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Once upon a time I knew the answer, or could just try it. But with distance I forgot. When I use an open collector, I do it to drive some load. So High means current is flowing. But when it is used as a signal voltage output along with a pull-up resistor, then High means current is not flowing and the pull-up resistor pulls the output pin high.

So if I want to actually run current through it in the High state of the input, do I need an open-collector buffer or an inverter? To add to the confusion, some parts descriptions say "octal buffer/line driver, inverting outputs" (74LS756). Why not call it an inverter then?

If you held a pistol at me to guess, I would guess I need an inverter, because in TTL speak Low means sinking current, and High means potentially sourcing current. But since open collectors are used to drive current, I could just as well be wrong.

What is the right logical argument to imprint into my mind to never again forget this coin-toss answer?

Below is a schematic.

TTL IN H -> LED lights up, TTL IN LOW -> LED off.

schematic

simulate this circuit – Schematic created using CircuitLab

In this schematic I represent my gate as a buffer, not an inverter, and my question is whether a buffer or an inverter needs to be used.

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    \$\begingroup\$ A part that's called a 'buffer/line driver' has high current outputs, much higher than normal logic, so this is what's getting the headline in the description. \$\endgroup\$
    – Neil_UK
    Jul 31 at 9:45
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    \$\begingroup\$ You're confusing "on/off" (a semantic meaning) with logic high/low (a voltage level). Logic can be active-low for example, in which case, for TTL for example, a voltage of 0-0.8V would be "on" or "active". It's additional meaning we give the signal, not the signal itself. Given this clarification, which case do you want for the input, and what for the output -- and what is the output wired as? Open-collector/drain usually implies a low-side switch and high-side load. \$\endgroup\$ Jul 31 at 9:47
  • \$\begingroup\$ Please can you edit your question to add the new info' and add a schematic, don't post new info' in comments. Thanks. \$\endgroup\$
    – TonyM
    Jul 31 at 9:51
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    \$\begingroup\$ You haven't added a schematic showing what your buffer or whatever is connected from and to, just some diagrams from datasheets that most of us know anyway. All that stuff can be removed. Instead, please add a schematic that shows all of what you're trying to connect up so we understand the now-confusing question clearly. \$\endgroup\$
    – TonyM
    Jul 31 at 11:28
  • \$\begingroup\$ @TonyM I added the data sheet schematics to develop my answer since no full and credible answer has been forthcoming so far. All these questions aren't zeroing in on the answer if H -> sink current through the open collector gate requires a buffer or an inverter. \$\endgroup\$ Jul 31 at 12:23

2 Answers 2

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Yes, sounds like you need an open-collector inverter. Or just a plain old transistor:

MOSFET switch

(V2 can be most any logic level signal, and M1 must be a logic-level type MOSFET. V1 and Load are unimportant, so long as M1's ratings are respected.)

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  • \$\begingroup\$ so, basically you are telling me to forget about open-collector and just use an IRLZ44N MOSFET. This is not a bad suggestion, because I need to get into MOSFETs at some point. Funny enough, while they should be simpler to use than BJPs I have not been very successful. Partly because I just don't know what specific MOSFET I should get. You proposed a part number, much appreciated. \$\endgroup\$ Jul 31 at 12:02
  • \$\begingroup\$ on the other hand, this MOSFET is big, a TO220 package. Is there a TO92? Or bus integrated chip? Sure with lower ratings. But that would be fine. \$\endgroup\$ Jul 31 at 12:04
  • \$\begingroup\$ @GuntherSchadow What load are you switching? What signal source? \$\endgroup\$ Jul 31 at 12:05
  • \$\begingroup\$ It doesn't matter. this question is about the principle. So let's just stick to low power LEDs just to clarify the principle. Nothing more. \$\endgroup\$ Jul 31 at 12:14
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    \$\begingroup\$ Ok, well ignore the MOSFET type as it is similarly not important; the diagram is just a likely suspect I found on EESE. Or if you must, a BSS138 will drive an LED just fine. \$\endgroup\$ Jul 31 at 12:28
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OK, I just answer it myself. The clue is to look at the schematics of 74LS05 hex inverter with open collector. Can't get much easier than that:

enter image description here

So when input is Low, it sinks current. That means it pulls the base of the first transistor low, so the first transistor is shut off. When that first transistor is shut off, then the following output transistor is also shut off. So that means no current is sunk through its open collector.

Now, if this is called an "inverter", it means an inverter with open collector is a gate which when given a Low input causes the output transistor to shut off, i.e., when used with a pull-up resistor on the output, it would cause the output to be High.

Conversely, a High input would cause the first transistor to be on, and the output transistor to also be on and therefore sink current.

Therefore, if what I want is to sink a current (e.g., to turn on an LED) on High with an open-collector, then indeed I need an inverter! But the state of the output transistor sinking current is Low. So a Low state on an open-collector output means current is flowing.

Now, to never again forget that, the thing to remember is that in all TTL (and I suppose CMOS logic) for a pin to be Low means it sinks current. For a pin to be high means it does not sink current, and optionally emits some current on its high voltage. This is the same regardless whether its a High driven gate or an open collector gate. Low = sink current. High = output voltage is optional.

I will add the use case that I had in distant memory was those Nixie tube drivers, that say H H L H H H H H H H for "2" in their truth table, meaning low sinks current, output transistor on, so that the 60V or whatever high voltage can be sunk driven by 5V TTL input signal that is high. I think this is also why 3 to 8 line selectors (74LS138) are output active low and this, in turn, I believe, is why all memory chips have come to use active low signals of /CS, /OE, and /WE. One would really wonder otherwise why else the first maker of memory had decided to just flip a coin to make his control inputs active low, he didn't say "let me confuse people in posterity by saying 'if you want to turn it on, give it a zero and not a one!' [evil laughter]", no it's because the selectors were intended to primarily sink current to drive Nixie tubes on higher voltage.

This is also why I think it is misleading to teach that TTL "High means high voltage" it is better to say "Low" means lower potential so that current flows into the gate onto lower ground. "High" means that the potential is high so nothing flows down. Most standard gate outputs can sink way more current in Low state than they can spit out in high state.

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  • \$\begingroup\$ Yeah sure, this answer is so very bad, it needs to be downvoted, by someone who feels his sense of authority being challenged. If it was about the subject, such person would have commented. \$\endgroup\$ Jul 31 at 18:39

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