You can do this, but normally you wouldn't, as doing so will increase the power consumption if the inputs are switching.
Then just tie the inputs to a rail and leave the outputs floating, that is standard practice.
That's because it's not standard practice.
Dave :)
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David L. Jones
You can often do lots of useful stuff with non-standard practices! :->
BTW, I hope you didn't mean just connecting the inputs together and
*not* connecting them to a rail as well? That's bad.
Actually, it can be good practice to connect the inputs via high value resistors (say 1M) to ground (or Vcc) instead of just tying them hard to the rail. In this case it means that you can use the gates at a later stage without cutting any PCB tracks, simply wire your signal up with some mod wire. If cost and board space etc isn't an issue then it can come in handy. Oh, and if you are going to hard tie the rails, then leave some track length so you can cut the tracks if you need to. If they are connected via tracks under the chip then you have to start lifting IC legs to isolate the pin if needed - yuck.
Dave :)
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Winfield Hill
Rob said he didn't need extra output, so not paralleling gates is good advice. But your warning against paralleling to solve drive-current requirements should be taken with a grain of salt.
In practice propagation delays between gates within one chip are pretty well matched. There's always a supply-rail shoot-through spike current that occurs when the complementary output FETs of any gate are both conducting. The increase in the spike current caused by whatever degree the paralleled delays are mismatched is modest, because the FETs are only partially on at that moment, and because the time is very short, e.g. one ns. A modest spike increase is an acceptable price for the needed drive capability, which is why this technique is indeed a common practice. When compared to all the other supply-rail current spikes in a full logic circuit, a little more can hardly be of any consequence.
Thanks,
- Win
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mc
I've paralleled 4049 or 4050 outputs many times. My understanding was that the resistance of the CMOS outputs (which is nonzero, I thought it was about
300 ohms) provided protection against heavy current flowing in the case that one of the gates was momentarily out of sync with the others.
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OBones
Why do you want to connect them together ? Unused outputs are left floating, this is not a problem in regular applications. And if you still wan't to connect them together, you need a way to prevent the current from entering the output, or you will most likely "burn" it. Unless it's an open collector device, or it has some kind of protection, of course.
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Winfield Hill
I picked out that comment, which is a good one. If the input is not from a logic-gate output, and the signal transitions may be slow, one absolutely cannot count on the propagation delays among gates within a chip to be matched, because the various gates may have slightly different logic thresholds.
Thanks,
- Win
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Andy
Hi,
No, just connect all the unused inputs to ground.
-- Andy
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Andy
Hi Rob, one should not parallel the buffers/invertors and such, as the skew between the transitions of the output signals of the different buffers/invertors is not specified, so there is potential for creating transitional current spikes across the output elements and for deteriorating significantly the form of the output signal. But in some lax practical cases such paralleling might work.
-- Andy
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Rob
Hi all, I'm using a 74HCT240 tristate inverting buffer, is it ok to connect outputs together? - the inputs are also connected together.
I don't need extra drive, I just have some spare gates.
I could not see anything in either the device or HCT family data sheets in relation to doing this.
TIA rob
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Adrian Tuddenham
If it's a tri-state array, I suppose it might be possible to momentarily open all the outputs just before they changed state. If they came back on with an unequal time delay after the change, at least they would either all be pulling in the same direction or not pulling at all.
I wouldn't like to guarantee it, though.
~ Adrian Tuddenham ~
(Remove the ".invalid"s and add ".co.uk" to reply)
www.poppyrecords.co.uk
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Rob
connect
in
Hi Andy - do you mean "no - you shouldn't parallel HCT buffers" or just that the usual approach is to tie unused i/p's low and leave o/p's floating. thanks rob
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Rob
that
Thanks for the help Andy. rob
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Rob
connect
in
Thanks for the help Dave.
I'm a bit of a master of "non-standard" practices.... : ) rob
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David L. Jones
It's around 0.1uA max for HC stuff, so 1M is fine. Always check the datasheet if in doubt though. 100K is a popular value for this task for CMOS logic.
1K is way too low! If you do the mod (which is why you put the resistor there in the first place) and drive the input from another gate with 5V logic, that's 5mA you are throwing down the drain. These "large" current spikes could have follow-on effects into other circuitry too. If you did use 1K then you'd remove the resistor. Often you'd pick a resistor value that you already had in your BOM, so it doesn't mean another line item.
Dave :)
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Andy
Thanks for that. As with any non-standard practice, one has to consider additional factors. I myself have paralleled such gates, with success, but I did not need clean output edges, the input signal transitions were fast, and these gates were minority in the circuit.
-- Andy
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Tim Hubberstey
It is a good idea, but 1M is too high a value for the resistor. You have to consider the voltage drop developed across any pull up/down caused by the input leakage current. For HC/HCT, this is (from memory) 10 uA, so
100k is the largest value I would consider. Besides, the "harder" you pull the input to the rail, the less chance there is that the leakage voltage at the input will put any of the transistors in the gate into a partially conducting state. I usually use 1k for this since most assembly houses have reels of this value in stock and they tend to be dirt-cheap.
Thats a good idea using a p/up p/down resistor, I should do that. cheers rob
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Rob
Thanks Win, I've used the // gate set up before, just wasn't too sure if HCT was any different. Seems like the best route for this particular job is to stick with the individual gate & tie the inputs for the spare gates low. thnx rob
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Tim Hubberstey
Actually, we were both wrong. I just looked up a couple of data sheets and it's 1 uA max, across temperature. I remembered the 100k correctly though. This gives a maximum voltage of 100 mV, well within the 800 mV worst case for a "0" and low enough to keep "off" transistors off.
We have different opinions on how to do mods. I assumed that the resistor will be removed when the board is modified, then the resistor value has no impact on current drain. Even with a 1M, you are still throwing away up to 5 uA per input if you don't remove the resistor. Besides, this way you have a nice pad to solder your jumper wire to.