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ADP2389ACPZ-R7 数据表(PDF) 17 Page - Analog Devices

部件名 ADP2389ACPZ-R7
功能描述  18 V, 12 A Step-Down Regulator with Programmable Current Limit
PDF  23 Pages
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制造商  AD [Analog Devices]
网页  http://www.analog.com
标志 AD - Analog Devices

ADP2389ACPZ-R7 数据表(HTML) 17 Page - Analog Devices

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Data Sheet
ADP2389/ADP2390
DESIGN EXAMPLE
This section provides the procedures of selecting the external
components based on the example specifications listed in Table 7.
The schematic of this design example is shown in Figure 32.
Table 7. Step-Down DC-to-DC Regulator Requirements
Parameter
Specification
Input Voltage
12.0 V ± 10%
Output Voltage
1.2 V
Output Current
12 A
Output Voltage Ripple
12 mV
Load Transient
±5%, 3 A to 9 A, 2 A/µs
Switching Frequency
500 kHz
OUTPUT VOLTAGE SETTING
Select a 10 kΩ resistor as the top feedback resistor (RTOP) and
calculate the bottom feedback resistor (RBOT) using the
following equation:
0.6
0.6
BOT
TOP
OUT
R
R
V
=
×
To set the output voltage to 1.2 V, the resistors values are RTOP =
10 kΩ and RBOT = 10 kΩ.
FREQUENCY SETTING
Use the following equation to calculate the value of RT:
67,000
(k )
– 12
(kHz)
T
SW
R
f
Ω=
Thus, when fSW = 500 kHz, the value of RT = 122 kΩ.
Select the standard resistor value of 121 kΩ for RT.
INDUCTOR SELECTION
The peak-to-peak inductor ripple current, ∆IL, is set to 33% 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 = 12.0 V
VOUT = 1.2 V
D = 10%
∆IL = 4 A
fSW = 500 kHz
This results in L = 0.54 µH. Select the standard inductor value
of 0.68 µH.
Calculate the peak-to-peak inductor ripple current using the
following equation:
(
)
SW
OUT
IN
L
f
L
D
V
V
I
×
×
=
This results in ∆IL = 3.176 A.
Calculate the peak inductor current with the following equation:
2
L
PEAK
OUT
I
II
=
+
This results in IPEAK = 13.588 A.
Calculate the rms current flowing through the inductor by the
following equation:
2
2
12
L
RMS
OUT
I
II
=
+
This results in IRMS = 12.035 A.
According to the calculated current value, select an inductor
with a minimum rms current rating of 12.035 A and a minimum
saturation current rating of 13.588 A.
However, to protect the inductor from reaching its saturation
point under a current-limit condition, the inductor must be
rated for at least a 20 A saturation current for reliable operation.
Based on these requirements, select a 0.68 µH inductor, such as
the 7443330068 from Würth Elektronik, which has 1.35 mΩ dc
resistance (DCR) and a 38 A saturation current.
OUTPUT CAPACITOR SELECTION
The output capacitor must meet both the output voltage ripple
requirement and load transient response.
To meet the output voltage ripple requirement, use the following
equation to calculate the ESR and capacitance value 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 = 66 μF and RESR = 3.78 mΩ.
To meet the ±5% overshoot and undershoot transient
requirements, use the following equations to calculate the
capacitance:
(
)
OUT
OV
OUT
OUT
STEP
OV
OV
OUT
V
V
V
L
I
K
C
×
×
×
=
_
2
_
2
(
)
UV
OUT
OUT
IN
STEP
UV
UV
OUT
V
V
V
L
I
K
C
_
2
_
2
×
×
×
×
=
where:
KOV = KUV = 2, and are the coefficients for estimation purpose.
∆ISTEP = 6 A, and is the load transient step.
∆VOUT_OV = 5% × VOUT and, is the overshoot voltage.
∆VOUT_UV = 5% × VOUT, and is the undershoot voltage.
This results in COUT_OV = 332 µF and COUT_UV = 38 µF.
Rev. 0 | Page 17 of 23



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