ISL12008IB8Z【PDF 10/19 页】IC RTC I2C LO-POWER 8-SOIC

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Sheet目录2000

6页 7页 8页 9页 [10页]11页 12页 13页 14页

FN6690.1

September 26, 2008

Following are some examples with equations to assist with

calculating backup times and required capacitance for the

ISL12008 device. The backup supply current plays a major

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

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)

BACKUP

0.47

2.9

() 4.38E 7

3.107E6s

–

(EQ. 11)

BAT

0.70

35.96

25.2

days

CBAT = dT*I/dV

(EQ. 12)

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

(EQ. 13)

(EQ. 14)

BAT

1.3

0.784

1.02F

ISL12008

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