FN6690 datasheet(18/19 Pages) RENESAS

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ISL12008 I2C Real Time Clock with Battery Backup

FN6690 Rev 1.00

Page 18 of 19

Sep 26, 2008

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 18, use CBAT = 0.47F and VCC = 5V. With VCC = 5V,

the voltage at VBAT will approach 4.7V as the diode turns off

completely. The ISL12008 is specified to operate down to VBAT =

1.8V. The capacitance charge/discharge in Equation 5 is used to

estimate the total backup time as follows:

Rearranging gives Equation 6:

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 ISL12008 (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” on page 6). 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

shown in Equation 7:

Using Equation 7 to solve for the average current given 2

voltage points gives Equation 8:

Combining with Equation 6 gives the equation for backup time

in Equation 9:

where:

CBAT = 0.47F

VBAT2 = 4.7V

VBAT1 = 1.8V

ILKG = 0 (assumed minimal)

Solving Equation 8 for this example (IBATAVG = 4.387E-7A)

yields Equation 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:

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

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 6 to

solve for capacitance in Equation 12:

Using the terms previously described, Equation 12 becomes

Equation 13:

where:

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

IBATAVG = 4.387 E-7A (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:

FIGURE 18. SUPERCAPACITOR CHARGING CIRCUIT

2.7V TO 5.5V

VCC

VBAT

GND

1N4148

CBAT

I = CBAT*dV/dT

(EQ. 5)

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

(EQ. 6)

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

(EQ. 7)

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

(EQ. 8)

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

(EQ. 9)

seconds

(EQ. 10)

tBACKUP

0.47

2.9

 4.38E 7

3.107E6s

–





(EQ. 11)

CBAT

0.70 35.96

25.2



days

CBAT = dT*I/dV

(EQ. 12)

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

(EQ. 13)

(EQ. 14)

CBAT

1.3 0.784

1.02F



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