LM2575
Procedure (Adjustable Output Version: LM2575−Adj) (continued)
Procedure
4. Inductor Selection (L1)
A.
Use the following formula to calculate the inductor Volt x
microsecond [V x
ms]
constant:
V out
6
x 10 [V x
ms]
E x T
+
V – V out
in
F[Hz]
V on
B.
Match the calculated E x T value with the corresponding
number on the vertical axis of the Inductor Value Selection
Guide shown in Figure 21. 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.
C.
Next step is to identify the inductance region intersected by
the E x T value and the maximum load current value on the
horizontal axis shown in Figure 21.
D.
From the inductor code, identify the inductor value. Then
select an appropriate inductor from the Table 1 or 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
Ioad
.
The inductor current rating can also be determined by
calculating the inductor peak current:
I
p(max)
+
I
Example
4. Inductor Selection (L1)
A.
Calculate E x T [V x
ms]
constant:
E x T
+
(12 – 8.0) x 8.0 x 1000
+
51 [V x
ms]
52
12
B.
E x T = 51 [V x
ms]
C.
I
Load(max)
= 1.0 A
Inductance Region = L220
D.
Proper inductor value = 220
mH
Choose the inductor from the Table 1 or Table 2.
Load(max)
)
V
in
– V out t on
2L
where t
on
is the “on” time of the power switch and
t on
+
V out
V
x
1
f osc
in
For additional information about the inductor, see the
inductor section in the “External Components” section of
this data sheet.
5. Output Capacitor Selection (C
out
)
A.
Since the LM2575 is a forward−mode switching regulator
with voltage mode control, its open loop 2−pole−2−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:
V
in(max)
[µF]
Cout
w
7.785
V out x L [µH]
B.
Capacitor values between 10
mF
and 2000
mF
will satisfy
the 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.
C.
Due to the fact that the higher voltage electrolytic capacitors
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 8V is appropriate, and a 10 V
or 16 V rating is recommended.
5. Output Capacitor Selection (C
out
)
A.
Cout
w
7.785 12
+
53
µF
8.220
To achieve an acceptable ripple voltage, select
C
out
= 100
mF
electrolytic capacitor.
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