Datasheet

MAX16932/MAX16933
2.2MHz, 36V, Dual Buck
with 20µA Quiescent Current
19Maxim Integrated
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 1200µS (typ),
and R
OUT,EA
is the output resistance of the error ampli-
fier, which is 30MI(typ) (see the Electrical Characteristics
table.)
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
). Select a value
for f
C
in the range:
<< ≤
SW
pMOD C
f
ff
5
At the crossover frequency, the total loop gain must be
equal to 1. So:
CC
FB
MOD(f ) EA(f )
OUT
V
GAIN GAIN 1
V
×× =
C
EA(f ) m,EA C
GAIN g R= ×
C
pMOD
MOD( f ) MOD(dc)
C
f
GAIN GAIN
f
= ×
Therefore:
C
FB
MOD(f ) m,EA C
OUT
V
GAIN g R 1
V
× × ×=
Solving for R
C
:
C
OUT
C
m,EA FB MOD(f )
V
R
g V GAIN
=
××
Set the error-amplifier compensation zero formed by R
C
and C
C
at the f
pMOD
. Calculate the value of C
C
as follows:
C
1
C
2f R
pMOD C
=
π× ×
If f
zMOD
is less than 5 x f
C
, add a second capacitor C
F
from COMP to AGND. The value of C
F
is:
F
1
C
2f R
zMOD C
=
π× ×
As the load current decreases, the modulator pole also
decreases; however, the modulator gain increases
accordingly and the crossover frequency remains the
same.
Below is a numerical example to calculate the compen-
sation network component values of Figure 2:
A
V_CS
= 11V/V
R
DCR
= 15mI
g
mc
= 1/(A
V_CS
x R
DC
) = 1/(11 x 0.015) = 6.06
V
OUT
= 5V
I
OUT(MAX)
= 5.33A
R
LOAD
= V
OUT
/I
OUT(MAX)
= 5V/5.33A = 0.9375I
C
OUT
= 2x47µF = 94µF
ESR = 9mI/2 = 4.5mI
f
SW
= 26.4/65.5kI = 0.403MHz
=×=
MOD(dc)
GAIN 6.06 0.9375 5.68
pMOD
1
f 1.8kHz
2 94µF 0.9375
= ≈
π× ×
<< ≤
SW
pMOD C
f
ff
5
C
1.8kHz f 80.6kHz<< ≤
select f
C
= 40kHz
zMOD
1
f 376kHz
2 4.5m 94µF
= ≈
π× Ω×
since f
zMOD
>f
C
:
R
C
≈ 16kI
C
C
≈ 5.6nF
C
F
≈ 27pF