参数资料
型号: LT1376HVCS8
厂商: Linear Technology
文件页数: 25/28页
文件大小: 0K
描述: IC REG BUCK ADJ 1.5A 8SOIC
标准包装: 100
类型: 降压(降压)
输出类型: 可调式
输出数: 1
输出电压: 2.42 V ~ 21.5 V
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: 管件
供应商设备封装: 8-SOIC
产品目录页面: 1327 (CN2011-ZH PDF)
LT1375/LT1376
APPLICATIO N S I N FOR M ATIO N
INDUCTOR VALUE
Minimum inductor discontinuous mode:
( f )( I )
Unlike buck converters, positive-to-negative converters
cannot use large inductor values to reduce output ripple
voltage. At 500kHz, values larger than 25 μ H make almost
no change in output ripple. The graph in Figure 19 shows
peak-to-peak output ripple voltage for a 5V to – 5V con-
L MIN =
2 ( V OUT )( I OUT )
2
P
verter versus inductor value. The criteria for choosing the
Minimum inductor continuous mode:
( )( )
V V
? I ? I ? 1 +
? P OUT ?
?
) ? ? ? ?
? ?
inductoristhereforetypicallybasedonensuringthatpeak
switch current rating is not exceeded. This gives the
lowest value of inductance that can be used, but in some
cases (lower output load currents) it may give a value that
creates unnecessarily high output ripple voltage. A com-
promise value is often chosen that reduces output ripple.
L MIN =
2 ( f ) ( V IN + V OUT
)
? ?
IN OUT
?
(
V OUT + V F
V IN
? ?
As you can see from the graph, large inductors will not
give arbitrarily low ripple, but small inductors can give
high ripple.
150
For the example above, with maximum load current of
0.25A:
( ) ( ) . 1 5
120
90
5V TO –5V CONVERTER
OUTPUT CAPACITOR
ESR = 0.1 ?
I CONT =
2 2
5
4 ( 5 + 5 ) ( 5 + 5 + 0 . 5 )
= 0 . 37 A
I LOAD = 0.25A
60
I LOAD = 0.1A
This says that discontinuous mode can be used and the
minimum inductor needed is found from:
( )( )
? ( )
? 500 ? 10 ? 1 . 5
3
30
0
0
5
10 15
INDUCTOR SIZE ( μ H)
20
25
L MIN =
?
2 5 0 . 25
2
= 2 . 2 μ H
1375/76 F19
Figure 19. Ripple Voltage on Positive-to-Negative Converter
The difficulty in calculating the minimum inductor size
needed is that you must first know whether the switcher
will be in continuous or discontinuous mode at the critical
point where switch current is 1.5A. The first step is to use
the following formula to calculate the load current where
the switcher must use continuous mode. If your load
current is less than this, use the discontinuous mode
formula to calculate minimum inductor needed. If load
current is higher, use the continuous mode formula.
Output current where continuous mode is needed:
In practice, the inductor should be increased by about
30% over the calculated minimum to handle losses and
variations in value. This suggests a minimum inductor of
3 μ H for this application, but looking at the ripple voltage
chart shows that output ripple voltage could be reduced by
a factor of two by using a 15 μ H inductor. There is no rule
of thumb here to make a final decision. If modest ripple is
needed and the larger inductor does the trick, go for it. If
ripple is noncritical use the smaller inductor. If ripple is
extremely critical, a second filter may have to be added in
any case, and the lower value of inductance can be used.
Keep in mind that the output capacitor is the other critical
( ) ( ) I
V
4 ( ) ( V + V
I CONT =
V IN + V OUT IN OUT + V F
IN P
2 2
)
factor in determining output ripple voltage. Ripple shown
on the graph (Figure 19) is with a capacitor ESR of 0.1 ? .
This is reasonable for an AVX type TPS “D” or “E” size
13756fd
25
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