参数资料
型号: LT3512EMS#PBF
厂商: Linear Technology
文件页数: 16/26页
文件大小: 0K
描述: IC REG FLYBK ISO ADJ .42A 16MSOP
标准包装: 37
类型: 回扫,隔离
输出类型: 可调式
输出数: 1
输出电压: 可调
输入电压: 4.5 V ~ 100 V
PWM 型: 电流模式,混合
频率 - 开关: 40kHz ~ 650kHz
电流 - 输出: 420mA
同步整流器:
工作温度: -40°C ~ 125°C
安装类型: 表面贴装
封装/外壳: 16-TFSOP(0.118",3.00mm),12 引线
包装: 管件
供应商设备封装: 16-MSOP
LT3512
APPLICATIONS INFORMATION
N PS <
N PS <
t OFF(MIN) ? N PS ? ( V OUT + V F )
L PRI ≥
Step1:Selectthetransformerturnsratio.
V SW(MAX) – V IN(MAX) – V LEAKAGE
V OUT + V F
V SW(MAX) = Max rating of internal switch = 150V
V LEAKAGE = Margin for transformer leakage spike = 40V
V F = Forward voltage of output diode = assume approxi-
mately ~ 0.5V
Example:
150V – 72V – 40V
15V + 0.5V
N PS < 2.45
N PS = 2
The choice of turns ratio is critical in determining output
power as shown earlier in the Output Power section. At
this point, a third winding can be added to the transformer
to drive the BIAS pin of the LT3512 for higher efficiencies.
Choose a turns ratio that sets the third winding voltage
to regulate between 3.3V and 6V for maximum efficiency.
Choose a third winding ratio to drive BIAS winding with
5V. (Optional)
Example:
D = 0.46
P OUT(VIN(MIN)) = 3W
I OUT(VIN(MIN)) = P OUT(VIN(MIN)) /V OUT = 0.2A
The chosen turns ratio satisfies the output current re-
quirement of 200mA. If the output current was too low,
the minimum input voltage could be adjusted higher. The
turns ratio in this example is set to its highest ratio given
switch voltage requirements and margin for leakage in-
ductance voltage spike.
Step 3: Determine primary inductance, switching
frequency and saturation current.
Primary inductance for the transformer must be set above
a minimum value to satisfy the minimum off and on time
requirements.
I PEAK(MIN)
t OFF(MIN) = 400ns
I PEAK(MIN) = 100mA
= THIRD =
L PRI ≥
Example:
N THIRD
N S
V
V OUT
5V
15V
= 0.33
t ON(MIN) ? V IN(MAX)
I PEAK(MIN)
t ON(MIN) = 100ns
( V OUT + V F ) ? N PS
( V OUT + V F ) ? N PS + V IN(MIN)
400ns ? 2 ? ( 15 + 0.5 )
L PRI ≥
L PRI ≥
The turns ratio of the transformer chosen is as follows
N PRIMARY : N SECONDARY : N THIRD = 2:1:0.33.
Step 2: Calculate maximum power output at minimum
V IN .
P OUT(VIN(MIN)) = η ? V IN(MIN) ? I IN = η ? V IN(MIN) ? D ?
I PEAK ? 0.5
D =
I PEAK(MIN) = 100mA
Example:
0.1
L PRI ≥ 124μH
100ns ? 72
0.1
L PRI ≥ 72μH
η = Efficiency = ~83%
I PEAK = Peak switch current = 0.44A
3512fb
16
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