参数资料
型号: LTM4603IV-1#PBF
厂商: Linear Technology
文件页数: 20/26页
文件大小: 0K
描述: IC DC/DC UMODULE 6A 118-LGA
产品培训模块: Power Module vs. Discrete DC to DC
软件下载: LTM4603-1 Spice Model
产品目录绘图: LTM Series_15x15x2.8
设计资源: LTM4603IV-1#PBF Footprint.bxl
特色产品: μModule Products
标准包装: 119
系列: µModule®
类型: 非隔离(POL)
输出数: 1
电压 - 输入(最小): 4.5V
电压 - 输入(最大): 20V
Voltage - Output 1: 0.6 ~ 5 V
电流 - 输出(最大): 6A
安装类型: 表面贴装
封装/外壳: 118-LGA
尺寸/尺寸: 0.59" L x 0.59" W x 0.11" H(15.0mm x 15.0mm x 2.8mm)
包装: 托盘
工作温度: -40°C ~ 85°C
效率: 93%
产品目录页面: 2711 (CN2011-ZH PDF)
LTM4603/LTM4603-1
APPLICATIONS INFORMATION
Example for 5V Output
LTM4603 minimum on-time = 100ns
t ON = [(V OUT ? 10pF)/I fSET ], for V OUT > 4.8V use 4.8V
LTM4603 minimum off-time = 400ns
t OFF = t – t ON , where t = 1/Frequency
Duty Cycle = t ON /t or V OUT /V IN
Equations for setting frequency:
I fSET = (V IN /(3 ? R fSET )), for 20V operation, I fSET = 201μA, t ON
= [(4.8 ? 10pF)/I fSET ], t ON = 239ns, where the internal R fSET
is 33.2k. Frequency = (V OUT /(V IN ? t ON )) = (5V/(20 ? 239ns))
~ 1MHz. The inductor ripple current begins to get high at
the higher input voltages due to a larger voltage across the
inductor. This is noted in the Inductor Ripple Current vs
Duty Cycle graph at ~5A at 25% duty cycle. The inductor
ripple current can be lowered at the higher input voltages by
adding an external resistor from f SET to ground to increase
the switching frequency. A 3A ripple current is chosen, and
the total peak current is equal to 1/2 of the 3A ripple current
plus the output current. The 5V output current is limited
to 5A, so total peak current is less than 6.5A. This is below
the 7A peak specified value. A 150k resistor is placed from
f SET to ground, and the parallel combination of 150k and
33.2k equates to 27.2k. The I fSET calculation with 27.2k
and 20V input voltage equals 245μA. This equates to a t ON
of 196ns. This will increase the switching frequency from
1MHz to ~1.28MHz for the 20V to 5V conversion. The
minimum on time is above 100ns at 20V input. Since
the switching frequency is approximately constant over
input and output conditions, then the lower input voltage
range is limited to 10V for the 1.28MHz operation due to
the 400ns minimum off time. Equation: t ON = (V OUT /V IN )
? (1/Frequency) equates to a 382ns on time, and a 400ns
off time. The V IN to V OUT Step-Down Ratio curve reflects
an operating range of 10V to 20V for 1.28MHz operation
with a 150k resistor to ground, and an 8V to 16V operation
for f SET floating. These modifications are made to provide
wider input voltage ranges for the 5V output designs while
limiting the inductor ripple current, and maintaining the
400ns minimum off time.
20
Example for 3.3V Output
LTM4603 minimum on-time = 100ns
t ON = [(V OUT ? 10pF)/I fSET ]
LTM4603 minimum off-time = 400ns
t OFF = t – t ON , where t = 1/Frequency
Duty Cycle (DC) = t ON /t or V OUT /V IN
Equations for setting frequency:
I fSET = [V IN /(3 ? R fSET )], for 20V operation, I fSET = 201μA,
t ON = [(3.3 ? 10pF)/I fSET ], t ON = 164ns, where the internal
R fSET is 33.2k. Frequency = [V OUT /(V IN ? t ON )] = [3.3V/
(20 ?164ns)] ~ 1MHz. The minimum on-time and minimum
off-time are within specification at 164ns and 836ns.
However, the 4.5V input to 3.3V output circuit will not meet
the minimum off-time specification of 400ns (t ON = 733ns,
Frequency = 1MHz, t OFF = 267ns).
Solution
Lower the switching frequency at lower input voltages to
allow for higher duty cycles, and meet the 400ns minimum
off-time at 4.5V input voltage. The off-time should be about
500ns with 100ns guard band included. The duty cycle
for (3.3V/4.5V) = ~73%. Frequency = (1 – DC)/t OFF , or
(1 – 0.73)/500ns = 540kHz. The switching frequency needs
to be lowered to 540kHz at 4.5V input. t ON = DC/frequency,
or 1.35μs. The f SET pin voltage is 1/3 of V IN , and the I fSET
current equates to 45μA with the internal 33.2k. The I fSET
current needs to be 24μA for 540kHz operation. A resistor
can be placed from V OUT to f SET to lower the effective I fSET
current out of the f SET pin to 24μA. The f SET pin is 4.5V/3
=1.5V and V OUT = 3.3V, therefore 82.5k will source 21μA
into the f SET node and lower the I fSET current to 24μA.
This enables the 540kHz operation and the 4.5V to 20V
input operation for down converting to 3.3V output. The
frequency will scale from 540kHz to 1.2MHz over this
input range. This provides for an effective output current
of 5A over the input range.
4603fb
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