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| 型号： | LT1956IGN#PBF |
| 厂商： | Linear Technology |
| 文件页数： | 25/28页 |
| 文件大小： | 0K |
| 描述： | IC REG BUCK ADJ 1.5A 16SSOP |
| 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： | 1.2 V ~ 45 V |
| 输入电压： | 5.5 V ~ 60 V |
| PWM 型： | 电流模式 |
| 频率 - 开关： | 500kHz |
| 电流 - 输出： | 1.5A |
| 同步整流器： | 无 |
| 工作温度： | -40°C ~ 125°C |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 16-SSOP（0.154"，3.90mm 宽） |
| 包装： | 管件 |
| 供应商设备封装： | 16-SSOP |
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LT1956/LT1956-5
APPLICATIO S I FOR ATIO
? I P – 2 ( V
OUT + V IN )( f )( L ) ?
? ( V OUT )( V IN – 0 . 3 )
currentinL2andC4flowsviathecatchdiodeD3,charging
the negative output capacitor C6. If the negative output is
not loaded enough, it can go severely unregulated (be-
come more negative). Figure 14b shows the maximum
I MAX
=
? ( V IN )( V OUT ) ?
?
( V OUT + V IN – 0 . 3 )( V OUT + V F )
allowable –5V output load current (vs load current on the
5V output) that will maintain the –5V output within 3%
tolerance. Figure 14c shows the –5V output voltage regu-
lation vs its own load current when plotted for three
separate load currents on the 5V output. The efficiency of
the dual output converter circuit shown in Figure 14a is
given in Figure 14d.
POSITIVE-TO-NEGATIVE CONVERTER
The circuit in Figure 15 is a positive-to-negative topology
using a grounded inductor. It differs from the standard
approach in the way the IC chip derives its feedback signal
because the LT1956 accepts only positive feedback sig-
nals. The ground pin must be tied to the regulated negative
output. A resistor divider to the FB pin, then provides the
proper feedback voltage for the chip.
The following equation can be used to calculate maximum
load current for the positive-to-negative converter:
D2
MMSD914TI
I P = maximum rated switch current
V IN = minimum input voltage
V OUT = output voltage
V F = catch diode forward voltage
0.3 = switch voltage drop at 1.5A
Example: with V IN(MIN) = 5.5V, V OUT = 12V, L = 15 μ H,
V F = 0.63V, I P = 1.5A: I MAX = 0.36A.
INDUCTOR VALUE
The criteria for choosing the inductor is typically based on
ensuring that peak 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.
The difficulty in calculating the minimum inductor size
needed is that you must first decide whether the switcher
will be in continuous or discontinuous mode at the critical
point where switch current reaches 1.5A. The first step is
to use the following formula to calculate the load current
V IN
0.1 μ F
+
( V IN ) ( I P ) 2
I CONT >
V IN
12V
C3
2.2 μ F
25V
BOOST
SW
LT1956
FB
GND V C
C C
C F
R C
C2
L1*
7 μ H
R1
36.5k
D1
10MQO60N
R2
4.12k
C1
100 μ F
20V TANT
OUTPUT**
–12V, 0.25A
above which 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:
2
4 ( V IN + V OUT )( V IN + V OUT + V F )
* INCREASE L1 TO 10 μ H OR 18 μ H FOR HIGHER CURRENT APPLICATIONS.
SEE APPLICATIONS INFORMATION
** MAXIMUM LOAD CURRENT DEPENDS ON MINIMUM INPUT VOLTAGE
1956 F15
Minimum inductor discontinuous mode:
AND INDUCTOR SIZE. SEE APPLICATIONS INFORMATION
Figure 15. Positive-to-Negative Converter
L MIN =
2( V OUT )(I OUT )
( f )( I P ) 2
1956f
25
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