ISL1218IUZ【PDF 11/21 页】IC RTC LP BATT BACKED SRAM 8MSOP

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Sheet目录2001

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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 CBAT = 0.47F and VCC = 5.0V. With VCC =

5.0V, the voltage at VBAT will approach 4.7V as the diode

turns off completely. The ISL1218 is specified to operate

down to VBAT = 1.8V. The capacitance charge/discharge

equation is used to estimate the total backup time:

Rearranging gives

CBAT is the backup capacitance and dV is the change in

voltage from fully charged to loss of operation. Note that

ITOT is the total of the supply current of the ISL1218 (IBAT)

plus the leakage current of the capacitor and the diode, ILKG.

In these calculations, ILKG 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 IBAT changes with VBAT almost linearly (see

Typical Performance Curves). This allows us to make an

approximation of IBAT, using a value midway between the

two endpoints. The typical linear equation for IBAT vs VBAT

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:

CBAT = 0.47F

VBAT2 = 4.7V

VBAT1 = 1.8V

ILKG = 0 (assumed minimal)

Solving equation 4 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 20 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 2 to

solve for capacitance:

Using the terms described above, this equation becomes:

where:

TBACKUP = 60 days * 86,400 sec/day = 5.18 E6 sec

IBATAVG = 4.387 E-7 A (same as Example 1)

ILKG = 0 (assumed)

VBAT2 = 4.7V

VBAT1 = 1.8V

Solving gives:

CBAT = 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:

CBAT = 1.3 *.784 = 1.02F

FIGURE 20. SUPERCAPACITOR CHARGING CIRCUIT

2.7V to 5.5V

VCC

VBAT

GND

1N4148

CBAT

I = CBAT * dV/dT

(EQ. 1)

dT = CBAT * dV/ITOT to solve for backup time.

(EQ. 2)

IBAT = 1.031E-7*(VBAT) + 1.036E-7 Amps

(EQ. 3)

IBATAVG = 5.155E-8*(VBAT2 + VBAT1) + 1.036E-7 Amps

(EQ. 4)

TBACKUP = CBAT * (VBAT2 - VBAT1) / (IBATAVG + ILKG)

(EQ. 5)

seconds

CBAT = dT*I/dV

(EQ. 6)

CBAT = TBACKUP * (IBATAVG + ILKG)/(VBAT2 – VBAT1)

(EQ. 7)

ISL1218

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