RoHS Compliant in Accordance with EU Directive 2011/65/EU
- Lead-Free Termination Finish
- Exemption 7(c)-I, Electrical and electronic components containing
lead [Pb] in glass
Applications
Telecom Infrastructure
Optical Networking
Wireless Networks
Edge Routers
Internet/Network Security
Storage Area Networks
Network Attached Storage
Switches
RAID Controllers
Description
767 Series Resistor Networks are single packaged devices containing an array of homogeneous resistor elements.
CTS network designs provide a smaller circuit footprint, excellent reliability, improved TCR tracking and resistor
tolerance matching; while helping to save placement costs by reducing application component count.
Ordering Information
Model
767
Number
of Pins
16
Schematic
3
Resistor
Code
103
Resistor
Tolerance
G
RoHS
Compliant
P
Packaging
TR13
Code
14
16
Pin Count
14 Pins
16 Pins
Code Resistor Value *
10k ohm
103
See Addendum for
Standard EIA Values
Code
P
Compliance
RoHS
Packing
Code
Slide Pack
Blank
TR13 Tape & Reel, 13"
Schematic Type
Code
Bussed Circuit
1
Isolated Circuit
3
5 Dual Terminator Circuit
Ladder Circuit
7
Notes:
1. No dashes or spaces to appear in part number.
Schematic Types 1 & 3
Tolerance
Code
±5%
J
±2% [std]
G
±1%
F
±0.5%
D
Schematic Type 5
Tolerance
Code
±2%
Blank
Schematic Type 7
Tolerance
Code
±1 LSB
F
±0.5 LSB
D
[LSB = Least Significant Bit]
Not all performance combinations and frequencies may be available.
Contact your local CTS Representative or CTS Customer Service for availability.
This product is specified for use only in standard commercial applications. Supplier disclaims all express and implied warranties and liability in connection with any use of this
product in any non-commercial applications or in any application that may expose the product to conditions that are outside of the tolerances provided in its specification.
C++ 属于面向对象的编程语言,OOP的思想不必多说,特别对于复杂的软件工程来说,利用OOP绝对是事半功倍,相对于传统的C来说; 当然用C来写单片机程序无可厚非,已经延续了一个传统,从大学时学的开始到工作岗位,好多人都是一直用C来做,但是既然Keil支持C++编译, 可以用C++来编写你的代码,可以利用高级语言来结构化,清晰化你的程序,为嘛不用呢!哈哈,个人看法!下面进入正题: C+...[详细]