Placement of End Termination

My upcoming article for EDN (March 2, 1998)(How Close is Close Enough?) discusses the placement of series terminating resistors, and it gives a simple expression for estimating effectiveness of the resistor compared to its placement. Bob Haller saw a preview of this article and submitted the following question.

I agree that series termination is a very effective way to eliminate SI problems on networks, and generating a simple expression to handle effectiveness based on placement is a great idea. I wonder if you can also handle stub length of parallel, terminated nets also? Both "in -line" and "down stream" parallel termination.

When I am performing parallel termination it is often difficult to far end (or down stream) terminate. I have found that if stub length is kept very short, and edge rates are not excessively fast, terminating "far end - in line" driver, termination, receiver can be as effective and save significant routing channels when surface mounted components are utilized (especially high pin count BGA's.)

Best Regards,
Bob Haller

Thanks for your interest in High-Speed Digital Design.

Great Idea!

When working with very fast edge rates, the sequencing of the end-terminator and its associated load can make a measurable difference in signal quality. We can use the "short stub" analysis idea to predict the effect. If you hook up a net in this sequence: driver, long line, terminator, short stub, receiver, the additional short stub (which is OPEN-circuited at both ends), will act as a small lumped-element capacitor. This small capacitance, along with the parasitic input capacitance of the receiver pin, creates an imperfection in the termination network. When the first incident wave arrives from the driver, part of that wave, a small pulse, bounces off the imperfection and returns to the driver.

The small reflected pulse travels backwards along the line to the driver, where it bounces again (off the low impedance of the driver output) and returns, one roundtrip-time later, to the receiver. What we observe at the receiver is an initial rising edge, followed one roundtrip time later by a secondary pulse. If the initial reflected pulse is sufficiently small, all tertiary and subsequent events will be of negligible amplitude.

Assuming the delay of the short stub is less than 1/3 of a risetime, we can model the amplitude of the initial reflected pulse as:

AMPLITUDE = (1/2)(Z0*[CSTUB+CGATE])/RISETIME

Where:

  • Z0 is the line impedance,
  • CSTUB is the stub impedance in pF, = STUBDELAY/STUBIMPEDANCE,
  • STUBDELAY is the one-way propagation delay of the stub, in pS,
  • STUBIMPEDANCE is the characteristic impedance of the stub, in ohms,
  • CGATE is the parasitic input capacitance of the receiver, in pF, and
  • RISETIME is the 10-90% risetime of the driver.

The accuracy of this formula, if it gives you an answer less than about 0.3, is approximately +/- 20 percent. If it gives an answer greater than 0.3, then you know two things: (1) you are going to get a whopping huge reflection, and (2) I don't care exactly how big the reflection is because the system probably isn't going to function.

Hint: if the stub capacitance is significant compared to CGATE, then you've unnecessarily increased the size of the reflected pulse. The best results are obtained by making the stub impedance very HIGH (skinny trace), reducing its effective capacitance.

Best Regards,
Dr. Howard Johnson