参数资料
型号: LT1375CS8-5#TRPBF
厂商: Linear Technology
文件页数: 10/28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
I OUT(MAX) = ( V OUT ) ( V IN ? V OUT )
2 ( L )( f )( V )
I P ?
( 5 ) ( 8 ? 5 )
()
2 ? ? 10 ? ? ?? 500 ? 10 ? ? 8
pinwhenoutputvoltageislow.Theequivalentcircuitryis
shown in Figure 2. Q1 is completely off during normal
operation. If the FB pin falls below 1V, Q1 begins to
conduct current and reduces frequency at the rate of
approximately 5kHz/ μ A. To ensure adequate frequency
foldback (under worst-case short-circuit conditions), the
external divider Thevinin resistance must be low enough
to pull 150 μ A out of the FB pin with 0.6V on the pin (R DIV
≤ 4k). The net result is that reductions in frequency and
current limit are affected by output voltage divider imped-
ance. Although divider impedance is not critical, caution
should be used if resistors are increased beyond the
suggested values and short-circuit conditions will occur
with high input voltage. High frequency pickup will in-
crease and the protection accorded by frequency and
current foldback will decrease.
formula assumes continuous mode operation, implying
that the term on the right is less than one-half of I P .
Continuous Mode
IN
For the conditions above and L = 10 μ H,
I OUT ( MAX ) = 1 . 44 ? ? 5 3
= 1 . 44 ? 0 . 19 = 1 . 25 A
At V IN = 15V, duty cycle is 33%, so I P is just equal to a fixed
2 ? 10 ? ?? 500 ? 10 ? ( 15 )
MAXIMUM OUTPUT LOAD CURRENT
Maximum load current for a buck converter is limited by
the maximum switch current rating (I P ) of the LT1376.
This current rating is 1.5A up to 50% duty cycle (DC),
decreasing to 1.35A at 80% duty cycle. This is shown
1.5A, and I OUT(MAX) is equal to:
1 . 5 ? ( 5 ) ( 15 ? 5 )
? 5 3
= 1 . 5 ? 0 . 33 = 1 . 17 A
graphically in Typical Performance Characteristics and as
shown in the formula below:
I P = 1.5A for DC ≤ 50%
I P = 1.65A – 0.15 (DC) – 0.26 (DC) 2 for 50% < DC < 90%
DC = Duty cycle = V OUT /V IN
Example: with V OUT = 5V, V IN = 8V; DC = 5/8 = 0.625, and;
I SW(MAX) = 1.64 – 0.15 (0.625) – 0.26 (0.625) 2 = 1.44A
Current rating decreases with duty cycle because the
Note that there is less load current available at the higher
input voltage because inductor ripple current increases.
This is not always the case. Certain combinations of
inductor value and input voltage range may yield lower
available load current at the lowest input voltage due to
reduced peak switch current at high duty cycles. If load
current is close to the maximum available, please check
maximum available current at both input voltage ex-
tremes. To calculate actual peak switch current with a
given set of conditions, use:
LT1376 has internal slope compensation to prevent cur-
rent mode subharmonic switching. For more details, read
Application Note 19. The LT1376 is a little unusual in this
regard because it has nonlinear slope compensation which
gives better compensation with less reduction in current
I SW ( PEAK ) = I OUT +
V OUT ( V IN ? V OUT )
2 ( L )( f )( V IN )
limit.
Maximum load current would be equal to maximum
For lighter loads where discontinuous operation can be
used, maximum load current is equal to:
( ) ( )( )( )
I f L V
) ( V ? V )
2 ( V
switch current for an infinitely large inductor, but with
finite inductor size, maximum load current is reduced by
one-half peak-to-peak inductor current. The following
I OUT(MAX) =
Discontinuous mode
2
P OUT
OUT IN OUT
13756fd
10
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