参数资料
型号: MIC2155YML TR
厂商: Micrel Inc
文件页数: 22/35页
文件大小: 0K
描述: IC REG CTRLR BUCK PWM VM 32MLF
特色产品: MIC2155/6 Buck Control IC
标准包装: 1
PWM 型: 电压模式
输出数: 1
频率 - 最大: 550kHz
占空比: 80%
电源电压: 4.5 V ~ 14.5 V
降压:
升压:
回扫:
反相:
倍增器:
除法器:
Cuk:
隔离:
工作温度: -40°C ~ 125°C
封装/外壳: 32-VFQFN 裸露焊盘,32-MLF?
包装: 标准包装
产品目录页面: 1091 (CN2011-ZH PDF)
其它名称: 576-3527-6
Micrel, Inc.
For this example, assume the output transient loading is
small and the filter design is based on output ripple
voltage requirement.
The inductance value is calculated by the equation
below:
MIC2155/2156
The peak inductor current in each channel is equal to the
average output current plus one half of the peak to peak
inductor ripple current:
I PK = I OUT + 0 . 5 × I PP = 15 A + 0 . 5 × 3 A = 16 . 5 A
L =
V OUT × ( η × V IN(MAX ) ? V OUT )
η × V IN ( MAX ) × f S × 0 . 2 × I OUT
The RMS inductor current is used to calculate the I 2 × R
losses in the inductor:
1 ? I PP
12 ? ? I OUT
? ?
Where:
f S is the switching frequency
0.2 is the ratio of AC ripple current to DC output
I INDUCTOR ( RMS ) = I OUT × 1 +
?
?
?
2
1 ? 3 A ?
current
V IN(MAX) is the maximum input voltage
I OUT is output current of the each channel or ? of the
total output current
I INDUCTOR ( RMS ) = 15 A × 1 +
2
? ? = 15 . 1 A
12 ? 15 ?
η is the converters efficiency
Maximizing efficiency requires the proper selection of
core material and minimizing the winding resistance. The
L =
For this example:
1 . 8 V × ( 0 . 88 × 12 V ? 1 . 8 V )
0 . 88 × 12 V × 500 kHz × 0 . 2 × 15 A
= 1 μ H
high frequency operation of the MIC2155 requires the
use of ferrite materials for all but the most cost sensitive
applications. Lower cost iron powder cores may be used
but the increase in core loss will reduce the efficiency of
the power supply. This is especially noticeable at low
If another inductor value is used, the ripple current for
each channel is calculated from the formula below:
output power. The inductor winding resistance
decreases efficiency at the higher output current levels.
The winding resistance must be minimized although this
I PP =
V OUT × ( η × V IN(MAX ) ? V OUT )
η × V IN ( MAX ) × f S × L
usually comes at the expense of a larger inductor.
The power dissipated in the inductor is equal to the sum
of the core and copper losses. At higher output loads,
I PP =
1 . 8 V × ( 0 . 88 × 12 V ? 1 .8 V )
0 . 88 × 12 V × 500 kHz × 1 μ H
= 3 A
the core losses are usually insignificant and can be
ignored. At lower output currents the core losses can be
a significant contributor. Core loss information is usually
The output capacitors see less ripple current than each
channel because they are out of phase.
The normalizing factor is:
available from the magnetics vendor.
For this example a Cooper HCF1305-1R0 inductor was
chosen. Core loss for this application was taken from
the data sheet and is 15mW. Winding resistance is
1.9m ? s
V OUT
f S × L OUT
=
1 . 8 V
500 kHz × 1 μ H
= 3 . 6
Copper loss in the inductor is calculated by the equation
below:
The output ripple current in the two-phase configuration
is approximately:
P INDUCTOR(COPPER) = (I INDUCTOR(RMS) ) 2 x R WINDING =
15.12 x 1.9m ? = 0.43W
0 . 65 ×
V OUT
f S × L OUT
= 0 . 65 ×
1 . 8 V
500 kHz × 1 μ H
= 2 . 3 A
For the input and output voltage in this application, going
to a two-phase design decreased the total output ripple
current from 3A PP to 2.3A PP .
November 2009
22
M9999-111209-B
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