LM2575 datasheet(18/28 Pages) MOTOROLA

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LM2575

MOTOROLA ANALOG IC DEVICE DATA

Packages on a Heatsink

If the actual operating junction temperature is greater than

the selected safe operating junction temperature determined

in step 3, than a heatsink is required. The junction

temperature will be calculated as follows:

TJ = PD (R

θJA + RθCS + RθSA) + TA

where

θJC is the thermal resistance junction–case,

θCS is the thermal resistance case–heatsink,

θSA is the thermal resistance heatsink–ambient.

If the actual operating temperature is greater than the

selected safe operating junction temperature, then a larger

heatsink is required.

Some Aspects That can Influence Thermal Design

It should be noted that the package thermal resistance and

the junction temperature rise numbers are all approximate,

and there are many factors that will affect these numbers,

such as PC board size, shape, thickness, physical position,

location, board temperature, as well as whether the

surrounding air is moving or still.

Other factors are trace width, total printed circuit copper

area, copper thickness, single– or double–sided, multilayer

board, the amount of solder on the board or even colour of

the traces.

The size, quantity and spacing of other components on

the board can also influence its effectiveness to dissipate

the heat.

Figure 25. Inverting Buck–Boost Regulator Using the

LM2575–12 Develops –12 V @ 0.35 A

1N5819

100

µH

Output

Feedback

Unregulated

DC Input

12 V to 25 V

Cin

100

µF

/50 V

3ON/OFF

Gnd

+Vin

Regulated

Output

–12 V @ 0.35 A

Cout

1800

µF

/16 V

LM2575–12

ADDITIONAL APPLICATIONS

Inverting Regulator

An inverting buck–boost regulator using the LM2575–12 is

shown in Figure 25. This circuit converts a positive input

voltage to a negative output voltage with a common ground

by bootstrapping the regulators ground to the negative output

voltage. By grounding the feedback pin, the regulator senses

the inverted output voltage and regulates it.

In this example the LM2575–12 is used to generate a

–12 V output. The maximum input voltage in this case

cannot exceed +28 V because the maximum voltage

appearing across the regulator is the absolute sum of the

input and output voltages and this must be limited to a

maximum of 40 V.

This circuit configuration is able to deliver approximately

0.35 A to the output when the input voltage is 12 V or higher.

At lighter loads the minimum input voltage required drops to

approximately 4.7 V, because the buck–boost regulator

topology can produce an output voltage that, in its absolute

value, is either greater or less than the input voltage.

Since the switch currents in this buck–boost configuration

are higher than in the standard buck converter topology, the

available output current is lower.

This type of buck–boost inverting regulator can also

require a larger amount of startup input current, even for light

loads. This may overload an input power source with a

current limit less than 1.5 A.

Such an amount of input startup current is needed for at

least 2.0 ms or more. The actual time depends on the output

voltage and size of the output capacitor.

Because of the relatively high startup currents required by

this inverting regulator topology, the use of a delayed startup

or an undervoltage lockout circuit is recommended.

Using a delayed startup arrangement, the input capacitor

can charge up to a higher voltage before the switch–mode

regulator begins to operate.

The high input current needed for startup is now partially

supplied by the input capacitor Cin.

Design Recommendations:

The inverting regulator operates in a different manner than

the buck converter and so a different design procedure has to

be used to select the inductor L1 or the output capacitor Cout.

The output capacitor values must be larger than is

normally required for buck converter designs. Low input

voltages or high output currents require a large value output

capacitor (in the range of thousands of

µF).

The recommended range of inductor values for the

inverting converter design is between 68

µH and 220 µH. To

select an inductor with an appropriate current rating, the

inductor peak current has to be calculated.

The following formula is used to obtain the peak inductor

current:

where ton +

in )

fosc

, and fosc + 52 kHz.

peak [

Load

in )

)

xton

Under normal continuous inductor current operating

conditions, the worst case occurs when Vin is minimal.

Note that the voltage appearing across the regulator is the

absolute sum of the input and output voltage, and must not

exceed 40 V.

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