ADP1821
Rev. C | Page 15 of 24
For initial practical designs, a good choice for the crossover
frequency is 1/10 of the switching frequency; first calculate
10
SW
CO
f
f =
(19)
This gives sufficient frequency range to design a compensation
that attenuates switching artifacts, yet also gives sufficient
control loop bandwidth to provide good transient response.
The output LC filter is a resonant network that inflicts two poles
upon the response at a Frequency f
LC
, so next calculate
LCπ
f
LC
2
1
=
(20)
The LC corner frequency is about two orders of magnitude
below the switching frequency, and therefore about one order of
magnitude below crossover. To achieve sufficient phase margin
at crossover to guarantee stability, the design must compensate
for the two poles at the LC corner frequency with two zeros to
boost the system phase prior to crossover. The two zeros require
an additional pole or two above the crossover frequency to
guarantee adequate gain margin and attenuation of switching
noise at high frequencies.
Depending on component selection, one zero might already be
generated by the ESR of the output capacitor. Calculate this zero
corner frequency,
f
ESR
, as
OUT
ESR
ESR
CRπ
f
2
1
= (21)
This zero is often near or below crossover and is useful in
bringing back some of the phase lost at the LC corner.
Figure 16 shows a typical bode plot of the LC filter by itself.
The gain of the LC filter at crossover can be linearly
approximated from
Figure 16 as
ESRLC
FILTER
AAA +=
×
×=
ESR
CO
LC
ESR
FILTER
f
f
f
f
A logdB20logdB40
(22)
If f
ESR
≈ f
CO
, then add another 3 dB to account for the local
difference between the exact solution and the preceding linear
approximation.
To compensate the control loop, the gain of the system must be
brought back up so that it is 0 dB at the desired crossover
frequency. Some gain is provided by the PWM modulation
itself, and it is given by
=
RAMP
IN
MOD
V
V
A log20
(23)
For systems using the internal oscillator, this becomes
=
V25.1
log20
IN
MOD
V
A
(24)
0dB
GAIN
FREQUENCY
LC FILTER BODE PLOT
PHASE
f
LC
f
ESR
f
CO
f
SW
A
FILTER
–40dB/dec
Φ
FILTER
–90°
–180°
–20dB/dec
05310-015
Figure 16. LC Filter Bode Plot
Note that if the converter is being synchronized, the ramp
voltage, V
RAMP
, is lower than 1.25 V by the percentage of
frequency increase over the nominal setting of the FREQ pin
=
SYNC
FREQ
RAMP
f
f
V V25.1
(25)
The rest of the system gain is needed to reach 0 dB at crossover.
The total gain of the system therefore, is given by
A
T
= A
MOD
+ A
FILTER
+ A
COMP
(26)
where:
A
MOD
is the gain of the PWM modulator.
A
FILTER
is the gain of the LC filter including the effects of the
ESR zero.
A
COMP
is the gain of the compensated error amplifier.
Additionally, the phase of the system must be brought back up
to guarantee stability. Note from the bode plot of the filter that
the LC contributes −180° of phase shift. Additionally, because
the error amplifier is an integrator at low frequency, it contrib-
utes an initial −90°. Therefore, before adding compensation or
accounting for the ESR zero, the system is already down −270°.
To avoid loop inversion at crossover, or −180° phase shift, a
good initial practical design is to require a phase margin of 60°,
which gives an overall phase loss of −120° from the initial low
frequency dc phase. The goal of the compensation is to boost
the phase back up from −270° to −120° at crossover.
ADP1821
Rev. C | Page 16 of 24
Two common compensation schemes are used, which are
sometimes referred to as Type II or Type III compensation,
depending on whether the compensation design includes two
or three poles. (Dominant pole compensations, or single pole
compensation, is referred to as Type I compensation, but it is
not very useful for dealing successfully with switching regulators.)
If the zero produced by the ESR of the output capacitor provides
sufficient phase boost at crossover, Type II compensation is
adequate. If the phase boost produced by the ESR of the output
capacitor is not sufficient, another zero is added to the compen-
sation network, and thus Type III is used.
In
Figure 17, the location of the ESR zero corner frequency gives
significantly different net phase at the crossover frequency.
GAIN
FREQUENCY
PHASE
LC FILTER BODE PLOT
PHASE CONTRIBUTION AT CROSSOVER
OF VARIOUS ESR ZERO CORNERS
f
SW
f
CO
f
ESR3
f
ESR2
f
ESR1
0dB
f
LC
–40dB/dec
–20dB/dec
–90°
–180°
Φ
1
Φ
2
Φ
3
05310-016
Figure 17. LC Filter Bode Plot
Use the following guidelines for selecting between Type II and
Type III compensators.
If
2
CO
ESRZ
f
f
, use Type II compensation.
If
2
CO
ESRZ
f
f >
, use Type III compensation.
The following equations were used for the calculation of the
compensation components as shown in
Figure 18 and Figure 19:
IZ
Z
CR
f
π
2
1
1
=
(27)
)(2
1
2
FFTOPFF
Z
RRC
f
+
=
π
(28)
HFI
HFI
Z
P
CC
CC
R
f
+
=
π
2
1
1
(29)
FFFF
P
CR
f
π
2
1
2
=
(30)
where:
f
Z1
is the zero produced in the Type II compensation.
f
Z2
is the zero produced in the Type III compensation.
f
P1
is the pole produced in the Type II compensation.
f
P2
in the pole produced in the Type III compensation.
Type II Compensator
G
(dB)
PHASE
–180°
–270°
f
Z
f
P
0V
VRAMP
C
HF
C
I
R
Z
R
TOP
R
BOT
FROM
V
OUT
VREF
EA
COMP
TO PWM
1
S
L
O
P
E
1
S
L
O
P
E
0
5310-017
Figure 18. Type II Compensation
If the output capacitor ESR zero frequency is sufficiently low (≤ ½
of the crossover frequency), use the ESR to stabilize the regulator.
In this case, use the circuit shown in
Figure 18. Calculate the
compensation resistor, Rz, with the following equation:
2
LCIN
COESRRAMPTOP
Z
fV
ffVR
R =
(31)
where:
f
CO
is chosen to be 1/10 of f
SW.
V
RAMP
is 1.25 V.
Next, choose the compensation capacitor to set the compensation
zero,
f
Z1
, to the lesser of ¼ of the crossover frequency or ½ of the
LC resonant frequency
IZ
SWCO
Z
CR
ff
f
π
2
1
404
1
===
(32)
or
IZ
LC
Z
CR
f
f
π
2
1
2
1
==
(33)
Solving for C
I
in Equation 32 yields
SWZ
I
fR
C
π
20
=
(34)
ADP1821
Rev. C | Page 17 of 24
Solving for C
I
in Equation 33 yields
LCZ
I
fR
C
π
1
=
(35)
Use the larger value of C
I
from Equation 34 or Equation 35.
Because of the finite output current drive of the error amplifier,
C
I
needs to be less than 10 nF. If it is larger than 10 nF, choose a
larger R
TOP
and recalculate R
Z
and C
I
until C
I
is less than 10 nF.
Next, choose the high frequency pole
f
P1
to be ½ of f
SW
.
SWP
ff
2
1
1
=
(36)
Since
C
HF
<< C
I
, Equation 29 is simplified to
HFZ
P
CR
f
π
2
1
1
=
(37)
Solving for C
HF
in Equation 36 and Equation 37 yields
ZSW
HF
Rf
C
π
1
=
(38)
Type III Compensator
0V
VRAMP
G
(dB)
PHASE
–270°
–90°
f
Z
f
P
C
HF
C
I
R
Z
C
FF
R
TOP
R
BOT
FROM
V
OUT
VREF
EA
COMP
TO PWM
R
FF
1
S
L
O
P
E
1
S
L
O
P
E
+
1
S
LO
P
E
0
5310-018
Figure 19. Type III Compensation
If the output capacitor ESR zero frequency is greater than ½ of
the crossover frequency, use Type III compensator as shown in
Figure 19. Set the poles and zeros as follows:
SWPP
fff
2
1
21
==
(39)
IZ
SWCO
ZZ
CR
ff
ff
π
2
1
404
21
====
(40)
or
IZ
LC
ZZ
CR
f
ff
π
2
1
2
2
1
===
(41)
Use the lower zero frequency from Equation 40 or Equation 41.
Calculate the compensator resistor, R
Z
, by
2
1
LCIN
COZRAMPTOP
Z
fV
ffVR
R =
(42)
Next calculate C
I
1
2
1
ZZ
I
fR
C
π
=
(43)
Because of the finite output current drive of the error amplifier,
C
I
needs to be less than 10 nF. If it is larger than 10 nF, choose a
larger R
TOP
and recalculate R
Z
and C
I
until C
I
is less than 10 nF.
Since
C
HF
<< C
I
, combining Equation 29 and Equation 39 yields
ZSW
HF
Rf
C
π
1
=
(44)
Next calculate the feedforward capacitor C
FF.
Assume R
FF
<< R
TOP
,
then Equation 28 is simplified to
TOPFF
Z
RC
f
π
2
1
2
=
(45)
Solving
C
FF
in Equation 45 yields
2
2
1
ZTOP
FF
fR
C
π
=
(46)
where
f
Z2
is obtained from Equation 40 or Equation 41.
The feedforward resistor,
R
FF
, can be calculated by combining
Equation 30 and Equation 39
SWFF
FF
fC
R
π
1
=
(47)
Check that the calculated component values are reasonable.
For instance, capacitors smaller than about 10 pF should be
avoided. In addition, the ADP1821 error amplifier has finite
output current drive, so R
Z
values less than 3 kΩ and C
I
values
greater than 10 nF should be avoided. If necessary, recalculate
the compensation network with a different starting value of R
TOP
. If
C
HF
is too small, start with a smaller value of R
TOP
. If R
Z
is too
small and C
I
is too big, start with a larger value of R
TOP
.
In general, aluminum electrolytic capacitors have high ESR,
therefore, a Type II compensation is adequate. However, if
several aluminum electrolytic capacitors are connected in
parallel, producing a low effective ESR, then Type III compensation
is needed. In addition, ceramic capacitors have very low ESR, on
the order of a few milliohms, requiring Type III compensation for
ceramic output capacitors. Type III compensation offers better
performance than Type II in terms of more low frequency gain,
more phase margin, and less high frequency gain at the
crossover frequency.

ADP1821ARQZ-R7

Mfr. #:
Manufacturer:
Analog Devices / Linear Technology
Description:
Switching Controllers 20 Amp Buck Controller in QSOP
Lifecycle:
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