LTC1702A View Datasheet(PDF) - Linear Technology
Part Name
Description
MFG CO.
'LTC1702A' PDF : 36 Pages View PDF
LTC1702A
APPLICATIONS INFORMATION
10A
0
6.8A
0
–3.2A
10A
0
3.6A
0
–6.4A
32%
68%
32%
68%
32% 18% 32% 18%
32% 18% 32% 18%
Q1 CURRENT, SIDE 1 ONLY
(FOR 1-PHASE, 2 SIDES:
MULTIPLY CURRENT BY 2)
CURRENT IN CIN, SIDE 1 ONLY
ICIN = 4.66ARMS, (1-PHASE,
2 SIDES: ICIN = 9.3ARMS)
Q11 CURRENT
Q21 CURRENT
BOTH SIDES EQUAL LOAD
2-PHASE OPERATION
CURRENT IN CIN,
BOTH SIDES EQUAL LOAD
ICIN = 4.8ARMS
1702A F07
Figure 7. RMS Input Current
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 0.01Ω 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 0.0048Ω. This can be met with three 0.014Ω,
470µF low ESR tantalum capacitors in parallel.
20
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:
( ) IRIPPLE
=
tON(Q2)
L
VOUT
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 = ( tON(Q2) VOUT ) = (1.2µs)(1.6V) = 0.5µH
IRIPPLE
4A
with
tON(Q2)
=
⎛⎝⎜1−
1.6V
5V
⎞⎠⎟
/
550kHz
=
1.2µs
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 IRIPPLE,
or 17A without saturating.
FEEDBACK LOOP/COMPENSATION1
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
1The 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.
1702afa
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