参数资料
型号: LM2575TV-12
厂商: MOTOROLA INC
元件分类: 稳压器
英文描述: 3.2 A SWITCHING REGULATOR, 52 kHz SWITCHING FREQ-MAX, PZFM5
封装: 314B-05
文件页数: 10/28页
文件大小: 380K
代理商: LM2575TV-12
LM2575
18
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
R
θJC is the thermal resistance junction–case,
R
θCS is the thermal resistance case–heatsink,
R
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
D1
1N5819
L1
100
H
Output
2
4
Feedback
Unregulated
DC Input
12 V to 25 V
Cin
100
F
/50 V
1
5
3ON/OFF
Gnd
+Vin
Regulated
Output
–12 V @ 0.35 A
Cout
1800
F
/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
F).
The recommended range of inductor values for the
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.
The following formula is used to obtain the peak inductor
current:
where ton +
|V
O
|
V
in )
|V
O
|
x
1
fosc
, and fosc + 52 kHz.
I
peak [
I
Load
(V
in )
|V
O
|)
V
in
)
V
in
xton
2L
1
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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