MIC2155_0911 Datasheet, PDF (22/35 Pages) Micrel Semiconductor – Two-Phase, Single-Output, PWM Synchronous Buck Control IC

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:
L = ^{VOUT × (η × VIN(MAX) − VOUT )}
η × V_IN(MAX) × f_S × 0.2 × I_OUT
Where:
f_S is the switching frequency
0.2 is the ratio of AC ripple current to DC output
current
V_IN(MAX) is the maximum input voltage
I_OUT is output current of the each channel or ½ of the
total output current
η is the converters efficiency
For this example:
L = ^{1.8V × (0.88 ×12V − 1.8V)} = 1μH
0.88 ×12V × 500kHz × 0.2 ×15A
If another inductor value is used, the ripple current for
each channel is calculated from the formula below:
I_PP
=
V_OUT × (η × V_IN(MAX) − V_OUT )
η × V_IN(MAX) × f_S × L
I_PP
=
1.8V × (0.88 ×12V − 1.8V)
0.88 ×12V × 500kHz ×1μH
=
3A
The output capacitors see less ripple current than each
channel because they are out of phase.
The normalizing factor is:
V_{OUT =}
1.8V
= 3.6
f_S × L_OUT 500kHz ×1μH
The output ripple current in the two-phase configuration
is approximately:
0.65 × ^VOUT = 0.65 × ^1.8V
= 2.3A
f_S × L_OUT
500kHz ×1μH
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.
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 = 15A + 0.5 × 3A = 16.5A
The RMS inductor current is used to calculate the I² × R
losses in the inductor:
I_{INDUCTOR(RMS)} = I_OUT ×
1+
1
12
⎜⎜⎝⎛
I_PP
I_OUT
⎟⎟⎠⎞²
I_{INDUCTOR(RMS)} = 15A ×
1+ ¹ ⎜⎛ ^3A ⎟⎞² = 15.1A
12 _⎝ 15 _⎠
Maximizing efficiency requires the proper selection of
core material and minimizing the winding resistance. The
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
output power. The inductor winding resistance
decreases efficiency at the higher output current levels.
The winding resistance must be minimized although this
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,
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
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
1.9mΩs
Copper loss in the inductor is calculated by the equation
below:
P_{INDUCTOR(COPPER)} = (I_{INDUCTOR(RMS)})² x R_WINDING =
15.12 x 1.9mΩ = 0.43W
November 2009
22
M9999-111209-B