LTC3108 Linear Technology, LTC3108 Datasheet - Page 15

• Manufacturer Part Number: LTC3108
• Description: Ultralow Voltage Step-Up Converter and Power Manager
• Manufacturer: Linear Technology

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APPLICATIONS INFORMATION
the capacitor. A method for calculating the maximum rate
at which the load pulses can occur for a given output cur-
rent from the LTC3108 will also be shown.
In this example, V
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 V
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 150μF or larger could be used for C
in this case. Note that the load current is the total current
draw on V
these outputs must come from V
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 V
being used. This number is best found empirically, since
there are many factors affecting the effi ciency of the
converter. Also determine what the total load current is
on V
that this must include any losses, such as storage ca-
pacitor leakage.
Assume, for instance, that the charge current from the
LTC3108 is 50μA and the total current drawn on V
the sleep state is 17μA, including capacitor leakage. In
C
OUT
OUT
OUT
( )
is:
µF
during the sleep state (between bursts). Note
OUT
, V
≥
40
OUT2
mA
OUT
0 33
OUT
.
and VLDO, since the current for all of
•
is set to 3.3V, and the maximum
pin given the input voltage source
V
1
ms
=
OUT
121
during a burst. Current
µF
OUT
OUT
in
addition, use the value of 150μF for the V
The maximum transmit rate (neglecting the duration of
the transmit burst, which is typically very short) is then
given by:
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 10μA and
the sleep current was only 5μA, 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 V
the charge current available from the LTC3108 must be
known. For this calculation, it is assumed that the LDO
output capacitor is 2.2μF .
If there were 50μA of charge current available and a 5μA
load on the LDO (when the processor is sleeping), the time
for the LDO to reach regulation would be 107ms.
If V
was 150μF , the time for V
If there were 50μA of charge current available and 5μA of
load on V
the initial application of power would be 11.1 seconds.
t
t
t
OUT
LDO
VOUT
=
OUT
150
(
were programmed to 3.3V and the V
50
=
OUT
capacitor the fi rst time, from 0V. Here again,
=
µA
2 2
µF
I
.
CHG
3 3
I
, the time for V
CHG
.
−
• .
V
0 33
17
V
• .
−
2 2
−
•
µA
I
150
LDO
I
VOUT
V
)
µF
=
OUT
µF
1 5
. sec
+
to reach regulation would be:
OUT
t
LDO
to reach regulation after
or f
MAX
LTC3108
OUT
=
OUT
0 666
.
capacitor.
capacitor
15
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3108p