# Pole-Zero Analysis of Multi-Stage Amplifiers: A Tutorial ```Pole-Zero Analysis of Multi-Stage
Amplifiers: A Tutorial Overview
Punith
R.
Surkanti, Annajirao Garimella and Paul
M.
Furth
Klipsch School of Electrical and Computer Engineering,
New Mexico State University, Las Cruces, NM
88003, USA
Email: [email protected], [email protected], [email protected]
Abstract- Analyzing pole-zero locations of an amplifier is
essential to
1)
understand the characteristics of a circuit in
the frequency domain, and
2)
choose appropriate frequency
compensation techniques to guarantee the stability of a circuit
over a specified range of load resistance and capacitance. The
objective of this paper is to provide tutorial treatment of the
steps for analyzing poles and zeros in multi-stage amplifiers.
These techuiques can be equally applied for the analysis of power
management circuits such as low-dropout voltage regulators
(LDOs) and controllers for DC-DC converters.
I. INTRODUCTION
Fig. 1.
Finding the analytical equations of poles and zeros is often
1) choose appropriate frequency compensation
techniques to stabilize an amplifier and 2) understand which
Schematic of two-stage amplifier with cascode compensation [I].
3 of . The first stage is a PMOS
essential to
folded cascode differential amplifier and the second stage
parameters have more impact than others, aiding in tuning
output resistance and lumped parasitic capacitance of the first
is a PMOS common-source amplifier. The transconductance,
the circuit for desired frequency response. In this tutorial, we
and second stage are represented by
explain the process of performing small-signal analysis and
C1
deriving the equations of poles and zeros in two-stage and
three-stage amplifiers. Section II outlines the steps for finding
and
COUT
gmb gm2, R1, ROUT,
respectively. The first stage is inverting so as
to ensure that the overall gain from
VI N + to
VOUT
is non&shy;
inverting. Equations for the small-signal parameters of the
poles and zeros. Section III illustrates pole/zero analysis in
amplifier are given in Table I and the small-signal model is
a two-stage folded-cascode op-amp. Section IV shows an
shown in Fig.
2.
example of analyzing poles and zeros for a more complex
three-stage amplifier. SPICE simulations are performed to
compare hand-computed pole/zero locations with AC analysis.
II. STEPS FOR FINDING POLES/ZEROS
The steps for obtaining the DC gain, poles and zeros:
1) Draw the small-signal model
2) Apply Kirchhoff's Current Law (KCL) at each node
3) Solve the KCL equations using symbolic manipulation
Fig. 2. Small-signal model of two-stage amplifier with cascode compensation.
software to obtain the transfer function
4) Set 8=0 to find the DC gain of the amplifier
5) Factor and solve the numerator to obtain zeros
6) Applying the assumption of widely-separated poles,
solve the denominator to obtain the poles, OR applying
the assumption of widely-separated poles with complex
pole pair, solve the denominator to obtain the poles
III. POLE-ZERO ANALYSIS OF TWO-STAGE AMPLIFIER
Compensation networks such as Miller compensation &shy;
, cascode compensation , -, nested Miller compen&shy;
sation (NMC) , , and reverse NMC (RNMC) , 
are widely used techniques to stabilize multi-stage amplifiers.
The transistor-level schematic of a two-stage op-amp using
cascode compensation is shown in Fig.
1. This circuit is
The first-stage is represented with voltage-controlled current
gMIVS, where Vs
Vs == VI N + VI N
source (VCCS)
age, given by
is
VI .
-
_.
is the differential input volt&shy;
The output of the first-stage
The second-stage is represented with VCCS
The output of the second-stage is
capacitor
and
vx,
transistor
Cc
and resistor
where
Vx
Rc
VOUT.
gM2Vl.
The compensation
are connected between
VOUT
is the source node of the common-gate
M6. Transistor M6 acts as a positive current buffer
Vx and VI represented with VCCS gmBvx. Transistor
resistor Rc and compensation capacitance Cc form the
between
M6,
cascode compensation. The impedance looking into the source
M6 is
1/ gmB, connected between nodes Vx and
Rc can be used to increase the impedance
1
looking into the source to __ + Rc, as illustrated in , .
9&quot;,B
terminal of
ground. Resistor
978-1-61284-857-0/11/\$26.00 &lt;92011 IEEE
TABLE
I
(WPI
9m of M9
.
ro9lirolOIIRL
Gqd9 + GqdlO + GL
*
.
9m of M6
.
b1
�
.
ic
WPI.
WP2
(1), we apply KCL at every node in Fig. 2. Let
o
VOUT to node vx.
-
=
The set of KCL equations are
=
:!!.QJl.X...
(7)
(2) so as to eliminate
(1) using symbolic manipulation software. The transfer
WZI
) ( sJ )
(1+ +
8
wPl
s, S2
(2)
VI , Vx and ic
to obtain the transfer function G(s) == VOUT(S)jvS(s) shown
Solve the equations in
(1+
The coefficients for
of (8) are given by
9mBVX 9mlVS + &quot;*&quot; + vlsGl
(V OUT - V X )
..,....::!.X..&shy;
t. c - Rc
+(l/sCc) 119m B
ic + 9m2 Vl + RV OUT + VOUTSGOUT
B. Transfer Function
in
G(s) =
be the current flowing through the compensation network
from node
the
3) One Real Pole &amp; Two Complex Poles: The generic
transfer function with one real pole
one complex pole
pair
with a quality factor Q and a real zero
is
In order to obtain the transfer function of the amplifier
shown in
wPl
__
OmItting bulk capacitances for simplicity
A. Applying Kirchhoff's current la w (KCL)
WP3),
1
_1_ ::::}
WPI� -b 1
1_ ::::}
b2 �
WP2 � �b 2
WPlwP2
1
::::} WP3 �
b3 �
�
��
b3
WPlWP2WP3
ro4119m6ro6{rosllro2)
*
*
s,
9m of MbM2
9m1
R1
G1
9m2
RouT
GOUT
9mB
WP2
Assuming widely separated poles
&laquo;:
&laquo;:
2
3
coefficients of
s and s can be approximated as
SMALL-SIGNAL PARAMETERS OF TWO-STAGE AMPLIFIER (FIG. 2)
QWP2
(8)
2
8
�
'
3
and s in the expanded denominator
_1_
_1 _
b1 - WPI + QWP2
1
1
b2 - QWPIWP2 +�
(9)
b3 =�
WPIWp
2
(WPI
WP2 ),
Assuming widely separated poles
&laquo;:
3
cient of
and s can be approximated as
s, S2
the coeffi&shy;
function of a two-stage amplifier with three left-half-plane
(LHP) poles and one LHP zero has a general form given by
G(s) (1 + bls + b2 s2 + b3s3)
Gain: ADc is the DC gain of the
(10)
(3)
1) DC
amplifier, given
by the product of the gains of the two stages as
(4)
2) Three Real Poles: The generic transfer function with
three real LHP poles
is
WZI
WPI . WP2 , WP3
and single LHP zero
C. Calculating Pole Locations
The two-stage amplifier in Fig.
the output capacitor
1) Three Real Poles: Assuming real and widely separated
poles, the equation for the dominant pole
is given by
WPI
2
3
The coefficients for
s and s in the expanded denominator
of the transfer function in (5) are compared with those of the
transfer function in (3) as shown below:
s,
WPI=
__
1
= - 1(
/ RouTCe+ReCe+ROUTCOUT
b1
+Ce/ gmb +R1 C1 +RICegm2RoUT)
(6)
__
Applying
gm1 Rl , gm2ROUT &raquo; 1
:::
ROUT
(
&quot;
1+S
(aa[
RO
: .�J)
)
�
��
�[ - - - - - -------+-8
------- - - - - - -- - - - - - - - - - � '
- - - - -+- c- - g- �
�
1
ROUTCC + R
CC c
C �k + ROUTC OUT + RIC l + RlCC g m 2Ro uTl+
bl
82 R
9�kCCI +
[ C CC ROUTC OUT + RIC IROUTCC + RIC IR
C CC + ROUTC OUT9�k CC + RIC IROUTC OUT + RIC 1
b2
3
+ RIC IROUTC OUT g �kCCl
. [RI C IROUTC OUTR
C CC
b3
(12)
(12), that is, neglecting smaller terms,
WPI� - 1/RICegm2RouT .
&quot;-.R'
(11)
In order to simplify this expression, we make the following
assumptions concerning small-signal parameters
R1 , RouT &raquo; Re and
__1_ +_1_ +_1_
b1 wp2
WPI
WP3
1_
1_ +
1_ +
b2 = w w
p
wp
w
p3
wp3wPI
Pl
2
2
1
b3 - wPlwp2wp3
G(s) = -
2 has three real poles if
is very large; otherwise, it has a
complex pole pair.
(5)
__
GOUT
(13)
(1)
(7), the equation of first non-dominant
WPI
Substituting the
in
pole
is given by
WP2
WP2
=
WP
b;1
=
Equation
9m 1 R 1 9m2 ROUT
1
R, Ceo a ? RnlTT
ROUTCOUTCC +ROUTCOUT9mBRc Cc +
WP 1
R 1 C1 9mBRoUTCC +R 1 C 1 9mBRc Cc +
R 1 C 1 9mBRoUTCOUT).
WP2
WP2 ::::;-
(12),
WP2
( cg�?)
(119m2 )C 1 1+
WP2
in
in
(13)
(15). As the value of
compensation capacitor Cc increases, the dominant pole
WP2
moves to higher frequencies.
Substituting
and
in (7), the high frequency pole
is given by
WPI
WP3
WP3
=
WP2
Cc +COUT
- -:::---::--=--,=-....::....::....:..,----,CCCOUT(Rc +119mB)
WP2
and
WP3
were very close to each other, violating
the assumption that the real poles are widely separated. Hence,
we anticipate the existence of a complex pole pair.
2) One Real Pole and One Complex Pole Pair:
For
moderately-sized CouT,the two non-dominant poles converge
and form a single complex pole pair. The dominant pole
WPI
Q
9mB9m2
C 1 COUT
::::;
Cc
( Cc +COUT)
0.78
24MHz
1 / 9&quot;,B)
9mI/CC
( Rc+
14MHz
gml
=
Cc
(19)
.
The phase margin for a system with three real poles and
single zero is computed using the equation given by
(
tan- 1
PM::::;900-tan- 1 ( a2 ) -tan- 1 ( a3 ) +
WCBW/WP2
i:&yen;3
WGBW
WZ 1
)
, (20)
WCBW/WP3'
where
==
and
==
If there exist
a complex pole pair, the phase margin becomes
PM::::;90&deg; -tan- 1
(
)
a2
tan- 1
+
Q(1- a� )
(
WGBW
WZ 1
)
(21)
.
4 shows the AC analysis simulated result of two-stage
1. The hand-calculated gain and pole&shy;
Fig_
amplifier shown in Fig.
zero locations of Table II are very close to the simulated
values.
&quot; F=======:::::-r--&quot;&quot;&quot;&quot;&quot;---&quot;-&quot;&quot;&quot;'----=1
WP2
::::;
V
1
is unchanged and is given by (13).
Applying the assumptions from (12), the non-dominant
complex pole
and quality factor Q are given by
WP2
- Cc
)
The gain-bandwidth product is given by
i:&yen;2
signal parameters of Table I, we found that the calculated val&shy;
(C
C+C
28MHz
9rn 2CQlJT
9m BCl
Bandwith and Phase Margin
(16)
In a design example, when we input realistic values for small&shy;
V
CCOUT .
WGBW
WPI
moves to lower frequencies, whereas the non-dominant pole
••
WZ 1
E.
WPI
50dB
40kHz
9mB9&quot;, 2
CICOUT
Q
(15)
.
Observe the equations of the dominant pole
and first non-dominant pole
(14)
is approximated as
1
ValuelFrequency
Parameter
-(9m2RoUT9mBR 1 Cc)/(R 1 C 1 CC +
Applying the assumptions in
ues for
TABLE II
POLE/ZERO EQUATIONS OF FIG I AND THEIR ApPROXIMATE VALUES
&quot;
(17)
Gain=.51.OdB
GBP=II.6r..&middot;IHz
PM=.59 . .52&deg;
-2S
,
,
,
.
&quot;
9m2 COUT
9mB C 1
D. One Real Zero
The equation of the LHP zero is given by
WZ 1
=
1
.
Cc(Rc +119mB)
-
(18)
-180
Table II summarizes the derived equations of the poles and
10
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - .,
'
'
10
J
10
10
�
Frtqlltu('�'(Hl)
'
10
10
'
'
10
-
'
10
zeros. The third column in the table represents the hand&shy;
calculated values. Fig.
3 illustrates the pole/zero locations
before and after compensation.
Two-stage op-amp
Uncompensated
Fig. 4. Simulated loop gain and phase response of the two-stage amplifier
with the theoretical pole and zero locations indicated.
IV. POLE-ZERO ANALYSIS OF THREE-STAGE AMPLIFIER
5
. The first stage is a
The three-stage pseudo class-AB amplifier shown in Fig.
�)2
&amp;�)2
Two-Stage Cascode
Compensation
COZ]
Complex
pole-pair
Fig. 3.
Diagram illustrating pole/zero locations (not to scale).
is the NMOS version of the circuit in
differential amplifier and the next two stages are common&shy;
source amplifiers. The PMOS transistor
MlO, which is an
inverting common-source amplifier, creates a feed-forward
path
gmF from the intermediate node to the output node. The
last stage of the amplifier is a non-inverting common-source
amplifier. It comprises an NMOS current mirror formed by
Mg and M11 of dimensions in I:K ratio, as shown in Fig. 5.
by proper placement of the two LHP zeros created by RNMC.
The fourth pole is at very high frequency. As a result, the
overall amplifier response can be designed to behave as a two&shy;
pole system with widely-separated poles.
As a design example, the simulated frequency response of
the three-stage amplifier is shown in Fig.
7.
100
==
2-
50
t3
0
=
Fig. 5.
.......
. . . . . . . . . . . . . . . • .1
I
-50
'
10
RNMC with nulling resistor is adopted to stabilize the
10
'
10i
I
amplifier for a wide range of capacitive loads. The small-signal
�
.....,...
8
ICI
Fig. 6.
�
&pound;
vOUT
C,
Small-signal model of three-stage amplifier shown in Fig. 5.
Using notations similar to the previous example, iC I and
iC2 are the currents flowing through the compensation network
from node VOUT to node VI and from node V2 to node VI,
respectively. Apply KCL at every node in Fig.
6, to obtain the
set of equations:
iC I + iC2 gmlvS + * + VI SCI
iC I + iC2 V ��V !
iC I v�gc��x
.
tC 2 �
I/SCC2
o
iC2 + gm2VI + � + V2SC2
V
gm3V2 iCI + gmFVI + :!!&pound;l
+ VOUTSCOUT
R
llI...
OUT
=
=
=
(22)
=
=
=
Solving these equations as illustrated in previous example, so
as to eliminate
Vb V2 , iC I
and
iC2
the DC gain, poles and
zeros are derived as shown in Table III
TABLE III
POLE/ZERO EQUATIONS OF THREE-STAGE AMPLIFIER WITH RNMCR
Value
Parameter
Equation
9ml RI(9m2R2 9m3ROUT + 9mFROUT)
WPI
W
P2
W
P3
ZI
W
Z2
W
WGBW
-
1
RlCCI9=2R29=3ROUT
2TTti,lcCJ
CC2(CCl+COUT)
- 9=3 CCI+COUT
CCICCOUT
1
RC(CCl+CC2)
- 9=29=3 CCI+CC2
(9=2+9=3 CCI CC2
9mI/CC
The amplifier has four poles.
CC I
..
.
I
.
10
'
I
�
R,
I
. . ... . ....... .......
'
10
FrE'qut'u(y (Hz)
10
'
91dB
122Hz
2.1MHz
32MHz
3.3MHz
29MHz
4.7MHz
creates the dominant pole.
The effect of the next two non-dominant poles can be nullified
10
'
10
I
I
'
I
I
6
CCI
v,
I
I
Schematic of three-stage amplifier based on [12).
model of this three-stage amplifier is shown in Fig.
I
I
-45
-90
-135
8
[PI
-1 0
'
10
10
'
10
'
10
'
'
10
Frt&quot;qu('ufY (Hz)
10
'
Fig. 7. Simulated loop gain and phase response of the three-stage amplifier
with the theoretical pole and zero locations indicated.
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