I have a data line, to which 4 smaller devices (which each write) and a "master" (which reads from the line) are attached. Only one device should be writing (e. g. connected) at a time to the data line, so I thought to use an IC with a transmission gate in it to leave the line disconnected if the current device is not active (there are other lines which determine the active device, clock and other data). Since the devices are only writing to the wire, I think that the same could be achieved using a multiplexer, but that would require each device to have a separate cable to the multiplexer, which I am trying to avoid. So my question is: Is this the right way to make only one device write to the line? And following up: I have not found any IC containing only transmission gates (using a quick google search), are they sold under a separate name/only included in complexer ICS? I'm quite new to this

EDIT: I think I choose a confusing wording here: So by "devices"/"master" I meant some AND, NOR, Flipflops etc. assembled onto a breadboard. The boxes represent the breadboards ("master" and "devices", to make the drawing simpler I added only two "devices") with gates (shift registers, counters, and-gates etc. but nothing "sophisticated" like Arduinos).

Build option 1 This is how I currently plan to make it: The "master" breadboard (upper middle) either enables "breadboard 1" or the other. If "breadboard 1" is enabled it can set the output (e. g. the "data line") to 1 or 0, while "breadboard 2" has to make it's own output 0 (so that the or-gate takes only the output of "breadboard 1". If "breadboard 2" is enabled, the opposite is the case.

Option 2 The second option uses switches (or what I intended to be transmission gates). If "breadboard 1" is enabled, S1 is closed and S2 open. The downside is that if neither breadboard is enabled, the line is left floating. But on the other hand, adding new breadboards is easy, and I do not have to use an OR gate. So to rephrase my question: is the second option a good option, and what kind of IC contain this logic (e. g. can you give me example?)

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    \$\begingroup\$ They would be called analog switches which are a bit more than just generic transmission gates. But it is unclear why you think a mux is extra work. You would need a transmission gate or switch for every IC anyways. Outputs can be back driven and fight. Absence of activity does not necessarily mean the line is not being driven. If you don't disconnect them then you need to be able to set them to be high impedance, and if you can do this then switches or muxes are not needed at all. Does "not active" = high impedance? Provide a block diagram of your proposal. Not clear what you have in mind. \$\endgroup\$
    – DKNguyen
    Aug 10, 2022 at 14:14
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    \$\begingroup\$ I currently do not have time to do some drawing, I will add a diagram in the evening \$\endgroup\$ Aug 10, 2022 at 14:23
  • \$\begingroup\$ Basically, I can't tell if you want to disconnect the ICs from the line and each other or if you want to disconnect the ICs from the line together (that won't work). Unless you have something else in mind entirely. This sentence is the one that has me confused: "would require each device to have a separate cable to the multiplexer". Not untrue but how does that differ from a transmission gate? Is this perhaps not on a PCB and you're talking about spider-web of cables? \$\endgroup\$
    – DKNguyen
    Aug 10, 2022 at 14:25
  • \$\begingroup\$ Also consider established standards for IC - IC communication, such as I2C, SPI, RS-485, etc. Might not have to reinvent the wheel. \$\endgroup\$
    – rdtsc
    Aug 10, 2022 at 14:27
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    \$\begingroup\$ On a breadboard? Then I'll wait for a diagram. Sounds like you're misunderstanding something. \$\endgroup\$
    – DKNguyen
    Aug 10, 2022 at 14:45

1 Answer 1


So diagram 1 is not ideal. You shouldn't go with that especially when you have so much control over the way the output of the breadboard works.

With regards to your second diagram, do you plant for SW1 and SW2 to be on their respective breadboards? If so, that's the only way I can see how transmission gates would save things over a mux (i.e. the mux would have to sit on a fourth breadboard). If SW1 and SW2 do not sit on their respective breadboards, nothing is saved over the mux.

In that respect, if the transmission gate was on it's respective PCB and disconnected the breadboard from the output when it was commanded to be disabled, this is identical to the output going high impedance.

You can use "analog switches" for that (which contain transmission gates inside but packaged to function as a switch than a bare bones transmission gate).

You can just add a pull-up or pull-down resistor to the output to prevent it from floating if all boards are disabled.

  • \$\begingroup\$ Yes, SW1 and SW2 are on the breadboards with the rest of the gates, so I'll go with option 2. Thanks for the hint :) \$\endgroup\$ Aug 11, 2022 at 12:10

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