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LTC1702A
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APPLICATIONS INFORMATION
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from
C
IN
(time point A). 50% of the way through, TG2
turns on and the total current is 13A (time point B).
Shortly thereafter, TG1 turns off and the current drops to
10A (time point C). Finally, TG2 turns off and the current
spends a short time at 0 before TG1 turns on again (time
point D).
IA A
AA A
AVG
=
( )
+
( )
+
( )
+
( )
=
305 13016
10 016 0 018 518
•. •.
•. •. .
Now we can calculate the RMS current. Using the same
waveform we used to calculate the average DC current,
subtract the average current from each of the DC values.
Square each current term and multiply the squares by the
same period percentages we used to calculate the aver-
age DC current. Sum the results and take the square root.
The result is the approximate RMS current as seen by the
input capacitor with both sides of the LTC1702A at full
load. Actual RMS current will differ due to inductor ripple
current and resistive losses, but this approximate value is
adequate for input capacitor calculation purposes.
TIME
0ABCD
50% 16% 16% 18%
5.2
AC INPUT CURRENT (A)
0
2.2
4.8
7.8
1702A SB2
Figure SB2. AC Current Calculation
I
A
RMS
RMS
=
( )
+
( )
+
( )
+
( )
=
–. . . .
.•. ..
.
218 05 782 016
482 016 518 018
455
22
22
If the circuit is likely to spend time with one side operating
and the other side shut down, the RMS current will need
to be calculated for each possible case (side 1 on, side 2
off; side 1 off, side 2 on; both sides on). The capacitor
must be sized to withstand the largest RMS current of the
three—sometimes this occurs with one side shut down!
Side only
IA A A
IA
Side only
IAA A
I
AVE
RMS RMS
AVE
RMS
1
3 0 67 0 0 33 2 01
1 0 67 2 0 33 1 42
2
10 032 0 068 32
68 032 32 068
1
1
22
2
2
22
:
•. •. .
•. •. .
:
•. •. .
.•. –..
=
( )
+
( )
=
=
( )
+
( )
=
=
( )
+
( )
=
=
( )
+
( )
=4466 455..AA
RMS RMS
>
Consider the case where both sides are operating at the
same load, with a 50% duty cycle at each side. The RMS
current with both sides running is near zero, while the
RMS current with one side active is 1/2 the total load
current of that side. The 2-phase, 5V to 2.5V circuit in the
applications section takes advantage of this phenom-
enon, allowing it to supply 40A of output current with only
120µF of input capacitance (and only 40µF of output
capacitance!).
the capacitor we chose for the single side application can
support the slightly higher 4.8A
RMS
current, we can add
the second channel without changing the input capacitor
at all. As a general rule, an input bypass capacitor capable
of supporting the larger output current channel can sup-
port both channels running simultaneously (see the
2-Phase Operation section for more details).
Tantalum capacitors are a popular choice as input capaci-
tors for LTC1702A applications, but they deserve a special
caution here. Generic tantalum capacitors have a destruc-
tive failure mechanism when they are subjected to large
RMS currents (like those seen at the input of a LTC1702A).
At some random time after they are turned on, they can
blow up for no apparent reason. The capacitor manufac-
turers are aware of this and sell special “surge tested”
tantalum capacitors specifically designed for use with
switching regulators. When choosing a tantalum input
capacitor, make sure that it is rated to carry the RMS
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LTC1702A
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current that the LTC1702A will draw. If the data sheet
doesn’t give an RMS current rating, chances are the
capacitor isn’t surge tested. Don’t use it!
OUTPUT BYPASS CAPACITOR
The output bypass capacitor has quite different require-
ments from the input capacitor. The ripple current at the
output of a buck regulator like the LTC1702A is much
lower than at the input, due to the fact that the inductor
current is constantly flowing at the output whenever the
LTC1702A is operating in continuous mode. The primary
concern at the output is capacitor ESR. Fast load current
transitions at the output will appear as voltage across the
ESR of the output bypass capacitor until the feedback loop
in the LTC1702A can change the inductor current to match
the new load current value. This ESR step at the output is
often the single largest budget item in the load regulation
calculation. As an example, our hypothetical 1.6V, 10A
switcher with a 0.01 ESR output capacitor would expe-
rience a 100mV step at the output with a 0 to 10A load
step—a 6.3% output change!
Usually the solution is to parallel several capacitors at the
output. For example, to keep the transient response inside
of 3% with the previous design, we’d need an output ESR
better than 0.0048. This can be met with three 0.014,
470µF low ESR tantalum capacitors in parallel.
APPLICATIONS INFORMATION
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INDUCTOR
The inductor in a typical LTC1702A circuit is chosen
primarily for value and saturation current. The inductor
value sets the ripple current, which is commonly chosen
at around 40% of the anticipated full load current. Ripple
current is set by:
I
tV
L
RIPPLE
ON Q OUT
=
()
()2
In our hypothetical 1.6V, 10A example, we'd set the ripple
current to 40% of 10A or 4A, and the inductor value would
be:
L
tV
I
sV
A
H
with t
V
V
kHz s
ON Q OUT
RIPPLE
ON Q
=
()
=
µ
()()
=
()
()
..
.
.
/.
2
2
12 16
4
05
1
16
5
550 1 2
The inductor must not saturate at the expected peak
current. In this case, if the current limit was set to 15A, the
inductor should be rated to withstand 15A + 1/2 I
RIPPLE
,
or 17A without saturating.
FEEDBACK LOOP/COMPENSATION
1
Feedback Loop Types
In a typical LTC1702A circuit, the feedback loop consists
of the modulator, the external inductor and output capaci-
tor, and the feedback amplifier and its compensation
network. All of these components affect loop behavior and
need to be accounted for in the loop compensation. The
modulator consists of the internal PWM generator, the
output MOSFET drivers and the external MOSFETs them-
selves. From a feedback loop point of view, it looks like a
linear voltage transfer function from COMP to SW and has
a gain roughly equal to the input voltage. It has fairly
benign AC behavior at typical loop compensation frequen-
cies with significant phase shift appearing at half the
switching frequency.
The external inductor/output capacitor combination makes
a more significant contribution to loop behavior. These
components cause a second order LC roll-off at the
1
The information in this section is based on the paper “The K Factor: A New Mathematical Tool for
Stability Analysis and Synthesis” by H. Dean Venable, Venable Industries, Inc. For complete paper,
see “Reference Reading #4” at www.linear-tech.com.
0
10A
32%
68%
0
10A
32% 18%
18%
18%32%
3.2A
0
6.8A
32%
68%
Q1 CURRENT, SIDE 1 ONLY
(FOR 1-PHASE, 2 SIDES:
MULTIPLY CURRENT BY 2)
CURRENT IN C
IN
, SIDE 1 ONLY
I
CIN
= 4.66A
RMS
, (1-PHASE,
2 SIDES: I
CIN
= 9.3A
RMS
)
CURRENT IN C
IN
,
BOTH SIDES EQUAL LOAD
I
CIN
= 4.8A
RMS
Q11 CURRENT
Q21 CURRENT
BOTH SIDES EQUAL LOAD
2-PHASE OPERATION
6.4A
0
3.6A
32% 18%
1702A F07
32%
Figure 7. RMS Input Current
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LTC1702A
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APPLICATIONS INFORMATION
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output, with the attendant 180° phase shift. This roll-off is
what filters the PWM waveform, resulting in the desired
DC output voltage, but the phase shift complicates the
loop compensation if the gain is still higher than unity at
the pole frequency. Eventually (usually well above the LC
pole frequency), the reactance of the output capacitor will
approach its ESR, and the roll-off due to the capacitor will
stop, leaving 6dB/octave and 90° of phase shift (Figure 8).
OUT
IN
R1
C2
C1
R2
R
B
1702A F10a
V
REF
+
GAIN
(dB)
PHASE
(DEG)
1702A F10b
00
–90
180
270
PHASE
GAIN
–6dB/OCT
6dB/OCT
Figure 10a. Type 2 Amplifier Schematic Diagram
Figure 10b. Type 2 Amplifier Transfer Function
So far, the AC response of the loop is pretty well out of the
user’s control. The modulator is a fundamental piece of the
LTC1702A design, and the external L and C are usually
chosen based on the regulation and load current require-
ments without considering the AC loop response. The
feedback amplifier, on the other hand, gives us a handle
with which to adjust the AC response. The goal is to have
180° phase shift at DC (so the loop regulates) and some-
thing less than 360° phase shift at the point that the loop
gain falls to 0dB. The simplest strategy is to set up the
feedback amplifier as an inverting integrator, with the 0dB
frequency lower than the LC pole (Figure 9). This “type 1”
configuration is stable but transient response will be less
than exceptional if the LC pole is at a low frequency.
Figure 10 shows an improved “type 2” circuit that uses an
additional pole-zero pair to temporarily remove 90° of
phase shift. This allows the loop to remain stable with 90°
more phase shift in the LC section, provided the loop
reaches 0dB gain near the center of the phase “bump.”
GAIN
(dB)
PHASE
(DEG)
1702A F08
A
V
00
–90
180
6dB/OCT
PHASE
GAIN
–12dB/OCT
Figure 8. Transfer Function of Buck Modulator
OUT
IN
R1
C1
R
B
1702A F09a
V
REF
+
GAIN
(dB)
PHASE
(DEG)
1702A F09b
00
–90
180
270
GAIN
PHASE
6dB/OCT
Figure 9a. Type 1 Amplifier Schematic Diagram
Figure 9b. Type 1 Amplifier Transfer Function

LTC1702ACGN#PBF

Mfr. #:
Manufacturer:
Analog Devices / Linear Technology
Description:
Switching Voltage Regulators 2x 550kHz Sync 2-PhSw Reg Cntr
Lifecycle:
New from this manufacturer.
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