ISL12008IB8Z Intersil, ISL12008IB8Z Datasheet - Page 18

• Manufacturer Part Number: ISL12008IB8Z
• Description: IC RTC I2C LO-POWER 8-SOIC
• Manufacturer: Intersil
• Type: Clock/Calendarr

Specifications of ISL12008IB8Z

• Time Format: HH:MM:SS (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-SOIC (3.9mm Width)
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Memory Size: -

Following are some examples with equations to assist with
calculating backup times and required capacitance for the
ISL12008 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 18, use C
the voltage at V
completely. The ISL12008 is specified to operate down to
V
is used to estimate the total backup time as follows:
Rearranging gives Equation 6:
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” on page 6). This allows us to
make an approximation of I
between the two endpoints. The typical linear equation for
I
Using Equation 7 to solve for the average current given 2
voltage points gives Equation 8:
dT = C
I
I = C
I
TOT
BAT
BAT
BATAVG
BAT
BAT
FIGURE 18. SUPERCAPACITOR CHARGING CIRCUIT
vs V
2.7V TO 5.5V
BAT
is the total of the supply current of the ISL12008 (I
= 1.031E-7*(V
= 1.8V. The capacitance charge/discharge in Equation 5
is the backup capacitance and dV is the change in
BAT
*dV/dT
BAT
= 5.155E-8*(V
BAT
*dV/I
is shown in Equation 7:
changes with V
BAT
TOT
BAT
will approach 4.7V as the diode turns off
BAT
to solve for backup time.
LKG
= 0.47F and V
) + 1.036E-7A
V
BAT2
CC
is assumed to be extremely small
BAT
1N4148
18
+ V
GND
BAT
, using a value midway
BAT1
almost linearly (see
V
BAT
CC
) + 1.036E-7A
= 5V. With V
C
BAT
CC
(EQ. 5)
(EQ. 6)
(EQ. 7)
(EQ. 8)
= 5V,
BAT
LKG
ISL12008
)
.
Combining with Equation 6 gives the equation for backup
time in Equation 9:
where:
Solving Equation 8 for this example (I
yields Equation 10:
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 represented in Equation 11:
EXAMPLE 2: CALCULATING A CAPACITOR VALUE FOR
A GIVEN BACKUP TIME
Referring to Figure 18 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 6 to
solve for capacitance in Equation 12:
Using the terms previously described, Equation 12 becomes
Equation 13:
where:
If the 30% tolerance is included for tolerances, then worst
case capacitor value would be:
C
t
seconds
C
C
t
C
BACKUP
BACKUP
CC
BAT
BAT
BAT
BAT
C
V
V
I
t
I
I
V
V
C
LKG
BACKUP
BATAVG
LKG
BAT2
BAT1
BAT2
BAT1
BAT
BAT
= 5.0V. As in Example 1, the V
= dT*I/dV
= t
=
=
= 0 (assumed minimal)
= 0 (assumed)
= 0.47F
= 5.18 E6*(4.387 E-7)/(2.9) = 0.784F
1.3 0.784
0.70 35.96
BACKUP
= 4.7V
= 1.8V
= 4.7V
= 1.8VSolving gives
= C
=
= 4.387 E-7A (same as Example 1)
= 60 days*86,400 sec/day = 5.18 E6 seconds
•
0.47
•
BAT
•
*(V
*(I
(
2.9
=
BATAVG
=
BAT2
1.02F
) 4.38E 7
25.2
⁄
- V
days
+ I
BAT1
–
LKG
=
) / (I
BAT
)/(V
3.107E6s
BATAVG
BATAVG
BAT2
voltage will vary from
– V
= 4.387E-7A)
September 26, 2008
+ I
BAT1
LKG
(EQ. 10)
)
(EQ. 13)
(EQ. 14)
(EQ. 11)
(EQ. 12)
)
(EQ. 9)
FN6690.1