Datasheet

LTC3109
18
3109fb
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applicaTions inForMaTion
The time for V
OUT
to charge and reach regulation can be
calculated by the formula below, which assumes V
OUT
is
programmed to 3.3V and C
OUT
is 330µF:
t
VOUT
=
3.3V • 330µF
I
CHG
–I
VOUT
–I
LDO
+ t
LDO
With 150µA of charge current available and 5µA of load on
both V
OUT
and VLDO, the time for V
OUT
to reach regula-
tion after the initial application of power would be 7.81
seconds.
DESIGN EXAMPLE 2
In most pulsed-load applications, the duration, magnitude
and frequency of the load current pulses are known and
fixed. In these cases, the average charge current required
from the LTC3109 to support the average load must be
calculated, which can be easily done by the following:
I
CHG
≥ I
Q
+
I
PULSE
• t
PULSE
T
where I
Q
is the sleep current supplied by V
OUT
and V
LDO
to the external circuitry in-between load pulses, including
output capacitor leakage, I
PULSE
is the total load current
during the pulse, t
PULSE
is the duration of the load pulse
and T is the pulse period (essentially the time between
load pulses).
In this example, I
Q
is 5µA, I
PULSE
is 100mA, t
PULSE
is 5ms
and T is one hour. The average charge current required
from the LTC3109 would be:
I
CHG
≥ 5µA +
100mA • 0.005sec
3600sec
= 5.14µA
Therefore, if the LTC3109 has an input voltage that allows
it to supply a charge current greater than just 5.14µA, the
application can support 100mA pulses lasting 5ms every
hour. It can be seen that the sleep current of 5µA is the
dominant factor in this example, because the transmit
duty cycle is so small (0.00014%). Note that for a V
OUT
of 3.3V, the average power required by this application is
only 17µW (not including converter losses).
Keep in mind that the charge current available from the
LTC3109 has no effect on the sizing of the V
OUT
capacitor,
and the V
OUT
capacitor has no effect on the maximum
allowed pulse rate.