1
\$\begingroup\$

I have a circuit I traced out off an old, dead PCB that came from the early 90s. It was using TTL chips back then. One thing I noticed is that it has a dual flip-flop. One flip-flop is being used as a reset switch and the other is just unused.

I know with CMOS, all inputs need to be tied to something, so I was wondering what I tie the PRE, D, CLK, and CLR to (high or low) on the unused flip flop, if I decided to go to a 74HC74 instead?

Also, do I need to do anything with the PRE and D on the used flip-flop, if I go to a 74HC74?

Lastly, if I wanted to convert the 74LS193 to a 74HC193, do I need to do anything pullup or pulldown with the A-D inputs, the DN (Down) or LOAD inputs?

Also, I am intending to convert the whole circuit to CMOS HC logic to avoid the whole "input voltage issues".

enter image description here

\$\endgroup\$
1
  • \$\begingroup\$ Whatever is connected to P17 and U6.8 might have stability issues. TTL have different input thresholds than CMOS. \$\endgroup\$
    – Lior Bilia
    Oct 16, 2023 at 18:04

3 Answers 3

3
\$\begingroup\$

With bipolar TTL such as 74LS, unconnected inputs will appear as a High, so for the CMOS parts, I would add pull-up resistors or direct connections to Vcc on the inputs that were not connected in the original circuit - including the inputs on the unused section of the 74xx74.

You should also add pull-up resistors on any TTL outputs that are feeding CMOS inputs, otherwise a TTL High level may not be recognized as a High by the CMOS part.

\$\endgroup\$
2
\$\begingroup\$

TTL (including LSTTL) inputs default to a logic high when not connected. However, it's bad practice to leave them that way as they are susceptible to noise. LSTTL inputs can be tied directly to Vcc or GND. Typical practice held over from the 74xx and 74Sxx days is to use a 10k pull-up or 100 ohm pull-down, a strategy that persists as it helps testability.

More here: 10k Resistor Pull Up/Down Standard for 74 series chips

Now you mentioned you're considering converting the circuit to CMOS. There's two issues with that:

  • Unused inputs need to be tied off. It's ok to tie CMOS inputs directly to Vcc or GND, no resistors needed, unless you want them for testability.
  • CMOS (AC, HC) input-high levels aren't compatible with TTL: CMOS needs 2/3 Vcc (3.33V) as Vih, which is higher than what LSTTL guarantees (2.7V).

Related: In general, is 5V CMOS compatible with 5V TTL?

The first trouble spot in your diagram as shown is where U6D (74LS00) is driving U1A (74LS74). That's not going to work with an 'HC74 or 'AC74 part: the 'LS00 output-high level isn't enough. You can fix this one of two ways:

  • Use 74HCT74 or ACT74 device which have TTL-compatible input levels, or

  • Add a pull-up (about 2k or so) to the 'LS00 to bring bring its output-high up to 'HC or 'AC CMOS level.

Then, tie the unused CMOS inputs to Vcc or GND as needed.

Another trouble spot: if you replace the 74LS193 with an HC type, make sure the UP pin gets a high enough level (2/3 Vcc or better), or use an HCT device. And again, tie off the unused inputs as you need them.

And finally, a really janky trouble spot: the U7 reset chip is open-collector, yet there is no pull-up on its output. Instead, it is relying on the ‘LS74 "R" input for its pull-up, and it’s tied together with the ‘LS163 "CO" through a resistor as a rather stupid kind of wire-or. That’s terrible design, even with LS, and won't work with a CMOS 74HC or HCT74. Instead, replace ROFF4 with a Schottky diode (BAT54 or similar) and add a pull-up to the U7 reset chip output.

\$\endgroup\$
0
\$\begingroup\$

So, if I go to change out all the LS TTL chips to HC CMOS chips, for the unused 74HC74, I need the PRE, D, and CLR pulled high, while the CLK needs to be pulled low. On the 74HC74 that is being used as a reset switch via simple 2-state FSM — "run" and "stop", I need to pull up the PRE and D.

On the 74HC193, I need the A, B, C, and D inputs pulled low. But the DN and /LOAD need to be pulled high.

Circuit with inputs properly set

\$\endgroup\$

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.