文件名称：差分阻抗定义
文件大小：17KB
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更新时间：2020-10-09 11:29:53
差分
Just when you thought you had mastered Zo, the characteristic impedance of a PCB trace, along comes a data sheet that tells you to design for a specific differential impedance. And to make things tougher, it says things like: “… since the coupling of two traces can lower the effective impedance, use 50 Ohm design rules to achieve a differential impedance of approximately 80 Ohms!” Is that confusing or what!! This article shows you what differential impedance is. But more than that, it discusses why it is, and shows you how to make the correct calculations. Single Trace: Figure 1(a) illustrates a typical, individual trace. It has a characteristic impedance, Zo, and carries a current, i. The voltage along it, at any point, is (from Ohm’s law) V = Zo*i. General case, trace pair: Figure 1(b) illustrates a pair of traces. Trace 1 has a characteristic impedance Z11, which corresponds to Zo, above, and current i1. Trace 2 is similarly defined. As we bring Trace 2 closer to Trace 1, current from Trace 2 begins to couple into Trace 1 with a proportionality constant, k. Similarly, Trace 1’s current, i1, begins to couple into Trace 2 with the same proportionality constant. The voltage on each trace, at any point, again from Ohm’s law, is: V1 = Z11 * i1 + Z11 * k * i2 Eqs. 1 V2 = Z22 * i2 + Z22 * k * i1 Now let’s define Z12 = k*Z11 and Z21 = k*Z22. Then, Eqs. 1 can be written as: V1 = Z11 * i1 + Z12 * i2 Eqs. 2 V2 = Z21 * i1 + Z22 * i2 This is the familiar pair of simultaneous equations we often see in texts. The equations can be generalized into an arbitrary number of traces, and they can be expressed in a matrix form that is familiar to many of you. Special case, differential pair: Figure 1(c) illustrates a differential pair of traces. Repeating Equations 1: V1 = Z11 * i1 + Z11 * k * i2 Eqs. 1 V2 = Z22 * i2 + Z22 * k * i1 Now, note that in a carefully designed and balanced situation, Z11 = Z22 = Zo, and i2 = -i1 This leads (with a little manipulation) to: V1 = Zo * i1 * (1-k) Eqs. 3 V2 = -Zo * i1 * (1-k) Note that V1 = -V2, which we already knew, of course, since this is a differential pair. Effective (odd mode) impedance: The voltage, V1, is referenced with respect to ground. The effective impedance of Trace 1 (when taken alone this is called the “odd mode” impedance in the case of differential pairs, or “single mode” impedance in general) is voltage divided by current, or: Zodd = V1/i1 = Zo*(1-k) And since (from above) Zo = Z11 and k = Z12/Z11, this can be rewritten as: Zodd = Z11 - Z12 which is a form also seen in many textbooks. The proper termination of this trace, to prevent reflections, is with a resister whose value is Zodd. Similarly, the odd mode impedance of Trace 2 turns out to be the same (in this special case of a balanced differential pair).