Datasheet

Iripple

Ipk+

Io Buck

Ipk-

Iripple

1/Fs

L1 Current

Io/(1-D) B-B

6.4 x 10

- 3.02 x 10

LM5118, LM5118-Q1

SNVS566H –APRIL 2008–REVISED JANUARY 2014

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APPLICATION INFORMATION

The procedure for calculating the external components is illustrated with the following design example. The

designations used in the design example correlate to Figure 28. The design specifications are:

• VOUT = 12 V

• VIN = 5 V to 75 V

• f = 300 kHz

• Minimum load current (CCM operation) = 600 mA

• Maximum load current = 3 A

R7 = R

RT sets the oscillator switching frequency. Generally speaking, higher operating frequency applications will use

smaller components, but have higher switching losses. An operating frequency of 300 kHz was selected for this

example as a reasonable compromise for both component size and efficiency. The value of R

can be calculated

as follows:

(9)

therefore, R7 = 18.3 kΩ

Figure 19. Inductor Current Waveform

Inductor Selection

The inductor value is determined based upon the operating frequency, load current, ripple current and the input

and output voltages. Refer to Figure 19 for details.

To keep the circuit in continuous conduction mode (CCM), the maximum ripple current I

RIPPLE

should be less

than twice the minimum load current. For the specified minimum load of 0.6 A, The maximum ripple current is 1.2

A p-p. Also, the minimum value of L must be calculated both for a buck and buck-boost configurations. The final

value of inductance will generally be a compromise between the two modes. It is desirable to have a larger value

inductor for buck mode, but the saturation current rating for the inductor must be large for buck-boost mode,

resulting in a physically large inductor. Additionally, large value inductors present buck-boost mode loop

compensation challenges which will be discussed in the Error Amplifier Configuration section. For the design

example, the inductor values in both modes are calculated as:

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