参数资料
型号: LT3748HMS#PBF
厂商: Linear Technology
文件页数: 14/30页
文件大小: 0K
描述: IC REG CTRLR FLYBK ISO CM 16MSOP
产品培训模块: LT3748 100V Flyback Controller
标准包装: 37
PWM 型: 电流模式
输出数: 1
电源电压: 5 V ~ 100 V
降压:
升压:
回扫:
反相:
倍增器:
除法器:
Cuk:
隔离:
工作温度: -40°C ~ 150°C
封装/外壳: 16-TFSOP(0.118",3.00mm),12 引线
包装: 管件
配用: 732-3308-ND - BOARD EVAL FOR LT3748
732-3305-ND - BOARD EVAL FOR LT3748
相关产品: 732-2669-2-ND - TRANS FLYBACK LT3748 20UH SMD
732-2669-6-ND - TRANS FLYBACK LT3748 20UH SMD
732-2668-6-ND - TRANS FLYBACK LT3748 14UH SMD
732-2667-6-ND - TRANS FLYBACK LT3748 500UH SMD
732-2666-6-ND - TRANS FLYBACK LT3748 12UH SMD
732-2665-6-ND - TRANS FLYBACK LT3748 12UH SMD
732-2664-6-ND - TRANS FLYBACK LT3748 15UH SMD
732-2663-6-ND - TRANS FLYBACK LT3748 15UH SMD
732-2662-6-ND - TRANS FLYBACK LT3748 8UH SMD
732-2661-6-ND - TRANS FLYBACK LT3748 8UH SMD
更多...
LT3748
APPLICATIONS INFORMATION
the output voltage multiplied by the windings ratio plus
25
some amount of overshoot caused by leakage inductance.
Second, increasing the turns ratio will increase the peak
current seen on the output diode generally increasing the
20
N PS = 2:1
I LIM = 3A
RMS diode current thereby lowering the efficiency. This
efficiency limitation is worse at lower output voltages when
the diode forward voltage is significant compared to the
output voltage. In a typical application such as the 5V, 2A
output shown on the back page, the diode losses dominate
15
10
5
N PS = 6:1
I LIM = 1A
N PS = 3:1
I LIM = 2A
all the other losses, as shown in Figure 4. To calculate
RMS diode current, two equations are needed—the first
0
0
20
40
60
80
100
for calculating duty cycle, D, and the second to calculate
the RMS current of a triangle waveform:
INPUT VOLTAGE (V)
3748 F03
Figure 3. Maximum Output Power at 12V Out Using Three
D =
( V OUT + V F(DODE) ) ? N PS
V IN + ( V OUT + V F(DIODE) ) ? N PS
Transformers with Equal Peak Output Current and Secondary
Inductance
100
V IN = 12V
( I LIM ? N PS ) ? ( ) D
I DIODE(RMS)
=
3
2
1–
95
90
D OUT
For a more general analysis, Figure 5 illustrates a sweep
85
of windings ratio on the x-axis while comparing output
power and estimated efficiency for a 5V output using a
48V input. If the desired application required 20W, the
80
75
f SW ? Q G + I Q
FET R DS(ON)
maximum power curve indicates that a winding ratio of
12:1 would be sufficient at a current limit of 2A (R SENSE =
0.05Ω), while a winding ratio of 5:1 would deliver the same
power at 3A. However, when examining the corresponding
efficiency at max load for those two windings ratios and
TRANSFORMER I ? R + LEAKAGE
70
0.2A MIN 2A MAX
I OUT (A)
3748 F03
Figure 4. Sources of Loss In 5V, 2A Out Typical Application
current limits, the 5:1, 3A selection is clearly the superior
solution with an estimated efficiency of 85% compared to
78% for the 12:1, 2A application.
There are several caveats to this evaluation. First, as the
diode forward voltage becomes a smaller percentage of
total loss at higher output voltages (>12V) the RMS current
becomes less of a concern and minimizing it will have a
much smaller impact on efficiency. More significantly, if
a lower turns ratio forces the use of a diode with a larger
100
95
90
85
80
75
70
65
I LIM = 3A
I LIM = 2A
OUTPUT
POWER
EFFICIENCY
32
28
24
20
16
12
8
4
forward drop to obtain a higher reverse voltage rating,
any gains from minimizing current might be lost. For low
60
0
3
6
9
12
15
18
0
output voltages (3.3V or 5V) or high input voltages (>48V),
a turns ratio greater than one can be used with multiple
primary windings relative to the secondary to maximize
the transformer’s current gain.
N PS
3748 F05
Figure 5. Estimated Efficiency and Output Power at 5V OUT from
48V IN vs Windings Ratio, N PS , at 2A and 3A Current Limits
3748fa
14
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