LM2574DW-ADJR2G ON Semiconductor, LM2574DW-ADJR2G Datasheet - Page 12

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LM2574DW-ADJR2G

Manufacturer Part Number
LM2574DW-ADJR2G
Description
IC REG SW 0.5A ADJ OUT 16-SOICW
Manufacturer
ON Semiconductor
Type
Step-Down (Buck)r
Datasheet

Specifications of LM2574DW-ADJR2G

Internal Switch(s)
Yes
Synchronous Rectifier
No
Number Of Outputs
1
Voltage - Output
1.23 ~ 37 V
Current - Output
500mA
Frequency - Switching
52kHz
Voltage - Input
4.75 ~ 40 V
Operating Temperature
-40°C ~ 125°C
Mounting Type
Surface Mount
Package / Case
16-SOIC (0.300", 7.5mm Width)
Mounting Style
SMD/SMT
Lead Free Status / RoHS Status
Lead free / RoHS Compliant
Power - Output
-
Lead Free Status / Rohs Status
Lead free / RoHS Compliant

Available stocks

Company
Part Number
Manufacturer
Quantity
Price
Part Number:
LM2574DW-ADJR2G
Manufacturer:
ON
Quantity:
4 500
Company:
Part Number:
LM2574DW-ADJR2G
Quantity:
2 200
Procedure (Adjustable Output Version: LM2574−ADJ)
4. Inductor Selection (L1)
5. Output Capacitor Selection (C
A. Use the following formula to calculate the inductor Volt x
B. Match the calculated E x T value with the corresponding
C. Next step is to identify the inductance region intersected by
D. From the inductor code, identify the inductor value. Then
A. Since the LM2574 is a forward−mode switching regulator with
B. Capacitor values between 10 mF and 2000 mF will satisfy the
C. Due to the fact that the higher voltage electrolytic capacitors
E x T + (V
microsecond [V x ms] constant:
number on the vertical axis of the Inductor Value Selection
Guide shown in Figure 23. This E x T constant is a measure
of the energy handling capability of an inductor and is
dependent upon the type of core, the core area, the number
of turns, and the duty cycle.
the E x T value and the maximum load current value on the
horizontal axis shown in Figure 27.
select an appropriate inductor from Table 2. The inductor
chosen must be rated for a switching frequency of 52 kHz
and for a current rating of 1.15 x I
rating can also be determined by calculating the inductor
peak current:
where t
For additional information about the inductor, see the inductor
section in the “External Components” section of this data
sheet.
voltage mode control, its open loop 2−pole−1−zero frequency
characteristic has the dominant pole−pair determined by the
output capacitor and inductor values.
For stable operation, the capacitor must satisfy the following
requirement:
loop requirements for stable operation. To achieve an
acceptable output ripple voltage and transient response, the
output capacitor may need to be several times larger than the
above formula yields.
generally have lower ESR (Equivalent Series Resistance)
numbers, the output capacitor’s voltage rating should be at
least 1.5 times greater than the output voltage. For a 5.0 V
regulator, a rating of at least 8.0 V is appropriate, and a 10 V
or 16V rating is recommended.
I
p
(
on
max
C out w 13, 300
is the “on” time of the power switch and
in
)
* V out )
+ I
t
on
Load
+
Procedure
V out
(
V
max
V
V
V out x L mH
in
out
in
V
)
in
x 10
)
x 1.0
out
(
max
F [ Hz ]
f
)
V
osc
Load
6
in
)
* V
V x ms
. The inductor current
mF
2L
out
t
on
LM2574, NCV2574
http://onsemi.com
12
4. Inductor Selection (L1)
5. Output Capacitor Selection (C
A.
B.
C. I
D. Proper inductor value = 1000 mH
A.
Calculate E x T V x ms constant :
Inductance Region = 1000
Choose the inductor from Table 2.
To achieve an acceptable ripple voltage, select
C
E x T + (40 * 24) x 24
E x T + 185 V x ms
Load(max)
out
C out w 13, 300 x
= 100 mF electrolytic capacitor.
= 0.4 A
Example
40
24 x 1000
x 1000
40
52
out
)
+ 22.2 mF
+ 105 V x ms

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