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设计资源库

单端时钟驱动多路负载匹配电阻的计算

A single end clock drives multiple seperate loads

 

Figure 1 Multiple loads schematics

How the value of series termination resistor, Rs can be calculated in a system where a single clock output is used to drive multiple separate loads?
Consider point A in the figure. To its right, point A sees a resistance of Zo. To its left, point A
observes a resistance consisting of:
Rs + (Rd || (Rs + Zo) || (Rs +Zo))                               Equation  1
Equating both sides of point A yields the following formula:
Zo = Rs + (Rd || (Rs + Zo) || (Rs +Zo))                       Equation 2
If we substitute Zo = 50 and Rd =25 into Equation 2 and solve for Rs, we will find that Rs is
equal to 38.4.

Note that Equation 2 only covers a system with 3 separate loads. Use Equation
3 for a system with N parallel loads.
Zo = Rs + (Rd || Rx)                                                    Equation 3
where Rx = (Rs +Zo) / (N-1)

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何宾

数字系统设计专家,EDA技术领域畅销书作者。

从业:15年 领域:FPGA,EDA 器件:XiLinx,STC 不对路,寻找其他 诊断专家》
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