4 active low inputs to 7 active high outputs decoder

Discussion in 'Homework Help' started by 0Wafi0, May 15, 2015.

  1. 0Wafi0

    Thread Starter New Member

    May 15, 2015
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    0
    hello there!
    This is m first post here. I'm Wafi and I'm a second year aerospace avionics major. I'm posting my first question about active low and active high inputs and I've struggling to understand this for a while now!
    Any help or push in the right direction would be greatly appreciated.
     
  2. MrCarlos

    Active Member

    Jan 2, 2010
    400
    134
    Hello 0Wafi0

    The first thing you should do is fill in the truth table in columns a to g.
    For example, in the first row, from A to D contain 0’s. It means that no input is selected.

    The next line says: 0001, means that the switch has enabled the input D. So that segments that will be active high Are: b c d e g.
    And so on.

    Does not look complete truth table but should be considered, when more than one entry is Active Low. In this case should light segments to form the letter "F".
     
  3. 0Wafi0

    Thread Starter New Member

    May 15, 2015
    25
    0
    awesome thanks...I knew about the truth table but I was confused about this active low inputs with active high outputs...It's easy just work through that...but I wanted to know how the active low input affects this whole thing with the active high output...
     
  4. MrCarlos

    Active Member

    Jan 2, 2010
    400
    134
    Hello 0Wafi0

    Ok. Normally in positive logic:
    "Active Low" is a voltage level close to zero volts,
    while:
    "Active High" is a voltage close to the bias voltage (VCC).

    So:
    Active Low = 0
    Active High = 1

    Note that the truth table "A", "B", "C" and "D" column are represented in the other direction, it means that:
    1 Yes, yes it is active.
    0 No, it is not active.
     
  5. bertus

    Administrator

    Apr 5, 2008
    15,648
    2,347
    Hello,

    @MrCarlos : may I remind you of the fact that this is an ENGLISH ONLY forum.
    Please translate the post into english.

    Bertus
     
    MrCarlos likes this.
  6. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    The normal convention is that 0 means LO (low voltage) and 1 means HI (high voltage). Active-LO means LO is TRUE (hence, "active-LO) so that 0 means TRUE and 1 means FALSE. In this case, inputs 0 0 0 0 means that ALL inputs are being asserted, while if the switch is open-circuited it means that the inputs are 1 1 1 1 (as a result of the pullup resistors).
     
  7. 0Wafi0

    Thread Starter New Member

    May 15, 2015
    25
    0
    awesome guys!!! thanks for the help!!! so basically in the truth table I would have to flip everythin for ABCD? correct?
     
  8. MrCarlos

    Active Member

    Jan 2, 2010
    400
    134
    Hello 0Wafi0

    Yes you are correct.

    Change the zeros by ones, and ones with zeros
     
  9. 0Wafi0

    Thread Starter New Member

    May 15, 2015
    25
    0
    so after that i do everything else normally? like just follow the new flipped inputs?
     
  10. MrCarlos

    Active Member

    Jan 2, 2010
    400
    134
    Hello 0Wafi0

    Yes, of course

    Note now that the inputs A, B, C and D are as mentioned in the phrase that we see in the image that you enclose in your post #1.

    Now 0 represents the letter is active.

    In that image there is a box to the right of the 7-Segment Display.
    Under the "Sw Input" column are the letters when they are "Active Low" light segments named in the "Segs" column.

    For example: for A, the segments should light are abcefg.
    They will be "active high".

    Remember:
    When no input (ABCD) is activated, the display must be turned off.
    When more than one input are active the display should show a F.
     
  11. 0Wafi0

    Thread Starter New Member

    May 15, 2015
    25
    0
    awesome thanks a lot!!!
     
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