0
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After I synthesize it, the error occured like this:

Multi-source in Unit <BCDcountmod> on signal <BCD0<3>>; this signal is connected to multiple drivers.>

Any solution? (Here's below my code)

module BCDcountmod(
  input Clock, Clear, up, down,
  output [3:0] BCD1_1, BCD0_0 );
reg [3:0] BCD1, BCD0;
//reg [3:0] BCD1_1, BCD0_0;
always @(posedge Clock) begin
  if (Clear) begin
    BCD1 <= 0;
    BCD0 <= 0;
    end
end


 always @(posedge up) begin
      if (BCD0 == 4'b1001) begin
        BCD0 <= 0;
        if (BCD1 == 4'b1001)
          BCD1 <= 0;
        else
          BCD1 <= BCD1 + 1;
      end
      else
        BCD0 <= BCD0 + 1;
    end


always @(posedge down) begin
      if (BCD0 == 4'b0000) begin
        BCD0 <= 4'b1001;
        if (BCD1 == 4'b1001)
          BCD1 <= 4'b1001;
        else
          BCD1 <= BCD1 - 1;
      end
      else
        BCD0 <= BCD0 - 1;
    end

 assign BCD1_1 = BCD1;
 assign BCD0_0 = BCD0;

endmodule
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2
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Here, the signal BCD0 is being assigned values under 3 separate 'always' processes. Hence the synthesis is returning multiple driver error. Try moving all assignments of BCD0 signal under a single 'always' process. That will solve the issue. You could use different conditions under the same always block itself to define the behaviour of the signal.

The BCD counter which you are trying to implement is sensitive to clock edges, clear, up as well as down signals. So the process should be sensitive to clock edges and all other signals can be sensed w.r.t clock edges. One way of rewriting the code is like:

always @(posedge Clock) begin  
  if (Clear) begin  
    BCD1 <= 0;  
    BCD0 <= 0;  
  else if (up == 1) begin  
    if (BCD0 == 4'b1001) begin  
        BCD0 <= 0;  
        if (BCD1 == 4'b1001)  
          BCD1 <= 0;  
        else  
          BCD1 <= BCD1 + 1;  
        end  
    else  
        BCD0 <= BCD0 + 1;  
    end  
  else if (down == 1) begin  
    if (BCD0 == 4'b0000) begin  
        BCD0 <= 4'b1001;  
        if (BCD1 == 4'b1001)  
          BCD1 <= 4'b1001;  
        else  
          BCD1 <= BCD1 - 1;  
        end  
    else  
        BCD0 <= BCD0 - 1;  
    end  
  end  
end  

Here, all sensitivity conditions of both BCD0 and BCD1 signal are captured under a single 'always' block which will synthesize it properly.

\$\endgroup\$
  • \$\begingroup\$ could you please elabore it? it's not easy for me to separate it that easy. not only BCD0 is being multi-source but BCD1 either. \$\endgroup\$ – user2020598 Dec 17 '13 at 9:57
  • \$\begingroup\$ Ok. i will edit my comment above \$\endgroup\$ – Avin Dec 17 '13 at 11:10
  • \$\begingroup\$ @user2020598 Please have a look at the above answer and tell me if it helps. \$\endgroup\$ – Avin Dec 17 '13 at 11:30
  • \$\begingroup\$ @user2020598 Were you able to get the code synthesized? \$\endgroup\$ – Avin Dec 18 '13 at 14:45

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