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ADP2381AREZ-R7 数据表(PDF) 20 Page - Analog Devices

部件名 ADP2381AREZ-R7
功能描述  20 V, 6 A Synchronous Step-Down
PDF  28 Pages
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制造商  AD [Analog Devices]
网页  http://www.analog.com
标志 AD - Analog Devices

ADP2381AREZ-R7 数据表(HTML) 20 Page - Analog Devices

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ADP2381
Data Sheet
Rev. 0 | Page 20 of 28
DESIGN EXAMPLE
This section provides the procedures of selecting the external
components based on the example specifications listed in Table 10.
The schematic of this design example is shown in Figure 36.
Table 10. Step-Down DC-to-DC Regulator Requirements
Parameter
Specification
Input Voltage
VIN = 12.0 V ± 10%
Output Voltage
VOUT = 3.3 V
Output Current
IOUT = 6 A
Output Voltage Ripple
∆VOUT_RIPPLE = 33 mV
Load Transient
±5%, 1 A to 5 A, 2 A/μs
Switching Frequency
fSW = 500 kHz
OUTPUT VOLTAGE SETTING
Choose a 10 kΩ resistor as the top feedback resistor (RTOP) and
calculate the bottom feedback resistor (RBOT) by using the
following equation:
6
.
0
6
.
0
OUT
TOP
BOT
V
R
R
To set the output voltage to 3.3 V, the resistors values are
RTOP = 10 kΩ, RBOT = 2.21 kΩ.
FREQUENCY SETTING
Connect a 100 kΩ resistor from RT pin to GND to set the
switching frequency at 500 kHz.
INDUCTOR SELECTION
The peak-to-peak inductor ripple current, ΔIL, is set to 30% of
the maximum output current. Use the following equation to
estimate the inductor value:
SW
L
OUT
IN
f
I
D
V
V
L
)
(
where:
VIN
= 12 V.
VOUT
= 3.3 V.
D
= VOUT/VIN = 0.275.
ΔIL = 1.8A.
fSW
= 500 kHz.
This results in L = 2.659 μH. Choose the standard inductor
value of 2.2 μH.
The peak-to-peak inductor ripple current can be calculated by
the following equation:
SW
OUT
IN
L
f
L
D
V
V
I
This results in ΔIL = 2.18 A.
The peak inductor current can be calculated using the following
equation:
2
L
OUT
PEAK
I
I
I
This results in IPEAK = 7.09 A.
The rms current flowing through the inductor can be calculated
by the following equation:
12
2
2
L
OUT
RMS
I
I
I
This results in IRMS = 6.03 A.
According to the calculated rms and peak inductor current
values, select an inductor with a minimum rms current rating of
6.03 A and a minimum saturation current rating of 7.09 A.
To protect the inductor from reaching its saturation limit, the
inductor should be rated for at least 9.6 A saturation current for
reliable operation.
Based on these requirements, select a 2.2 μH inductor, such as
the FDVE1040-2R2M from Toko, which has 6.8 mΩ DCR and
11.4 A saturation current.
OUTPUT CAPACITOR SELECTION
The output capacitor is required to meet both the output voltage
ripple requirement and the load transient response.
To meet the output voltage ripple requirement, use the
following equation to calculate the ESR and capacitance of the
output capacitor:
RIPPLE
OUT
SW
L
RIPPLE
OUT
V
f
I
C
_
_
8
L
RIPPLE
OUT
ESR
I
V
R
_
This results in COUT_RIPPLE = 16.5 μF and RESR = 15.1 mΩ.
To meet the ±5% overshoot and undershoot transient
requirements, use the following equations to calculate the
capacitance:
UV
OUT
OUT
IN
STEP
UV
UV
OUT
OUT
OV
OUT
OUT
STEP
OV
OV
OUT
V
V
V
L
I
K
C
V
V
V
L
I
K
C
_
2
_
2
2
_
2
_
)
(
2
)
(
where:
KOV = KUV = 2, the coefficients for estimation purposes.
ΔISTEP = 4 A, the load transient step.
ΔVOUT_OV = 5%VOUT, the overshoot voltage.
ΔVOUT_UV = 5%VOUT, the undershoot voltage.
This results in COUT_OV = 63.1 μF and COUT_UV = 24.5 μF.
According to the preceding calculation, the output capacitance
must be larger than 63 μF, and the ESR of the output capacitor
must be smaller than 15 mΩ. It is recommended that one 100
μF, X5R, 6.3 V ceramic capacitor and one 47 μF, X5R, 6.3 V
ceramic capacitor be used, such as the GRM32ER60J107ME20
and GRM32ER60J476ME20 from Murata with an ESR = 2 mΩ.



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