ADP2386-EVALZ Analog Devices, ADP2386-EVALZ Datasheet - Page 18

• Manufacturer Part Number: ADP2386-EVALZ
• Description: Power Management IC Development Tools 20V 6A Evaluation Board
• Manufacturer: Analog Devices
• Type: DC/DC Converters, Regulators & Controllersr
• Series: ADP2386r

Specifications of ADP2386-EVALZ

• Rohs: yes
• Tool Is For Evaluation Of: ADP2386
• Input Voltage: 4.5 V to 20 V
• Factory Pack Quantity: 1

ADP2386
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
Input Voltage
Output Voltage
Output Current
Output Voltage Ripple
Load Transient
Switching Frequency
OUTPUT VOLTAGE SETTING
Choose a 10 kΩ resistor as the top feedback resistor (R
and calculate the bottom feedback resistor (R
following equation:
To set the output voltage to 3.3 V, the resistors values are as
follows: R
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
the maximum output current. Use the following equation to
estimate the inductor value:
where:
V
V
D = 0.275.
∆I
f
SW
IN
OUT
L
= 600 kHz.
= 12 V.
= 1.8 A.
R
L =
= 3.3 V.
BOT
(
= R
TOP
V
IN
∆
TOP
= 10 kΩ, and R
−
I
L
V
×
×
OUT

 

f
V
SW
)
OUT
×
0
D
6 .
− 6 .
V
0
IN
BOT
= 12V

 

10µF
= 2.21 kΩ.
Specification
V
V
I
∆V
±5%, 1 A to 5 A, 2 A/μs
f
25V
C
OUT
SW
IN
OUT
IN
OUT_RIPPLE
= 12.0 V ± 10%
= 600 kHz
= 6 A
C
22nF
100kΩ
= 3.3 V
C
1µF
VREG
R
SS
T
= 33 mV
PVIN
EN
PGOOD
SYNC
RT
VREG
SS
BOT
L
, is set to 30% of
) by using the
ADP2386
GND
Figure 33. Schematic for Design Example
TOP
PGND
),
Rev. A | Page 18 of 24
COMP
BST
SW
FB
C
4.7pF
CP
0.1µF
C
BST
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:
This calculation results in ∆I
Use the following equation to calculate the peak inductor current:
This calculation results in I
Use the following equation to calculate the rms current flowing
through the inductor:
This calculation results in I
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.
R
44.2kΩ
2.2µF
C
1.2nF
C
L1
C
R
10kΩ
ΔI
I
I
PEAK
RMS
TOP
1%
L
=
=
= I
R
2.21kΩ
1%
C
100
(
TOP
V
OUT1
6.3V
OUT
I
IN
µ
OUT
F
L
+
−
×
2
V
I ∆
C
OUT
+
f
2
OUT2
47
6.3V
SW
L
∆
µ
12
F
)
I
×
L
2
D
PEAK
RMS
V
OUT
L
= 1.81 A.
= 6.023 A.
= 6.905 A.
= 3.3V
Data Sheet