LTC1702ACGN#TR-SMI Linear Technology, LTC1702ACGN#TR-SMI Datasheet - Page 18

• Manufacturer Part Number: LTC1702ACGN#TR-SMI
• Manufacturer: Linear Technology

Specifications of LTC1702ACGN#TR-SMI

• Lead Free Status / RoHS Status: Not Compliant

APPLICATIONS
LTC1702A
implies a duty cycle of 32%, which means QT is on 32%
of each switching cycle. During QT’s on-time, the current
drawn from the input equals the load current and during
the rest of the cycle, the current drawn from the input is
near zero. This 0A to 10A, 32% duty cycle pulse train adds
up to 4.7A
last about 1.8µs—most system logic supplies have no
hope of regulating output current with that kind of speed.
A local input bypass capacitor is required to make up the
difference and prevent the input supply from dropping
drastically when QT kicks on. This capacitor is usually
chosen for RMS ripple current capability and ESR as well
as value.
The input bypass capacitor in an LTC1702A circuit is
common to both channels. Consider our 10A example
case with the other side of the LTC1702A disabled. The
input bypass capacitor gets exercised in three ways: its
ESR must be low enough to keep the initial drop as QT
turns on within reason (100mV or so); its RMS current
capability must be adequate to withstand the 4.6A
ripple current at the input and the capacitance must be
large enough to maintain the input voltage until the input
supply can make up the difference. Generally, a capacitor
that meets the first two parameters will have far more
capacitance than is required to keep capacitance-based
droop under control. In our example, we need 0.01Ω ESR
to keep the input drop under 100mV with a 10A current
step and 4.6A
heating the capacitor. These requirements can be met with
multiple low ESR tantalum or electrolytic capacitors in
parallel, or with a large monolithic ceramic capacitor.
The two sides of the LTC1702A run off a single master
clock and are wired 180° out of phase with each other to
significantly reduce the total capacitance/ESR needed at
the input. Assuming 100mV of ripple and 10A output
current, we needed an ESR of 0.01Ω and 4.7A ripple
current capability for one side. Now, assume both sides
are running simultaneously with identical loading. If the
two sides switched in phase, all the loading conditions
would double and we’d need enough capacitance for
9.4A
phase, the input current is 4.8A
the single case (Figure 7)! The peak current deltas are still
only 10A, requiring the same 0.01Ω ESR rating. As long as
18
RMS
and 0.005Ω ESR. With the two sides out of
RMS
RMS
at the input. At 550kHz, switching cycles
ripple current capacity to avoid over-
U
INFORMATION
U
RMS
W
—barely larger than
U
RMS
Calculating RMS Current in C
A buck regulator like the LTC1702A draws pulses of
current from the input capacitor during normal opera-
tion. The input capacitor sees this as AC current, and
dissipates power proportional to the RMS value of the
input current waveform. To properly specify the capaci-
tor, we need to know the RMS value of the input current.
Calculating the approximate RMS value of a pulse train
with a fixed duty cycle is straightforward, but the
LTC1702A complicates matters by running two sides
simultaneously and out of phase, creating a complex
waveform at the input.
To calculate the approximate RMS value of the input
current, we first need to calculate the average DC value
with both sides of the LTC1702A operating at maximum
load. Over a single period, the system will spend some
time with one top switch on and the other off, perhaps
some time with both switches on, and perhaps some
time with both switches off. During the time each top
switch is on, the current will equal that side’s full load
output current. When both switches are on, the total
current will be the sum of the two full load currents, and
when both are off, the current is effectively zero. Multiply
each current value by the percentage of the period that
the current condition lasts, and sum the results—this is
the average DC current value.
As an example, consider a circuit that takes a 5V input
and generates 3.3V at 3A at side 1 and 1.6V at 10A at
side 2. When a cycle starts, TG1 turns on and 3A flows
Figure SB1. Average Current Calculation
5.2
13
10
3
0
0
50%
A
TIME
16% 16% 18%
IN
B
C
D
1702A SB1
I
AVE
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