Bi-directional Terminations

There are many books talking about how to do termination in high-speed digital circuits. They usually consider termination topologies of driver- receiver (unidirectional) pairs such as source or end terminations.

If we use the bi-directional transceivers, how does one do correct termination? Because both buffers may function as source or receiver and source or end termination just functions properly for one direction, does there exist any termination scheme for bi-directional transceivers? Thanks.

Thanks for your interest in High-Speed Digital Design.

I assume you are talking about a bi-directional line in which only one party or the other may successfully transmit at any given time.

Let's work with two transceivers, labeled A and F, configured like this:

     [XCVR  A]--[resistor B]--[C--long line--D]--[resistor E]--[XCVR F]        

From transceiver A, the signal passes first through series terminating resistor B, and then to the left end (C) of a long line. The long line runs from C to D.

From the right end (D) of the long line, the signal goes through another series terminating resistor E, and from there into transceiver F.

Resistor B, plus the natural output impedance of driver A when it is engaged and transmitting, should equal the Z0 of the line.

Resistor E, plus the natural output impedance of driver F when it is engaged and transmitting, should equal the Z0 of the line.

When driver A is working, driver F should be in the tri-state (open-circuited) condition. This provides a normal series-terminated configuration when A is driving.

When driver F is working, driver A should be in the tri-state (open-circuited) condition. Now the line is source-terminated at F.

The only peculiar thing you will notice about this setup is that the rise time at the receiving device may be somewhat degraded, and a little delayed, from the normal situation you would expect in a series terminated line. This is due to the effect of the idle (unused) series terminator at each end of the line. The received signal has to go through this extra resistor into the receiver. In the process of doing so, we get an R-C lowpass filter effect with a time constant equal to:

T(Bidirectional,0-63) = (Z0+Rs)*Cout


  • Z0 is the line impedance,
  • Rs is the value of series terminator, and
  • Cout is the output capacitance of the transceiver when in the tri-state (off) condition.

With a normal series-terminated condition, the R-C time constant would have only been:

T(RegularSeriesTerminated,0-63) = (Z0)*Cout

Best regards,
Dr. Howard Johnson