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參數(shù)資料
| 型號(hào)： | ISL12082IB8Z |
| 廠商： | Intersil |
| 文件頁數(shù)： | 17/26頁 |
| 文件大?。?/td> | 0K |
| 描述： | IC RTC I2C LO-POWER 8-SOIC |
| 標(biāo)準(zhǔn)包裝： | 980 |
| 類型： | 時(shí)鐘/日歷 |
| 特點(diǎn)： | 警報(bào)器，閏年，監(jiān)視計(jì)時(shí)器 |
| 時(shí)間格式： | HH:MM:SS:hh（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 |
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24
FN6731.3
November 24, 2008
margin should be included if periods of very warm
temperature operation are expected.
Example 1. Calculating Backup Time Given
Voltages and Capacitor Value
VCC = 5.0V, the voltage at VBAT will approach 4.7V as the
diode turns off completely. The ISL12082 is specified to
operate down to VBAT = 1.8V. The capacitance
charge/discharge equation is used to estimate the total
backup time as shown in Equation 10:
Rearranging gives Equation 11.
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 ISL12082 (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 7). This allows us to
make an approximation of IBAT, using a value midway
between the two endpoints. The typical linear equation for
Using this equation to solve for the average current given 2
voltage points gives Equation 13:
Combining with Equation 11 gives the equation for backup
time in Equation 14:
where:
CBAT = 0.47F
VBAT2 = 4.7V
VBAT1 = 1.8V
ILKG = 0 (assumed minimal)
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:
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 11 to
solve for capacitance in Equation 16:
Using the terms previously, this Equation 16 becomes
Equation 17:
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 as shown in
Equation 18.
FIGURE 20. SUPERCAPACITOR CHARGING CIRCUIT
2.7V TO 5.5V
VCC
VBAT
GND
1N4148
CBAT
I = CBAT*dV/dT
(EQ. 10)
dT = CBAT*dV/ITOT to solve for backup time.
(EQ. 11)
IBAT = 1.031E-7*(VBAT) + 1.036E-7A
(EQ. 12)
IBATAVG = 5.155E-8*(VBAT2 + VBAT1) + 1.036E-7A
(EQ. 13)
TBACKUP = CBAT*(VBAT2 - VBAT1)/(IBATAVG + ILKG)
(EQ. 14)
seconds
(EQ. 15)
C
BAT
0.70
35.96
25.2
=
days
=
CBAT = dT*I/dV
(EQ. 16)
CBAT = TBACKUP*(IBATAVG + ILKG)/(VBAT2 - VBAT1)
(EQ. 17)
CBAT = 1.3*0.784 = 1.02F
(EQ. 18)
ISL12082
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