![]() ![]() It is during this transition time, as the edge passes by, that the voltage is changing and current flows through the initial capacitor. As the signal is launched into the transmission line, the voltage across the signal- and return-path conductors ramps up. The only way current flows through a capacitor is if the voltage across it changes. As we described in Chapter 5, “The Physical Basis of Capacitance,” if the voltage across the initial capacitor is constant, there will be no current flow through this capacitor. As the signal launches into the line, it sees the first capacitor. The best way of thinking about it is by going back to the zeroth-order model, which describes the line as a bunch of tiny capacitors. When do we see the current come out the return path? Does it take 2 seconds-1 second to go down and 1 second to come back? What would happen then if the far end were really open? If there is insulating dielectric material between the signal and return conductors, how could the current possibly get from the signal to the return conductor, except at the far end? But how long does this take? The current flow in a transmission line is very subtle. If current travels in loops and must return to the source, eventually we’d expect to see the current travel to the end of the line and flow back down the return path. When does the current actually exit the return path? Figure 7-18 Current injected into the signal path of a transmission line and the current distribution after a long time. ![]()
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