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Homework Help: DEMUX (Help me see if its correct!)

  1. Oct 13, 2009 #1

    I have this diagram and I assume it works like below, I drew a Truth Table, so please reassure my assumption whether its correct or not.


    2 Tables, top for Enabled, below for Disabled.

    1) When the DEMUX is enabled and a Binary Counter is continously pulsing through it, all the LEDs will be Light Up and the Output will be the LED that is Not Light Up and running through all 8 Output LEDs. This is because the DEMUX has active low outputs where Logic 1 does not light up the LED.

    2) When the toggled is switch on, the DEMUX is disabled. All 8 LEDs are Light Up. However, as the Binary Counter is continously pulsing through it, the Enable/Disable state switches instantly. So the running LED continues to run, but skips when it (For example we use Switch I0) reaches I0 and LED I0 never switches off because when the DEMUX is enabled, it is Switch On and the running LED never reaches it because when it reaches its turn and when the DEMUX is disabled, it also remain Switch On.

    Am I right? I just need reassurance?

    And also 1 more thing I want to confirm is for active low activation, a 1 is still required for activation right? Its just that a 0 is needed to be pulse in for it to be activated. But the 0 is than conveted into a 1 and activates.

    For example for an active low enabled, E = 0, so E* = 1 so it activated

    E = 1, so E* = 0, so it does not activate right?
  2. jcsd
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