Electronic – using ram verilog instantiation

ramverilog

I have implemented ram Verilog module. I made its instantiation and read the data from some address .After that I want to write the data to other address.
looks that I don`t need to made the instantiation of the ram twice,since each ram defines different registers array.The question is how (should I?) reuse the same ram instance?

module ram(we,addr,data);

input we;
input[DATA_WIDTH - 1:0] addr,
inout[DATA_WIDTH - 1:0] data;

reg [DATA_WIDTH - 1:0] mem[0:ADDR_DEPTH - 1];
always @ ( posedge clk )
begin
if(we)
mem[addr] <= data;
else
data <= mem[addr];
end
endmodule

module top (add1,add2);
wire [DATA_WIDTH - 1:0] w_data;
ram i_ram1(1b'0,add1,w_data);
ram i_ram2(1b'1,add2,w_data);//<- other mem array is used in this instance

Best Answer

You'll need to use the clock to break up the write and read operations. Here is an example of a test bench that writes data to two addresses then reads them back.

wire [DATA_WIDTH-1:0] data = we ? wdata : {DATA_WIDTH{1'bz}}; // tri-state write output driver
ram i_ram0(clk,we,addr,data);

initial begin
  clk   = 1'b1;
  data1 = 'h0F;
  data2 = 'hF0;
  addr1 = 0;
  addr2 = 2;
  we    = 1'b1;
  addr  = addr1;
  wdata = data1;
  @(posedge clk) #1;
  addr  = addr2;
  wdata = data2;
  @(posedge clk) #1;
  we    = 1'b0;
  addr  = addr1;
  @(posedge clk) #1;
  if(data !== data1) $display("ERROR data:%h != data1:%h @ addr:%h", data,data1,addr);
  addr  = addr2;
  @(posedge clk) #1;
  if(data !== data2) $display("ERROR data:%h != data2:%h @ addr:%h", data,data2,addr);
  @(posedge clk) #1;
  $finish;
end

FLY if data is an inout, then you need a tri-state driver in your ram module:

assign data = we ? {DATA_WIDTH{1'bz}} : rdata; // tri-state read output driver

Full working example here: http://www.edaplayground.com/s/6/221

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Electronic – Verilog – Referencing Flattened Busses in Module Instantiation

It seems like you are running into Verilog's Escape identifiers gotcha:

§2.7.1 Escaped identifiers

Escaped identifiers shall start with the backslash character () and end with white space (space, tab, newline). They provide a means of including any of the printable ASCII characters in an identifier (the decimal values 33 through 126, or 21 through 7E in hexadecimal).

Neither the leading backslash character nor the terminating white space is considered to be part of the identifier. Therefore, an escaped identifier \cpu3 is treated the same as a nonescaped identifier cpu3.

So not only you need to properly write down escaped port name, but you also need to make sure you don't forget to put a whitespace at the end. For example:

DES_SBOX1 feistel( .CLK(CLK_tb),
                 .\ptext[5] (ptext_tb[5]),
                 .\ptext[4] (ptext_tb[4]),
  ...

Hope it helps. Good Luck!

Electronic – Verilog asynchronous reads of regs – and design question

Your code simulates two multiplexers. These are actually asynchronous components. The fact that Verilog requires data1_temp and data2_temp to be declared as reg's is a quirk of Verilog syntax and your choice of coding style, and doesn't mean these signals would be the outputs of storage elements in a physical implementation.

If you want to capture these values in actual registers, you need to add those explicitly:

reg [7:0] data1, data2;
always @(posedge someclock) begin
    data1 <= data1_tmp;
    data2 <= data2_tmp;
end

But I would like to know what this mini register file would be made of in hardware. Particularly, the 4x8 bit array consisting of k0,k1,k2,k3.

You haven't shown how these variables are assigned, so it's not possible to say how they are implemented. As your code showed, just declaring them as reg doesn't guarantee they are implemented with actual storage elements. If you assign them inside a block that begins always @(posedge clk) then very likely they are flip-flops, but there are ways you could code them that would make them synthesize differently.

I thought when it came to registers and arrays like this, you need a clock to read out data, like RAM?

You need a clock to update a (physical) register. You can read it out at any time. For example:

wire [8:0] sum;
assign sum = k0 + k1;

is perfectly valid code. sum will change whenever any of its inputs changes. If k0 and k1 are the outputs of flip-flops, their values will only change when there is a clock edge.

For another example, you could equally well describe your multiplexers with code like this:

reg [7:0] k0, k1, k2, k3;
wire [7:0] data1_tmp;
reg [1:0] reg1;
// k<n> and reg1 are assigned elsewhere.
assign data1_tmp = (reg1 == 0) ? k0 :
                   (reg1 == 1) ? k1 :
                   (reg1 == 2) ? k2 : k3;

how do I read from this tag_array and do the comparison all within the same clock cycle?

Let me repeat a key point for emphasis: You need to use a clock to assign a new value to a register (an actual hardware register or group of flip-flops). It's output is available at any time.

RAMs are different and how you access the contents of a RAM will depend on details of the type of RAM you use.

I got confused because frankly I don't know enough about memory hardware and how that's possible.

Another key strategy: When you are learning digital logic, I recommend you learn about the physical hardware first, and then work out or study how to simulate it in HDL second. So first, learn what a physical flip-flop is, then learn the standard Verilog methods of describing a flip-flop. Especially if you are trying to write HDL for synthesis, trying to write good code before you learn the capabilities of the underlying hardware will lead you down a lot of dead-end paths.

Best Answer

Related Solutions

Electronic – Verilog – Referencing Flattened Busses in Module Instantiation

Electronic – Verilog asynchronous reads of regs – and design question

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