LM2576T-ADJ datasheet(12/28 Pages) ONSEMI

Part #	LM2576T-ADJ

Description	3.0 A, 15 V, Step?묭own Switching Regulator

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LM2576

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12

Procedure (Adjustable Output Version: LM2576−ADJ) (continued)

Procedure

Example

4. Inductor Selection (L1)

A. Use the following formula to calculate the inductor Volt x

microsecond [V x

ms] 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 22. 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 25.

D. From the inductor code, identify the inductor value. Then

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 ILoad.

The inductor current rating can also be determined by

calculating the inductor peak current:

where ton 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.

ExT

+ V

in

–Vout

Vout

V

in

x

106

F[Hz]

[V x

ms]

I

p(max)

+ I

Load(max) )

V

in

–Vout ton

2L

ton +

Vout

V

in

x 1.0

fosc

4. Inductor Selection (L1)

A. Calculate E x T [V x

ms] constant:

B. E x T = 80 [V x

ms]

C. ILoad(max) = 2.5 A

Inductance Region = H150

D. Proper inductor value = 150

mH

Choose the inductor from Table 2.

ExT

+ (25–8.0) x 8.0

25

x

1000

52

+ 80 [V x ms]

5. Output Capacitor Selection (Cout)

A. Since the LM2576 is a forward−mode switching regulator

with 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:

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 8.0 V is appropriate, and a 10 V

or 16 V rating is recommended.

C

out w

13, 300

V

in(max)

Vout xL [μH]

[

μF]

5. Output Capacitor Selection (Cout)

A.

To achieve an acceptable ripple voltage, select

Cout = 680 mF electrolytic capacitor.

C

out w

13, 300 x

25

8 x 150

+ 332.5 μF

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