ISL1208
Combining with Equation 5 gives the equation for backup
time in Equation 8:
TBACKUP = CBAT * (VBAT2 - VBAT1) / (IBATAVG + ILKG)
seconds
(EQ. 8)
where:
CBAT = 0.47F
VBAT2 = 4.7V
VBAT1 = 1.8V
ILKG = 0 (assumed minimal)
Solving Equation 7 for this example, IBATAVG = 4.387E-7 A
TBACKUP = 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:
CBAT = 0.70 * 35.96 = 25.2 days
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
VCC = 5.0V. As in Example 1, the VBAT voltage will vary from
4.7V down to 1.8V. We will need to rearrange Equation 5 to
solve for capacitance in Equation 9:
CBAT = dT*I/dV
(EQ. 9)
Using the terms described above, this equation becomes
Equation 10:
CBAT = TBACKUP * (IBATAVG + ILKG)/(VBAT2 â€“ VBAT1)
(EQ. 10)
where:
TBACKUP = 60 days * 86,400 sec/day = 5.18 E6 seconds
IBATAVG = 4.387 E-7 A (same as Example 1)
ILKG = 0 (assumed)
VBAT2 = 4.7V
VBAT1 = 1.8VSolving gives
CBAT = 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:
CBAT = 1.3 Ã— 0.784 = 1.02F
(EQ. 11)
21
FN8085.8
September 12, 2008