Datasheet

20
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
()()
=µ
= −
⎛
⎝
⎜
⎞
⎠
⎟
=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.
APPLICATIO S I FOR ATIO
WUUU