I apologize if my question sounds stupid, but I'm a novice and trying to explore all of my options. I currently have a 4/16 decoder, but I did not realize it was active-low and what I really need is an active-high decoder. How might I go about fixing this without putting an inverter on every output? Thanks!

(datasheet: https://datasheet.octopart.com/NTE74HC154-NTE-Electronics-datasheet-22129262.pdf)

  • \$\begingroup\$ I'm considering just buying a 16-gate inverter if that exists \$\endgroup\$ Apr 22, 2018 at 13:31
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    \$\begingroup\$ @BrianDrummond I was in a bit of a rush and this was the only decoder available, so I didn't check the datasheet when I got it (should have). Downstream is just some leds with a pull-down resistor, so I can't think of a way to modify this to also be active low. Any ideas? (Might just go with 7404s) \$\endgroup\$ Apr 22, 2018 at 13:51
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    \$\begingroup\$ Is that all? Run the LEDs with pull UP resistors to +V. That's how it was always done in the TTL (good at pulling low) days. \$\endgroup\$
    – user16324
    Apr 22, 2018 at 13:52
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    \$\begingroup\$ @BrianDrummond Would that look something like this? (4/16 decoder output pin)------(resistor to V+)-------(led) \$\endgroup\$ Apr 22, 2018 at 14:02
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    \$\begingroup\$ @CalebWhittington yes, although it's more traditional to put the resistor on the VCC side of the LED. \$\endgroup\$
    – Alnitak
    Jun 15, 2018 at 15:56

1 Answer 1


Just invert the logic of the downstream circuitry.

To drive a LED, do this:


simulate this circuit – Schematic created using CircuitLab

(BUF1 is just any of your decoder's output pins) The resistor and LED can be interchanged.
That's the way TTL circuitry (good at pulling low) was always used.


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