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Verilog HDL Coding Practice Question

 Hello Dear Readers, 

Today first time we will go to the pages part in which I will post differents types of practice questions that are more frequently asked by the interviewer.

Q-1:

See the verilog code I and II and explain what kind of dissimilarity is in between it and which code removes the glitches at the output and why.

Code-I:

module MUX2_1(input a,b,sel, output out);

wire sel_b,a1,b1;

not g1(sel_b,sel);

and g2(a1,a,sel_b);

and g3(b1,b,sel);

or  g4(out,a1,b1);

endmodule


Code-II:

module MUX2_1(input a,b,sel, output out);

wire sel_b,a1,b1,ab;

not g1(sel_b,sel);

and g2(a1,a,sel_b);

and g3(b1,b,sel);

and g4(ab,a1,b1);

or  g5(out,a1,b1,ab);

endmodule

Write down your answer in the comments and we will discuss it.

Q-2: In which conditions latches are created during RTL coding.


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Comments

  1. Both code are same !!!

    ReplyDelete
    Replies
    1. Yes sorry now it has updated and thanks for feedback

      Delete
    2. Ok sir now second code I think cancel the glitches.

      Delete
    3. Answer of the second question is I think incomplete sensitivity list,case statement if we forgot any conditions etc ...

      Delete
  2. First code because it is contain last or gate logic so output is avoid static hazards.

    ReplyDelete
  3. First code is cancelled glitches

    ReplyDelete
  4. First code avoid glitches

    ReplyDelete
  5. Second code glitches free because of last AND gate logic.

    ReplyDelete
  6. First code cancel glitches and for second question if we not used all possibility inside case statements.

    ReplyDelete
  7. Latches are created during if any case is not declared and not even matches with default stmt

    ReplyDelete

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