参数资料
型号: LT1375CS8-5#TR
厂商: Linear Technology
文件页数: 24/28页
文件大小: 0K
描述: IC REG BUCK 5V 1.5A 8SOIC
标准包装: 2,500
类型: 降压(降压)
输出类型: 固定
输出数: 1
输出电压: 5V
输入电压: 5 V ~ 25 V
PWM 型: 电流模式
频率 - 开关: 500kHz
电流 - 输出: 1.5A
同步整流器:
工作温度: 0°C ~ 125°C
安装类型: 表面贴装
封装/外壳: 8-SOIC(0.154",3.90mm 宽)
包装: 带卷 (TR)
供应商设备封装: 8-SOIC
LT1375/LT1376
APPLICATIO N S I N FOR M ATIO N
POSITIVE-TO-NEGATIVE CONVERTER
Maximum load current:
? I P ?
= ?
( )( ) ( ) (
)
?
? V OUT V IN ? 0 . 5
The circuit in Figure 18 is a classic positive-to-negative
topology using a grounded inductor. It differs from the
standard approach in the way the IC chip derives its
feedback signal, however, because the LT1376 accepts
only positive feedback signals, the ground pin must be tied
to the regulated negative output. A resistor divider to
I MAX
?
?
V IN V OUT
?
2 ( V OUT + V IN )( f )( L ) ?
( V OUT + V IN ? 0 . 5 ) ( V OUT + V F )
ground or, in this case, the sense pin, then provides the
proper feedback voltage for the chip.
D1
1N4148
I P = Maximum rated switch current
V IN = Minimum input voltage
V OUT = Output voltage
V F = Catch diode forward voltage
INPUT
4.5V TO
20V
C3
10 μ F TO
50 μ F
+
V IN
BOOST
V SW
LT1376-5
SENSE
GND V C
C C
R C
C2
0.1 μ F
D2
1N5818
L1*
5 μ H
+
C1
100 μ F
10V TANT
OUTPUT**
0.5 = Switch voltage drop at 1.5A
Example: with V IN(MIN) = 4.7V, V OUT = 5V, L = 10 μ H, V F =
0.5V, I P = 1.5A: I MAX = 0.52A. Note that this equation does
not take into account that maximum rated switch current
(I P ) on the LT1376 is reduced slightly for duty cycles
above 50%. If duty cycle is expected to exceed 50% (input
voltage less than output voltage), use the actual I P value
from the Electrical Characteristics table.
–5V, 0.5A
* INCREASE L1 TO 10 μ H OR 20 μ H FOR HIGHER CURRENT APPLICATIONS.
SEE APPLICATIONS INFORMATION
Operating duty cycle:
** MAXIMUM LOAD CURRENT DEPENDS ON MINIMUM INPUT VOLTAGE
AND INDUCTOR SIZE. SEE APPLICATIONS INFORMATION
Figure 18. Positive-to-Negative Converter
1375/76 F18
DC =
V OUT + V F
V IN ? 0 . 3 + V OUT + V F
(This formula uses an average value for switch loss, so it
Inverting regulators differ from buck regulators in the
basic switching network. Current is delivered to the output
may be several percent in error.)
With the conditions above:
as square waves with a peak-to-peak amplitude much
greater than load current. This means that maximum load
current will be significantly less than the LT1376’s 1.5A
DC =
5 + 0 . 5
4 . 7 ? 0 . 3 + 5 + 0 . 5
= 56 %
maximum switch current, even with large inductor values.
The buck converter in comparison, delivers current to the
output as a triangular wave superimposed on a DC level
equal to load current, and load current can approach 1.5A
with large inductors. Output ripple voltage for the positive-
to-negative converter will be much higher than a buck
converter. Ripple current in the output capacitor will also
be much higher. The following equations can be used to
This duty cycle is close enough to 50% that I P can be
assumed to be 1.5A.
OUTPUT DIVIDER
If the adjustable part is used, the resistor connected to
V OUT (R2) should be set to approximately 5k. R1 is
calculated from:
calculate operating conditions for the positive-to-negative
converter.
R 1 =
R 2 ( V OUT ? 2 . 42 )
2 . 42
13756fd
24
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