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| 型号： | LT1110CS8#PBF |
| 厂商： | Linear Technology |
| 文件页数： | 8/16页 |
| 文件大小： | 0K |
| 描述： | IC REG BUCK BOOST ADJ 0.4A 8SOIC |
| 标准包装： | 100 |
| 类型： | 降压（降压），升压（升压） |
| 输出类型： | 可调式 |
| 输出数： | 1 |
| 输出电压： | 可调 |
| 输入电压： | 1 V ~ 30 V |
| 频率 - 开关： | 70kHz |
| 电流 - 输出： | 400mA |
| 同步整流器： | 无 |
| 工作温度： | 0°C ~ 70°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 |
LT1110
A PPLICATI
S I FOR ATIO
Picking an inductor value of 47 μ H with 0.2 ? DCR results
in a peak switch current of
V OUT = output voltage
V IN = minimum input voltage
4 . 5 V ?
I PEAK =
? 1 ? e
1 . 0 W ? ?
? 1 .0 W ? 10ms ?
47 m H ? = 862 mA .
? ?
( 08 )
V SW is actually a function of switch current which is in turn
a function of V IN , L, time and V OUT . To simplify, 1.5V can
be used for V SW as a very conservative value.
Substituting I PEAK into Equation 05 results in
Once I PEAK is known, inductor value can be derived from
( )( )
47 μ H 0 . 862 A 2 = 17 . 5 μ J .
E L =
L =
? t ON
1
( 09 )
2
Since 17.5 μ J > 13.7 μ J, the 47 μ H inductor will work. This
trial-and-error approach can be used to select the opti-
V IN MIN – V SW – V OUT
I PEAK
where t ON = switch ON time (10 μ s).
( 11 )
mum inductor. Keep in mind the switch current maximum
rating of 1.5A. If the calculated peak current exceeds this,
an external power transistor can be used.
A resistor can be added in series with the I LIM pin to invoke
switch current limit. The resistor should be picked such
that the calculated I PEAK at minimum V IN is equal to the
Next, the current limit resistor R LIM is selected to give
I PEAK from the R LIM Step-Down Mode curve. The addition
of this resistor keeps maximum switch current constant as
the input voltage is increased.
As an example, suppose 5V at 250mA is to be generated
from a 9V to 18V input. Recalling Equation (10),
Maximum Switch Current (from Typical Performance
Characteristic curves). Then, as V IN increases, switch
current is held constant, resulting in increasing efficiency.
I PEAK =
2 ( 250 mA ) ? 5 + 0 . 5 ?
? ?
0 . 69 ? 9 – 1 . 5 + 0 . 5 ? = 498 mA . ( 12 )
Inductor Selection — Step-Down Converter
Next, inductor value is calculated using Equation (11)
L =
? 10 μ s = 50 μ H .
The step-down case (Figure 5) differs from the step-up in
that the inductor current flows through the load during
both the charge and discharge periods of the inductor.
Current through the switch should be limited to ~800mA
9 – 1.5 – 5
498 mA
Use the next lowest standard value (47 μ H).
( 13 )
in this mode. Higher current can be obtained by using an
external switch (see Figure 6). The I LIM pin is the key to
successful operation over varying inputs.
After establishing output voltage, output current and input
voltage range, peak switch current can be calculated by the
formula
Then pick R LIM from the curve. For I PEAK = 500mA,
R LIM = 82 ? .
Inductor Selection — Positive-to-Negative Converter
Figure 7 shows hookup for positive-to-negative conver-
sion. All of the output power must come from the inductor.
I PEAK =
2 I OUT ? V OUT + V D ?
? ?
DC ? V IN – V SW + V D ?
( 10 )
In this case,
P L = ( | V OUT | + V D )( I OUT ) .
( 14 )
where DC = duty cycle (0.69)
V SW = switch drop in step-down mode
V D = diode drop (0.5V for a 1N5818)
I OUT = output current
8
In this mode the switch is arranged in common collector
or step-down mode. The switch drop can be modeled as
a 0.75V source in series with a 0.65 ? resistor. When the
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