ISL1208_06 Datasheet, PDF (19/21 Pages) Intersil Corporation – Low Power RTC with Battery Backed SRAM

ISL1208
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 20. SUPERCAPACITOR CHARGING CIRCUIT
In Figure 20, use C_BAT = 0.47F and V_CC = 5.0V. With V_CC =
5.0V, the voltage at V_BAT will approach 4.7V as the diode
turns off completely. The ISL1208 is specified to operate
down to V_BAT = 1.8V. The capacitance charge/discharge
equation is used to estimate the total backup time:
I = C_BAT * dV/dT
(EQ. 1)
Rearranging gives
dT = C_BAT * dV/I_TOT to solve for backup time.
(EQ. 2)
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 ISL1208 (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). 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:
I_BAT = 1.031E-7*(V_BAT) + 1.036E-7 Amps
(EQ. 3)
Using this equation to solve for the average current given 2
voltage points gives:
I_BATAVG = 5.155E-8*(V_BAT2 + V_BAT1) + 1.036E-7 Amps
(EQ. 4)
Combining with Equation 2 gives the equation for backup
time:
T_BACKUP = C_BAT * (V_BAT2 - V_BAT1) / (I_BATAVG + I_LKG)
seconds
(EQ. 5)
where
C_BAT = 0.47F
V_BAT2 = 4.7V
V_BAT1 = 1.8V
I_LKG = 0 (assumed minimal)
Solving equation 4 for this example, I_BATAVG = 4.387E-7 A
T_BACKUP = 0.47 * (2.9) / 4.38E-7 = 3.107E6 sec
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_BAT = 0.70 * 35.96 = 25.2 days
Example 2. Calculating a capacitor value for a
given backup time
Referring to Figure 20 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 2 to
solve for capacitance:
C_BAT = dT*I/dV
(EQ. 6)
Using the terms described above, this equation becomes:
C_BAT = T_BACKUP * (I_BATAVG + I_LKG)/(V_BAT2 – V_BAT1)
(EQ. 7)
where
T_BACKUP = 60 days * 86,400 sec/day = 5.18 E6 sec
I_BATAVG = 4.387 E-7 A (same as Example 1)
I_LKG = 0 (assumed)
V_BAT2 = 4.7V
V_BAT1 = 1.8V
Solving 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 cap value would be
C_BAT = 1.3 *.784 = 1.02F
19
FN8085.5
August 23, 2006