参数资料
型号: LT1107CN8-5
厂商: Linear Technology
文件页数: 7/16页
文件大小: 0K
描述: IC REG BUCK BST INV 5V 0.4A 8DIP
标准包装: 50
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: 63kHz
电流 - 输出: 400mA
同步整流器:
工作温度: 0°C ~ 70°C
安装类型: 通孔
封装/外壳: 8-DIP(0.300",7.62mm)
包装: 管件
供应商设备封装: 8-PDIP
LT1107
A PPLICATI
S I FOR ATIO
Inductor Selection –– Step-Up Converter
As an example, suppose 12V at 60mA is to be generated
In a step-up, or boost converter (Figure 1), power gener-
ated by the inductor makes up the difference between
input and output. Power required from the inductor is
from a 3V to 6V input. Recalling equation (1),
P L = ( 12 V + 0 . 5 V ? 3 V )( 60 mA ) = 570 mW
( 6 )
P L = ? V OUT + V D ? V IN ( ) ? ( I OUT )
?
MIN ?
=
= 9 . 05 μ J
determined by:
( 1 )
where V D is the diode drop (0.5V for a 1N5818 Schottky).
Energy required by the inductor per cycle must be equal or
greater than:
Energy required from the inductor is:
P L 570 mW
( 7 )
f OSC 63 kHz
Picking an inductor value of 33 μ H with 0.2 ? DCR results
in a peak switch current of:
1 ? e 33 μ H ? = 850 mA
P L / f OSC ( 2 )
in order for the converter to regulate the output.
When the switch is closed, current in the inductor builds
I PEAK
=
?
3V ?
1 ? ?
?
? 1 ? ? 11 μ s ?
?
?
( 8 )
according to:
Substituting I PEAK into Equation 4 results in:
? 1 ? e L ?
R ′ ? ?
( )( ) 2 = 11 . 91 μ J
E L =
33 μ H 0 . 85 A
I L ( t ) =
V IN
? ? R t ?
? ?
( 3 )
1
( 9 )
2
Since 11.9 μ J > 9.05 μ J, the 33 μ H inductor will work. This
()
I L t =
V IN
LI 2 PEAK
E L =
DC ? ? V IN SW D ? ?
where R′ is the sum of the switch equivalent resistance
(0.8 ? typical at 25 ° C) and the inductor DC resistance.
When the drop across the switch is small compared to V IN ,
the simple lossless equation:
t ( 4 )
L
can be used. These equations assume that at t = 0,
inductor current is zero. This situation is called “discon-
tinuous mode operation” in switching regulator parlance.
Setting “t” to the switch ON time from the LT1107 speci-
fication table (typically 11 μ s) will yield I PEAK for a specific
“L” and V IN . Once I PEAK is known, energy in the inductor
at the end of the switch ON time can be calculated as:
1
( 5 )
2
E L must be greater than P L /f OSC for the converter to deliver
the required power. For best efficiency I PEAK should be
kept to 1A or less. Higher switch currents will cause
excessive drop across the switch resulting in reduced
efficiency. In general, switch current should be held to as
low a value as possible in order to keep switch, diode and
inductor losses at a minimum.
trial-and-error approach can be used to select the opti-
mum inductor.
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
Switch Current (from Typical Performance Characteristic
curves). Then, as V IN increases, peak switch current is
held constant, resulting in increasing efficiency.
Inductor Selection –– Step-Down Converter
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 through the switch should be limited to ~650mA
in this mode. Higher current can be obtained by using an
external switch (see LT1111 and LT1110 data sheets). 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:
2 I OUT ? V OUT + V D ?
I PEAK = ? ? ( 10 )
? V + V
1107fa
7
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