Ok thanks for the help, I really appreciate this. I was meticulous when soldering, I checked for both continuity and shorts once I'd cleared all the holes and again once I'd installed the socket. All the joints are neat and tidy and are in good order and I've confirmed everything at E6 is connected to where it needs to go. So I don't think it's a problem with the installation.

But on rechecking C2 I find that pin 8 is now low at 0.08V, regardless of whether E6 is installed or not. Perhaps I misread before. So for clarity:

C2 (without E6 installed)

pin 1 - 0.16V
pin 2 - 0.16V
pin 3 - 4.54V
pin 4 - 3.97V
pin 5 - 4.46V
pin 6 - 0V
pin 7 - 0V
pin 8 - 0.08V (this becomes 0.24V when E6 is installed)
pin 9 - 15.625KHz at 5.1V
pin 10 - 4.54V
pin 11 - jittery signal at around 1V
pin 12 - as pin 11
pin 13 - 4.54V
pin 14 - 5.11V

And for good measure I rechecked C5 and C6, which are as they were before, working fine. And A1 is still unchanged, no matter if E6 is installed or not.

One difference I have noted is that C7 now has a 15.625KHz signal at pin 1, whereas before it did not. I am wondering if this is why I now see a similar signal on the vertical sync pin.

C2 pin 8 being stuck low, which I think is caused by its input pin 10 being stuck high, brings me back again to B1 - a NAND gate with some of its inputs forever being low. And those inputs are all coming from A1.

I get the feeling I'm chasing something around the board and just narrowing things down as I go. I don't mind replacing ICs as they're cheap and it's easy to do, but I do hope it's leading somewhere! BTW I don't think this is connected but I've also noticed that one of the RAM chips isn't getting hot like the others, so I suspect that's dead. And some of the ROM chips are quite cool, so perhaps there's some issues there, but again I think the problem we're trying to solve is more fundamental.

Again:

C2 pin 10 is high.
C2 pin 9 is 15.625KHz @ 5v.

There is supposed to be a signal at the output pin 8 of the C2 NAND gate.

In the schematic, C2 pin 8 _only_ goes to E6 (74100) pin 12 and pin 23.

If there isn't a short circuit due to soldering, and if E6 isn't in the board,
and there still is no signal at C2 pin 8, replace C2 with a SN7400 or SN74LS00.

Ok thanks, you're like a teacher guiding a student I see where I'm going wrong, I'll replace C2. Some of these ICs are still sold new so I've ordered new.

PS attached a pic of my soldering on E6 just so people can see where I am

Soldering looks nice and clean.

SN74LS00 drains less current from the power supply compared to SN7400.

SN74100 and SN74177 are very exotic, but the rest of the TTL chips in that schematic still seems to be in production.


Don't worry: learning the basics isn't too difficult.
The datasheets usually nicely expain, what a TTL chip does.
If you had tinkered with Lego (TM) bricks as a kid, you are set...

I can find some 54100 chips still around and they're just higher tolerance versions of the 74100, so if the ones I bought don't work out I still have options. Breadboard might be a simpler way for me to test using equivalents if need be (before permanent installation).

Interestingly, the datasheet for C2 shows an expected output voltage of between 2.7-3.4V, which is well below the output I measured before it appeared to stop working. That makes me wonder if that's why E6 was only half working - simply not enough voltage on that clock.

For SN74\SN74LS\SN74HCT, the logic level threshold at the inputs is ca. 1.3V.
Means a voltage below 1.3V would be interpreted as LOW, and a voltage above 1.3V would be interpreted as HIGH.

In the datasheets, input voltage is specified <0.8V for LOW and >2.0V for HIGH.


SN74\SN74LS outputs are good at sinking current to GND,
but they are less good when it comes to sourcing current from +5V.

The HIGH voltage of a SN74\SN74LS output depends a lot on the resistive load to GND at the output.

TI SN74LS00 datasheet, PDF page 6: 6.6 Electrical characteristics SNx4LS00:
Vho=2.5V minimum, 3.4V typical when 0.4mA flows out of the output at VCC=4.5V.


In the good old times, when you ordered SN74 chips and the manufacturer had not enough of them, you sometimes got SN54 chips in ceramic package instead.
Hmm... to be on the safe side, thou shalt not export/sell SN54 chips to certain countries...

When not driving any load, or where the load tends to pull-up the line to +5V, you can get TTL digital outputs that are near to +5V...

Mark

The way the TTL circuit is arranged internally, the highest it can pull up will be more than two diode voltage drops below Vcc. However, it will let you pull the output up externally like with a pull-up resistor to Vcc. The problem with that is that the last part of the rising edge can be pretty slow. Using a matched transistor pair like the Nexperia BCM62B (thanks to Dr Jefyll for pointing it out) may be fast enough to be a good help. You'd connect it to act as a current mirror, so the resistance drops as the voltage does, to keep the pull-up current constant.

_________________
http://WilsonMinesCo.com/ lots of 6502 resources
The "second front page" is http://wilsonminesco.com/links.html .
What's an additional VIA among friends, anyhow?

Garth, My post that you quoted was in answer to the Pet 2001-8 guy‘s post:
And some of his other results where clearly the lines are static, but he is getting a voltage greater than 3.8V.

I agree with you on the slow rise time if the only circuit element that is providing or helping to provide a logic high to +5V is a resistor or similar. I was not suggesting a resistor be fitted or used.

Mark

My word! I quake in horror at what those darn Ruskies might learn from an SN54LS00, or what weapons they might build if they got enough of them....

_________________
Curt J. Sampson - github.com/0cjs

And for reference here is a typical TTL schematic.


When the output is logic high, Q1 and Q2 are off. Base current for Q3 is provided by resistor R2 causing Q3 to be on. As Q3 is in an emitter follower (common collector) configuration, it’s emitter will be around 0.6V below the voltage at it’s base. As the current to the base via R2 is a low current, there will not be much voltage drop across R2. So Q2’s base will be near to the supply voltage. So Q3’s emitter will be around 0.6V below the supply voltage.

Then you will also have the forward voltage drop of the diode (D1). At very low output currents, the volt drop across R3 is not significant. But at greater currents, R3 will proportionally drop a greater voltage as the current increases. Hence the data sheets stated output voltage range. Note however that these voltages are steady state / static values. For very fast changing signals, the switching on and more importantly the switching off times of the transistors will affect the operation.

Mark

Yes, but it's better to avoid scenarios where Prostetnic Vogon Jeltz might show up with a poetry book, asking if you had filled out all of the export forms correctly.

Dieter, is Vogon poetry actually fatal? Or does it only make you wish you were dead?

_________________
In 1988 my 65C02 got six new registers and 44 new full-speed instructions!
https://laughtonelectronics.com/Arcana/ ... mmary.html

Quite a few chips on the board have inputs tied to the 5V rail. In fact when I socketed C2 today and was checking for shorts, I noticed that C2 9 and 14 are shorted. I thought I'd made a mistake but no, the board is designed that way. Pin 14 goes straight onto the 5V rail, and pin 9 heads over to D8 where it too connects to 5V.