Implementing a 5-to-32 Line Decoder in Verilog

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  • เผยแพร่เมื่อ 28 ส.ค. 2024

ความคิดเห็น • 9

  • @dirkgerritoort4055
    @dirkgerritoort4055 3 ปีที่แล้ว +1

    hi, great video, but was this also an opportunity (perhaps on the end) to show how a generate block will work. something like:
    genvar idx;
    genvar idy;
    generate
    for (idx = 7; idx >=0; idx = idx-1)
    for (idy = 3; idy >=0; idy =idy-1)
    begin
    assign D[idx * 4 + idy] = W[idx + 4] & W[idy];
    end
    endgenerate

  • @ZingatSidLive
    @ZingatSidLive 6 หลายเดือนก่อน

    wolf
    man you are great

  • @kevinwood3131
    @kevinwood3131 3 ปีที่แล้ว

    what program are you using to draw the design in the beginning of this lesson?

    • @VSThrasher
      @VSThrasher 3 ปีที่แล้ว

      It is OneNote

  • @minhajsixbyte
    @minhajsixbyte 3 ปีที่แล้ว

    at 10:25 you did D3, D2, D1, D0 sequence but everything before was in opposite order (D0, D1, ...., D31)
    why :(

  • @venkateshjagannathan7682
    @venkateshjagannathan7682 2 ปีที่แล้ว

    hello, can I know where u have initialized W for wire in the code?

  • @andreavenegas5134
    @andreavenegas5134 3 ปีที่แล้ว

    in a quaternary to binary decoder, how can negative input be implemented in verilog?

  • @yung_broccolini
    @yung_broccolini 3 ปีที่แล้ว

    Don't know why he changed his initial "endian-ness" of the decoder. The LSB should be the 0-bit, e.g. for the 1-to-2 decoder if A = 0, then D0 = 1, and D1 = 0 (if D0 is the LSB) -- that is the correct convention

  • @minhajsixbyte
    @minhajsixbyte 3 ปีที่แล้ว

    what are "Rnumber"s?