参数资料
型号: LT3437EDD#TRPBF
厂商: Linear Technology
文件页数: 14/28页
文件大小: 0K
描述: IC REG BUCK ADJ 0.5A 10DFN
标准包装: 2,500
类型: 降压(降压)
输出类型: 可调式
输出数: 1
输出电压: 1.25 V ~ 54 V
输入电压: 3.3 V ~ 60 V
PWM 型: 电流模式,混合
频率 - 开关: 200kHz
电流 - 输出: 500mA
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: 10-WFDFN 裸露焊盘
包装: 带卷 (TR)
供应商设备封装: 10-DFN(3x3)
LT3437
APPLICATIO S I FOR ATIO
V OUT ( IN OUT )
( )( ) 5
2 ( 68 e – 6 )( 200 e 3 )( 8 )
( )( ) 5
2 ( 68 e – 6 )( 200 e 3 )( 15 )
V OUT ( IN OUT )
( )( )( )
V OUT
10mV/DIV
100 μ F TANTALUM
ESR 75m ?
V OUT
10mV/DIV
100 μ F CERAMIC
V SW
10V/DIV
V IN = 12V 1 μ s/DIV 3437 F03
V OUT = 3.3V
I LOAD = 500mA
L = 100 μ H
Figure 3. LT3437 Ripple Voltage Waveform
MAXIMUM OUTPUT LOAD CURRENT
Maximum load current for a buck converter is limited by
the maximum switch current rating (I PK ). The current
rating for the LT3437 is 500mA. Unlike most current mode
converters, the LT3437 maximum switch current limit
does not fall off at high duty cycles. Most current mode
converters suffer a drop off of peak switch current for duty
cycles above 50%. This is due to the effects of slope
compensation required to prevent subharmonic oscilla-
tions in current mode converters. (For detailed analysis,
see Application Note 19.)
The LT3437 is able to maintain peak switch current limit
over the full duty cycle range by using patented circuitry to
cancel the effects of slope compensation on peak switch
current without affecting the frequency compensation it
provides.
Maximum load current would be equal to maximum
switch current for an infinitely large inductor, but with
Discontinuous operation occurs when:
V – V
I OUT ( DIS ) ≤
2 ( L )( f )( V IN )
For V OUT = 5V, V IN = 8V and L = 68 μ H:
5 8 –
I OUT ( MAX ) = 0 . 5 –
= 0 . 5 – 0 . 069 = 0 . 431 A
Note that there is less load current available at the higher
input voltage because inductor ripple current increases. At
V IN = 15V, duty cycle is 33% and for the same set of
conditions:
5 15 –
I OUT ( MAX ) = 0 . 5 –
= 0 . 5 – 0 . 121 = 0 . 379 A
To calculate actual peak switch current in continuous
mode with a given set of conditions, use:
V – V
I SW ( PK ) = I OUT +
2 L f V IN
If a small inductor is chosen which results in discontinuous
mode operation over the entire load range, the maximum
load current is equal to:
( )( )( )
( V OUT )( V IN OUT ) = I PK – I P -P
2 ( )( )( V IN )
finite inductor size, maximum load current is reduced by
one-half peak-to-peak inductor current. The following
formula assumes continuous mode operation, implying
that the term on the right (I P-P /2) is less than I OUT .
– V
I OUT ( MAX ) = I PK –
L f 2
I OUT ( MAX ) =
2
I PK 2 f L V IN
2 ( V OUT )( V IN – V OUT )
3437fc
14
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