ISL1208IRT8Z-TK Intersil, ISL1208IRT8Z-TK Datasheet - Page 21

• Manufacturer Part Number: ISL1208IRT8Z-TK
• Description: IC RTC/CALENDER I2C 8-TDFN
• Manufacturer: Intersil
• Type: Clock/Calendar/NVSRAMr

Specifications of ISL1208IRT8Z-TK

• 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: 8-TDFN
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant

Combining with Equation 5 gives the equation for backup
time in Equation 8:
where:
Solving Equation 7 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
V
4.7V down to 1.8V. We will need to rearrange Equation 5 to
solve for capacitance in Equation 9:
Using the terms described above, this equation becomes
Equation 10:
where:
T
seconds
C
C
BAT
CC
BACKUP
BAT
BAT
C
V
V
I
T
T
I
I
V
V
C
If the 30% tolerance is included for tolerances, then worst
LKG
BATAVG
LKG
BACKUP
BACKUP
BAT2
BAT1
BAT2
BAT1
BAT
BAT
= 5.0V. As in Example 1, the V
case capacitor value would be:
C
= 0.70 * 35.96 = 25.2 days
= dT*I/dV
= T
BAT
= 0 (assumed minimal)
= 0 (assumed)
= 0.47F
= 5.18 E6 * (4.387 E-7)/(2.9) = 0.784F
= 4.7V
= 1.8V
= 4.7V
= 1.8VSolving gives
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 seconds
1.3
BAT
×
* (V
* (I
0.784
BATAVG
BAT2
=
1.02F
- V
21
+ I
BAT1
LKG
BAT
) / (I
)/(V
BATAVG
voltage will vary from
BATAVG
BAT2
= 4.387E-7 A
– V
+ I
BAT1
LKG
(EQ. 10)
(EQ. 11)
(EQ. 8)
(EQ. 9)
)
)
ISL1208
September 12, 2008
FN8085.8