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Altera_Forum
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Honored Contributormultiple driver error during simulation
Hi
I am doing a project and am stuck in the synthesis part in Quartus. The code was successfully compiled using icarus verilog software.But while compiling the code in Quartus , I get some error messages. The code is as follows with its description: /* Whenever positive edge of switch occurs , I need to take the incoming 8 bits (serial inputs on a single pin) and after that I have to discard all the bits, until the next posedge of switch. These bits are coming from a parallel in serial out register. These bits are stored in a reg called c and i is an integer which keeps the count. */ module test(si,po,clk,switch,flag); input si,clk,switch; output reg [7:0] po; reg [7:0] c,d; output reg flag; reg count; //reg flag; integer j,i; initial i=0; initial flag=0; initial count=0; always@(posedge clk) begin if (flag==1) begin c[i]=si; i=i+1; if(i==8) begin //count<=1; //po<=c; po<=(c[7:4]*10)+c[3:0]; i=0; flag<=0; end end end always@(posedge switch) begin flag<=1; end endmodule The error messges are as follows: Error: Can't elaborate top-level user hierarchy Error (10029): Constant driver at test.v(16) Error (10028): Can't resolve multiple constant drivers for net "flag" at test.v(38) Could anyone help in rectifying the mistake?17 Replies
Altera_ForumYour flag is driven in two processes. the second process sets it permanently to high...
You must have a clock input as fast as serial stream (at least) then you run a counter(integer 0 ~ 8) set at zero at MSB or LSB location then increment to 8 and reset zero.(vhdl clocked process) --detect edge of switch switch_d <= switch; if switch = '1' and switch_d = '0' then count <= 1; elsif count < 8 then count <= count + 1; end if; if count < 8 then output(count) <= input; end if; if count = 7 then valid <= '1'; else valid <= '0'; end if;- Altera_Forum
Hi Kaz
I have some doubts on your expalaination as follows: switch_d <= switch; if switch /= switch_d then count <= 1; else count <= count + 1; end if; 1)by this are you detecting positive edge of switch 2) output <= input(count); what does this do? - Altera_Forum
I just corrected the output assignment.
Yes I detect switch rising level, the count is set to 1 due to one clock latency. for safer design you may put this for edge detection: if switch = '1' and switch_d = '0' then instead - Altera_Forum--- Quote Start ---
--- Quote End --- As per the specifications the first 8 inputs are to be taken at the positive edge of switch. So why do you set count =1 at +ve edge and increase it by 1 when there's no posedge of switch? 2) if count < 8 then output(count) <= input; end if; what is this used for?(vhdl clocked process) --detect edge of switch switch_d <= switch; if switch = '1' and switch_d = '0' then count <= 1; elsif count < 8 then count <= count + 1; end if; if count < 8 then output(count) <= input; end if; if count = 7 then valid <= '1'; else valid <= '0'; end if; - Altera_Forum
You are right, I am still playing with this code for demo only as I am not aware of your inputs behaviour.
remember you can only detect switch edge one clock period after it rises. I am not sure if switch will rise once only or many times. you can run count 0~7 instead of 0 ~ 8 then it will set back to 0 ready waiting for switch to go up again. if switch = '1' and switch_d = '0' then count <= 1; else count <= count + 1; -- 0~7 free running end if; output(count) <= input; ...etc. - Altera_Forum--- Quote Start --- I am not sure if switch will rise once only or many times. ...etc. --- Quote End --- Yes switch is allowed transitions "if switch = '1' and switch_d = '0' then" If I am not wrong this is detecting the positive edge So my count needs to be increased only when there is a positive edge and inputs(serial 8 bits) need to be stored in a reg. Moreover, me being familiar with Verilog, not VHDL, the syntax output(count) <= input; is troubling me
- Altera_Forum
if your 8 bits are serial then they can't all come at rising edge of switch but one after the other as per input clock edge otherwise you need to clearify this point. I know RS232 uses no clock. Instead you run a counter to read bits by guess in effect... is that your case as well? is your clock as fast as serial bit stream?
- Altera_Forum
Do you mean one bit arrives per switch edge?
- Altera_Forum
Well what I intend to do is that the serial bits are coming at every clock edge(global clock) . i use the switch mechanism to detect that my inputs are stable and I am ready to take the incoming 8 bits serially. So the value of count needs to be increased only when I am taking the inputs.
Hope this would clarify my specifications. - Altera_Forum--- Quote Start --- Do you mean one bit arrives per switch edge? --- Quote End --- its per clock edge
