13
LTC1878
A second, more severe transient is caused by switching in
loads with large (>1µF) supply bypass capacitors. The
discharged bypass capacitors are effectively put in parallel
with C
OUT
, causing a rapid drop in V
OUT
. No regulator can
deliver enough current to prevent this problem if the load
switch resistance is low and it is driven quickly. The only
solution is to limit the rise time of the switch drive so that
the load rise time is limited to approximately (25 • C
LOAD
).
Thus, a 10µF capacitor charging to 3.3V would require a
250µs rise time, limiting the charging current to about
130mA.
PC Board Layout Checklist
When laying out the printed circuit board, the following
checklist should be used to ensure proper operation of the
LTC1878. These items are also illustrated graphically in
the layout diagram of Figure 7. Check the following in your
layout:
1. Are the signal and power grounds segregated? The
LTC1878 signal ground consists of the resistive
divider, the optional compensation network (R
C
and
C
C1
) and C
C2
. The power ground consists of the (–)
plate of C
IN
, the (–) plate of C
OUT
and Pin 4 of the
LTC1878. The power ground traces should be kept
short, direct and wide. The signal ground and power
ground should converge to a common node in a star-
ground configuration.
2. Does the V
FB
pin connect directly to the feedback
resistors? The resistive divider R1/R2 must be con-
nected between the (+) plate of C
OUT
and signal ground.
3. Does the (+) plate of C
IN
connect to V
IN
as closely as
possible? This capacitor provides the AC current to the
internal power MOSFETs.
4. Keep the switching node SW away from sensitive small
signal nodes.
Design Example
As a design example, assume the LTC1878 is used in a
single lithium-ion battery-powered cellular phone applica-
tion. The input voltage will be operating from a maximum
of 4.2V down to about 2.7V. The load current requirement
is a maximum of 0.3A but most of the time it will be in
standby mode, requiring only 2mA. Efficiency at both low
and high load currents is important. Output voltage is
2.5V. With this information we can calculate L using
equation (1),
L
fI
V
V
V
L
OUT
OUT
IN
=
()
∆
()
−
1
1
(3)
Substituting V
OUT
= 2.5V, V
IN
= 4.2V, ∆I
L
=120mA and
f = 550kHz in equation (3) gives:
L
V
kHz mA
V
V
H=−
=µ
25
550 120
1
25
42
15 3
.
()
.
.
.
A 15µH inductor works well for this application. For best
efficiency choose a 1A inductor with less than 0.25Ω
series resistance.
C
IN
will require an RMS current rating of at least 0.15A at
temperature and C
OUT
will require an ESR of less than
0.25Ω. In most applications, the requirements for these
capacitors are fairly similar.
For the feedback resistors, choose R1 = 412k. R2 can
then be calculated from equation (2) to be:
R
V
R k use
OUT
2
08
1 1 875 5 8=−
=
.
. ; 87k
Figure 8 shows the complete circuit along with its effi-
ciency curve.
APPLICATIO S I FOR ATIO
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