ISL12008 Datasheet, PDF (18/19 Pages) Intersil Corporation – I2C Real Time Clock with Battery Backup

ISL12008
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
1N4148
2.7V TO 5.5V
V_CC
V_BAT
GND
C_BAT
FIGURE 18. SUPERCAPACITOR CHARGING CIRCUIT
In Figure 18, use C_BAT = 0.47F and V_CC = 5V. With V_CC = 5V,
the voltage at V_BAT will approach 4.7V as the diode turns off
completely. The ISL12008 is specified to operate down to
V_BAT = 1.8V. The capacitance charge/discharge in Equation 5
is used to estimate the total backup time as follows:
I = C_BAT*dV/dT
(EQ. 5)
Rearranging gives Equation 6:
dT = C_BAT*dV/I_TOT to solve for backup time.
(EQ. 6)
C_BAT is the backup capacitance and dV is the change in
voltage from fully charged to loss of operation. Note that
I_TOT is the total of the supply current of the ISL12008 (I_BAT)
plus the leakage current of the capacitor and the diode, I_LKG.
In these calculations, I_LKG is assumed to be extremely small
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_BAT changes with V_BAT almost linearly (see
“Typical Performance Curves” on page 6). This allows us to
make an approximation of I_BAT, using a value midway
between the two endpoints. The typical linear equation for
I_BAT vs V_BAT is shown in Equation 7:
I_BAT = 1.031E-7*(V_BAT) + 1.036E-7A
(EQ. 7)
Using Equation 7 to solve for the average current given 2
voltage points gives Equation 8:
I_BATAVG = 5.155E-8*(V_BAT2 + V_BAT1) + 1.036E-7A
(EQ. 8)
Combining with Equation 6 gives the equation for backup
time in Equation 9:
t_BACKUP = C_BAT*(V_BAT2 - V_BAT1) / (I_BATAVG + I_LKG)
seconds
(EQ. 9)
where:
C_BAT = 0.47F
V_BAT2 = 4.7V
V_BAT1 = 1.8V
I_LKG = 0 (assumed minimal)
Solving Equation 8 for this example (I_BATAVG = 4.387E-7A)
yields Equation 10:
t_BACKUP = 0.47 • (2.9) ⁄ 4.38E – 7 = 3.107E6s
(EQ. 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:
C_BAT = 0.70 • 35.96= 25.2 days
(EQ. 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_CC = 5.0V. As in Example 1, the V_BAT voltage will vary from
4.7V down to 1.8V. We will need to rearrange Equation 6 to
solve for capacitance in Equation 12:
C_BAT = dT*I/dV
(EQ. 12)
Using the terms previously described, Equation 12 becomes
Equation 13:
C_BAT = t_BACKUP*(I_BATAVG + I_LKG)/(V_BAT2 – V_BAT1)
(EQ. 13)
where:
t_BACKUP = 60 days*86,400 sec/day = 5.18 E6 seconds
I_BATAVG = 4.387 E-7A (same as Example 1)
I_LKG = 0 (assumed)
V_BAT2 = 4.7V
V_BAT1 = 1.8VSolving gives
C_BAT = 5.18 E6*(4.387 E-7)/(2.9) = 0.784F
If the 30% tolerance is included for tolerances, then worst
case capacitor value would be:
C_BAT = 1.3 • 0.784 = 1.02F
(EQ. 14)
18
FN6690.1
September 26, 2008