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Taken from Practical Electronics for Inventors is the following diagram: enter image description here

I’m little confused here. The pinout diagram shows pin 1 with \$\bar E_a\$. The functional diagram also shows a bubble with the corresponding pin with the inner label being \$E_a\$. So does that mean that we will have to supply inverted input, i.e., \$\bar E\$ to the pin?

Similar questions for the output pins.

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    \$\begingroup\$ First thing to realize is that your excerpted picture is in error, as on the right it mistakenly gives pin 15 different treatment than pin 1, when actual data sheets and the diagram on the left show them to be the same. Next is that there's little reason reason to use this in a new design today, instead for a demultiplexing role you would fill this need and preceding or following functions in a cheap MCU, and things like parallel address bus decoders are mostly extinct in small systems except inside multifunction ICs. \$\endgroup\$
    – Chris Stratton
    Sep 3, 2019 at 17:32
  • \$\begingroup\$ it means that the input must be low for the device to activate ... it does not mean that you have to supply an inverted input \$\endgroup\$
    – jsotola
    Sep 3, 2019 at 17:44
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    \$\begingroup\$ Atom - Just FYI, that book (at least the earlier editions) had several mistakes. I suggest you search for {"Practical Electronics for Inventors" errata} (remove the {}) for whichever edition of the book you have. At least some errata lists are linked on the EEVBlog. That will help to avoid you getting confused by something which is actually a mistake in the book. \$\endgroup\$
    – SamGibson ♦
    Sep 3, 2019 at 19:40

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That notation is indicating the "Enable" pin expressed by "E" is active low. Meaning to enable each decoder, you must pull the "Enable" pin low. Check out the pin descriptions on page 3 of the datasheet here. This type of notation is common. You will see this type of thing implemented in several ways across different IC manufacturers. i.e. \$\text{Enable}\$, \$\overline{\text {Enable}} \$, \$\text{SD}\$ (Shutdown), and \$\overline{\text{SD}}\$ (barShutdown). You just have to the reason through the language. \$\text{Enable}\$ is active high enable, \$\overline{\text{Enable}}\$ is active low enable, \$\text{SD}\$ is active high shutdown, \$\overline{\text{SD}}\$ is active low shutdown.

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    \$\begingroup\$ Actually here, it is not the enable input as in multiplexers. Rather, it here stands for the input signal for the demultiplexer. \$\endgroup\$
    – Atom
    Sep 3, 2019 at 17:00
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    \$\begingroup\$ You may chose to use it for data, but it is effectively an enable. It's really a matter of your perspective. The enable role does help explain the depiction as active low. Many data sheets will call this a "Decoder/Demultiplexor" \$\endgroup\$
    – Chris Stratton
    Sep 3, 2019 at 17:13
  • \$\begingroup\$ Yup, both are perfectly reasonable ways to look at the functionality of said pin(s). \$\endgroup\$
    – AJbotic
    Sep 3, 2019 at 17:43
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The bubble means "invert the signal". It's the same bubble you see on the output of inverters, just put on the input. So the internal signal is considered to be active high, but because it's inverted, the external signal is most definitely active low.

Somewhere in that datasheet there should be a truth table, which makes it explicit. If it's not in that datasheet, there will be a better one from a different manufacturer.

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  • \$\begingroup\$ Part of the implication of the inversion at both input and output is that the deselected outputs which might by hunch or even convention be expected to be inactive or even "zero" are actually going to be electrically high on this part. \$\endgroup\$
    – Chris Stratton
    Sep 3, 2019 at 17:29
  • \$\begingroup\$ I was just addressing the input -- but yes, the outputs are also active-low, which fits the active-low enable inputs on most memory chips at the time that the '139 hit the market. \$\endgroup\$
    – TimWescott
    Sep 3, 2019 at 17:49

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