ISL1209IU10Z Intersil, ISL1209IU10Z Datasheet - Page 23

• Manufacturer Part Number: ISL1209IU10Z
• Description: IC RTC LP BATT BACK SRAM 10MSOP
• Manufacturer: Intersil
• Type: Clock/Calendar/NVSRAMr

Specifications of ISL1209IU10Z

• Memory Size: 2B
• Time Format: HH:MM:SS (12/24 hr)
• Date Format: YY-MM-DD-dd
• Interface: I²C, 2-Wire Serial
• Voltage - Supply: 2.7 V ~ 5.5 V
• Operating Temperature: -40°C ~ 85°C
• Mounting Type: Surface Mount
• Package / Case: 10-MSOP, Micro10™, 10-uMAX, 10-uSOP
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant

Below are some examples with equations to assist with
calculating backup times and required capacitance for the
ISL1209 device. The backup supply current plays a major
part in these equations, and a typical value was chosen for
example purposes. For a robust design, a margin of 30%
should be included to cover supply current and capacitance
tolerances over the results of the calculations. Even more
margin should be included if periods of very warm
temperature operation are expected.
Example 1. Calculating backup time given
voltages and capacitor value
In Figure 21, use C
5.0V, the voltage at V
turns off completely. The ISL1209 is specified to operate
down to V
equation is used to estimate the total backup time:
Rearranging gives
C
voltage from fully charged to loss of operation. Note that
I
plus the leakage current of the capacitor and the diode, I
In these calculations, I
and will be ignored. If an application requires extended
operation at temperatures over 50°C, these leakages will
increase and hence reduce backup time.
Note that I
Typical Performance Curves). This allows us to make an
approximation of I
two endpoints. The typical linear equation for I
is:
Using this equation to solve for the average current given 2
voltage points gives:
dT = C
I
I = C
I
TOT
BAT
BATAVG
BAT
FIGURE 21. SUPERCAPACITOR CHARGING CIRCUIT
2.7V to 5.5V
BAT
is the total of the supply current of the ISL1209 (I
= 1.031E-7*(V
is the backup capacitance and dV is the change in
BAT
= 5.155E-8*(V
* dV/dT
BAT
BAT
* dV/I
= 1.8V. The capacitance charge/discharge
changes with V
TOT
BAT
BAT
BAT
BAT
, using a value midway between the
to solve for backup time.
LKG
= 0.47F and VDD = 5.0V. With V
) + 1.036E-7 Amps
VDD
BAT2
will approach 4.7V as the diode
is assumed to be extremely small
1N4148
23
+ V
GND
BAT
BAT1
almost linearly (see
V
BAT
) + 1.036E-7 Amps
BAT
C
BAT
vs. V
(EQ. 1)
(EQ. 2)
(EQ. 3)
(EQ. 4)
BAT
DD
LKG
BAT
)
=
ISL1209
.
Combining with Equation 2 gives the equation for backup
time:
where
Solving equation 4 for this example, I
Since there are 86,400 seconds in a day, this corresponds to
35.96 days. If the 30% tolerance is included for capacitor
and supply current tolerances, then worst case backup time
would be:
C
Example 2. Calculating a capacitor value for a
given backup time
Referring to Figure 21 again, the capacitor value needs to be
calculated to give 2 months (60 days) of backup time, given
VDD = 5.0V. As in Example 1, the V
from 4.7V down to 1.8V. We will need to rearrange Equation
2 to solve for capacitance:
Using the terms described above, this equation becomes:
where
Solving gives
If the 30% tolerance is included for tolerances, then worst
case cap value would be
T
seconds
C
C
BAT
BACKUP
BAT
BAT
C
V
V
I
T
T
I
I
V
V
C
C
LKG
BATAVG
LKG
BAT2
BAT1
BACKUP
BACKUP
BAT2
BAT1
BAT
BAT
BAT
= 0.70 * 35.96 = 25.2 days
= dT*I/dV
= T
= 0 (assumed minimal)
= 0 (assumed)
= 0.47F
= 5.18 E6 * (4.387 E-7)/(2.9) = 0.784F
= 1.3 *.784 = 1.02F
= 4.7V
= 1.8V
= 4.7V
= 1.8V
BACKUP
= C
= 4.387 E-7 A (same as Example 1)
= 0.47 * (2.9) / 4.38E-7 = 3.107E6 sec
= 60 days * 86,400 sec/day = 5.18 E6 sec
BAT
* (V
* (I
BATAVG
BAT2
- V
+ I
BAT1
LKG
) / (I
BAT
)/(V
BATAVG
BATAVG
voltage will vary
BAT2
= 4.387E-7 A
– V
October 17, 2006
+ I
BAT1
LKG
(EQ. 5)
(EQ. 7)
(EQ. 6)
FN6109.4
)
)