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参数资料
| 型号： | LT1108CS8-5#TR |
| 厂商： | Linear Technology |
| 文件页数： | 7/12页 |
| 文件大小： | 0K |
| 描述： | IC REG BUCK BST INV 5V .4A 8SOIC |
| 标准包装： | 2,500 |
| 类型： | 降压（降压），升压（升压），反相 |
| 输出类型： | 固定 |
| 输出数： | 1 |
| 输出电压： | 5V |
| 输入电压： | 2 V ~ 30 V |
| 频率 - 开关： | 19kHz |
| 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： | 带卷 (TR) |
| 供应商设备封装： | 8-SOIC |
LT1108
A PPLICATI
S I FOR ATIO
Energy required from the inductor is
where DC = duty cycle (0.60)
P L
f OSC
=
315mW
19 kHz
= 16 . 6 μ J
( 07 )
V SW = switch drop in step-down mode
V D = diode drop (0.5V for a 1N5818)
I OUT = output current
Picking an inductor value of 100 μ H with 0.2 ? DCR results
in a peak switch current of
V OUT = output voltage
V IN = minimum input voltage
?
100 μ H ? = 605 mA
I PEAK =
2 V
1 . 0 ?
?
?
1 – e
– 1. 0 ? × 36 μ s ?
?
( 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 04 results in
Once I PEAK is known, inductor value can be derived from
E L =
1
2
( 100 μ H )( 6 . 605 A ) 2 = 18 . 3 μ J
( 09 )
L =
V IN MIN ? V SW ? V OUT
I PEAK
× t ON
( 11 )
Since 18.3 μ J > 16.6 μ J, the 100 μ H inductor will work. This
trial-and-error approach can be used to select the optimum
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 so the
calculated I PEAK at minimum V IN is equal to the Maximum
where t ON = switch-ON time (36 μ s).
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 300mA is to be generated
from a 12V to 24V input. Recalling Equation (10),
Switch Current (from Typical Performance Characteristic
curves). Then, as V IN increases, switch current is held
constant, resulting in increasing efficiency.
I PEAK =
2 ( 300 mA ) ? 5 + 0 . 5 ?
? ?
0 . 60 ? 12 – 1 . 5 + 0 . 5 ? = 500 mA
( 12 )
Step-Down Converter
Next, inductor value is calculated using Equation (11)
The step-down case (Figure 2) 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
L =
12 – 1.5 – 5
500 mA
36 μ s = 396 μ H
( 13 )
through the switch should be limited to ~650mA in this
mode. Higher current can be obtained by using an external
switch (see Figure 3). 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:
Use the next lowest standard value (330 μ H).
Then pick R LIM from the curve. For I PEAK = 500mA,
R LIM = 220 ? .
Positive-to-Negative Converter
Figure 4 shows hookup for positive-to-negative conver-
sion. All of the output power must come from the inductor.
? V + V ?
? IN SW D ?
I PEAK =
2 I OUT
DC
? V OUT + V D ?
– V
( 10 )
In this case,
P L = ( ? V OUT ? + V D )( I OUT )
(14)
7
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