19
LTC1703
1703fa
Calculating RMS Current in C
IN
A buck regulator like the LTC1703 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 LTC1703
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 LTC1703 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
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
current that the LTC1703 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 LTC1703 is much lower
than at the input, due to the fact that the inductor current
is constantly flowing at the output whenever the LTC1703
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
LTC1703 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 tantalum capacitors in parallel.
INDUCTOR
The inductor in a typical LTC1703 circuit is chosen prima-
rily 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 QB OUT
=
()
()
TIME
0ABCD
50% 16% 16% 18%
I
AVE
0
INPUT CURRENT (A)
5.2
3
10
13
1703 SB1
Figure SB1. Average Current Calculation
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LTC1703
1703fa
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 LTC1703 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.
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
>
C
onsider 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.
TIME
0ABCD
50% 16% 16% 18%
5.2
AC INPUT CURRENT (A)
0
2.2
4.8
7.8
1703 SB2
Figure SB2. AC Current Calculation
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 QB OUT
RIPPLE
ON QB
=
()
=
µ
()()
=
()
()
..
.
.
/.
12 16
3
064
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.
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LTC1703
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FEEDBACK LOOP/COMPENSATION
1
Feedback Loop Types
In a typical LTC1703 circuit, the feedback loop consists of
the modulator, the external inductor and output capacitor,
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 themselves. 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 frequencies 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 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).
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
LTC1703 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.
GAIN
(dB)
PHASE
(DEG)
1703 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
1703 F09a
V
REF
+
GAIN
(dB)
PHASE
(DEG)
1703 F09b
00
–90
180
270
GAIN
PHASE
6dB/OCT
Figure 9a. Type 1 Amplifier Schematic Diagram
Figure 9b. Type 1 Amplifier Transfer Function
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.”
Type 2 loops work well in systems where the ESR zero in
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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.

LTC1703IG#TRPBF

Mfr. #:
Manufacturer:
Analog Devices / Linear Technology
Description:
Switching Voltage Regulators 2x 550kHz Sync 2-PhSw Reg Cntr w/ 5-B VI
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New from this manufacturer.
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