LTC3109
17
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applicaTions inForMaTion
DESIGN EXAMPLE 1
This design example will explain how to calculate the
necessary reservoir capacitor value for V
OUT
in pulsed-
load applications, such as a wireless sensor/transmitter.
In these types of applications, the load is very small for a
majority of the time (while the circuitry is in a low power
sleep state), with pulses of load current occurring periodi
-
cally during a transmit burst.
The reservoir capacitor on V
OUT
supports the load during
the transmit pulse; the long sleep time between pulses
allows the LTC3109 to accumulate energy and recharge
the capacitor (either from the input voltage source or the
storage capacitor). A method for calculating the maximum
rate at which the load pulses can occur for a given output
current from the LTC3109 will also be shown.
In this example, V
OUT
is set to 3.3V, and the maximum
allowed voltage droop during a transmit pulse is 10%, or
0.33V. The duration of a transmit pulse is 5ms, with a total
average current requirement of 20mA during the pulse.
Given these factors, the minimum required capacitance
on V
OUT
is:
C
OUT
µF
( )
≥
= 303µF
Note that this equation neglects the effect of capacitor ESR
on output voltage droop. For ceramic capacitors and low
ESR tantalum capacitors, the ESR will have a negligible
effect at these load currents. However, beware of the voltage
coefficient of ceramic capacitors, especially those in small
case sizes. This greatly reduces the effective capacitance
when a DC bias is applied.
A standard value of 330µF could be used for C
OUT
in
this case. Note that the load current is the total current
draw on V
OUT
, V
OUT2
and VLDO, since the current for all
of these outputs must come from V
OUT
during a pulse.
Current contribution from the capacitor on VSTORE is not
considered, since it may not be able to recharge between
pulses. Also, it is assumed that the harvested charge
current from the LTC3109 is negligible compared to the
magnitude of the load current during the pulse.
To calculate the maximum rate at which load pulses can
occur, you must know how much charge current is avail
-
able from the LTC3109 V
OUT
pin given the input voltage
source being used. This number is best found empirically,
since there are many factors affecting the efficiency of the
converter. You must also know what the total load cur
-
rent is on V
OUT
during the sleep state (between pulses).
Note that this must include any losses, such as storage
capacitor leakage.
Let’s assume that the charge current available from the
LTC3109 is 150µA and the total current draw on V
OUT
and
VLDO in the sleep state is 17µA, including capacitor leakage.
We’ll also use the value of 330µF for the V
OUT
capacitor.
The maximum transmit rate (neglecting the duration of
the transmit pulse, which is very short compared to the
period) is then given by:
T =
150µA – 17µA
= 0.82sec or f
MAX
= 1.2Hz
Therefore, in this application example, the circuit can sup-
port a 5ms transmit pulse of 20mA every 0.82 seconds.
It can be seen 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 or standby current is low. Even if the available
charge current in the example above was only 21µA, if the
sleep current was only 5µA, it could still transmit a pulse
every seven seconds.
The following formula will allow you to calculate the time
it will take to charge the LDO output capacitor and the
V
OUT
capacitor the first time, from zero volts. Here again,
the charge current available from the LTC3109 must be
known. For this calculation, it is assumed that the LDO
output capacitor is 2.2µF:
t
LDO
=
I
CHG
– I
LDO
If there was 150µA 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 only 33ms.
