signal transfer function. As a
result, the denominator of the voltage transfer functions
AJs) and A(s) are equal to each other and hence the
PSRR is given by the ratio of the signal transconductance
(GM) and the power supply transconductance (GM,) or
( 5 )
In order to find the power supply transconductance of the
internal nodes of the circuit, the following technique is
used.
The power supply under study is enclosed by a curve,
whereby this curve cuts only once every branch coming
from that power supply. The gain node is connected to
the acground. As a result, the current flow into the gain
node due to the inputoutput transfer path can be calcu
lated separately from the one due to any impedance
between the gain node and any power supply line. The
following are properties of this technique.
1) For every cut, each branch is divided in two cutsides.
There is always at one cutside a current transfer function
to the gain node. The current transfer function is given by
the ratio of the current that flows into the gain node (into
the acground because the gain node is acgrounded) and
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STEYAERT AND \ANSEN' PSKK OTA'S
I
gm 4.7 4.65 3.5 3 . 5 3
g0
8 . 4 8.6
CP 191 191
14
64
14.6 7 . 2
71 93
1079
PA/Y
nA/Y
fF
I
Fig 4
Curves C1 and C2 for cdlculating the power supply
transconductance
PSRR, vdd
lowfreq.
f3dB
77dB
12kHz
PSR 53dB
PSRR,vss
lowfreq.
f3dB
P SR
47dB
2lkHz
28dB
the current that is applied to that cutside. If there is no
current transfer function for either of the cutsides, then
the PSRR of the subcircuits is not influenced by that
branch.
2) If there exists a current transfer function for one of
the cutsides, the power supply transconductance to the
gain node is given by the admittance seen into the other
cutside (y,), multiplied by the ratio y 1 /(yl + y,) (with y 1
the admittance seen into the cutside where there exists a
current transfer function), and multiplied by the current
transfer function (or current gain). This product gives the
contribution of that branch to the PSRR in the subcircuit.
The summation of the power supply transconductances of
each branch results in the total power supply transcon
ductance of the subcircuit. In a normal circuit design the
ratio y , / y , is much smaller than 1 ( y l /(yl + y2) = 1,
which is further assumed in this text). Because the PSRR
is given by the ratio of the signal transconductance and
the power supply transconductance (5), a high PSRR can
be obtained by designing the admittance y , as small as
possible. This means also that the ratio of the impedances
on both cutsides must be made as high as possible.
As an example, the PSRR of an OTA is analyzed. The
structure is presented in Fig. 4. First the PSRR, cdd is
studied. Therefore the power supply is enclosed by curve
C1. Let us now discuss cut a. For this cut there exists a
current transfer function to the gain node at the cutside
of transistor M8: gm7/gmX. So the contribution of this
branch to the power supply transconductance is given by
the product of gm7/gInx and the admittance seen into the
other cutside. In this example, this is the admittance seen
into the drain of M,: go, + s.C,,. Hence the contribution
of this branch is (go, + s.Cp5).gm7/gm8. The same can
be done for the other cuts, which results in a total GM,
( = io,, /
given by
~3~~)
76dB
9kHz
49dB
47dB
2OkHz
27dB
with C, the draingate plus the drainbulk capacitance
of the transistors. For the case C,,, (cut a), this C,, is a
drainsubstrate capacitance if a pwell process is used. In
the case of an nwell, Cps is a drainwell capacitance.
TABLE 1
AND SPICE SIMULATIONS
ON AN OTA
HAND CALCULAT~ONS
OTA small signal spec's out of SPICE level 2
However, in both cases C,, is the capacitance between
that node and the positive power supply rail.
Equation (6) can be simplified if the current mirrors
used have a current gain factor of one (gm7 = gm8, g,,
gm5, gm, = gm6), and if the contribution of cuts b and c is
neglected (sol = go,, CPl = C,,>, to
=
(7)
This last assumption means that the two sides of the input
stage balance. As a result, the vdd signal appears on
nodes b and c as a common mode signal and is as a result
rejected in the circuit.
The signal transconductance of this amplifier is given
by GM = (gml + gm2)/2 = gm1. So the PSRR is given by
(using relationships 5 and 7)
Analyzing the PSRR of the V,,, curve C2 in Fig. 4 is
used. The contribution of each branch in the transcon
ductances is calculated and results in (GM, = io,, / Vss)
It has to be remarked that in this case mismatches in the
transconductances of the input transistors (gml and gm2)
have been included in the calculations. As can be seen
they can decrease the PSRR. If g,, = g,,
the input stage balances), the PSRR, uSs can be simpli
fied, resulting in
(assuming that
In Table I the relationships of the PSRR,udd and
PSRR, c , ,
are compared with SPICE level 2 simulations
and they fit very well.
From this example it can be concluded that a high
PSRR, cdd can be realized by designing a very symmetrical
OTA. On the other hand, the phase of the PSRR, c~~ can
be at low frequencies (0" or 180") depending on the
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1080
gut 4.63 4.60 3.5 3.5
g0
0.5 9.3 15
CP
191 220
14
74
80
IEEE TRANSACTIONS ON CIRCUITS AND SYSTEMS, VOL. 37, NO. 9, SEPTEMBER. 1990
6.7 P A / ' ?
22.0 nA/V
779 fF
c 1
PSRR,Vdd
lowfreq.
f3dB
P SR
PSRR .vss
lowfreq.
f3dB
PSR
...............................................................
.3rve
:$zkinM.."
46dB
14kHz
23dB
46 dB
14 kHz
23dB
91dB
4.6kHz
58dB
90dB
5kHz
58dB
............................
P .................................
Fig. 5. A sourcecoupled input stage with active load.
mismatch in the output conductance (go) of transistor M6
and M5. Also the pole of the PSRR transfer function can
be situated in the left or righthalf plane, depending on
which parasitic capacitance is the largest. For the
PSRR,u,, it can be concluded that the output conduc
tance (8,)
and the sum of the draingate and the
drainbulk capacitances of transistors M5, M6, and M7
have to be made very small in order to realize a high
PSRR,v,,.
V. COMPARISON
AMPLIFIER
OF DIFFERENT
STRUCTURES
A. The SourceCoupled Input Stage with Active Load
A commonly used input stage of an amplifier is a
sourcecoupled differential pair with an active load, as
presented in Fig. 5. Studying the PSRR,udd, cut C1 is
used. It is clear that go, introduces a current with the
value go4'udd. On the other hand, go, (via cut a) intro
duces a current into the common mode node V, with a
value gol'udd. This current can only flow into the sources
of M1 and M2. It means that one half of that current
flows into the output node via the source of M 2 and the
other half via the current mirror M3 and M4. The total
current (iddl introduced by a variation on the udd in the
gain node is then given by
' dd
G M =  = g
'dd
+ g
01 04'
The transconductance of this stage is given by GM=
(gml + gm2)/2 = gml, or the PSRR, vdd results in, using
(9,
Using the same technique for the parasitic capacitors, the
relationship becomes
TABLE I1
AND SPICE SIMULATIONS
ON A SOURCECOUPLED
INPUT STAGE
ACTIVE LOAD
HAND CALCULATIONS
WITH
OTA small signal spec's out of SPICE level 2
parasitic capacitances (if the cascode transistors have the
same dimensions as the drive transistors). Hence only the
low frequency PSRR,udd can be improved by using cas
code transistors, but the high frequency PSRR, vdd
(f3 dB < f < GBW) will not be improved at all.
The PSRR,u,, can be calculated similar to the input
stage of an OTA (see (9)), which results in a PSRR, U,, of
(14)
gm 1
gm1  gm2 *
gm1+ gm2
PSRR, U,, =
(go9 + s.c,9).
In Table 11, the hand calculations are compared with the
SPICE level2 simulations. However, it has to be re
marked that in this case the transistors are designed in a
separated pwell ( = no bulk effect). The capacitance C , ,
is, however, the total capacitance wellsubstrate of M1
and M 2 plus drainsubstrate of M9. So in this case the
PSRR is mainly affected by the wellsubstrate capaci
tance of the input devices. If the structure is designed in
an nwell process, the input transistors can not be de
signed in a separated well and they are affected by the
bulk modulation. Due to this bulk effect an extra current
(id) flows into the gain node, given by
.
(13)
(16)
gm 1
PSRR, vdd =
So to improve the PSRR,Vss, matched input devices are
required and the drainbulk capacitance of M9 must be
minimized.
go1 + go4 + S.(C,l+ cp4)
A comparison of the hand calculations and the SPICE
simulations on this structure is presented in Table 11.
It can be concluded that in order to improve the high
frequency PSRR, mainly the parasitic drainbulk capaci
tances of transistors M1 and M4 have to be made very
small. Inserting cascode transistors increases the output
conductances go, and go4, but it does not decrease the
A
A widely used transconductance amplifier is the two
stage amplifier presented in Fig. 6. This amplifier is
internally compensated with a Miller capacitance (Cc).
OTA
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STEYAERT AND SANSEN: PSRR OTA’S
1081
gm 4.67 4.7
go 8 . 5 8.5 15
Cp
191
3.5 3 . 5 35
15
80
80
145
211
191
St.9.
1
stapc 2
. .
35
105 22.8 nA/V
79 779 fF
6 . 1 U I V
Fig. 6. A twostage Miller transconductance amplifier.
PSRR, vdd
lowfreq.
f3dB
P SR
PSRR.vss
lowfreq.
f3dB
P SR
92dB
4.9Hz
OdB
96dB
680HZ
47dB
The first stage (Ml M4) consists of a sourcecoupled
input structure with active load; the second one (M5M6)
consists of an inverter structure. Between the two gain
nodes a compensation capacitor is placed (Cc).
The study of the low frequency PSRR,u,, is worked
out by dividing the circuit in two subcircuits as is shown in
Fig. 6. The equivalent circuit at low frequencies (i.e., the
circuit without any capacitances) is given in Fig. 7, with
A1 and A2 the openloop gains of the two gain stages,
each stage having a high output impedance. In this equiv
alent network the openloop gain of the amplifier is given
by A =  Al.A2, and the power supply gain by A, =
A,, + A2.(1 A,,), with A,, ( = U1 /Udd)
( = U,,,* / udd when both inputs, inverting and noninvert
ing, are acgrounded) the power supply gains of each
stage. The power supply is then
and Ap2
93dB
4Hz
OdB
97dB
650HZ
47dB
output
QX2
Fig. 7. Equivalent network of the Miller OTA for calculating
PSRR, ~,j,j.
output
Fig. 8. Equivalent network, including the capacitors, of the Miller
OTA for calculating PSRR, udd.
/’
1 1
1 1
+
_ 
=
PSRR PSRR2.Al A1 PSRR1’ (18) The comparison between hand calculations and SPICE
simulation on this circuit is given in Table 111. Thus it can
be concluded that the PSRRp,,, of the Miller OTA is
almost given by the PSRR of the second stage multiplied
by the gain of the first stage.
The Miller OTA at higher frequencies is analyzed using
the equivalent network of Fig. 7 and adding the capaci
tors of the circuit. The result is represented in Fig. 8,
where Cc stands for the compensation capacitance, CL
for the load capacitance, C i for the parasitic, to the
ground, and C1 for the gatesource capacitance of tran
sistor m5. For frequencies up to the GBW (and thus
lower than the bandwidth of the second stage), the invert
ing input of the second stage always follows the noninvert
ing input due to the feedback by Cc. Hence the capacitor
C1 sees no acvoltage drop, which means that the value of
this capacitor is of no importance in this analysis (e.g., in
a SPICE simulation where C1 was doubled, the PSRR
decreased only 0.1 dB, which can be neglected). This is in
contrast with the usually suspected signal feedthrough
through C1Cc for the degradation of the PSRR. Let us
further simplify the structure by assuming an infinite
PSRR in the first stage. Hence Fig. 8 can be redrawn into
In order to calculate the PSRR, udd, the openloop gain of
the first stage (Al), the PSRR of the first stage (PSRRl),
and the one of the second stage (PSRR21, must be calcu
lated. The gain of the first stage is given by A1 =
 g,,
first part of this section (see (12), whereby the inverting
and the noninverting input nodes are interchanged), or
/(go* + go4). The PSRRl was calculated in the
( 12b)
g m 1
go1 + go4
PSRRl =  ~.
The PSRR2 is calculated using cut b at the output of
stage 2 (see Fig. 61, which results in PSRR2 =  g,,
So, if the relationships are substituted in (181, it becomes
/go5.
(19)
1
(go2 + go4) *go5
go2  go1
+
gm 1
 
PSRR
grnl.grn5
or with go, =: go, = go, =: go4, it results in