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IBM 5110/5100 video Display card repair (possible diode fix)

For those 7417's there is a D, DR, N variant; "PDIP (N)" vs "SOIC (D)" ? Well, they're out of stock from TI of course. So I have some 7417D's on order from 2nd hand, since those may take awhile to show up (if we decide that's the thing to try).
 
A few more thoughts now. Absent a 'scope that can resolve things a little better than your current setup, it's a bit harder to tell what's going on. Can you share information on what you're using? The clock recordings you've posted recently do not look very good, to be honest. What we want to see are nice square waves going from 0V to +5V or so, and we know they must be there. The fact that you're getting wiggles around +2.5V suggest that those square waves are getting averaged. Details and trouble signs we might be looking for in other parts of the circuit, like glitches, could be invisible.

Is there a chance that your oscilloscope software might have some setting that allows better resolution on the time axis? Maybe there's something like a "high resolution" mode? Or maybe your 'scope can only store/display a certain number of samples, and by setting a narrower time interval over which it stores, you can get higher resolution on the time axis? What relevant-looking options are available?

The D02 waviness is still fishy and worth investigating. But I think that right now we are not quite ready to replace the IC we've been zeroing in on. Note that the 7417 can't be the Character Counter all by itself --- as a driver IC, all it does is "retransmit" the signals that the chips doing the counting are emitting, and that's happening elsewhere.

Before we do any replacing, I think we want to keep investigating in order to minimise rough treatment of the board. At some point we will probably want to probe points on the board itself, including parts not accessible through A1 board pins.

More in a few hours, including the reason it's too early to get discouraged by that pin 3 connection to the big IBM metal can chip.
 
Hantek 1008A USB scope, outputs on USB 2.0 (older discontinued model, yep on the cheap side). One spec it has is "2.4MSa/s" - 2.4 million samples per second? But I'm not sure how much of that makes it across USB, or that ends up to the application under Windows (as for sure, Win10 is no RTOS).

There isn't much in settings - it has a Calibration, which I did run through that. There is some MATH to mix singals and a sort of Trigger mode, but nothing that led me to squaring up those clock inputs.

Some errands to do for a while.... // if I do replace a chip, I wonder if I could/should just use a socket to make it easier to replace in the future? But I suppose sockets, mounted vertically, have some risk of coming out on their own? I won't mind it being obvious that some alternation has happened on this card.
 
Thanks for identifying your 'scope. I found the manual, and like manuals for many lower-end oscilloscopes (mine included!), it's a bit thin. The only thing that stands out to me right now is that the acquisition rate is qualified as "single channel". I think that if multiple channels are active (even if they are not displaying), then your horizontal resolution is worsened by a factor of 8. If you're probing these signals one-at-a-time, using only one channel, can you make sure that all the other channels are disabled --- not just hidden, but off? If you're not doing this already, I think it will make a difference.


An earlier post seemed uncertain about how to identify pin locations on a DIP. Here's an image from Wikipedia:

1656019781548.png
It's always clockwise from Pin 1, which is at top left. Which end is the top? It's often identified by a little U-shaped notch, although sometimes Pin 1 is marked instead or additionally by a little round recess or a painted dot next to Pin 1. Sometimes it's a stripe. Here are some examples where Pin 1 is always at bottom left:

1656020008019.png
1656020051137.png
1656020112186.png

If you'd like to test your skill, try the image where these examples came from: it's jam-packed with DIPs. All oriented in the same way, as far as I can tell.

if I do replace a chip, I wonder if I could/should just use a socket to make it easier to replace in the future?
It wouldn't bother me and would probably be what I'd do, provided it fit. Use a good quality socket: cheap ones are less reliable.

But I have no insight as to how those IC DIP chips can fail. I mean, a power surge or short causes "random" failures (typically starting with the "weakest" component?).
That can happen --- remember me warning you about not having the -12V power supply? :) Depending on the chip, bad voltages are an example of something that can challenge the chip's integrity. ESD from handling parts without proper precautions can also be a factor. But also, things can just happen. Chips are highly ordered, intricate devices and sometimes entropy wins. Legendary Commodore engineer Bil Herd has a talk that reminds us how all things will fail someday, even ICs (and why): it's worth a watch. It speaks to exactly this question (although it talks about more than ICs).

Haven't found it yet. I tried against the other ICs, nothing obvious stood out. Then I realized, maybe I better first trace this using the GOOD Display card instead of the BAD and tried again. Still no luck. It's a lot of pads to try from the backside. Then I remembered I had one of those "electronics" cameras -- basically a web cam that can be mounted on a steady frame to look like a microscope.
Tracing connections is a pain, and the only sure way to do it is to have a continuity tester with one probe on (in this case) pin 3 of that driver chip and then to probe every single other component leg one by one. After you do it for a while, you get a knack for dragging the second probe over the legs fairly quickly. There are tricks you can try like clipping the second probe to a piece of aluminium foil and then dragging the foil around the PCB to try multiple places at once, but it's not so reliable.

Well, bad news... If we're looking at the right pins, that traces directly over to one of the IBM 22 tin cans (the 2nd one from the edge, where I marked "A")
Don't despair (especially if this was the wrong pin you mentioned earlier, in which case this measurement is irrelevant --- but I'll assume it was correct and carry on). We know the Character Counter has several outputs, only one of which is D02/+Printer Clock Pwrd. We don't know if the can is the source of the signal going into the driver IC, or if it is just another destination for that signal.

Just to reassure ourselves, though, let's assume the worst and say that the signal does come from the can. Well, the driver IC could still be what's wrong here. One way to tell for sure would be to get an oscilloscope trace on Pin 3: if the signal there looks healthy, then that's an important sign. Of course, how you probe that pin while the machine is running is a good question: you may need to solder on a little tag wire that snakes outside of the card cage, as we've discussed earlier.

...

It's important to remember that we don't know why this driver IC is here. It's not marked at all on the IBM logic diagram. It could be that it's there just to give the signal an extra boost on its journey out to the printer --- call this Theory 1. If this is its only use, then replacing it very likely won't do us any good, even if it's truly broken: you don't have a printer.

But it could also be that the driver boosts the signal for a different reason: maybe there are lots of ICs on the display card that "listen" to the signal, and that means that the signal needs to be boosted. (A chip can't provide infinite power on one of its outputs: it will only have enough power to talk to a limited number of other chips. This limit is called the maximum fan-out. If the Character Counter needs to talk to more chips than its maximum fan-out allows, then you'd use a driver to boost its signal.) If this is the case, then a broken driver could truly be the culprit. Call this Theory 2.

Hopefully further investigation will tell the story. We've already discussed getting an oscilloscope on Pin 3. Here's another thing you could check: if you found that Pin 4 (that's right, pin 4 this time, the driver output) only connects to D02, then this suggests that Theory 1 is correct. But if you find that it connects to more things besides D02, then maybe Theory 2 is right.


Good luck, and I hope you can bump up the temporal resolution of your 'scope --- it would be a big help!
 
If you're probing these signals one-at-a-time, using only one channel, can you make sure that all the other channels are disabled --- not just hidden, but off? If you're not doing this already, I think it will make a difference.

Check, all but CH1 set to OFF and disconnected. I think I've found a "reasonably priced" 200MHz TDS350 Tektronix scope (meaning used, but I'll take my chances). I'll give it a shot, probably next week.

I looked at that "test your skill" board you linked - very nice, indeed they all seem to be oriented in the same direction :) There are some white-color chips on the left side labeled BECKMAN - I feel like I've seen that brand somewhere, maybe in a Commodore system (maybe in the power connector internals)?


And the Bil Herd discussion is great, I hadn't seen that before. Absolutely the engineering perspective is different than the collector's perspective. Taken to the extreme, entropy eats the entire universe, trillion-trillion years from now (or the other model is this infinite oscillation between expansion and collapse, but the jury is still out on that). But on a smaller timescale - take the fantastic marble statues of the Greeks, or those detailed columns or triumph arches that they left out in the weather. Sun and rain are brutal on those detailed carvings, after centuries of exposure. Though that's one thing I noticed about some of the Egyptians columns (obelisk) - whatever material they used (red granite?), their etchings still seem as crisp as new. Indeed, materials matter, and the atmospheric conditions.

Anyhow, recognizing that hardware will inevitably fail at the fundamental component level -- even if just sitting there (like an old car engine -- cylinders can fuse to blocks, maybe; or at least the rings on the cylinders might) -- that gives greater importance/priority on getting the 5100 emulation up and going (that's my version of a Sudoku or brain-teaser type mental puzzles, for certain days :D ).


There are tricks you can try like clipping the second probe to a piece of aluminium foil and then dragging the foil around the PCB to try multiple places at once, but it's not so reliable.
Ha! The idea of using something like a foil crossed my mind - to just at least isolate it down to a quadrant of the board. But then I noticed with good enough eyes, you can follow the traces on those boards. But right, once it crossed to the other side, then holding a probe on one side securely while searching the otherside wasn't the most graceful things to do (hunting 1 pin is fine, but scaling it up to reverse out the whole board is daunting)



Just to reassure ourselves, though, let's assume the worst and say that the signal does come from the can. Well, the driver IC could still be what's wrong here. One way to tell for sure would be to get an oscilloscope trace on Pin 3: if the signal there looks healthy, then that's an important sign. Of course, how you probe that pin while the machine is running is a good question: you may need to solder on a little tag wire that snakes outside of the card cage, as we've discussed earlier.

Well... Like anything, one needs proper tools. My issue in soldering is I can glob together some metal - I did that for a quad I printed, wired up, and successfully flew. But for precise control, I often glob too much solder, spill over to another connector, and it's a huge mess to clean up. I've tried finer tips -- it's probably that I need more precise control on the temperature.

I did come across some 8481 Premium Carbon Conductive Grease - maybe an alternative to any conductive glue material. It's an option, but I don't think I'll use it (yet). The other option is the 96 long jumper cables (as in 2+ feet long) and sort-of rising the entire card out. And I've realized I don't have to run those jumper wires from the inside of the A1 board - I could just use the pins on the backside (except some of those pins have so much wire-wrap already, even on the Display card, that there isn't much pin left to work with).

Alright, I understand what's being requested and why, and will strategize. If I run with the really longer jumper cables, I'll have to do that very carefully, and I won't be able to move the machine (or swap in the GOOD Display - not easily) while setup for that. And I'll have to be even more careful about bumping things, to avoid jostling any of those cables. But if I think we'll need to explore things further, that's what is needed to be done (which was essentially suggested near the start of this thread :) ) since I can probe both sides of the board very easily.


I may pause here just to wait for the new (to me) oscope - and verify my D02 observations with that equipment (of the SAWTOOTH signal with the BAD Card, and nothing with the GOOD Card). Meanwhile, I may physically trace (as far as I can) both pin 3 and 4 on that chip.

Specifically, I said D02 goes to this PIN 4 (as noted below). Well, there is another trace connected to that pin, shown here on the bottom ("back') side of the board). One of these (should) go to D02 at the base of the card. Where does the other one go? Then for PIN 3, yes it traced to underneath "the chip next door" and then off to that "tin can" IBM chip - as shown below, PIN 3 just has 1 trace (on this side; perhaps its possible it has another trace on the other side at this same pad) -- but the 1 pad that it traces to, I need to verify that (on both sides of the board) that all the intermedate pads also just have 1 trace.

1656053970992.png


And just to throw this out there - what about the IR camera? For example, take PIN3 in the above image. If I let the board completely cool (i.e. take it out of the system for a day or two), then if I just apply 5V to PIN3 -- won't just the parts of the board connected to it warm up? And would it be enough of a difference to observe on the IR camera? Might not even need 5V - say just start with 1V? 2V? just that, I don't have super high resolution on the IR camera, so it may be too ambiguous - or I'd have to leave that voltage for quite awhile as I pan the camera around.


BTW - I'm motivated to solve this because... One thing on my to-do list is to just emulate the entire Tape Control Card behavior. Forget the physical tape parts, the system just cares about the Data In and Out from the Tape Control Card. So the learning being done here hopefully applies to figuring out the tape stuff later (except it has some 12V pins that may make it "interesting"). But that's for the future ;)
 
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Worth noting that if you are dealing with open collector buffer IC's, you won't see anything on their outputs with the scope unless there is a pullup resistor. Looking at the card photo, the objects in the rectangular packages with the dots on one end are likely resistor arrays used as pullups, likely with the common connection tied to the + supply. So if that connection fails, or the connection to the individual pullup resistor fails, you won't see any signal/s on the output pins of the buffers.

Very important to have a scope that is guaranteed to to be able to resolve fast rising edge waveforms. Many digital scopes can but if you want to be 100% sure you are not getting misleading results from the setup on a digital scope, you could simply get a 20MHz to 100MHz bandwidth analog scope and there would be no ambiguity in the recording between what you are actually seeing and what is actually there on the point being tested. Also, when the scope is used it is important to have it set on DC coupling and note the position of the beam(or line) with zero volts and know the vertical scale in volts/cm, that way you can check if the levels you are seeing are proper logic levels. One way a fault can occasionally be found is if there is an intermediate or ambiguous voltage that is between logic high & low.

If there is a suspicion that an open collector output is not being pulled up you can simply connect a resistor such as a 2.2k from the output pin to the IC's + supply pin and see if a signal appears. For most TTL IC (not all) that is pin 14 on the 14 pin package and pin 16 on the 16 pin package. And the grounds are on pin 7 and pin 8 respectively.

One classic TTL IC with the power supply pins elsewhere is the 7490, and this is one way to identify that IC, say if it had another number on it.

I would be very reluctant to remove an IC from a board like this unless there was very solid evidence it was defective.
 
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Worth noting that if you are dealing with open collector buffer IC's, you won't see anything on their outputs with the scope unless there is a pullup resistor. Looking at the card photo, the objects in the rectangular packages with the dots on one end are likely resistor arrays used as pullups, likely with the common connection tied to the + supply.

We're pretty sure that those are parallel resistor arrays. The mystery output we're seeing (assuming it isn't caused by odd aliasing) is a strange (relatively) low-frequency 5V p/p sawtooth wave.

I looked at that "test your skill" board you linked - very nice, indeed they all seem to be oriented in the same direction :) There are some white-color chips on the left side labeled BECKMAN - I feel like I've seen that brand somewhere, maybe in a Commodore system (maybe in the power connector internals)?

Those are DIPs but they are not chips: they're resistor packs. Of course, resistor pacts often have polarities too, and if they do, they should have a marking for Pin 1, just like an IC.

But then I noticed with good enough eyes, you can follow the traces on those boards. But right, once it crossed to the other side, then holding a probe on one side securely while searching the otherside wasn't the most graceful things to do (hunting 1 pin is fine, but scaling it up to reverse out the whole board is daunting)

With the Display Card circuit board, I'm pretty sure there's a lot more than what meets the eye: I think these boards have more than two layers. The traces you can see are helpful, but there may be others that are invisible. I think the only way to be sure that two components are connected or not is to measure between them. Luckily the components are all through-hole on the Display Card, so in theory you can do all of your probing work on the solder side, with the components facing the tabletop. It's still a major pain; no way around it.

Well... Like anything, one needs proper tools. My issue in soldering is I can glob together some metal - I did that for a quad I printed, wired up, and successfully flew. But for precise control, I often glob too much solder, spill over to another connector, and it's a huge mess to clean up. I've tried finer tips -- it's probably that I need more precise control on the temperature.

Practice, good equipment, and good materials are hard to find a substitute for --- unless you have a friend who can do the work for you. For some people, practice won't overcome hand tremors and the like, or they don't have time, or it's just hard; not much to be done. If you lived in London, I'd help...

I've realized I don't have to run those jumper wires from the inside of the A1 board - I could just use the pins on the backside (except some of those pins have so much wire-wrap already, even on the Display card, that there isn't much pin left to work with).

This strategy might work, but as you point out, it's really tricky. You want to make sure you have good connections, and you need to be certain that all 90+ are correct. Putting power into the wrong pin could be disastrous. Even just one weak connection could be hard to spot and might cause bizarre problems.

It would not be the hardest thing in the world to design an extender card based on the substitute SLT plugs and sockets I designed. But it would be fiddly to put it together, and you'd still have the issue of holding the card extremely still: my connectors are more wobbly than the IBM originals.

I may pause here just to wait for the new (to me) oscope - and verify my D02 observations with that equipment (of the SAWTOOTH signal with the BAD Card, and nothing with the GOOD Card). Meanwhile, I may physically trace (as far as I can) both pin 3 and 4 on that chip.

I think this could be a good decision. Having dependable measurements will help us make good choices about next steps.

And just to throw this out there - what about the IR camera? For example, take PIN3 in the above image. If I let the board completely cool (i.e. take it out of the system for a day or two), then if I just apply 5V to PIN3 -- won't just the parts of the board connected to it warm up? And would it be enough of a difference to observe on the IR camera? Might not even need 5V - say just start with 1V? 2V? just that, I don't have super high resolution on the IR camera, so it may be too ambiguous - or I'd have to leave that voltage for quite awhile as I pan the camera around.

Let's consider this option. This will take a little while, but I think it might be worth reading. Physics lets us run this experiment in our head before we try it on sensitive hardware, provided you do the maths right. Knowledge is power... did someone say power?

Heat comes from power. Forget that the chip is a chip, let's assume it's just a resistor. You want to turn electrical power into heat, and that's what resistors do. If all you care about is heat, anything that conducts is a resistor, and things that conduct only marginally --- like semiconductors --- are even better at getting hot.

What value should the resistor be? Well, the input to a driver IC is designed to be sensitive: it's not meant to demand a lot of power from the signal going into it --- after all, it's kind-of like an amplifier. It wants to turn a weak digital signal into a strong one. "Not meant to demand a lot of power" is in this case like saying it has a high resistance, or a high impedance to use a familiar term. You might want to think of this pin as a sensitive voltage sensor. Your voltmeter is also a sensitive, high-impedance voltage sensor for the same reason: it's not supposed to affect the circuit it's measuring very much. That also means the same thing as "not meant to demand a lot of power from the signal." All the same general idea.

So our thought experiment has to use a fairly high-value resistor to emulate the chip. Let's say 500Kohm. I could be off by an order of magnitude, who knows. 500K, got it, let's set that aside.

OK, you want to heat an IC DIP; that's a chunk of plastic. How much energy do you need? Well, a plastic chip is, what, a gram maybe? I'm from America, who knows. It's probably a couple grams; let's just say a gram. Let's say you need to raise the temperature of the gram of plastic by ten degrees celsius for your thermal camera to pick it up. We can use the specific heat of the plastic material to calculate an amount of energy. This page says that plastics have a specific heat of around 1000 joules per kilogram per degree c. You've got a .001 Kg chip that you want to raise by 10 degrees c, so that's .001 * 10 * 1000 = 10 joules. (We're assuming that you only try to heat one DIP and that none of the energy manages to escape the chip, which are simplifying assumptions that are kind to you.)

Energy is power times time, so how much power do you need and for how long? Well, you can't just trickle a low current in very slowly, since then you really will have to worry about the heat leaking out of your chip, and you may not get it to heat up at all. So let's say you only want to juice the chip for one minute to pump in those ten joules: sixty seconds. 10 joules divided by 60 seconds is .166 W = 166 milliwatts, and you may start to sense a bit of a problem at this point. Let's press on:

We have the information we need now to compute how many amps you'll be pushing into this pin, since power is current squared times resistance. Do a little algebra and: current = sqrt(power / resistance) = sqrt(.166 / 500000) = .00058 A = six hundred microamps. Hey, that doesn't sound too bad, does it? Hardly any current at all.

Well, the problem is that 500 Kohm is a pretty slender straw to push 600 uA through. You're going to need to push pretty hard. How hard? Ohm's law says V = current * resistance = .00058 * 500000 = 290 V. That is probably not very healthy for a sensitive voltage detector that was never designed to see more than +6V or so on its inputs.

Now remember that you're interested in heating up more than one chip to detect continuity, and you'll also want to work in the real world and not this spherical cow world where the chips can never cool down. So, you'll need more power and more voltage in real life than what we've calculated here. In any case, I would not advise pursuing this approach.

BTW - I'm motivated to solve this because... One thing on my to-do list is to just emulate the entire Tape Control Card behavior. Forget the physical tape parts, the system just cares about the Data In and Out from the Tape Control Card. So the learning being done here hopefully applies to figuring out the tape stuff later (except it has some 12V pins that may make it "interesting"). But that's for the future ;)

I'm glad you're pushing ahead with this and hope that my long messages have been helpful. Even though I've been long-winded and didactic, I'm learning a lot too thanks to your brave explorations. I'm keeping hopeful that we'll work out a fix, and even if we don't, there's still more to learn...
 
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I think these boards have more than two layers

Oh? Here's the side of the Display card (I keep the tray off for now). Maybe it's 3-5 layers, but with traces on the intermediate layers? I'm skeptical.

1656140463919.png

But if there are "unseen" traces within these layers... Well, I didn't bring it up earlier: there is a "hole" in the BAD Display card. I didn't mention earlier since it didn't appear there were any traces (top or bottom) associated with it. Here is the only shot I have of it: (solid hole, light goes clear thru). But there are no apparent copper traces coming out of it, on either side of the board. It's along the perimeter of this tin-can chip, not underneath of it.

1656141281661.png
 
290 V. That is probably not very healthy for a sensitive voltage detector that was never designed to see more than +6V or so on its inputs.

Fair analysis. I also bench tested it - 5v thru a wire, couple resistor packs, and to an LED. Just did a single LED (out of set of 9, where all the other 8 LEDs had the same wiring and no voltage flowing). IR showed no evidence of any difference in the line with the 5v. At least not in cables with insulation still on them. The thought was to heat up the exposed traces between the chips, not so much the chip themselves - but no way around that, they're all connected. Like, maybe a soldering iron with a very clean tip on it, could induce some heat thru the line? Also, how much "heat" you need will be relative to the sensitivity of the IR equipment?

Even if you were off by a factor of 10, that's still --- 29V ? Way too much.
 
Fair analysis. I also bench tested it
Ha, brilliant. There's no substitute for empirical testing!

Unfortunately I don't think there's a practical substitute for physically contacting all the pins when you are reverse engineering a netlist. Maybe there's some strange method where you could inject a very weak high frequency AC signal and detect EM radiation from connected traces, but I definitely don't have the physics knowledge to give that a theoretical whirl! Nor the equipment for an empirical attempt.

Oh? Here's the side of the Display card (I keep the tray off for now). Maybe it's 3-5 layers, but with traces on the intermediate layers? I'm skeptical.
I still think so :) My main reason for suspecting it is stuff like this in your photos:

1656152328789.png

You can tell the difference between a trace meant to carry signals (very narrow) and a trace meant to carry power (the thick one). (NB: besides the trace, one additional reason we think this is a power pin is because DIPs often --- not always!! --- have power pins at the corners of the package.) I think that if the board were a two-layer board, we'd see thick power traces in the pictures that don't just go one "grid step" like this --- instead, they'd go all the way to the power pins on the board's edge connectors.

Since we don't see this, my hunch is that the power supply traces tend to stick to intermediate layers. This is a common design: top and bottom layers for signals, and two middle layers for a power plane and a ground plane. But sometimes you still sneak a few little stretches of signal line into a power layer just to connect some awkward places together that would be difficult to reach otherwise. Maybe IBM did differently, but I wouldn't trust my eyes to tell the whole story.


As for the hole .... I'm not sure what to make of it! It could just be a manufacturing hiccup. Let's keep it in mind, it might be relevant later?
I imagine that the PCB uses plated VIAs (i.e. metal on all sides of the hole), so if it needs to carry current through, it doesn't matter whether there's solder in the hole or not.
 
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It is quite plausible it is a 4 layer pcb, with the gnd and +supply voltage planes in the central two layers. Dead easy to tell, just check the gnd and power supply pins of the DIL IC's. (I noted where these were for most IC's on a previous post) You will soon be able to figure out if this is the case. If you can trace tracks to all of those pins on both visible sides of the pcb, then it will only be a 2 sided board.
 
Not much to report, waiting for some equipment. I Do want to give Thanks to all the ideas and feedback - even if I don't seem to respond to each one, I am reading and digesting what I can. I nearly considered a 1977 Huntron - to play a "could this have been repaired with equipment available in 1978" kind of game :) But I think I may try the 20mhz signal generator approach instead, eventually.


During first part of the thread, I said I didn't think this was a resistor component issue. I wanted to circle back on that, and verify it.... So here's the backside of the IBM 5110 Display board.


IMG_2724B.jpg

I mocked this up in Excel, and started probing here and there.. (across both the "BAD" and "GOOD" Display cards, to check for consistency). I marked a few traces that I started exploring (purplish color), but that'll be a difficult thing to overlay into this same diagram.

There are actually a few "holes" in the 1978 Display card (vs the 1979 Display card, to what I have) - there is a cluster of them at the top left corner, then two others: AA20 (near center), and BI36 (bottom right, center of one of the tin-cans -- since it is below that large flat component, no light goes through making this one harder to spot). Tentatively, I don't think these holes are any kind of issue, but it is a physical difference between the GOOD/BAD Display cards to keep in mind.

1656235969186.png

There are a couple observations of interest:

*1) the resistance reported in this resistor pack varies depending on how the "font" Jumpers are set. Like if you just change Jumper 1 at the top, then only the resistance between pins 1 and 2 is reduced from the 750ohm (but the others remain at 750).


*2) these are at the "top" of the board and "look like" the other resistor packs, but I don't think these actually are resistor packs. These are part 2392699 instead of 2392712. I don't think x2699 is resistor packs, because I get fairly different values depending on how I connect the DMM leads. Hopefully the following will make sense - as a measurement of resistance with the BLACK lead on PIN 1 versus the RED lead on PIN 1, and versus both the GOOD and BAD Display. The resistance on the GOOD Display card is "half" (4k vs 2k) in a lot of cases, but when reversing the leads, the values are fairly consistent between the two. So whatever this component is, I don't think it's a major issue (but could be wrong, just the measurements are roughly equivalent between them).

Code:
"BAD"                                          
                    BLK/RED        BLK/RED              
7.23k    4.0k    4.3k    4.1k    7.2k    1        1    4.5k    4.4k    4.4k    4.4k    7.2k
3.9k    4.2k    4.3k    4.2k    3.9k    1        1    4.4k    4.1k    3.8k    4.5k    7.26k
                    RED/BLK        RED/BLK              
7.14k    1.1k    1.1k    1.1k    7.1k    1        1    1.1k    1.1k    1.1k    1.1k    7.26k
1.0k    1.0k    1.0k    1.0k    1.0k    1        1    1.1k    1.1k    1.1k    1.1k    7.3k
                                           
                                           
"GOOD"                                          
                    BLK/RED        BLK/RED              
7.33k    1.97k    1.85k    1.82k    7.3k    1        1    1.76k    2.14k    2.0k    1.8k    7.38k
1.93k    1.98k    1.91k    1.74k    2.0k    1        1    2.0k    2.1k    2.0k    2.0k    7.3k
                    RED/BLK        RED/BLK              
7.24k    0.93k    0.93k    0.92k    7.2k    1        1    0.92k    0.95k    0.93k    0.92k    7.24k
.93k    .93k    .92k    .90k    .94k    1        1    0.93k    0.94k    0.93k    0.93k    7.3k


#3) Recall there is one "mystery diode" on the board, sort of near the center (sell AJ34 in the diagram above). On the BAD Display, this reported 450ohm then 500ohm with the leads the other way. On the GOOD Display (card) was reported 350 vs 344 ohm. Probably no issue on this component, IMO.


All the actual resistors had comparable resistance in both directions (marked as R1 to R5).


The "black monolith" capacitors were also measured for resistance. Not that that's an effective reference, but still, the values were comparable: ( the component is identified as "951", but is installed in different directions across the board, so I tried to mark the negative side with "(-)".

Code:
                        "BAD"            "GOOD"                    "BAD"        "GOOD"
9(A)    51(-)        382    424        458    500            R1(A)    35.4        35
9(-)    51(B)        422    382        500    455            R2(B)    27        27
9(C)    51(-)        382    423        456    500            R3(C)     81        81
9(-)    51(D)        423    381        500    455            R4(D)    240        260
9(E)    51(-)        218    260        139    148            R5(Z)    214        207


#4) As mentioned, there may be a middle "power/ground" layer of the board. The numbers (in the diagram) for the black DIP/IC chips is correct, but the numbers for the "IBM tin can" components are all made-up and tentative (as far as order/sequence of pins).
 
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These charts are fantastic! Do you think you could label the DIPs with their identifying number? So 2392122 for the hex driver we've spent so much time looking at, for example?

Nearly all of the capacitors on this board are likely to be "bypass" or "decoupling" capacitors (same thing, different word). The deal with these is that they're like local reservoirs of energy for nearby ICs. When circuits inside an IC switch, the IC's power demand spikes for a moment, and without the decoupling cap there, the little "tug" on the power line could disrupt other ICs or potentially the switching IC itself.

In order for the decoupling cap to be a little power reservoir, this means that one pin will sit on +5V and the other will sit on ground. The resistance you're measuring is therefore not the resistance internal to the cap itself, but instead the resistance of the complicated network of components connected between power and ground; all the ICs on this board through different lengths of trace and so on. For this reason it's hard to interpret the values you've recorded or to know whether they are significant in some way.

In general, measurements for components in-circuit (except for simple things like continuity) is always a little difficult to interpret! (This may be part of the trouble with the 2392699 parts.) It is still worthwhile in case it turns up a clue (a dead short would be a really important sign of course!).

The numbers (in the diagram) for the black DIP/IC chips is correct, but the numbers for the "IBM tin can" components are all made-up and tentative (as far as order/sequence of pins).

There's a chance that there is a convention for this --- or probably several. I know there are a couple of "official" pin numbering schemes for the smaller 16-pin SLT etc. cans. Hours of poking around in IBM docs on Bitsavers might have the answer, but I'm not sure we have much use for that information at the moment...
 
Here is an updated chart.

The numbers at the top right (near "*2") are somewhat meaningless, since I don't think those are resistor packs -- but the values correspond to what the "GOOD" Display measured.

And as before, "*1" is just a note that the resistor values in this pack do correspond to the adjustments of the Jumpers right above that pack.

Throughout the board, when you hold it up to a light, there are these kind of single pad "islands" here and there. These are "square only" pads without the usual circular solder pad, and as you can see, it turns out they corresponded to either GND or 5V spots. Elsewhere I may annotate GND with "(-)" or 5V with "(+)".

1656308504267.png
NOTE: To be clear, the above diagram is from the perspective of the back side, *not* the component side (so component side would be opposite left/right).

The purplish-colored spots are my marking notes while trying to trace out all the pins of the 239 2122 chip. I started marking these with letters (A, B, C, D, ...). A tentative summary of that tracing is as follows (where the letters K, J, A, B, etc. correspond to my markings in the diagram above -- sometimes I use PRIME, like D', to indicate it continues on past a component like a resistor):

1656308989901.png
 
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The big news is that I managed to monitor that PIN 3 on the 239 2122 chip, on both cards (per the notes above, that's an input from the IBM 2462313 tincan).

Long story short: it's SAWTOOTH ! This is still with my cheaper scope - it shows a minor difference between the signal, but still: this is just a practice run, which I'll do again when the new equipment shows up.

1656309375153.png


I've probed around for more places where PIN 3 or PIN 4 might be connected to, but still no luck (just might not be any other traces).

So it's still a mystery while D02 looks dead silent on the "GOOD" Display card and not on the "BAD" Display card (if the chip should basically just be relaying this input signal). But also as mentioned, it may not matter anyway, since we don't really know what this Printer Clock Pwrd output is used for (if it is actually Printer related, that's a device we're not concerned about - maybe the Display somehow sets the "pace" of the printer, by relaying a portion of the clock? Or the word Printer here means something else? Or it's a mis-print in the SLM documentation, which seems not likely).

EDIT: Well, another possibility is we have the pinouts of the 239 2122 wrong - but that too seems not likely. I am going by the TI SN7417 PDF that @stepleton was able to cross-reference to.
1656310103683.png
 
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(Barely any time to read vcfed this evening!)

Huh, interesting. So at this point it looks like the input to the driver IC is the same for both cards. (How did you manage to record from Pin 3?) This is in some ways a relief, since it diminishes the likelihood that the problem isn't in the can IC, at least based on this evidence.

I suspect that our pinout assumptions are correct, but cannot commit to 100% certainty yet.

I'm suspicious of this sawtooth shape and wonder if it's an artifact of your scope. A different scope will let us know for sure. It may well resemble a square wave when we measure it with a higher performance instrument.

Thanks for plotting out the complete connectivity of that chip, and that very neat diagram. I think this is info we want to know. Based on all you've shown here, I'm surprised we don't get the sawtooth D02 output on the good board? It seems like we should see it if Pin 3 has the same input and pin 4 has the 750 ohm pull-up.

Some musings:
Pins 8 and 9 are typical for an unused driver input/output pair: ground the input.
Pin 2: a 108-ohm pull-up with no apparent output is odd. That's a very small resistor for a pull-up, too!
Pin 6: wonder if there's a missing pull-up somewhere?
Pin 10: I am surprised to find the +6V reference connected to the driver IC --- it's not really connected to anything on the schematic...
 
The digital scope recording looks like an R-C integrated square wave, which happens when the scope's bandwidth is too low for the signal frequency being measured.

Looking back at the CRT screen image you posted, it looks like there is loss of H synchronization and the VDU's H scan rate is crawling with respect to the screen data.

The first thing to check when you get a scope with a better bandwidth, is the video signal feeding the VDU, it may have corrupt or absent H syncs and that may lead directly to where to look for the fault on the pcb.
 
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