LTC3564 LINER [Linear Technology], LTC3564 Datasheet - Page 15

• Manufacturer Part Number: LTC3564
• Description: 2.25MHz, 1.25A Synchronous Step-Down Regulator
• Manufacturer: LINER [Linear Technology]

APPLICATIO S I FOR ATIO
PC Board Layout Checklist
When laying out the printed circuit board, the following
checklist should be used to ensure proper operation of the
LTC3564. These items are also illustrated graphically in
Figures 4 and 5. Check the following in your layout:
1. The power traces, consisting of the GND trace, the SW
2. Does the V
3. Does the (+) plate of C
4. Keep the switching node, SW, away from the sensitive
5. Keep the (–) plates of C
Design Example
As a design example, assume the LTC3564 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 1.25A but most of the time it will be in
standby mode, requiring only 2mA. Efficiency at both low
trace and the V
wide.
resistors? The resistive divider R1/R2 must be con-
nected between the (+) plate of C
possible? This capacitor provides the AC current to the
internal power MOSFETs.
V
TO 4.2V
FB
2.7V
V
IN
node.
C
22μF
CER
IN
FB
**
Figure 6a. Typical Application
IN
** TAIYO YUDEN JMK316BJ226ML
IN
*TOKO A915AY-1R1M (D53LC SERIES)
pin connect directly to the feedback
3
5
U
trace should be kept short, direct and
will be operating from a maximum of
V
RUN
IN
LTC3564
GND
IN
IN
2
U
and C
SW
V
connect to V
FB
4
1
OUT
316k
1.1μH*
3564 F06a
22pF
W
1M
OUT
as close as possible.
and ground.
IN
as closely as
U
C
22μF
CER
OUT
**
V
2.5V
OUT
and high load currents is important. Output voltage is
2.5V. With this information we can calculate L using
equation (1),
Substituting V
f = 2.25MHz in equation (3) gives:
A 1μH or 1.1μH inductor works well for this application.
For best efficiency choose a 1.5A or greater inductor with
less than 0.1Ω series resistance.
C
I
of less than 0.125Ω. In most cases, a ceramic capacitor
will satisfy this requirement.
For the feedback resistors, choose R1 = 316k. R2 can
then be calculated from equation (2) to be:
Figure 6 shows the complete circuit along with its effi-
ciency curve.
LOAD(MAX)
IN
L =
R
L
will require an RMS current rating of at least 0.6A ≅
2
=
=
( )
2.25MHz(500mA)
f
⎛
⎜
⎝
( )
V
100
/2 at temperature and C
1
Δ
0 6
95
90
85
80
75
70
65
60
OUT
Figure 6b. Efficiency vs Output Current
.
I
0.1
L
V
2.5V
OUT
OUT
V
−
OUT
= 2.5V
1 1 1000
= 2.5V, V
⎞
⎟
⎠
R
1
OUTPUT CURRENT (mA)
⎛
⎜
⎝
1
=
−
V
⎛
⎝ ⎜
V
OUT
10
1
IN
IN
−
k
= 4.2V, ΔI
2 5
4 2
⎞
⎟
⎠
.
.
OUT
V
V
100
V
V
V
⎞
⎠ ⎟
IN
IN
IN
=
will require an ESR
= 2.7V
= 3.6V
= 4.2V
3564 F06b
0 9
L
.
1000
LTC3564
= 500mA and
μH H
15
3564f
(3)