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參數(shù)資料
| 型號： | ISL12008IB8Z |
| 廠商： | Intersil |
| 文件頁數(shù)： | 10/19頁 |
| 文件大小： | 0K |
| 描述： | IC RTC I2C LO-POWER 8-SOIC |
| 標(biāo)準(zhǔn)包裝： | 980 |
| 類型： | 時鐘/日歷 |
| 特點： | 警報器，閏年 |
| 時間格式： | HH:MM:SS:hh（24 小時） |
| 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 |
18
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
the voltage at VBAT will approach 4.7V as the diode turns off
completely. The ISL12008 is specified to operate down to
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
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)
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)
t
BACKUP
0.47
2.9
() 4.38E 7
3.107E6s
=
–
=
(EQ. 11)
C
BAT
0.70
35.96
25.2
=
days
=
CBAT = dT*I/dV
(EQ. 12)
CBAT = tBACKUP*(IBATAVG + ILKG)/(VBAT2 – VBAT1)
(EQ. 13)
(EQ. 14)
C
BAT
1.3
0.784
1.02F
=
=
ISL12008
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