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| 型号： | LTC4311IDC#TRMPBF |
| 厂商： | Linear Technology |
| 文件页数： | 12/12页 |
| 文件大小： | 0K |
| 描述： | IC ACCELERATOR SMBUS DL 6-DFN |
| 产品培训模块： | LTC4310 - Hot-Swappable I²C Isolators |
| 标准包装： | 1 |
| 类型： | SMBus 加速器 |
| Tx/Rx类型： | I²C Logic |
| 电源电压： | 1.6 V ~ 5.5 V |
| 电流 - 电源： | 300µA |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 6-WFDFN 裸露焊盘 |
| 供应商设备封装： | 6-DFN-EP（2x2） |
| 包装： | 标准包装 |
| 产品目录页面： | 1355 (CN2011-ZH PDF) |
| 其它名称： | LTC4311IDC#TRMPBFDKR |
LTC4311
9
4311fa
APPLICATIONS INFORMATION
This is the lowest resistor value that may be chosen and
still meet VOL. Next calculate the largest value of RP that
will satisfy SR, the minimum slew rate requirement. Us-
ing VOL = 0.4V and SR = 0.5V/μs calculate the value of
RP with equation 2.
R
P =
3.3V
0.4V
600pF
0.5V / s
R
P = 9.667k
(11)
This is approximately the largest value of RP that will satisfy
the minimum slew rate requirement. Since RP is larger
than 2.175k the VOL will be below 0.4V, and the slew rate
will actually be faster than calculated. Choosing RP = 10k ,
VOL and SR are recalculated.
V
OL =
300
Ω3.3V
300
Ω+10kΩ
=96mV
SR
=
3.3V – 96mV
10k
Ω600pF
=0.534V / s
(12)
The rise and fall times need to be veried using equations
3 and 4.
t
r = –10kΩ 600pF
In
2.31V – 3.3V – 10k
Ω 2.5mA
0.99V – 3.3V – 10k
Ω 2.5mA
= 0.297s
(13)
t
f = 291Ω 600pF
In
2.31
3.3V
(10kΩ+ 300Ω)– 300Ω
0.99V
3.3V
(10kΩ+ 300Ω)– 300Ω
= 0.158s
(14)
Both the rise and fall times meet the 0.3μs I2C requirement
and the VOL is satised, while meeting the minimum slew
rate requirement, so RP is chosen to be 10k.
If tr is not met, RP should be decreased and if tf is not met
then RP should be increased.
SMBus Design Example
Given the following conditions and requirements for a low
power SMBus system:
V
CC = 3.3V NOMINAL
V
OL = 0.4V MAXIMUM
C
BUS = 400pF
V
ILMAX = 0.8V, VIHMIN = 2.1V
t
r = 1s MAXIMUM, t f = 0.3s MAXIMUM
(15)
If an RS of 100Ω is used and the max RON of the driver is
200Ω, then RL = 200Ω + 100Ω = 300Ω. Use equation 1
to nd the required RP to meet VOL.
R
P =
300
Ω(3.3V –0.4V)
0.4V
R
P = 2.175k
(16)
Calculate Maximum RP from equation 2.
R
P =
3.3V – 0.4V
400pF
0.5V / s
R
P = 14.5k
(17)
Choose RP = 13k and recalculate VOL and SR.
V
OL =
300
Ω3.3V
300
Ω+ 13kΩ
= 74mV
SR
=
3.3V – 74mV
13k
Ω 400pF
= 0.62V / s
(18)
The rise and fall times need to be veried using equations
5 to 8.
t
1 = –13kΩ 400pF
ln
0.9V–3.3V
0.8V–0.15V–3.3V
= 0.515s
(19)
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