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Divide by 3 counter with 75 % duty cycle

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nisheethg

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hello
i am preparing for interviews and have come across a question which is
Divide by 3 counter with 75 % duty cycle

For 75 % duty cycle i need to count 1/4th of a cycle...how do i do it??

Any clues...
 

nand_gates

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its very simple just build divide by 3 counter and invert any one of its output
to get 75% duty cycle!
Here is one more way! example in verilog!
Code:
module div3_75 (
   // Outputs
   clk_out,
   // Inputs
   clk, reset_n
   );
   input clk, reset_n;
   output clk_out;
   reg [1:0] count;
   assign    clk_out = count[1];
   always @(posedge clk or negedge reset_n) begin
      if (!reset_n)
        count <= 0;
      else
        count <= { ~count[0], ~count[0] & count[1]};
   end
endmodule
 

vcnvcc

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Hi nand_gate this is not 75% duty cycle.

it is 66% duty cycle, as output is high for 2 clok cycles and low for 1 clock cycles..

how to do 75%, could u tell me plz.
 

nand_gates

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OK!
What you need is 2x clock multiplier.

Code:
                                   +------+
                   +----------+    |      |
           +----+  | +----+   |    | div  |
 clk in    |    |  +-|   q|---(----|> by  |----+
  ----+----| 2x |----0>   |   |    |  3   |    |
      |    |    |    |  qn|---+    |      |    |  ___
      |    +----+    +----+        +------+    +--\  \
      |                                            )  )--- div by 3
      |                            +------+    +--/__/   75% duty cycle
      |                            |      |    |
      |                            | div  |    |
      +----------------------------|> by  |----+
                                   |  3   |    
                                   |      |    
                                   +------+
 

arvkum02

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first make a divide by 3 with 50% DC . Then delay this o/p by 25% . OR the two signals you will get the desired o/p.
 

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