19
FN6313.0
June 22, 2006
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 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 ISL1218 is specified to operate
down to V
BAT
= 1.8V. The capacitance charge/discharge
equation is used to estimate the total backup time:
Rearranging gives
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 ISL1218 (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:
Using this equation to solve for the average current given 2
voltage points gives:
Combining with Equation 2 gives the equation for backup
time:
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:
Using the terms described above, this equation becomes:
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
FIGURE 20. SUPERCAPACITOR CHARGING CIRCUIT
2.7V to 5.5V
V
CC
V
BAT
GND
1N4148
C
BAT
I = C
BAT
* dV/dT
(EQ. 1)
dT = C
BAT
* dV/I
TOT
to solve for backup time.
(EQ. 2)
I
BAT
= 1.031E-7*(V
BAT
) + 1.036E-7 Amps
(EQ. 3)
I
BATAVG
= 5.155E-8*(V
BAT2
+ V
BAT1
) + 1.036E-7 Amps
(EQ. 4)
T
BACKUP
= C
BAT
* (V
BAT2
- V
BAT1
) / (I
BATAVG
+ I
LKG
)
(EQ. 5)
seconds
C
BAT
= dT*I/dV
(EQ. 6)
C
BAT
= T
BACKUP
* (I
BATAVG
+ I
LKG
)/(V
BAT2
– V
BAT1
)
(EQ. 7)
ISL1218
