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Differential-to-common-mode conversion差模转共模转换损耗

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发表于 2016-12-8 13:14:04 | 显示全部楼层 |阅读模式

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Differential-to-common-mode conversion

by Dr. Howard Johnson. First printed in EDN magazine, October 17, 2002

Any unbalanced circuit element within an otherwise well-balanced transmission channel creates a region of partial coupling between the differential and common modes of transmission at that point. The coupling can translate part of a perfectly good differential signal into a common-mode signal, or vice versa.
Such differential-to-common-mode-conversion problems frequently arise in the design of LAN adapters. For example, assume the output winding of the transformer in Figure 1 has equal capacitances (C1) connected from point A to ground and from point B to ground. If the capacitances are exactly equal (and the cable and transformer perfectly symmetrical), the differential signal on the cable forces equal but opposite currents through these two capacitances. In the product chassis, the two currents perfectly cancel. The perfect cancellation implies that no current circulates between the twisted-pair cable and the surrounding chassis. In practice, however, one capacitance is always a little larger than the other.
Assume that capacitor C2 in Figure 1 represents the small amount of physical imbalance (2 pF) between the parasitic capacitances associated with circuit nodes A and B. Calculate the current flowing through this capacitor, see where it flows, and then decide whether it causes any problems.
Using Ethernet 10BaseT for this example, the drive amplitude is approximately 2V p-p on each wire, at a switching time of 25 nsec. The current forced through capacitor C2 is:
difftocommonEqu.gif
This current flows through capacitor C2 to the product chassis. It couples from the product chassis to the earth (either through the green-wire ground or through the capacitance between the product chassis and the earth). From the earth it couples capacitively to the cabling, along which it travels as a common-mode signal riding on the twisted-pair cable back to the transformer, completing the loop.
A balanced load comprising equal-valued capacitors from A to ground and from B to ground does not generate any common-mode currents, because the currents through the two capacitors cancel, leaving nothing to exit the system in common-mode format. In this example, it is the imbalance in capacitive loading that generates the common-mode current.
A capacitive imbalance even as small as 2 pF causes a big problem in this example, because the existence of 160 μA of high-frequency common-mode current on an exposed cable easily violates US and international emissions regulations.

difftocommonFig1.jpg

发表于 2018-5-9 08:40:58 | 显示全部楼层
Howard Johnson医生。2002年10月17日首次刊登在EDN杂志上
在另一个平衡良好的传输信道内的任何不平衡电路元件在该点处产生差分和公共传输模式之间的部分耦合区域。耦合可以将一个完全好的差分信号的一部分转换成共模信号,反之亦然。
这样的差分到共模转换问题在局域网适配器的设计中经常出现。例如,假设图1中变压器的输出绕组具有从A点接地和从B点接地的相等电容(C1)。如果电容完全相等(电缆和变压器完全对称),则电缆上的差分信号通过这两个电容产生相等但相反的电流。在产品底盘中,两个电流完全抵消。完美的取消意味着没有电流在双绞线电缆和周围的底盘之间循环。然而,在实践中,一个电容总是比另一个大一点。
假设图1中的电容器C2代表与电路节点A和B相关联的寄生电容之间的少量物理不平衡(2 PF),计算流过该电容器的电流,看它流过的位置,然后决定它是否引起任何问题。
在这个示例中,使用以太网10BaseT,在每根导线上的驱动幅度大约为2V P P,在25毫秒的切换时间。强制通过电容器C2的电流是:


该电流通过电容器C2流向产品底盘。它从产品底盘耦合到地球(通过绿线接地或通过产品底盘和地球之间的电容)。从地球上,它电容性地耦合到电缆,沿着它作为共模信号行进,在双绞线电缆上回到变压器,完成环路。
包括等值电容器的从A到地和从B到地的均衡负载不会产生任何共模电流,因为通过两个电容器的电流被取消,没有留下任何以共模格式退出系统的电流。在这个例子中,电容负载的不平衡产生共模电流。
在这个例子中,电容不平衡甚至小到2 pF会导致一个大问题,因为暴露在电缆上的高频共模电流的160μA的存在很容易违反美国和国际排放法规。

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