Looking at the internals, sorry but no.

The following posts prove I should not have posted.

To me looking at the internals they don't connect in the same order, I still look at the pics over and over until I move on.

My apologies for misleading the OP.

Learning from the continuing posts

Can you elaborate? The 74HCT240 and 74LS240 are functionally (logic) the same. They may not be interchangeable for this board, but there would have to be another reason.

Screen Shot 2023-03-04 at 5.44.20 PM (resized).png

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https://datasheetspdf.com/pdf/1392859/nexperia/74HCT240/1

https://datasheetspdf.com/datasheet/74LS240.html

Would need to see the schematic.

74LS drives high to 3v. 74HC(T) drives high to 5v. 74HC (and 4000 series) needs 5v level input. 74HCT was designed to work with a LS parts and works with a 3v level input.

Certain cases like a HCT240 is going to a mosfet gate that needs a 5v level to turn it on, then would be a no on a LS part. If all the LS240 outputs are going to other HCT or LS inputs it will be OK. If the LS240 is going to 4000 series or 74HC inputs, then it may not work right without pull up resistors.

FWIW, I found this Rottendog MPU080.v1 schematic posted in another thread.

pinside.ff507a935375b3db78d98268c66ecb2d85d0e410.pdf

That chip is strobing the switch matrix. The original board had plain 7400 series stuff in the switch strobe and returns. The rottendog has 74HCT parts. I think it will be OK with a LS240 inverter doing the strobes since the return is a HCT part, LS output can interface with HCT input safely (but not LS -> HC without pull up).

You could probably find some opinions/theories about which logic family is best to drive and return the switch matrix, but I don't see any reason why it wouldnt work with the LS240 part.

That was my initial thought too, but I am not an expert in this area. I hope the OP sees these responses.

In the old days we called that fanout. It's whether the output device can source or sink enough current to provide logic low or high to the input(s) of the next device. If I remember correctly the CMOS logic chips were very easy to drive because of the very high input impedance.

TTL chips require higher current to drive.

Correct -- Fan-in and fan-out. For this part, inputs are 1 to 1 from a 6532 so fan-in is not a problem.

My usual over-analysis --
To eliminate fan-out as a problem:
If no column switches are closed in a single row then there is zero load on the 74x240
If all column switches are closed in a single row then there are eight loads on the 74x240
Therefore the 74x240 must sink current for eight loads on a single output.

Each load is one NAND gate input plus one 4.7K pull up resistor.
The 74HCT00 NAND gate has zilch for low level current input load so the resistor is the actual load for the 74x240.
The return line (NAND gate input) must be below 0.8V at each Z13/Z14 NAND gate input.
Using that target voltage - there is a (5V - 0.8V) = 4.2V drop across each resistor in SIP4.
4.7K resistor with 4.2V drop ==> 893uA thru each resistor.
Up to eight of these can be 'switched' at any one time so that means the 74x240 column driver must sink 7.1mA.
If the output device is a 74LS240, the 74LS240 can sink up to 24mA on each output so no problem.
If the output device is a 74HCT240, the 74HCT240 can sink up to a specified 6mA on each output. In theory, the 74HCT40 will not work based on 74HCT240 vague specs and worst case conditions. There are several factors built into the specs for a 74HCT240 such as operating voltage and how many outputs are driven at the same time, operating temperature, etc - so the acceptable sink current can be much higher and there is no easy answer on this one.

In real life, the odds of having all eight columns switched to a single row are slim to none ... but is still possible.
Boards should always be designed for worst case. If I were replacing that single part, I would go to a 74LS240 instead of the 74HCT240.

Williams presumably ran into this when a bunch of drop targets where down (black knight?) and modified series resistors on the driver board down to zero ohm to make it work.

I don't think fanout is a problem in the direction he is going as you can drive pretty much as many 74HCT inputs from a 74 or 74LS output as you like, because they essentially draw zero current. The book says a 74HCT output will reliably drive about three 74 plain inputs and about ten 74LS inputs.
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Fanout normally isn't a problem especially if you are sourcing to HCT parts which are only a 1uA load. The problem occurs when there are pullup or pulldown resistors connected to the load. Each of these row drivers ("strobes") have eight columns ("Returns") tied to them. And each column has one 4.7K pullup connected. The pullup resistors in this design can amount to a large current draw if multiple switches are engaged at the same time. For full load (all switches closed for a row) - the pullup resistors alone will add 7.1mA of sink current (I-OL) to the 74HCT240.
74HCT240 outputs are spec'd at 6mA for both high (source) and low (sink) level outputs and the outputs on this board have a 7.1mA load. I would say this has a fan-out problem. Incidentally, driving ten 74LS loads (with no pullup/pulldown resistors) = 4mA and three 74 loads would be 4.8mA. So, yeah, it can drive ten 74LS loads or three 74 loads.
Although having eight simultaneous switches closed on one output is unlikely to ever happen, it is still possible.

However, I can see an even bigger problem with the designer's choice of 74HCT32 for switch decoder (Z15). There is a far greater chance of having eight loads on each row selection from the 74HCT32 yet the 74HCT32 has a much lower current sinking capability at 4mA. The load on each 74HCT32 output can be nearly 2x the rated output load.

Two other places where pullup resistors became an issue:
1 = Williams games that used the QVE11233.00whateveritwas opto. They used such a low value of pullup on this that the opto had problems sinking enough current which resulted in a too high of value for opto output voltage.
2 = Bally -35 boards and a 150 ohm pullup resistor on output of U15D. For this reason plus a decent number of loads, they had to use an MC3459 (high current driver). Bally could have easily used a more common 74S00 (or 74F00) for that part and omitted that hefty pullup resistor.

Interesting that Williams went to zero ohm resistors. Those must have been mounted in series with the load.