ISL1219IUZ Intersil, ISL1219IUZ Datasheet - Page 23

• Manufacturer Part Number: ISL1219IUZ
• Description: IC RTC LP BATT BACK SRAM 10MSOP
• Manufacturer: Intersil
• Type: Clock/Calendar/NVSRAMr

Specifications of ISL1219IUZ

• 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
• Clock Format: HH
• Clock Ic Type: RTC
• Interface Type: I2C, Serial
• Memory Configuration: 2 X 8
• Supply Voltage Range: 2.7V To 5.5V
• Digital Ic Case Style: MSOP
• Rohs Compliant: Yes
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant

best choice. These devices are available from such vendors
as Panasonic and Murata. The main specifications include
working voltage and leakage current. If the application is for
charging the capacitor from a +5V ±5% supply with a signal
diode, then the voltage on the capacitor can vary from ~4.5V
to slightly over 5.0V. A capacitor with a rated WV of 5.0V
may have a reduced lifetime if the supply voltage is slightly
high. The leakage current should be as small as possible.
For example, a Super Capacitor should be specified with
leakage of well below 1µA. A standard electrolytic capacitor
with DC leakage current in the microamps will have a
severely shortened backup time.
Below are some examples with equations to assist with
calculating backup times and required capacitance for the
ISL1219 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 23, use C
5.0V, the voltage at V
off completely. The ISL1219 is specified to operate down to
V
used to estimate the total backup time. (Equation 2 and 3):
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
dT = C
I = C
TOT
BAT
BAT
FIGURE 23. SUPERCAPACITOR CHARGING CIRCUIT
2.7V to 5.5V
BAT
is the total of the supply current of the ISL1219 (I
= 1.8V. The capacitance charge/discharge equation is
is the backup capacitance and dV is the change in
BAT
* dV/dT
BAT
* dV/I
changes with V
TOT
BAT
BAT
BAT
, using a value midway between the
to solve for backup time.
LKG
= 0.47F and V
VDD
will approach 4.7V as the diode turns
is assumed to be extremely small
1N4148
23
GND
BAT
almost linearly (see
V
BAT
DD
= 5.0V. With V
C
BAT
DD
(EQ. 2)
(EQ. 3)
BAT
LKG
=
)
ISL1219
.
two endpoints. The typical linear equation for I
is shown in Equation 4:
Using this equation to solve for the average current given 2
voltage points gives (Equation 5):
Combining with Equation 3 gives the equation for backup
time (Equation 6):
where:
Solving Equation 5 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 23 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 3 to
solve for capacitance (Equation 7):
Using the terms described above, this equation becomes
(Equation 8):
where:
Solving gives:
If the 30% tolerance is included for tolerances, then worst
case cap value would be:
C
I
T
seconds
C
C
I
BAT
BATAVG
DD
BAT
BAT
BACKUP
BAT
BAT
C
V
V
I
T
T
I
I
V
V
C
LKG
BATAVG
LKG
BAT2
BAT1
BACKUP
BACKUP
BAT2
BAT1
BAT
BAT
= 5.0V. As in Example 1, the V
= 1.031E-7*(V
= 0.70 * 35.96 = 25.2 days
= 1.3 *.784 = 1.02F
= dT*I/dV
= T
= 0 (assumed minimal)
= 0 (assumed)
= 0.47F
= 5.18 E6 * (4.387 E-7)/(2.9) = 0.784F
= 4.7V
= 1.8V
= 4.7V
= 1.8V
= 5.155E-8*(V
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
BAT
BATAVG
BAT2
) + 1.036E-7 Amps
BAT2
- V
+ V
+ I
BAT1
LKG
BAT1
BAT
) / (I
)/(V
BATAVG
) + 1.036E-7 Amps
voltage will vary from
BATAVG
BAT2
= 4.387E-7 A
BAT
– V
+ I
BAT1
vs. V
LKG
July 15, 2010
(EQ. 6)
(EQ. 8)
(EQ. 4)
(EQ. 7)
FN6314.2
(EQ. 5)
)
)
BAT