Datasheet

LTC3112
23
3112fc
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both higher order poles (f
POLE2
and f
POLE3
) occur at the
common frequency, f
P
. In most cases this is a reasonable
assumption since the zeros are typically located between
1kHz and 10kHz and the poles are typically located near
each other at much higher frequencies. Given this as
-
sumption, the
maximum phase boost, f
MAX
, provided by
the compensated error amplifier is determined simply by
the amount of separation between the poles and zeros as
shown by the following equation.
f
MAX
= 4tan
–1
f
P
f
Z






– 270°
A reasonable choice is to pick the frequency of the poles,
f
P
, to be about 50 times higher than the frequency of the
zeros, f
Z
, which provides a peak phase boost of approxi-
mately f
MAX
= 60° as was assumed previously. Next, the
phase boost must be centered so that the peak phase
occurs at the target crossover frequency. The frequency
of the maximum phase boost, f
CENTER
, is the geometric
mean of the pole and zero frequencies as shown below.
f
CENTER
= f
P
f
Z
= 50 • f
Z
≅ 7f
Z
Therefore, in order to center the phase boost given a factor
of 50 separation between the pole and zero frequencies,
the zeros should be located at one seventh of the cross
-
over frequency and the poles should be located at seven
times the crossover frequency as given by the fol lowing
equations.
f
Z
=
1
7
f
C
=
1
7
35kHz
( )
= 5kHz
f
P
= 7f
C
= 7 35kHz
( )
= 250kHz
This placement of the poles and zeros will yield a peak phase
boost of 60° that is centered at the cross over frequency,
f
C
. Next, in order to produce the desired target crossover
frequency, the gain of the compensation network at the
point of maximum phase boost, G
CENTER
, must be set to
−7dB. The gain of the compensated error amplifier at the
point of maximum phase gain is given by the following
equation.
G
CENTER
= 10log
2πf
P
2πf
Z
( )
3
R
TOP
C
FB
( )
2








dB
Assuming a multiple of 50 separation between the pole
frequencies and zero frequencies this can be simplified
to the following expression.
G
CENTER
= 20log
50
2πf
C
R
TOP
C
FB






dB
This equation completes the set of constraints needed to
determine the compensation component values. Specifi-
cally, the two zeros, f
ZERO1
and f
ZERO2
, should be located
near 5kHz. The two poles, f
POLE2
and f
POLE3
, should be
located near 250kHz and the gain should be set to provide
a gain at the crossover frequency of G
CENTER
= –7dB.
The first step in defining the compensation component
values is to pick a value for R
TOP
that provides an accept-
ably low quiescent current through the resistor divider. A
value
of R
TOP
= 845kΩ is a reasonable choice and is used
in several applications circuits. Next, the value of C
FB
can
be found in order to set the error amplifier gain at the
crossover frequency to −7dB as follows.
G
CENTER
= –7dB = 20log
50
2π 35kHz
( )
845kΩ
( )
C
FB






C
FB
=
50
0.185• 10
12
• antilog
–7
20






≅ 680pF
The compensation poles can be set at 250kHz and the
zeros at 5kHz by using the expressions for the pole and
zero frequencies given in the previous section. Setting the
frequency of the first zero f
ZERO1
, to 5kHz results in the
following value for R
FB
.
R
FB
=
1
2π 680pF
( )
5kHz
( )
≅ 45kΩ
A 33kΩ was selected to split the two zeros slightly apart,
giving a higher zero frequency of 7kHz. This leaves the
free parameter, C
POLE
, to set the frequency f
POLE1
to the
common pole frequency of 250kHz.
applicaTions inForMaTion