2.14 Pulling Energy from a Tapered Transmission Line (TTL)
In Section 2.12, we discussed wave action where energy flowed from an ideal voltage source through a short transmission line and was dissipated in a resistor. An increment of energy was transferred for each round trip of the wave in the transmission line. In that discussion, reflections occurred at abrupt changes to the characteristic impedance. In the case of a TTL, the reflection process still takes place but on a continuous basis. A linear change in characteristic impedance results in a continuous reflection process. Of course, there is a reflection at the edges of a board, but this occurs at a much later time. The phenomena we consider takes place before the wave action reaches the edge of the board.
To get some idea of how this TTL works, consider the ideal case where the capacitance of the tapered line is charged to a voltage V. Assume an ideal logic switch and a load of 5 ohm. When the switch closes, if the tapered entry point looks like 50 ohms, the voltage drops to 9.1% of V. We need to find out how long it takes for the voltage at the load to reach 95% of V.
At the moment of switch closure, the wave that propagates into the TTL is − 9.909 V. As the wave progresses into the tapered line, the continuous nature of the reflection reduces the magnitude of this wave. Stated another way, as the wave propagates radially, the wave amplitude decreases and the voltage across the leading edge rises. We are interested in the time t it takes for the wave amplitude to diminish so that the voltage across the planes is 95% of V. To obtain this time, the rate of energy transfer from the capacitance to the wave as the wave progresses out on the tapered line must be calculated.
If the wave amplitude progresses until the wave amplitude is − 0.95 V, the energy supplied to the wave per unit time depends on the capacitance at that point in its travel.
The energy stored in a capacitor between these two voltages is
2.18 
Using Equation 2.12 for capacitance, the increment of energy taken from an annular ring at a radius r in a distance dr is given by
The wave in the capacitor travels at the speed of light divided by the square root of the relative dielectric constant or v = c/( ε R)1/2. The power W that transfers to the wave at a distance r is given by dE/dt. Dividing both sides of Equation 2.19 by dt and noting that dr/dt is equal to the velocity v = c/( ε 1/2), we can write the power as
2.20 
The power required by the load is V2/R. The value of r where the power supplied to the wave equals the power required by the load is
2.21 
The time it takes a wave to travel this distance is given by r/v or
where c is the velocity of light and ε = 8.85 × 10−12 F/m.
The time t in Equation 2.22 is the time for the wave to travel out to get energy. The time required to obtain the energy and return to the entry point is double this value or
2.23 
If h is 10 mil and R = 5 ohm, then the time t is 0.4 ns. This time is too long for logic operating at 1 GHz. This delay is shown in Figure 2.10b.
The advantage of using a high dielectric constant is that the wave action stays closer to the point of demand. If several devices take energy from the planes then a higher dielectric constant can reduce cross talk. It may also be useful on circuit boards where the overall dimensions are small.
Vias and some trace length are needed to provide connections to the ground and power planes. The characteristic impedance of this connection is apt to be greater than 50 ohm. To provide a low impedance connection, assume that the connection was somehow coaxial. To achieve a low impedance connection, the ratio of conductor diameter to conductor spacing would have to be at least 100 : 1. This type of construction is not practical on a typical circuit board. Even if a low characteristic impedance connection could be constructed, how long would the connecting traces be? This argument shows that there are significant physical limitations to moving energy rapidly in or out of a ground/power plane structure.
The waves that propagate between conducting planes reduce in amplitude with increasing radius r. These waves reflect at the board edges and at any discontinuities along the way.
If energy is taken from the two planes near the board edge then the amplitude of the wave at the board edge will be high. Since the reflection at an open edge doubles the voltage, this increases the radiation from the edge. It is good practice to make rapid demands for energy at points that are removed from the perimeter of the board.