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I'm breadboarding a simple data bus terminator for a 1970s TTL computer bus that was normally unterminated. I'm trying both passive and active termination to see how they work, but have a question on how to construct it.

I know that in a logical schematic sense, the terminating connection/resistors can appear "anywhere" on the (say) D0 signal line, because whether they are at the "end" of the signal or somewhere "else", they're logically equivalent.

I want to understand how important the physical location of the terminating resistors are to be most effective. I can put a standalone terminating pack (etc) at the physical end of the bus, long beyond all other electronics. But if I want to combine a terminating network on the same board with a bit of other functionality using the data lines, do I need to literally extend the physical lines "beyond" the length of the real functionality in order to effectively terminate the lines at a definite "endpoint"?

Thanks for any insight. If this contains wildly wacky assumptions please do feel free to set me straight!

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    \$\begingroup\$ Keep in mind that GND and +5V isn't the same everywhere. I'd suggest that you determine where the terminator's gnd connection has most" GNDiness. Same with +5V. For example, a backplane bus suggests plug-in cards, which likely passes bus GND through a connector. So card GND might have more noise than backplane GND...I'd want to GND a passive terminator at the bus GND, not the card GND. \$\endgroup\$
    – glen_geek
    May 24, 2023 at 20:05
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    \$\begingroup\$ breadboarding a [transmission line] terminator is a contradiction in itself. If it normally went "unterminated", what prompted you to consider termination? (Brings back unhappy memories of frustration over a fast driver combined with a cable way out of spec causing massive data loss.) \$\endgroup\$
    – greybeard
    May 24, 2023 at 20:38
  • \$\begingroup\$ @glen_geek There are multiple cards but no backplane in this unusual system-- the boards are connected via a ribbon cable plugged into a DIP socket on each board. Including the power rails. I can bring +5 and GND in directly from the PSU if I need to instead of getting them from the ribbon cable rails. \$\endgroup\$
    – BZo
    May 24, 2023 at 20:56
  • \$\begingroup\$ @greybeard This is an early microcomputer (the obscure and unloved Sphere), the boards are connected with a ribbon cable and not a backplane, and it normally went unterminated but users at the time frequently complained of flakiness that could be improved by terminating the data bus. So I'm trying to do the same :) \$\endgroup\$
    – BZo
    May 24, 2023 at 20:57
  • \$\begingroup\$ (Found a hackaday image: Three flat ribbon cables, guessing 16 conductors, ~22 cm (I guess I'd try to use 5 cm and parallel vertical boards). No bus transceivers in sight, but a decent number of decoupling capacitors. Would have been easier with .1" pitch ID(C) connectors… if every second conductor was a static voltage as in SCSI-P. The usual place would be near the ends of the "cable"/backplane; single (120 Ω ballpark) resistors to 3…3.3 V overtaxing low drive even more severely than 220+330 Ω dividers, let alone two. "Dynamic termination" with a ~15 ns time constant seems saner.) \$\endgroup\$
    – greybeard
    May 24, 2023 at 21:49

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how important [is] the physical location of the [terminators?]

All important. Within limits:
A transmission line is a useful model when its properties are reasonably uniform along its length and it is long enough in relation to signal transition time that the signal state at the driver/transmitter can significantly differ from the state at the receiver.
A signal bus is a (topologically) linear connection of transmitters and receivers - at least three, otherwise communication is point-to-point.

(In the 70ies, drivers&receivers where LS TTL more likely than not, rise time on the order of 8 ns, fall time ~5 ns. Forth&back, that's 50 cm, give or take:
Signal problems you are looking at most probably are not caused by reflections.)

Driving a signal line anywhere but near one end actually starts two signals travelling it: one in each direction.
Think of it as energy:
It doesn't just disappear when reaching the end -
what isn't absorbed gets reflected.

The termination should be geometrically close to the end of the transmission line.
It does not matter if the terminator or the last transceiver is a fraction of a transition time closer to that end.

A terminator being effective depends on the similarity of the signal network to a transmission line and the impedance match between line and terminator - in the critical frequency range, obviously.
And the position of an ineffective terminator is pointless.

(I remember pairs twisted in alternating direction as a flat cable ~132 Ω, .05" flat cable ~100 Ω, coax at 50, 60, 75, 93.
There were a lot of problems (if not in the 70ies, then with the introduction of stronger drivers and faster logic families - AC(T), A(L)S, FAST… Manufacturers provided application notes mentioning termination issues, RS keeping at least one as of 2023/05), and a lot of snake oil.)

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  • \$\begingroup\$ Thank you for the writeup on this @greybeard, it's helpful, esp the observation about length and whether state is likely to be different at the opposite ends relative to frequency vs length. I'm obviously not an electrical engineer by trade so forgive the dumb followup: you suggest based on the lengths of cable I have that reflections are probably not the cause of signal degradation in this case. Does that mean that termination is actually not that important here, or won't accomplish much of value? Or does it have other benefits besides mitigating "reflection"? \$\endgroup\$
    – BZo
    May 26, 2023 at 16:23
  • \$\begingroup\$ One benefit of a 132 or 187 Ω pull-up to 3 V (or 3.3) is that it mitigates the unlike drive capabilities to low and high. Such must have been discussed at length in standards committee meetings for buses that had any. \$\endgroup\$
    – greybeard
    May 26, 2023 at 18:02
  • \$\begingroup\$ Thanks. When you say 3/3.3V are you referring to the result of a termination voltage divider for TTL 5V levels? Or is that referring to non-TTL levels to begin with? Thanks \$\endgroup\$
    – BZo
    May 26, 2023 at 19:33
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    \$\begingroup\$ I remember 5 V TTL levels (0.4/2.4 out, 0.8/2.0 in, about 1.2 threshold) - the only TTL 5 V level I remember is supply voltage. 3.3 V from 220&330 Ω, 3 V a possible voltage for "pull-up & active termination". If I was to trouble shoot a non-trivial digital circuit, I'd first try and figure out how to assess what's happening, then why, what to change, what difference that made. What's happening (beyond logic levels & some 10 M samples/s) looked hard back in the 80ies with a hobbyist's budget and know-how. \$\endgroup\$
    – greybeard
    May 26, 2023 at 20:37

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