LTC1876 View Datasheet(PDF) - Linear Technology
Part Name
Description
MFG CO.
LTC1876
Linear Technology
'LTC1876' PDF : 36 Pages View PDF
LTC1876
APPLICATIO S I FOR ATIO
Low VIN Applications
In applications where the input supply is low (<5V), the
LTC1876 auxiliary regulator can be used to step-up the
input to provide the gate drive to the external MOSFETs as
shown in Figure 9.
Shown in the Typical Application section of the data sheet
is a circuit (3.3VIN Dual-Phase High Efficiency Power
Supply) with input supply of 3.3V. The boost section of the
LTC1876 is set up to generate 5V and is used to provide the
gate drive to the external MOSFETs. The circuit provides
dual outputs, a 2.5V/15A and 1.8V/15A. Both drawing
power directly from VIN.
INPUT
SUPPLY
AUXVIN
AUXSW
LTC1876
BOOST
SECTION
AUXVFB
SGND
L1
VIN
D1
LTC1876
R8
+
STEP-DOWN
SECTION
COUT
R7
EXTERNAL
MOSFETs
1876 F09
Figure 9. Generating the Gate Drive
for Low Input Supply Applications
Single Output/High Current Applications
In applications that demand current much higher than a
single stage can supply (>20A), the LTC1876 can be
configured as a single output converter. Figure 10 shows
the block diagram of the configuration. Note that the
compensation pins (ITH1 and ITH2) of the two channels are
connected together, saving a set of passive components.
In addition, the output voltage sense pins (VOSENSE1 and
VOSENSE2) are shorted together, using only one resistor
divider to set the output voltage.
Although the output current requirement is high, the input
capacitors ripple current requirement is not much differ-
ent compared to the dual outputs circuit. This is attributed
to the fact that the current is shared between two channels
and an out-of-phase architecture is implemented for the
controllers (See Theory and Benefits of 2-Phase
Operation).
INPUT
SUPPLY
RC CC
R1
VIN
ITH1
ITH2
LTC1876
VOSENSE1
VOSENSE2
SGND
R2
TO SENSE1+
AND SENSE1–
L1
RS1
EXTERNAL
MOSFETs
L2
TO SENSE2+
AND SENSE2–
VOUT
RS2
+
1876 F10
Figure 10. Single Output Configuration
Auxiliary Regulator’s Inductor Value Calculation
Since the current limit for the auxiliary regulator is inter-
nally set at 1A, it makes the selection of components
easier. For the boost regulator, the duty cycle is given by:
Duty Cycle = 1– VIN
VOUT
Since energy is only transferred to the output capacitor(s)
during the off-time, the maximum output current that can
be supplied by the regulator without losing regulation is:
IOUT = 0.5(2 • IPK – ∆IL)(1 – Duty Cycle)
where IPK = peak inductor current and is internally set at
1A.
∆IL = inductor’s ripple current
With the required ripple current determined, the value of
the inductor is:
L = (VIN •Duty Cycle)
(f • ∆IL)
where f = operating frequency (1.2MHz)
In most cases, a larger value of inductance is used. This is
done to account for component variation. It also lowers
the inductor ripple current and results in lower core
losses. In addition, lower ripple also translates into lower
ESR losses in the output capacitors and smaller output
voltage ripple.
1876fa
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