LTC3108

15

3108p

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.

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

CµF

mA

ms

V

µF

•

.

≥

=

40

1

033

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.

in this case. Note that the load current is the total 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

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

that this must include any losses, such as storage ca-

pacitor leakage.

Assume, for instance, that the charge current from the

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

The maximum transmit rate (neglecting the duration of

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

given by:

t

µF

V

µA

µA

=

−

==

150

0 33

50

17

15

0666

•.

()

. sec

.

H

Hz

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 charge current available from the LTC3108 must be

known. For this calculation, it is assumed that the LDO

output capacitor is 2.2μF.

t

VµF

II

LDO

CHG

LDO

=

−

22

22

.• .

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.

t

VµF

II

t

VOUT

CHG

VOUT

LDO

=

−

+

3 3

150

.•

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

the initial application of power would be 11.1 seconds.

APPLICATIONS INFORMATION