# how to trigger another clk in mainclk (verilog)

I wrote somekind of prescaler in verilog to make sclk_adc signal from clk_i. by now my code looks like:

always @(posedge clk_i)
begin
end


now i wonder is there any probability to load shift registers on pos/negedge of generated clock in always @(posedge clk_i) block?

When i wrote this outside the main always block

always @(negedge sclk_adc)
begin
transdata = transtmp;
dataInCh[ch_cnt][15:0] = transdata;
end

begin
dout = shiftOut [ch_cnt] [15];
shiftOut [ch_cnt][15:0] = { shiftOut[ch_cnt] [14:0], 1'b0};
end


It is simulated well in the GTKwave, but quartus started to complain about multiply reference to dout (my output), deservedly, i think. So it seems like i have to load them in main always block, but when i add those lines there, shift registers start to load with clk_i, not with the sclk_adc what is logical, but how to avoid this? Please, provide me any clue, thanks

• You should share all the code you can. You haven't shared code that shows two references to dout, so it's hard for us to comment on how you can improve your code. – The Photon Jul 25 '18 at 5:11
• pastebin.com/NXUHLb3u – dshee Jul 25 '18 at 6:32
• now it is here (i am sorry it is my first verilog code. it is driver for adc circuit). But i faced another problem :c I almost get what i wanted link i needed dout loaded WITH posedge sclk_adc and transdata shift WITH negedge sclk. But they made it at another clk_i after event of pos/negedge, why? help me to explain this please – dshee Jul 25 '18 at 6:41

You can just update dout on the edge of sclk_adc:

always @(posedge sclk_adc)
begin
dout <= shiftOut [ch_cnt] [15];
end


But if you do this, you can only update dout in this always block. You can use an asynchronous reset if you are concerned that your reset signal won't be asserted long enough to see an edge of the slow clock:

always @(posedge sclk_adc or posedge reset) begin
if (reset) begin
dout <= 0;
else begin
dout <= shiftOut [ch_cnt] [15];
end
end


You can't reset dout in the always block timed by clk_i if you're also going to set it in the block timed by sclk_adc.

Alternatively, since you are generating sclk_adc using a counter timed by clk_i, you could just arrange to assert sclk_adc one cycle earlier than you are doing now. It's quite common when generating slow clocks to actually generate several clocks at the same frequency, but delayed from each other by 1 or more cycles of the master clock, to allow timing different events on different phases of the slow clock.

• According to you i will rewrite dout like this assign dout = shiftOut [15]; assign out_next = { shiftOut[14:0], 1'b0}; to not mention it in main block and will make another always with sclk_adc. but it still mystery why my pos and neg detectors did not work. Anyway, thank you a lot! – dshee Jul 26 '18 at 4:00

Well i beat this one. I just needed to built rising and falling edge detector

all i neeg is that few lines

reg    main_reg;
wire    pos_sclk, neg_sclk;
assign  pos_sclk = sclk_adc & ~main_reg;
assign  neg_sclk = ~sclk_adc & main_reg;


and in main clk block

main_reg <= sclk_adc;