LM2576 datasheet(23/46 Pages) TI

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1000

115 V

(10)

2. Use the E • T value from the previous formula and match it with the E • T number on the vertical axis of the

Inductor value selection guide shown in Figure 8-8.

E • T = 115 V • μs

3. On the horizontal axis, select the maximum load current.

ILOAD(Max) = 3 A

4. Identify the inductance region intersected by the E • T value and the maximum load current value, and note

the inductor code for that region.

Inductance Region = H150

5. Identify the inductor value from the inductor code, and select an appropriate inductor from the table shown in

Table 8-4. Part numbers are listed for three inductor manufacturers. The inductor chosen must be rated for

operation at the LM2576 switching frequency (52 kHz) and for a current rating of 1.15 × ILOAD. For additional

inductor information, see Section 8.1.2.

Inductor Value = 150 μH

Choose from AIE part #415-0936, Pulse Engineering part #PE-531115, or Renco part #RL2445.

8.2.2.2.3 Output Capacitor Selection (COUT)

1. The value of the output capacitor together with the inductor defines the dominate pole-pair of the switching

regulator loop. For stable operation, the capacitor must satisfy Equation 11:

IN Max

OUT

13,300

)

Equation 11 yields capacitor values between 10 μF and 2200 μF that satisfies the loop requirements for

stable operation. But to achieve an acceptable output ripple voltage, (approximately 1% of the output

voltage) and transient response, the output capacitor can need to be several times larger than Equation 11

yields.

COUT

13,300

22.2 )

150

However, for acceptable output ripple voltage select

COUT ≥ 680 μF

COUT = 680-μF electrolytic capacitor

2. The capacitor voltage rating must be at last 1.5 times greater than the output voltage. For a 10-V regulator,

TI recommends a rating of at least 15 V or more. Higher voltage electrolytic capacitors generally have lower

ESR numbers, and for this reason it can be necessary to select a capacitor rate for a higher voltage than can

normally be needed.

8.2.2.2.4 Catch Diode Selection (D1)

1. The catch-diode current rating must be at least 1.2 times greater than the maximum load current. Also, if the

power supply design must withstand a continuous output short, the diode must have a current rating equal

to the maximum current limit of the LM2576. The most stressful condition for this diode is an overload or

shorted output. See Table 8-3. For this example, a 3.3-A current rating is adequate.

2. The reverse voltage rating of the diode must be at least 1.25 times the maximum input voltage. Use a 30-V

31DQ03 Schottky diode, or any of the suggested fast-recovery diodes in Table 8-3.

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LM2576, LM2576HV

SNVS107G – JUNE 1999 – REVISED MARCH 2023

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