LTM4601HVIV#PBF Linear Technology, LTM4601HVIV#PBF Datasheet - Page 12

• Manufacturer Part Number: LTM4601HVIV#PBF
• Description: IC DC/DC UMODULE 12A 118-LGA
• Manufacturer: Linear Technology
• Series: µModuler
• Type: Point of Load (POL) Non-Isolatedr

Specifications of LTM4601HVIV#PBF

• Design Resources: LTM4601HV Spice Model
• Output: 0.6 ~ 5 V
• Number Of Outputs: 1
• Power (watts): 60W
• Mounting Type: Surface Mount
• Voltage - Input: 4.5 ~ 28 V
• Package / Case: 118-LGA
• 1st Output: 0.6 ~ 5 VDC @ 12A
• Size / Dimension: 0.59" L x 0.59" W x 0.11" H (15mm x 15mm x 2.8mm)
• Power (watts) - Rated: 60W
• Operating Temperature: -40°C ~ 85°C
• Efficiency: 95%
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• 3rd Output: -
• 2nd Output: -
• Other names: LTM4601HEIV#PBF; LTM4601HEIV#PBF

LTM4601HV
APPLICATIONS INFORMATION
Multiphase operation with multiple LTM4601HV devices in
parallel will lower the effective output ripple current due to
the interleaving operation of the regulators. For example,
each LTM4601HV’s inductor current of a 12V to 2.5V
multiphase design can be read from the Inductor Ripple
Current versus Duty Cycle graph (Figure 3). The large
ripple current at low duty cycle and high output voltage
12
12
10
8
6
4
2
0
0
Figure 3. Inductor Ripple Current vs Duty Cycle
Figure 4. Normalized Output Ripple Current vs Duty Cycle, Dlr = V
20
DUTY CYCLE (V
40
OUT
/V
1.00
0.95
0.90
0.85
0.80
0.75
0.70
0.65
0.60
0.55
0.50
0.45
0.40
0.35
0.30
0.25
0.20
0.15
0.10
0.05
IN
60
)
0
0.1 0.15 0.2 0.25
4601HV F03
80
2.5V OUTPUT
5V OUTPUT
1.8V OUTPUT
1.5V OUTPUT
1.2V OUTPUT
3.3V OUTPUT WITH
130k ADDED FROM
V
5V OUTPUT WITH
100k ADDED FROM
f
SET
OUT
0.3
TO GND
TO f
0.35
SET
0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9
DUTY CYCLE (V
can be reduced by adding an external resistor from f
ground which increases the frequency. If the duty cycle is
DC = 2.5V/12V = 0.21, the inductor ripple current for 2.5V
output at 21% duty cycle is ~6A in Figure 3.
Figure 4 provides a ratio of peak-to-peak output ripple cur-
rent to the inductor current as a function of duty cycle and
the number of paralleled phases. Pick the corresponding
duty cycle and the number of phases to arrive at the correct
output ripple current ratio value. If a 2-phase operation is
chosen at a duty cycle of 21%, then 0.6 is the ratio. This
0.6 ratio of output ripple current to inductor ripple of 6A
equals 3.6A of effective output ripple current. Refer to Ap-
plication Note 77 for a detailed explanation of output ripple
current reduction as a function of paralleled phases.
The output voltage ripple has two components that are
related to the amount of bulk capacitance and effective
series resistance (ESR) of the output bulk capacitance.
Therefore, the output voltage ripple can be calculated with
the known effective output ripple current. The equation:
ΔV
O
OUT(P-P)
/V
IN
)
O
T/L
≈ (ΔI
I
, Dlr = Each Phase’s Inductor Current
L
/(8 • f • m • C
1-PHASE
2-PHASE
3-PHASE
4-PHASE
6-PHASE
4601HV F04
OUT
) + ESR • ΔI
L
), where
SET
4601hvfa
to