LM2575D2T-12 ON Semiconductor, LM2575D2T-12 Datasheet - Page 18

• Manufacturer Part Number: LM2575D2T-12
• Description: EASY SWITCHERE 1.0 A STEP-DOWN VOLTAGE REGULATOR
• Manufacturer: ON Semiconductor

Packages on a Heatsink
the selected safe operating junction temperature determined
in step 3, than a heatsink is required. The junction
temperature will be calculated as follows:
where
selected safe operating junction temperature, then a larger
heatsink is required.
Some Aspects That can Influence Thermal Design
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.
area, copper thickness, single– or double–sided, multilayer
board, the amount of solder on the board or even colour of
the traces.
the board can also influence its effectiveness to dissipate
the heat.
ADDITIONAL APPLICATIONS
Inverting Regulator
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.
–12 V output. The maximum input voltage in this case
Unregulated
DC Input
12 V to 25 V
Figure 25. Inverting Buck–Boost Regulator Using the
18
If the actual operating junction temperature is greater than
If the actual operating temperature is greater than the
It should be noted that the package thermal resistance and
Other factors are trace width, total printed circuit copper
The size, quantity and spacing of other components on
An inverting buck–boost regulator using the LM2575–12 is
In this example the LM2575–12 is used to generate a
100 F
/50 V
C in
R JC is the thermal resistance junction–case,
R CS is the thermal resistance case–heatsink,
R SA is the thermal resistance heatsink–ambient.
LM2575–12 Develops –12 V @ 0.35 A
T J = P D (R JA + R CS + R SA ) + T A
+V in
1
3
LM2575–12
Gnd
5
ON/OFF
Output
4
2
Feedback
D1
1N5819
100 H
L1
–12 V @ 0.35 A
Regulated
C out
1800 F
/16 V
Output
LM2575
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.
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.
are higher than in the standard buck converter topology, the
available output current is lower.
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.
least 2.0 ms or more. The actual time depends on the output
voltage and size of the output capacitor.
this inverting regulator topology, the use of a delayed startup
or an undervoltage lockout circuit is recommended.
can charge up to a higher voltage before the switch–mode
regulator begins to operate.
supplied by the input capacitor C in .
Design Recommendations:
the buck converter and so a different design procedure has to
be used to select the inductor L1 or the output capacitor C out .
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 F).
inverting converter design is between 68 H and 220 H. To
select an inductor with an appropriate current rating, the
inductor peak current has to be calculated.
current:
conditions, the worst case occurs when V in is minimal.
absolute sum of the input and output voltage, and must not
exceed 40 V.
where t on
This circuit configuration is able to deliver approximately
Since the switch currents in this buck–boost configuration
This type of buck–boost inverting regulator can also
Such an amount of input startup current is needed for at
Because of the relatively high startup currents required by
Using a delayed startup arrangement, the input capacitor
The high input current needed for startup is now partially
The inverting regulator operates in a different manner than
The output capacitor values must be larger than is
The recommended range of inductor values for the
The following formula is used to obtain the peak inductor
Under normal continuous inductor current operating
Note that the voltage appearing across the regulator is the
I
peak
+
[
V
in
MOTOROLA ANALOG IC DEVICE DATA
I
Load
|V
)
O
|
|V
(V
O
V
in
|
in
x
)
f osc
1
|V
O
, and f osc
|)
)
V
in
2L 1
x t on
+
52 kHz.