Two men of average height meet in a crowded bar. They grasp hands. One man follows through with a vigorous up-and-down motion. The other hangs on, providing some resistance to the motion. Their meeting produces a normal, hearty shake about 32 in. above floor level. Now, suppose the second man were Yao Ming of the Houston Rockets. At 7 feet, 6 in., he ranks among the world's tallest basketball players. The same shake would occur, but slightly higher above the floor. On the other hand, if the second man were Danny DeVito, the diminutive Hollywood actor, the first must reach down to his level. Any handshake accommodates the characteristics of both men.
When a logic driver meets its load, it behaves in a similar way. Begin with the driver in Figure 1. Connect its output to a static load. Exercise the driver by swinging its output high and low. At the driver's output, you'll see something like the waveforms in Figure 1.
Figure 1 illustrates a Virtex-4 HSTL-I driver. The load comprises a single resistor of 50 Ω leading to an adjustable terminating voltage, VT, all simulated with Hyperlynx. It hardly matters what type of totem-pole output you choose or how exactly you load it—all totem-pole outputs exhibit the same general behavior. You can learn a lot about drivers looking at the details of this family of curves.
Start with the terminating voltage set to 0 V. This value produces the red (lowest) waveform in Figure 1. Driving high, IC1 fails to attain its rated VOH level. Why? At this setting, the load requires more current than the driver can provide. You can check that assumption by calculating the output current. The rule for output current is that it equals the voltage drop across RT divided by its value in ohms.
When the red waveform rises to 0.8V, then, because the terminating voltage for that waveform is zero, the output current is (0.8–0)/50=16 mA. Apparently, when sourcing 16 mA, this particular driver cannot pull all the way up to VOH. That's normal. Most totem-pole drivers can't pull hard enough to meet VOH when loaded with 50 Ω to ground. Only an exceptionally strong driver can achieve that goal.
Yao Ming is exceptionally tall and very strong. If Ming ever greets DeVito, the handshake will probably pull the actor's arm practically out of its socket.
Now go back to the chart, with VT still at zero (red waveform). Look at the low-side output voltage. It drops to precisely 0 V. So does the current. Whenever the output voltage equals the terminating voltage, the output current vanishes. A driver forces an output greater than VT by sourcing current. It draws the output below VT by sinking current. At precisely VT, the driver does nothing.
Making the output voltage equal VT is easy for a driver. The terminating voltage is a “natural resting place.” If you disconnect the driver, the load immediately relaxes, all by itself, to VT.
When Yao Ming lets go, Danny DeVito feels relieved.
Now adjust DeVito's height. Put him on a ladder. Ming needn't reach down as far, so, if DeVito is still game for testing, the handshake occurs at a higher level. In the electrical world, raising VT always drags the output waveform higher. Lowering VT has the opposite effect. The trick in end-termination design is setting a value of VT high enough so the driver can pull above VOH but low enough so that the driver can sink below VOL, while all the time not causing any shoulder injuries (in other words, not exceeding the current capabilities of the driver or burning out the terminating resistor). The driver in Figure 1 meets those requirements with a terminating voltage of 0.75V, producing currents in the high and low states of 9 mA and –9 mA, respectively.