Figure 12.33 shows that in the first instant after the driver impresses a rising (or falling) edge into a source terminated configuration, before the signal has a chance to bounce off the far end and return, the input impedance equals twice the characteristic impedance of the line. Not only that, the drive current requirement drops to zero after 2 T seconds, lowering the average power drain. These facts may tempt you to assume that a single gate designed to drive a single endterminated configuration should be able to drive multiple (at least) sourceterminated lines. That assumption is incorrect.
A careful examination of initial conditions, however, reveals that the peak drive current i PEAK for the source and endterminated lines are the same. For example, in Figure 12.33 the i PEAK required to initiate a full sized rising edge into an impedance of 2 Z is V CC /2 Z .
Figure 12.33. The drive current required to initiate a fullsized rising edge at the end of the line is therefore V CC /2 Z .
An endterminated example, assuming the end terminator is symmetrically split , may be analyzed using the Thevenin equivalent circuit of Figure 12.23. In the Thevenin equivalent circuit, because the battery is biased at the midpoint , the driver need only provide enough current to change the voltage at the input to R T by half of V CC in either direction. The i PEAK required is thus (1/2;)V CC / Z , precisely the same peak current required by the series terminator.
If your driver cannot pump out the required amount of current, then your initial rising edge at the receiver won't be full sized. Note that in the seriesterminated case the driver does not have to meet V OH at the stipulated peak current, but you must know with some degree of precision what voltage your driver is guaranteed to produce when sourcing i PEAK . Given the voltage produced at the required i PEAK , the external seriesterminating resistor is then sized to produce a voltage drop precisely equal to the difference between the driver output voltage at current i PEAK and the initial required voltage on the line, which is half of V CC . If the resistor is sized properly, the voltagedoubling effect at the unterminated far end of the line will ultimately bring the initial rising edge of the received signal up to exactly full value.
Some driver circuits can easily source enough current to drive two sourceterminated lines. Is it possible to drive two or more sourceterminated lines from such a driver? Yes, but only under the limited conditions diagramed in Figure 12.34.
Figure 12.34. A single driver can drive multiple sourceterminated loads only under restricted conditions.
The trick to understanding this figure is to realize that the lines are coupled together into a jointly resonant structure. You cannot properly analyze just one line without seeing what happens to all the lines. The coupling happens because of the finite output impedance of the driver.
If the driver output impedance R S were zero (it never is), there would be no crosscoupling between lines, and you could simply use a separate seriesterminating resistor of value R 1 = Z on each line. Unfortunately, the reality of finite driver impedance forces us to contemplate joint resonance . The paragraphs below show how to jointly analyze the system.
Skipping ahead to the answer, multiple source termination with a nonzero driver impedance works only if the lines are equally long and the loads at each end are balanced . The sourcetermination resistors must equal
Equation 12.6
where 
R S = output resistance of driver, W , 
Z = transmission line characteristic impedance, W , 

R 1 = value of resistance added to each trace, and 

N = number of driven lines. 
When driving one line ( N = 1), [12.6] matches the total source impedance ( R S + R 1 ) to the characteristic impedance Z . This is a normal source termination. When driving multiple lines, [12.6] prescribes smaller sourceterminating resistors. With N too large, [12.6] goes negative, implying that no practical solution exists.
Let's analyze the lines in Figure 12.34 one at a time to see what happens. In Figure 12.34, a pulse travels down line A toward the load. This pulse reflects off the far end of line A, returning to the driver. In the usual application of source termination, the source termination matches the characteristic impedance of the line, eliminating the reflection at the driver. In Figure 12.34, however, the effective source impedance is not matched; it is set slightly lower than the characteristic impedance of the line. The returning pulse on line A therefore bounces off the driver, producing a negative reflection. So far, the negative reflection looks like a problem.
Another effect occurs at the same time. As current from the returning pulse on line A surges into the driver chip and through R S , it generates a voltage at the driver output pin. This voltage couples into line B. The polarity of the crosstalk pulse coupled onto line B is positive.
So far, the consequences of the returning reflected pulse on line A seem to include a negative reflection on line A and a positive crosstalk pulse on line B.
Now imagine what happens if the returning signals reflected off the far ends of lines A and B arrive at the same time. Each signal will induce on its own line a negative reflection and on the other line a positive amount of crosstalk. If you choose the resistor values carefully (according to [12.6]), you can get the negative reflection and positive crosstalk to cancel exactly. The result is a perfectly damped system.
The conditions under which perfect cancellation may be achieved are very restrictive :
Equation [12.6] sets the sourceterminating resistance so that line A experiences a negative reflection pulse exactly compensated for by the positive crosstalk pulse from line B. Equation [12.6] works with any number of lines, as long as they are equal in length and identically loaded.
Perfect balance rarely occurs in practice. If the lines are not perfectly balanced, the reflections and crosstalk from each line will not cancel. Incomplete cancellation makes the system ring.
POINT TO REMEMBER
12.9.1 To Tee or Not To Tee
POINTS TO REMEMBER
12.9.2 Driving Two Loads
POINT TO REMEMBER
Fundamentals
Transmission Line Parameters
Performance Regions
FrequencyDomain Modeling
Pcb (printedcircuit board) Traces
Differential Signaling
Generic BuildingCabling Standards
100Ohm Balanced TwistedPair Cabling
150Ohm STPA Cabling
Coaxial Cabling
FiberOptic Cabling
Clock Distribution
TimeDomain Simulation Tools and Methods
Points to Remember
Appendix A. Building a Signal Integrity Department
Appendix B. Calculation of Loss Slope
Appendix C. TwoPort Analysis
Appendix D. Accuracy of Pi Model
Appendix E. erf( )
Notes