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schematic

I've come across the above schematic in a datasheet for a 4x2:1 bus switch. What exactly does that triangular symbol on S mean? It looks a lot like a NOT gate, but the third leg is confusing me.

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  • \$\begingroup\$ That 3rd (bottom) leg is a copy of the input. Don't know if is buffered or not, though. \$\endgroup\$ – Dwayne Reid Aug 10 at 20:35
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    \$\begingroup\$ It is the S input but the symbol is trying to pass on the idea that it has the same delay as the not-output, thus preventing spikes when switching. \$\endgroup\$ – Oldfart Aug 10 at 20:36
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It is a gate with an inverted and a normal output. The idea is that the two outputs switch exactly at the same time. There is hardly any delay between them.

The symbol as shown in your diagram is rather awkwardly made. More often the following symbol is used for a combined buffer + inverted like that:

enter image description here

You will find these used with differential line drivers.

The following diagram has an issue that the S input bypasses the inverter.

enter image description here

If this logic was used as depicted, the gate connected directly to the S input would switch a fraction faster to the new state then the one which uses the S-NOT from the inverter.

Most manufacturers don't bother with that sort of details. Here is a typical diagram of a 4 output de-mux:

enter image description here

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That is a very ambiguous symbol. It probably means that the buffer has both an inverting and non-inverting output. As Tom Carpenter comments it goes to the input of another gate so it must be an output.

There are two common variants of buffers.

  1. A buffer with tri-state output and an enable pin.

  2. A buffer with true and complement outputs. The symbol is usually drawn symmetrically. The virtue of these devices is that there is low timing skew between the two outputs.

enter image description here

enter image description here

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    \$\begingroup\$ In this case it is unlikely because the enable signal is an input which is connected in the circuit to another input only. \$\endgroup\$ – Tom Carpenter Aug 10 at 22:03
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    \$\begingroup\$ I agree - that was why I said it is more likely to be a complement output. \$\endgroup\$ – Kevin White Aug 10 at 23:53

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