IRU3055CQ IRF [International Rectifier], IRU3055CQ Datasheet - Page 11

• Manufacturer Part Number: IRU3055CQ
• Description: POWER MANAGEMENT CHIPSET FOR 3-PHASE VRM 9.0 CONVERTERS
• Manufacturer: IRF [International Rectifier]

Synchronous-Rectifier Driver
Synchronous rectification reduces conduction losses in
the rectifier by shunting the normal Schottky diode or
MOSFET body diode with a low on-resistance MOSFET
switch. The synchronous rectification also ensures good
transient dynamic. For IRU3055, the 3-phase synchro-
nous rectifier MOSFET drivers are built inside. To drive
the high-side MOSFET, it is necessary to supply a gate
voltage at least 4V greater than the bus voltage. In
IRU3055, the driver supply voltage for high side MOSFET
driver is supplied through the V
input voltage for DC-DC converter is 5V, the V
V
charge pump configuration as shown in Figure 11.
If the voltage Vc1 and V
input voltage 12V, the voltage at V
charged up to almost twice the input voltage. The high
side driver can be enabled. A capacitor in the range of
0.1mF to 1mF is generally adequate for capacitor C2.
For high current applications, a large ceramic capacitor
such as 2.2mF is recommended. The diode can be a
Schottky diode such as BAT54S.
With the charge bump configuration, shown in Figure
11, the voltage at pins V
up. When the low side MOSFET is on, the capacitor C2
is charged to voltage Vc1. When the high side MOSFET
is ON, the energy in the capacitor C2 is discharged to
the bypass capacitor C1 next to pins V
The voltage at V
sum of the voltage Vc1 and V
signal should be at least 4V higher than the input volt-
age (V
the demo-board, Vc1 is equal to input voltage (V
If the low power dissipation of IC is preferred, especially
at higher frequency, Vc1 can be connected to 5V in-
stead.
Rev. 1.4
08/13/02
CH3
pins can be connected to 12V or supplied by using
IN
). The voltage Vc1 has to be 5V or higher. For
V
CH12
Figure 11 - Supply V
IRU3055
charge bump configuration.
CH12
V
C H 3
and V
IN
CH12
CH3
in Figure 11 is connected to
C1
and V
pins is approximately the
CH12
CH12
IN
. The high side driver
C2
CH12
Phase 1
, V
and V
CH3
L1
CH3
and V
can be boosted
with
CH12
CH3
C3
CH3
V
V
pins. If the
C1
IN
and V
pins are
CH12
IN
=12V).
www.irf.com
and
CH3
.
Component Selection Guide
Output Inductor Selection
The inductor is selected based on the inductor current
ripple, operation frequency and efficiency consideration.
In general, a large inductor results in small output ripple
and higher efficiency but big size. A small value inductor
causes large current ripple and poor efficiency but small
size. Generally, the inductor is selected based on the
output current ripple. The optimum point is usually found
between 20% and 50% ripple of output inductor current.
For each phase synchronous buck converter, the output
peak-to-peak current ripple is given by:
Assuming the output current is evenly distributed in each
phase, we can define the ratio of the ripple current and
nominal output current as:
Where LIR is typically between 20% to 50% and m is
the phase number. In this case m=3. Then the inductor
can be selected by:
For example, in the application circuit, the ripple is se-
lected as LIR=40%, the inductor is selected as:
The RMS current of the inductor will be approximately
equal to average current:
The peak inductor current is about:
Output capacitor selection
The voltage rating of the output capacitor is the same as
output voltage. Typical available capacitors on the mar-
ket are electrolytic, tantalum and ceramic. If electrolytic
or tantalum capacitors are employed, the criteria is nor-
mally based on the value of Effective Series Resistance
(ESR) of total output capacitor. In most cases, the ESR
of the output capacitor is calculated based on the follow-
ing relationship:
Where DV is the maximum allowed output voltage drop
during the transient and Di is the maximum output cur-
rent variation. In the worst case, Di is the maximum out-
put current minus zero.
Di
LIR = Di
L>V
L>1.53(12-1.5)/(150K312340%360A/3)=1.1mH
I
I
ESR < DV/Di
Select L=1mH
OUT
L(PEAK)
(PEAK - PEAK)
/m = 60/3 = 20A.
OUT
= (1+LIR/2)3I
3(V
(PEAK - PEAK)
IN
= (V
-V
OUT
IN
-V
/ I
)/(Fs3V
OUT
OUT
OUT
---(8)
)3V
/ m
/m = 1.2320 = 24A
IN
OUT
3LIR3I
/(L3Fs3V
IRU3055
OUT
/m)
IN
) ---(6)
---(7)
11