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