ISL1208IU8-TK【PDF 14/24 页】IC RTC/CALENDAR I2C 8-MSOP

收藏本站

您好，
买卖IC网欢迎您。
请登录
免费注册

买卖IC网

Sheet目录2001

10页 11页 12页 13页 [14页]15页 16页 17页 18页

FN8085.8

September 12, 2008

Combining with Equation 5 gives the equation for backup

time in Equation 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:

Using the terms described above, this equation becomes

Equation 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:

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

(EQ. 8)

seconds

CBAT = dT*I/dV

(EQ. 9)

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

(EQ. 10)

(EQ. 11)

BAT

1.3

0.784

1.02F

ISL1208

发布紧急采购，3分钟左右您将得到回复。

采购需求

(若只采购一条型号，填写一行即可)

*型号	*数量	厂商	批号	封装

添加更多采购

采购需求

发布成功！您可以继续发布采购。也可以进入我的后台，查看报价

发布成功！您可以继续发布采购。也可以进入我的后台，查看报价

我的联系方式