C
參數(shù)資料
型號: ISL1208IB8ZR5291
廠商: Intersil
文件頁數(shù): 14/24頁
文件大?。?/td> 0K
描述: IC RTC LP BATT BACKED SRAM 8SOIC
標(biāo)準(zhǔn)包裝: 100
類型: 時(shí)鐘/日歷
特點(diǎn): 警報(bào)器,閏年,SRAM
存儲容量: 2B
時(shí)間格式: HH:MM:SS(12/24 小時(shí))
數(shù)據(jù)格式: YY-MM-DD-dd
接口: I²C,2 線串口
電源電壓: 2.7 V ~ 5.5 V
電壓 - 電源,電池: 1.8 V ~ 5.5 V
工作溫度: -40°C ~ 85°C
安裝類型: 表面貼裝
封裝/外殼: 8-SOIC(0.154",3.90mm 寬)
供應(yīng)商設(shè)備封裝: 8-SOIC
包裝: 管件
21
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)
C
BAT
1.3
0.784
1.02F
=
×
=
ISL1208
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