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我需要对CAN网络中的120欧姆电阻有一些清晰性。所有节点之间的有效电阻是否应为120欧姆或60欧姆?如果电阻是ECU的一部分,或者连接后ECU应该是CAN总线的一部分?请告知。
以上来自于百度翻译 以下为原文 I need some clarity on the 120 ohm resistance in the can network. Whether the effective resistance between all the nodes should be 120 ohm or 60 ohm? Should the resistance be part of the ECU or after connecting all the ECUs should be part of the CAN lines? Please advise. |
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7个回答
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总线两端各有一个120欧姆电阻。你如何活动取决于你的设计。这个问题不是有效的直流电阻,而是交流网络的阻抗匹配。
以上来自于百度翻译 以下为原文 One 120 ohm resistor on each end of the bus. How you active that depends on you design. The issue is not the effective dc resistance, but impedance matching for the ac network. |
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谢谢你的邮件。请你详细解释一下好吗?什么是交流网络的阻抗匹配。我如何确保它是正确的,因为最近我努力使CAN网络上升,但是当我在CAN -H和CAN-L之间连接了60欧姆时,它就开始工作了,我不知道它工作的确切原因。请告知。
以上来自于百度翻译 以下为原文 Thank you for the mail. Can you please explain me in detail? What is impedance matching of the ac network. How do i make sure it is correct because recently i struggled a lot to make CAN network up, but when i connected 60 ohm between CAN -H and CAN-L it started working, i don't know the exact reason why it worked. Please advise. |
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两个120欧姆电阻与一个60欧姆相同。如果电缆长度比较长,那么电阻必须位于电缆的两端。这是为了防止反射。当然,如果没有电缆,它都是一个小板,电阻器的位置并不重要,但你确实需要整个60欧姆电阻。
以上来自于百度翻译 以下为原文 Two 120 Ohm resistors is the same as one 60 Ohm. If you have relatively long cable, then the resistors must be located on the ends of the cable. This is to prevent reflections. Of course, if there's no cable and it all is on a small board, the position of the resistors doesn't matter, but you do need the overall 60 Ohm resistance. |
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这个(60欧姆)与阻抗匹配是相同的吗?
以上来自于百度翻译 以下为原文 Is this (60 ohm) is the same as impedance matching or it is something else? |
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穿过导线的信号与任何其它信号相连。它具有交流和直流分量。在某些方面,行为就像一个无线电波。穿过器皿的信号可以反映出来。这可能导致原始信号的回波通过导线反射。它们可以作为数据出现在网络中的节点上。钻头可以拉伸。或插入错误位。这会破坏真实数据。电阻器A称为终端电阻器,因为终止传输线。这吸收端部的信号,使反射最小化。传输线和阻抗匹配是整个过程。还有护目镜。你增加了电阻器改善了情况。但这并不是完美的解决方案。对于短的长度和低波特率,不良影响不足以影响性能。运行时间越长或波特率越高,它就越重要。
以上来自于百度翻译 以下为原文 The signal traving through the wire is link any other signal. It has ac and dc components. And in some ways behaves like a radio wave. Signals traving through the ware can reflect. This can cause echos of the original signal to reflect through the wire. They can appear as data to nodes in the network. Bit can be Stretched.. or false bits inserted. This corrupts the true data. The resisters a called termination resistors because the terminate the transmission line. This absorbs the signals at the ends, minimizing reflections. Transmission lines and impedance matching are entire courses. And goggle-able. Your addition of the resistor improved the situation. But was not nessacarily the perfect solution. For short lengths and low baud rates the bad effects are not enough to affect the performance. The longer the run or higher the baud rate the more it matters. |
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由于CAN收发器只“创造”主导信号电平,必须有一种建立隐性电平的方法,这基本上是由一些漏电流引起的,但是终端电阻对建立隐性电平有很大的贡献。我有时有(小)CAN网络工作,没有任何终端电阻器,但通常至少1Ω电阻需要120欧姆稳定运行。
以上来自于百度翻译 以下为原文 As the CAN transceivers only "create" the dominant signal level, there has to be a means to establish the recessive level. This basically happens by some leakage currents, but the termination resistors significantly contribute in establishing the recessive level. I've sometimes had (small) CAN networks work without any termination resistor, but usually at least 1 120 Ohms resistor is required for stable operation. |
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