2
\$\begingroup\$

I am trying to bring back into operation a late-70s disk controller board (all 74xx TTL). I'm stuck trying to figure out why a small pull-up network is resulting in a far out of spec voltage not doing what (I believe) it's supposed to be.

There is a section of the larger circuit that is simplified in my sketch below. There is a 1K pull-up resistor connected to the 5V rail that is in turn connected to several input pins on four different ICs, with the apparent intention of pulling them all high.

enter image description here Typo: 74109 above is actually a 74107

But instead of a ~5V level on the other side of that resistor, I measure a signal that floats around 1V. Even wackier, is that as a means of debugging, I removed all four of these chips, leaving in theory only the resistor with just an isolated network of traces on the other side of it-- same low goofy voltage!

So although I've illustrated the actual chips and pin connections there, it doesn't seem to make any difference whether the chips are even present.

The resistor is a new 1K film resistor that measures correctly, and if I connect it to the 5V rail and nothing else, I see 5V on both sides of it, as I'd expect. So something beyond the resistor is causing the bad voltage, but I'm at a loss to figure out next steps in determining what it might be. This is a small area of the PCB, and I find no solder bridges or other obvious connectivity issues when following the traces. I'm not an engineer, just a tinkerer trying to learn, so either my expectations are wrong or I've run out of good ideas on what to look for. Can anyone offer some pointers or pro suggestions for how to systematically figure out what's going on? This has my stumped. Thank you!


(The full schematic for the board I'm working with is here. The pull up circuit is chopped up across the first few sheets, look for "P.U." on sheet 1 which shows the resistor, and then blocks on subsequent pages with connections marked "PU"-- chips C1, D1, D4, and C5)

\$\endgroup\$
11
  • 1
    \$\begingroup\$ What do you mean with "floating voltage" exactly? \$\endgroup\$ Commented Mar 31, 2023 at 7:02
  • 1
    \$\begingroup\$ The only possible answer is: the physical circuit isn't what you think, you just have to find out how and where. Can you post a photo of the resistor? Did you measure good 5V on the top side of R1, with and without chips present? \$\endgroup\$
    – jonathanjo
    Commented Mar 31, 2023 at 8:11
  • 2
    \$\begingroup\$ Something connected to this network must be pulling it down. That could be a broken chip, a shorted trace, or something that's supposed to pull it down that you didn't realize \$\endgroup\$ Commented Mar 31, 2023 at 8:52
  • 1
    \$\begingroup\$ If all the chips have been pulled, and pullup resistor removed, leaving the trace isolated, then you can measure the trace resistance.. An ohmmeter might be able to discriminate between an actual resistance and an active transistor junction...if the ohmmeter reading is the same when you switch lead polarity, and gives the same reading on a different scale - it is likely a resistance. Perhaps difficult for your eyeball to see. \$\endgroup\$
    – glen_geek
    Commented Mar 31, 2023 at 12:20
  • 1
    \$\begingroup\$ In the set of schematic sheets you are missing sheets 5 and 6 of the drawing set 200028-200. (You included duplicates of sheets 5 and 6 of the 200058-200 set). Is this group of schematics from one board or two? (I just noticed that this may be a problem directly from the Vintage Computer website. ) So there may be other chips that use this same PU connection, with potential problems or bad connections. \$\endgroup\$
    – Nedd
    Commented Mar 31, 2023 at 13:36

2 Answers 2

4
\$\begingroup\$

At this point you may have a major task ahead of you to repair this disk system board. The documentation on the Vintage Computer site you linked to seems to be missing a few pages, so there could be additional components on the suspect PU line. If you are not able to find the missing pages one alternative is to manually follow the PU trace along the complete board and start marking each component pin. To help with this task you could use a fine tip marker or small sticky labels to highlight each connection. If you are not sure of the connection use an ohm meter to verify the continuity. If you have to remove a chip be sure to mark the empty position with the IC type and the orientation. During the handling and repair of the board it would be best to use a static wrist strap to minimize the chances of static damage.

If you are very lucky while tracing the PU signal you might find a shorted pin or a shorted trace. If you find a component other than an IC on this trace you could lift a pin and retest the PU line by briefly powering up the board and checking the voltage on that line.

If you find that only ICs are connected to the PU line you can take a sort of shot-gun approach, purchase a group of original TTL parts and replace all of the suspect parts at one time. So far it seems that there is just a few 74123 and 74107 ICs, (but of course there are still some missing schematic pages).

Another option would be to track the PU trace and if you find a main branch off point (for example near the center of the board) make a cut in the trace and retest the board measuring the voltage on the PU line (at the 1K resistor). If you do this a few times you may be able to isolate the defective point.

Yet another option, if you don't mind making modifications on the board, would be to get a few 1k resistors and start placing them on the pins of each IC that uses the PU line (preferably on the bottom surface). You would first cut the original PU trace going to (and away) from each pin then solder a resistor from the VCC pin (usually pin 14) to each pin. If you later test the board under power you might find that one of these IC will still have a low voltage on the end of the new pull-up resistor, that could be the one IC that caused the problem, at that point you might only need to replace that one IC.

It looks like chip A7 (74123, sht 4 of 6) should also have it's pins 3 and 11 connected to the PU line, this might be a drawing typo or maybe the pin was left open.

Also see
E5 (74107), p13, sht 5 of 6 200058-200
E6 (74123), p11, sht 4 of 6 200058-200
G2 (74107), p13, sht 3 of 6 200058-200
H15 (74107), p13, sht 1 of 6 200058-200
H16 (7476), p8, sht 4 of 6 200058-200

Note that some parts of the above chips have the CLR pins used for other purposes, (be sure to verify this before making any modifications).

Possible other connections on the missing sheets 5 and 6 of 200028-200.

IC datasheets for reference:
74123
74107
7476

Some examples of these ideas are show below:

.

schematic

simulate this circuit – Schematic created using CircuitLab

.

.

schematic

simulate this circuit

.
.

.

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Thanks Nedd for the thoughtful response. I have already removed and replaced (adding sockets) the original TTL chips with fresh parts to no avail (that was my first step as I kept checking and assuming the chips in turn were bad and sinking the voltage), but your suggestion of starting to cut traces and test is a smart one and I'll do that next. I don't mind modifying the board at all if that's what I end up with-- what I want is for it to work :) Thank you! \$\endgroup\$
    – BZo
    Commented Mar 31, 2023 at 18:16
  • \$\begingroup\$ PS: To clarify your questions from the comments above: yes, this controller is really two boards connected by large ribbon cables, and the schematic set linked above is missing two of the sheets because the original paper set was wrong, too. Fortunately they made another product that had identical controller boards so I have the two "missing sheets". This pull up network is isolated to the first board and seems limited only to what can be seen in the sheets linked (but I will triple check again) so I didn't want to overcomplicate the question. \$\endgroup\$
    – BZo
    Commented Mar 31, 2023 at 18:20
  • 1
    \$\begingroup\$ @BZo - Do note that there are a few additional ICs (I think on both boards) that have (or may need) the PU line, so if you trace out the complete length of the PU line you may find those too. - Also, it may be that some of the CLR pins were intentionally left open, this is a bad practice but usually results in a High input signal for old standard TTL parts, in some cases this can lead to intermittent problems. Personally I suspect that you have a single bad IC or pin-pin short deep in the circuit that is causing the whole problem, because of the shared connection finding it is complicated. \$\endgroup\$
    – Nedd
    Commented Mar 31, 2023 at 18:55
  • 1
    \$\begingroup\$ @BZo - Replacing some of the suspect parts with sockets is a good idea. If you continue with that method you may not need to make too many trace cuts. Best of luck with the repair,,, \$\endgroup\$
    – Nedd
    Commented Mar 31, 2023 at 19:01
3
\$\begingroup\$

Before embarking on the hunt for the problem, measure the voltage across the 1k resistor. It will be 4 volts if one end is 5v and the other is 1v. And this means that the voltage drop really is due to current being drawn from somewhere on that bus. In fact, ohm's law says that it's 4 mA going through the resistor.

Your comment about "floating", plus my own suspicion of what I'm measuring always makes me think about what's going on. Is the voltmeter working right? Is the resistor really what it says? Is Ground really ground here? That test should clear up all that. Good luck in the hunt.

\$\endgroup\$
1
  • 2
    \$\begingroup\$ Just occurred to me that you can change the value of the 1k and see what happens. Make it 2k and if the voltage goes from 1 to .5 that means there's a resistance causing it. If you still get 1v there's probably a semiconductor interfering with it. \$\endgroup\$
    – gbarry
    Commented Mar 31, 2023 at 18:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.