Owner manual

MAX5099
Dual, 2.2MHz, Automotive Synchronous Buck
Converter with 80V Load-Dump Protection
18 ______________________________________________________________________________________
Inductor Selection
Three key inductor parameters must be specified for
operation with the MAX5099: inductance value (L),
peak inductor current (I
L
), and inductor saturation cur-
rent (I
SAT
). The minimum required inductance is a func-
tion of operating frequency, input-to-output voltage
differential, and the peak-to-peak inductor current (ΔI
L
).
A good compromise is to choose ΔI
L
equal to 30% of
the full load current. To calculate the inductance, use
the following equation:
where V
IN
and V
OUT
are typical values (so that efficien-
cy is optimum for typical conditions). The switching fre-
quency is set by R
OSC
(see the Setting the Switching
Frequency section). The peak-to-peak inductor current,
which reflects the peak-to-peak output ripple, is worse
at the maximum input voltage. See the Output
Capacitor section to verify that the worst-case output
ripple is acceptable. The inductor saturation current is
also important to avoid runaway current during output
overload and continuous short circuit. Select the I
SAT
to
be higher than the maximum peak current limits of 4.3A
and 2.6A for converter 1 and converter 2.
Input Capacitor
The discontinuous input current waveform of the buck
converter causes large ripple currents at the input. The
switching frequency, peak inductor current, and allow-
able peak-to-peak voltage ripple dictate the input
capacitance requirement. Note that the two converters
of the MAX5099 run 180° out-of-phase, thereby effec-
tively doubling the switching frequency at the input.
The input ripple waveform would be unsymmetrical due
to the difference in load current and duty cycle between
converter 1 and converter 2. The worst-case mismatch
is when one converter is at full load while the other is at
no load or in shutdown. The input ripple is comprised of
ΔV
Q
(caused by the capacitor discharge) and ΔV
ESR
(caused by the ESR of the capacitor). Use ceramic
capacitors with high ripple-current capability at the input
connected between DRAIN_ and PGND. Assume the
contribution from the ESR and capacitor discharge
equal to 50%. Calculate the input capacitance and ESR
required for a specified ripple using the following equa-
tions:
where
and
where
where I
OUT
is the maximum output current from either
converter 1 or converter 2, and D is the duty cycle for
that converter. f
SW
is the frequency of each individual
converter. For example, at V
IN
= 12V, V
OUT
= 3.3V at
I
OUT
= 2A, and with L = 3.3μH, the ESR and input
capacitance are calculated for a peak-to-peak input rip-
ple of 100mV or less, yielding an ESR and capacitance
value of 20mΩ and 6.8μF for 1.25MHz frequency. At low
input voltages, also add one electrolytic bulk capacitor
of at least 100μF on the converters’ input voltage rail.
This capacitor acts as an energy reservoir to avoid pos-
sible undershoot below the undervoltage-lockout thresh-
old during power-on and transient loading.
D
V
V
OUT
IN
=
C
IDD
Vf
IN
OUT
QSW
=
×
()
×
−1
Δ
ΔI
VV V
Vf L
L
IN OUT OUT
IN SW
=
()
×
××
−
ESR
V
I
I
IN
ESR
OUT
L
=
+
Δ
Δ
2
L
VVV
Vf I
OUT IN OUT
IN SW L
=
()
××
−
Δ
R
A
V
OUT_
V
OUT_
SOURCE_
FB_
V
OUT_
≥
0.8V
R
B
MAX5099
R
C
FB_
SOURCE_
BYPASS
V
OUT_
< 0.8V
R
A
MAX5099
Figure 2. Adjustable Output Voltage