Datasheet

ADP2386 Data Sheet
Rev. A | Page 18 of 24
DESIGN EXAMPLE
ADP2386
BST
FB
COMP
PGOOD
GND
RT
SYNC
VREG
SS
SW
PGND
EN
PVIN
V
IN
= 12V
C
SS
22nF
C
OUT1
100 µ F
6.3V
C
OUT2
47
µ
F
6.3V
C
VREG
1µF
C
BST
0.1µF
R
T
100kΩ
L1
2.2µF
V
OUT
= 3.3V
C
C
1.2nF
C
CP
4.7pF
R
C
44.2kΩ
R
TOP
2.21kΩ
1%
10211-033
R
TOP
10kΩ
1%
C
IN
10µF
25V
Figure 33. Schematic for Design Example
This section describes the procedures for selecting the external
components, based on the example specifications that are listed
in Table 8. See Figure 33 for the schematic of this design example.
Table 8. Step-Down DC-to-DC Regulator Requirements
Parameter Specification
Input Voltage
V
IN
= 12.0 V ± 10%
Output Voltage V
OUT
= 3.3 V
Output Current I
OUT
= 6 A
Output Voltage Ripple ∆V
OUT_RIPPLE
= 33 mV
Load Transient ±5%, 1 A to 5 A, 2 A/μs
Switching Frequency
f
SW
= 600 kHz
OUTPUT VOLTAGE SETTING
Choose a 10 kΩ resistor as the top feedback resistor (R
TOP
),
and calculate the bottom feedback resistor (R
BOT
) by using the
following equation:
R
BOT
= R
TOP
×








− 6
.0
6
.0
OUT
V
To set the output voltage to 3.3 V, the resistors values are as
follows: R
TOP
= 10 kΩ, and R
BOT
= 2.21 kΩ.
FREQUENCY SETTING
To set the switching frequency to 600 kHz, connect a 100 kΩ
resistor from the RT pin to GND.
INDUCTOR SELECTION
The peak-to-peak inductor ripple current, ∆I
L
, is set to 30% of
the maximum output current. Use the following equation to
estimate the inductor value:
L =
SWL
OUTIN
fI
DVV
×∆
×− )(
where:
V
IN
= 12 V.
V
OUT
= 3.3 V.
D = 0.275.
∆I
L
= 1.8 A.
f
SW
= 600 kHz.
This calculation results in L = 2.215 μH. Choose the standard
inductor value of 2.2 μH.
The peak-to-peak inductor ripple current can be calculated by
using the following equation:
ΔI
L
=
SW
OUTIN
fL
DVV
×
×− )(
This calculation results in ∆I
L
= 1.81 A.
Use the following equation to calculate the peak inductor current:
I
PEAK
= I
OUT
+
2
L
I∆
This calculation results in I
PEAK
= 6.905 A.
Use the following equation to calculate the rms current flowing
through the inductor:
I
RMS
=
12
2
2
L
OUT
I
I
∆
+
This calculation results in I
RMS
= 6.023 A.
Based on the calculated current value, select an inductor with
a minimum rms current rating of 6.03 A and a minimum
saturation current rating of 6.91 A.
However, to protect the inductor from reaching its saturation
point under the current-limit condition, the inductor should be
rated for at least a 9.6 A saturation current for reliable operation.
Based on the requirements described previously, select a 2.2 μH
inductor, such as the FDVE1040-2R2M from Toko, which has
a 6.8 mΩ DCR and a 11.4 A saturation current.