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LTC3108

3108fb

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(μF)

40mA • 1ms

0.33V

121μF

OUT

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 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

these outputs must come from VOUTduringaburst.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 VOUT 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. 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 50μA and the total current drawn on VOUT in

the sleep state is 17μA, including capacitor leakage. In

addition, use the value of 150μF for the VOUT capacitor.

The maximum transmit rate (neglecting the duration of

the transmit burst, which is typically very short) is then

given by:

−

150μF • 0.33V

(50μA 17μA)

1.5sec or f

0.666Hz

MAX

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 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 2.2μF.

−

2.2V • 2.2μF

LDO

CHG

LDO

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 VOUT were programmed to 3.3V and the VOUT capacitor

was 150μF, the time for VOUT to reach regulation would be:

−−

3.3V • 150μF

VOUT

CHG

VOUT

LDO

If there were 50μA of charge current available and 5μA 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

CHG

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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