Rainbow Electronics MAX5098A - Manual (Page 24)

MAX5098A
Dual, 2.2MHz, Automotive Buck or Boost
Converter with 80V Load-Dump Protection
24 ______________________________________________________________________________________
Target the unity-gain crossover frequency for:
2) Place a zero
where R F ≥ 10kΩ.
3) Calculate C
I
for a target crossover frequency, f
C
:
where
ω
C = 2π x f C :
4) Place a pole
5) Place the second zero
where
6) Place the second pole
the switching frequency.
Load-Dump Protection MOSFET
Select the external MOSFET with an adequate voltage
rating, V
DSS
, to withstand the maximum expected load-
dump input voltage. The on-resistance of the MOSFET,
R
DS(ON) , should be low enough to maintain a minimal
voltage drop at full load, limiting the power dissipation
of the MOSFET.
During regular operation, the power dissipated by the
MOSFET is:
P NORMAL = I LOAD
2 x R DS(ON)
where I LOAD is equal to the sum of both converters’
input currents.
The MOSFET operates in a saturation region during
load dump, with both high voltage and current applied.
Choose a suitable power MOSFET that can safely oper-
ate in the saturation region. Verify its capability to sup-
port the downstream DC-DC converters input current
during the load-dump event by checking its safe oper-
ating area (SOA) characteristics. Since the transient
peak power dissipation on the MOSFET can be very
high during the load-dump event, also refer to the ther-
mal impedance graph given in the data sheet of the
power MOSFET to make sure its transient power dissi-
pation is kept within the recommended limits.
Improving Noise Immunity
In applications where the MAX5098A is subject to noisy
environments, adjust the controller’s compensation to
improve the system’s noise immunity. In particular, high-
frequency noise coupled into the feedback loop causes
jittery duty cycles. One solution is to lower the crossover
frequency (see the
Compensation section).
C
C
f R C
CF
F
SW F F
=
Ч Ч Ч Ч
( )
− 2 0 5 1 π .
f
R C
at
P
F CF
2
1
2
1 2 =
Ч Ч π
/
R
f C
R
LC I
I
1
1
2
=
Ч Ч
−
π
f
R C
at f
Z
I
LC 2
1
2 1
=
Ч Ч π
.
R
f C
I
ZERO RHP I
=
Ч Ч
1
2
π
,
f
R C
at f
P
I I
ZERO RHP 1
1
2
=
Ч Ч π
.
,
C
V D L C
R V
I
OSC C OUT OUT
C F IN
=
( )
+
⎡
⎣⎢
⎤
⎦⎥
− 1
2 2 ω
ω
C
f R
F
LC F
=
Ч Ч Ч
1
2 0 75
π
.
f
R C
at f
Z
F F
LC 1
1
2
0 75
=
Ч Ч
Ч
π
. .
f
f
C
ZERO RHP ≤ ,
5