参数资料
型号: LT1976EFE
厂商: Linear Technology
文件页数: 15/28页
文件大小: 0K
描述: IC REG BUCK ADJ 1.5A 16TSSOP
标准包装: 95
类型: 降压(降压)
输出类型: 可调式
输出数: 1
输出电压: 1.2 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: 1.5A
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-TSSOP(0.173",4.40mm)裸露焊盘
包装: 管件
供应商设备封装: 16-TSSOP-EP
LT1976/LT1976B
APPLICATIO S I FOR ATIO
( 3 . 3 )( 12 – 3 . 3 )
( 12 ) ( 33 e ? 6 )( 200 e 3 )
= = 3 . 63 e 5
( 5 )( 8 – 5 )
2 ( 20 e – 6 )( 200 e 3 )( 8 )
( 5 )( 15 – 5 )
2 ( 20 e – 6 )( 200 e 3 )( 15 )
Example: with V IN = 12V, V OUT = 3.3V, L = 33μH, ESR =
0.08 Ω , ESL = 10nH:
I P-P = = 0 . 362 A
di 12
dt 3 . 3 e – 5
V RIPPLE = (0.362A)(0.08) + (10e – 9)(363e3)
= 0.0289 + 0.003 = 32mV P-P
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 LT1976 is 1.5A. Unlike most current mode
converters, the LT1976 maximum switch current limit
For V OUT = 5V, V IN = 8V and L = 20 μ H:
I OUT ( MAX ) = 1 . 5 –
= 1 . 5 – 0 . 24 = 1 . 26 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:
I OUT ( MAX ) = 1 . 5 –
= 1 . 5 – 0 . 42 = 1 . 08 A
To calculate actual peak switch current in continuous
mode with a given set of conditions, use:
(
( )( )( )
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-
I SW ( PK ) = I OUT +
V OUT V IN – V OUT
2 L f V IN
)
tions in current mode converters. (For detailed analysis,
see Application Note 19.)
The LT1976 is able to maintain peak switch current limit
If a small inductor is chosen which results in discontinous
mode operation over the entire load range, the maximum
load current is equal to:
I PK 2 2 ( )( )( V IN )
overthefulldutycyclerangebyusingpatentedcircuitryto
cancel the effects of slope compensation on peak switch
current without affecting the frequency compensation it
provides.
I OUT ( MAX ) =
f L
2 ( V OUT )( V IN – V OUT )
Maximum load current would be equal to maximum
switch current for an infinitely large inductor, but with
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 .
CHOOSING THE INDUCTOR
For most applications the output inductor will fall in the
range of 15 μ H to 100 μ H. Lower values are chosen to
reduce physical size of the inductor. Higher values allow
more output current because they reduce peak current
seen by the LT1976 switch, which has a 1.5A limit. Higher
( V OUT )( V IN – V OUT ) = I – I P -P
2 ( L )( f )( V IN )
2
I OUT ( MAX ) = I PK –
PK
values also reduce output ripple voltage and reduce core
loss.
When choosing an inductor you might have to consider
V OUT ( V IN – V OUT )
Discontinuous operation occurs when:
I OUT ( DIS ) ≤
2 ( L )( f )( V IN )
maximum load current, core and copper losses, allow-
able component height, output voltage ripple, EMI, fault
current in the inductor, saturation and of course cost.
The following procedure is suggested as a way of han-
dling these somewhat complicated and conflicting
requirements.
1976bfg
15
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