MAX5072EVKIT Maxim Integrated Products, MAX5072EVKIT Datasheet - Page 18

• Manufacturer Part Number: MAX5072EVKIT
• Description: EVAL KIT FOR MAX5072
• Manufacturer: Maxim Integrated Products

Specifications of MAX5072EVKIT

• Main Purpose: DC/DC, Step Up or Down
• Outputs And Type: 2, Non-Isolated
• Voltage - Output: 3.3V, 12V
• Current - Output: 2A, 220mA
• Voltage - Input: 5.5 ~ 16V
• Regulator Topology: Boost, Buck
• Frequency - Switching: 2.2MHz
• Board Type: Fully Populated
• Utilized Ic / Part: MAX5072
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Power - Output: -
• Lead Free Status / Rohs Status: Lead free / RoHS Compliant

2.2MHz, Dual-Output Buck or Boost
Converter with POR and Power-Fail Output
For output voltages below 0.8V, set the MAX5072 out-
put voltage by connecting a voltage-divider from the
output to FB_ to BYPASS (Figure 6). Select R
BYPASS resistor) higher than a 50kΩ range. Calculate
R
where V
Characteristics table), and V
V
Three key inductor parameters must be specified for
operation with the MAX5072: inductance value (L), peak
inductor current (I
(I
operating frequency, input-to-output voltage differential
and the peak-to-peak inductor current (∆I
allows for a lower inductor value while a lower ∆I
requires a higher inductor value. A lower inductor value
minimizes size and cost, improves large-signal transient
response, but reduces efficiency due to higher peak cur-
rents and higher peak-to-peak output ripple voltage for
the same output capacitor. On the other hand, higher
inductance increases efficiency by reducing the ripple
current. However, resistive losses due to extra wire turns
can exceed the benefit gained from lower ripple current
levels, especially when the inductance is increased with-
out also allowing for larger inductor dimensions. A good
compromise is to choose ∆I
load current. To calculate the inductance use the follow-
ing equation:
where V
is optimum for typical conditions). The switching frequen-
cy is set by R
Frequency section). The peak-to-peak inductor current,
which reflects the peak-to-peak output ripple, is worst at
the maximum input voltage. See the Output Capacitor
Selection section to verify that the worst-case output rip-
ple is acceptable. The inductor saturating current is also
important to avoid runaway current during the output
overload and continuous short circuit. Select the I
be higher than the maximum peak current limits of 4.5A
and 2.2A for converter 1 and converter 2.
18
FB_
SAT
A
with the following equation:
.
______________________________________________________________________________________
). The minimum required inductance is a function of
IN
FB
and V
= 0.8V, V
R
L
A
OSC
OUT
=
=
L
V
V
), and inductor saturation current
R
IN
OUT IN
are typical values (so that efficiency
(see the Setting the Switching
C
BYPASS
⎡
⎢
⎣
×
(
V
V
BYPASS
f
V
SW
FB
OUT_
L
−
= 2V (see the Electrical
−
V
× ∆
equal to 30% of the full
OUT
V
OUT
−
can range from 0V to
I
V
L
Inductor Selection
)
FB
⎤
⎥
⎦
L
). Higher ∆I
C
(FB to
SAT
to
L
L
The discontinuous input current waveform of the buck
converter causes large ripple currents at the input. The
switching frequency, peak inductor current, and the
allowable peak-to-peak voltage ripple dictate the input
capacitance requirement. Increasing the switching fre-
quency or the inductor value lowers the peak to aver-
age current ratio, yielding a lower input capacitance
requirement. Note that two converters of MAX5072 run
180° out-of-phase, thereby effectively 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 input ripple is com-
prised of ∆V
∆V
load converter dictates the ESR requirement, while the
capacitance requirement is a function of the loading
mismatch between the two converters. The worst-case
mismatch is when one converter is at full load while the
other is at no load or in shutdown. Use low-ESR ceramic
capacitors with high ripple-current capability at the
input. Assume the contribution from the ESR and capac-
itor discharge equal to 50%. Calculate the input capaci-
tance and ESR required for a specified ripple using the
following equations:
where
and
where
where I
converter 1 or converter 2, and D is the duty cycle for
that converter. f
converter. For example, at V
I
capacitance are calculated for a peak-to-peak input
OUT
ESR
= 2A, and with L = 3.3µH, the ESR and input
(caused by the ESR of the capacitor). A higher
OUT
Q
is the maximum output current from either
∆I
L
(caused by the capacitor discharge) and
ESR
C
SW
=
IN
(
IN
V
=
is the frequency of each individual
IN
D
V
I
=
OUT
IN
−
=
∆
⎛
⎜
⎝
V
I
V
OUT
×
V
OUT
Q
OUT
V
∆
×
IN
f
V
SW
×
IN
ESR
)
D
+
×
(
f
= 12V, V
∆
1
SW
2
×
I
−
L
V
D
OUT
⎞
⎟
⎠
L
)
Input Capacitors
OUT
= 3.3V at