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参数资料
| 型号： | LTC3108EDE#PBF |
| 厂商： | LINEAR TECHNOLOGY CORP |
| 元件分类： | 稳压器 |
| 英文描述： | 0.45 A SWITCHING REGULATOR, PDSO12 |
| 封装： | 4 X 3 MM, LEAD FREE, PLASTIC, MO-229WGED, DFN-12 |
| 文件页数： | 9/22页 |
| 文件大小： | 1300K |
| 代理商： | LTC3108EDE#PBF |
LTC3108
3108fa
applicaTions inForMaTion
at which the load pulses can occur for a given output cur-
rent from the LTC3108 will also be shown.
In this example, VOUT is set to 3.3V, and the maximum
allowed voltage droop during a transmit burst is 10%, or
0.33V. The duration of a transmit burst is 1ms, with a total
average current requirement of 40mA during the burst.
Given these factors, the minimum required capacitance
on VOUT is:
C
F
mA ms
V
F
OUT(
)
.
≥
=
40
1
0 33
121
Note that this equation neglects the effect of capacitor
ESR on output voltage droop. For most ceramic or low
ESR tantalum capacitors, the ESR will have a negligible
effect at these load currents.
A standard value of 150F or larger could be used for COUT
in this case. Note that the load current is the total current
draw on VOUT, VOUT2 and VLDO, since the current for all of
theseoutputsmustcomefromVOUTduringaburst.Current
contribution from the holdup capacitor on VSTORE is not
considered, since it may not be able to recharge between
bursts. Also, it is assumed that the charge current from
the LTC3108 is negligible compared to the magnitude of
the load current during the burst.
To calculate the maximum rate at which load bursts can
occur, determine how much charge current is available
from the LTC3108 VOUTpingiventheinputvoltagesource
being used. This number is best found empirically, since
there are many factors affecting the efficiency of the
converter. Also determine what the total load current is
on VOUT during the sleep state (between bursts). Note
that this must include any losses, such as storage ca-
pacitor leakage.
Assume, for instance, that the charge current from the
LTC3108 is 50A and the total current drawn on VOUT in
the sleep state is 17A, including capacitor leakage. In
addition, use the value of 150F for the VOUT capacitor.
The maximum transmit rate (neglecting the duration of
the transmit burst, which is typically very short) is then
given by:
t
F
V
A
or fMAX
=
=
150
0 33
50
17
1 5
0 666
.
(
)
. sec
.
HHz
Therefore, in this application example, the circuit can sup-
port a 1ms transmit burst every 1.5 seconds.
It can be determined that for systems that only need to
transmit every few seconds (or minutes or hours), the
average charge current required is extremely small, as
long as the sleep current is low. Even if the available
charge current in the example above was only 10A and
the sleep current was only 5A, it could still transmit a
burst every ten seconds.
The following formula enables the user to calculate the
time it will take to charge the LDO output capacitor and
the VOUT capacitor the first time, from 0V. Here again,
the charge current available from the LTC3108 must be
known. For this calculation, it is assumed that the LDO
output capacitor is 2.2F.
t
V
F
I
LDO
CHG
LDO
=
2 2
.
.
If there were 50A of charge current available and a 5A
load on the LDO (when the processor is sleeping), the time
for the LDO to reach regulation would be 107ms.
If VOUT were programmed to 3.3V and the VOUT capacitor
was 150F, the time for VOUT to reach regulation would be:
t
V
F
I
t
VOUT
CHG
VOUT
LDO
=
+
3 3
150
.
If there were 50A of charge current available and 5A of
load on VOUT, the time for VOUT to reach regulation after
the initial application of power would be 12.5 seconds.
Design Example 2
In many pulsed load applications, the duration, magnitude
and frequency of the load current bursts are known and
fixed. In these cases, the average charge current required
from the LTC3108 to support the average load must be
calculated, which can be easily done by the following:
I
t
T
CHG
Q
BURST
≥ +
Where IQ is the sleep current on VOUT required by the ex-
ternal circuitry in between bursts (including cap leakage),
IBURST is the total load current during the burst, t is the
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