LM2575
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11
Procedure (Adjustable Output Version:
LM2575Adj) (continued)
Procedure
Example
4. Inductor Selection (L1)
A.
Use the following formula to calculate the inductor Volt x
microsecond [V x s] constant:
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
:
where t
on
is the “on” time of the power switch and
For additional information about the inductor, see the
inductor section in the “External Components” section of
this data sheet.
E x T
Vin– Vout
Vout
Von
x
106
F[Hz][V x s]
Ip(max)
ILoad(max)
Vin– Voutton
2L
ton
Vout
Vin
x
1
fosc
4. Inductor Selection (L1)
A.
Calculate E x T [V x s] constant:
B.
E x T = 51 [V x s]
C.
I
Load(max)
= 1.0 A
Inductance Region = L220
D.
Proper inductor value = 220 H
Choose the inductor from the Table 1 or Table 2.
E x T
(12 – 8.0) x12x152
51 [V x s]
5. Output Capacitor Selection (C
out
)
A.
Since the LM2575 is a forwardmode switching regulator
with voltage mode control, its open loop 2pole2zero
frequency characteristic has the dominant polepair
determined by the output capacitor and inductor values
.
For stable operation, the capacitor must satisfy the
following requirement:
B.
Capacitor values between 10 F and 2000 F 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.
Cout
7.785
Vin(max)
Voutx L [
μ
H][
μ
F]
5. Output Capacitor Selection (C
out
)
A.
Cout
To achieve an acceptable ripple voltage, select
C
out
= 100 F electrolytic capacitor.
7.785
12
8.220
53
μ
F