LTC3406-1.2 Linear Technology, LTC3406-1.2 Datasheet - Page 10

• Manufacturer Part Number: LTC3406-1.2
• Description: 1.5MHz / 600mA Synchronous Step-Down Regulator in ThinSOT
• Manufacturer: Linear Technology

APPLICATIO S I FOR ATIO
LTC3406-1.2
The regulator loop then acts to return V
state value. During this recovery time V
tored for overshoot or ringing that would indicate a stability
problem. For a detailed explanation of switching control
loop theory, see Application Note 76.
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
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
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
LTC3406-1.2. These items are also illustrated graphically
in Figures 3 and 4. Check the following in your layout:
1. The power traces, consisting of the GND trace, the SW
2. Does the (+) plate of C
3. Keep the (–) plates of C
10
trace and the V
wide.
possible? This capacitor provides the AC current to the
internal power MOSFETs.
OUT
, causing a rapid drop in V
Figure 3. LTC3406-1.2 Layout Diagram
V
OUT
–
+
BOLD LINES INDICATE HIGH CURRENT PATHS
IN
C
U
trace should be kept short, direct and
OUT
L1
IN
1
2
3
U
IN
and C
RUN
SW
LTC3406-1.2
connect to V
GND
V
OUT
OUT
V
C
IN
IN
W
OUT
as close as possible.
5
4
. No regulator can
OUT
OUT
IN
340612 F03
can be moni-
to its steady-
as closely as
V
IN
U
LOAD
).
Design Example
As a design example, assume the LTC3406-1.2 is used in
a single lithium-ion battery-powered cellular phone
application. The V
4.2V down to about 2.7V. The load current requirement
is a maximum of 0.6A but most of the time it will be in
standby mode, requiring only 2mA. Efficiency at both low
and high load currents is important. With this informa-
tion we can calculate L using equation (1),
Substituting V
equation (3) gives:
A 2.2µH inductor works well for this application. For best
efficiency choose a 720mA or greater inductor with less
than 0.2Ω series resistance.
C
I
of less than 0.25Ω. In most cases, a ceramic capacitor will
satisfy this requirement.
LOAD(MAX)
IN
L
L
will require an RMS current rating of at least 0.3A ≅
=
=
V
OUT
1 5
( )
f
.
( )
MHz
/2 at temperature and C
Figure 4. LTC3406-1.2 Suggested Layout
1
∆
I
L
1 2
IN
.
L1
(
1 2 1
240
V
= 4.2V, ∆I
.
IN
V
mA
will be operating from a maximum of
SW
⎛
⎜
⎝
PIN 1
−
C
)
OUT
⎛
⎜
⎝
1 2
1
V
.
L
−
IN
V
= 240mA and f = 1.5MHz in
LTC3406-1.2
1 2
4 2
⎞
⎟
⎠
.
.
VIA TO V
GND
V
V
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OUT
⎞
⎟ =
⎠
IN
VIA TO V
will require an ESR
2 38
C
IN
.
OUT
µ
H
V
340612 F04
IN
340612f
(3)