Initial Condition

A symmetric end termination biases the line halfway.

The PCB (printed-circuit-board) transmission line in Figure 1a lacks an end termination. The next circuit has an end termination popularly called a split termination, or, in the special case where R1=R2, a symmetric end termination. The third diagram is the same as (b), except that it shows imaginary current sources replacing the action of switch S2. Let's analyze the circuits.

If you leave switch S1 closed for a long time, the line comes to rest in a state with precisely 0 V at all points. The transmission line in Figure 1b behaves similarly. With S2 closed, it also comes to rest in a state with 0 V at all points. The voltages in the two situations are the same, but the currents differ. In the first case, the line at rest carries no current. In the second case, the end termination supplies a substantial current as long as you hold the line in a low state. To make the numbers easy, assume 2-V logic, a perfect switch at the source, and values of 100 Ω for both R1 and R2. Those values produce a steady-state current of 20 mA for Figure 1b.

At time zero, with both lines in their respective steady-state conditions, open both switches. In the case of Figure 1a, just before you open the switch, no current flows through it. Opening S1 therefore changes nothing; it has no effect on the circuit. Opening S2 in Figure 1b has a different effect. In the steady-state condition, 20 mA spills continuously through the switch. When you interrupt that state of events by opening S2, the current at the left end of the line changes from 20 mA to 0A. You can emulate that effect with a superposition of two linear-current sources, IB and IC, which connect (Figure 1c).

Current source IB replaces the 20 mA of steady-state current flowing through S2 in Figure 1b. It sets the initial conditions before your switching event, and it perpetually sinks 20 mA. At time zero, a 20-mA step of current from source IC cancels the current from source IB, bringing the net current to 0A. The combination of two sources duplicates the conditions at the left of Figure 1b the moment S2 opens. The linear-current-source model clarifies the actions that occur at time zero. Directing a positive step of 20 mA into the line must create a positive-step-voltage waveform moving to the right with an amplitude of 20 mA×50Ω=1V. In a 2-V system, that scenario makes a half-sized step.

If your goal is to inject a total voltage step of 2V into the line, making a full-sized step, which initial state do you prefer? Starting with Figure 1a—that is, with no termination—you must do all the work with the top half of your totem-pole driver, sourcing a full 40 mA to create a full-sized signal. Most drivers can't source that much current.

On the other hand, a circuit with a symmetric end termination enjoys the benefit of sinking 20 mA the entire time it holds low. When the bottom half of the totem-pole driver lets go, the line voltage at the source automatically jumps up halfway. The top half of the totem-pole driver then needs only to source the other half of the current (20 mA), bringing the line up to full voltage. A symmetric end termination biases the line at a halfway voltage, so that the driver need source or sink only enough current to swing the line halfway either direction. That's why I like it.