参数资料
型号: LM2576TV-ADJ
厂商: MOTOROLA INC
元件分类: 稳压器
英文描述: 3 A SWITCHING REGULATOR, 63 kHz SWITCHING FREQ-MAX, PSFM5
封装: TO-220, 5 PIN
文件页数: 11/28页
文件大小: 415K
代理商: LM2576TV-ADJ
LM2576
19
MOTOROLA ANALOG IC DEVICE DATA
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 start–up input current, even for
light loads. This may overload an input power source with a
current limit less than 5.0 A.
Such an amount of input start–up 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 start–up currents required
by this inverting regulator topology, the use of a delayed
start–up or an undervoltage lockout circuit is recommended.
Using a delayed start–up 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 start–up is now partially
supplied by the input capacitor Cin.
It has been already mentioned above, that in some
situations, the delayed start–up or the undervoltage lockout
features could be very useful. A delayed start–up circuit
applied to a buck–boost converter is shown in Figure 27,
Figure 33 in the “Undervoltage Lockout” section describes an
undervoltage lockout feature for the same converter
topology.
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 what 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
1.0
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.
Figure 27. Inverting Buck–Boost Regulator
with Delayed start–up
D1
1N5822
L1
68
H
Output
2
4
Feedback
12 V to 25 V
Unregulated
DC Input
Cin
100
F
/50 V
1
3
5ON/OFF
Gnd
+Vin
–12 V @ 700 m A
Regulated
Output
Cout
2200
F
/16 V
LM2576–12
C1
0.1
F
R1
47 k
R2
47 k
Figure 28. Inverting Buck–Boost Regulator Shutdown
Circuit Using an Optocoupler
LM2576–XX
1
3
5
Gnd
ON/OFF
+Vin
R2
47 k
Cin
100
F
NOTE: This picture does not show the complete circuit.
R1
47 k
R3
470
Shutdown
Input
MOC8101
–Vout
Off
On
5.0 V
0
+Vin
With the inverting configuration, the use of the ON/OFF pin
requires some level shifting techniques. This is caused by the
fact, that the ground pin of the converter IC is no longer at
ground. Now, the ON/OFF pin threshold voltage (1.3 V
approximately) has to be related to the negative output
voltage level. There are many different possible shut down
methods, two of them are shown in Figures 28 and 29.
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