Datasheet

15Maxim Integrated
MAX16909
36V, 220kHz to 1MHz Step-Down Converter
with Low Operating Current
The feedback voltage-divider has a gain of GAIN
FB
=
V
FB
/V
OUT
, where V
FB
is 1V (typ).
The transconductance error amplifier has a DC gain of
GAIN
EA(DC)
= g
m,EA
x R
OUT,EA
, where g
m,EA
is the error-
amplifier transconductance, which is 900FS (typ), and
R
OUT,EA
is the output resistance of the error amplifier.
A dominant pole (f
dpEA
) is set by the compensa-
tion capacitor (C
C
) and the amplifier output resistance
(R
OUT,EA
). A zero (f
zEA
) is set by the compensation
resistor (R
C
) and the compensation capacitor (C
C
).
There is an optional pole (f
pEA
) set by C
F
and R
C
to
cancel the output capacitor ESR zero if it occurs near
the crossover frequency (f
C
, where the loop gain equals
1 (0dB)). Thus:
dpEA
C O U T,E A C
1
f
2 C (R R )
=
π× × +
zEA
CC
1
f
2C R
=
π× ×
pEA
FC
1
f
2CR
=
π× ×
The loop-gain crossover frequency (f
C
) should be set
below 1/5th of the switching frequency and much higher
than the power-modulator pole (f
pMOD
):
SW
pMOD C
f
ff
5
<< ≤
The total loop gain as the product of the modulator gain,
the feedback voltage-divider gain, and the error-amplifier
gain at f
C
should be equal to 1. So:
FB
MOD(fC) EA(fC)
OUT
V
GAIN GAIN 1
V
×× =
For the case where f
zMOD
is greater than f
C
:
GAIN
EA(fC)
= g
m,EA
× R
C
pMOD
MOD(fC) MOD(dc)
C
f
GAIN GAIN
f
= ×
Therefore:
FB
MOD(fC) m,EA C
OUT
V
GAIN g R 1
V
× × ×=
Solving for R
C
:
OUT
C
m,EA FB MOD(fC)
V
R
g V GAIN
=
××
Set the error-amplifier compensation zero formed by R
C
and C
C
(f
zEA
) at the f
pMOD
. Calculate the value of C
C
as follows:
C
pMOD C
1
C
2f R
=
π× ×
If f
zMOD
is less than 5 x f
C
, add a second capacitor,
C
F
, from COMP to GND and set the compensation pole
formed by R
C
and C
F
(f
pEA
) at the f
zMOD
. Calculate the
value of C
F
as follows:
F
zMOD C
1
C
2f R
=
π× ×
As the load current decreases, the modulator pole
also decreases; however, the modulator gain increases
accordingly and the crossover frequency remains the
same.
For the case where f
zMOD
is less than f
C
:
The power-modulator gain at f
C
is:
pMOD
MOD(fC) MOD(dc)
zMOD
f
GAIN GAIN
f
= ×
The error-amplifier gain at f
C
is:
zMOD
EA(fC) m,EA C
C
f
GAIN g R
f
= ××
Therefore:
zMOD
FB
MOD(fC) m,EA C
OUT C
f
V
GAIN g R 1
Vf
× × ×× =
Solving for R
C
:
OUT C
C
m,EA FB MOD(fC) zMOD
Vf
R
g V GAIN f
×
=
×× ×
Set the error-amplifier compensation zero formed by R
C
and C
C
at the f
pMOD
(f
zEA
= f
pMOD
) as follows:
C
pMOD C
1
C
2f R
=
π× ×
If f
zMOD
is less than 5 x f
C
, add a second capacitor C
F
from COMP to GND. Set f
pEA
= f
zMOD
and calculate C
F
as follows:
F
zMOD C
1
C
2f R
=
π× ×