I'm trying to implement an interface with modports. I have 2 modules inside a testbench.
module test;
test_intf intf();
clk_reset u_clk_reset (
intf.CLK_RESET
);
mem u_mem(
intf.MEM
);
....
endmodule
This is the interface:
interface test_intf;
logic reset_n;
logic aclk;
logic bclk;
logic areset_n;
logic wr_in;
logic rd_en;
logic [ 9:0] wr_addr;
logic [23:0] wr_data;
logic [ 9:0] rd_addr;
logic [23:0] rd_data;
modport CLK_RESET (output aclk, bclk, areset_n);
modport MEM (output rd_data, input clk, reset_n, wr_in, rd_en, wr_addr, wr_data, rd_addr);
endinterface
This is the clk_reset module:
module clk_reset(
test_intf intf
);
always #(5) intf.aclk = ~intf.aclk;
initial begin
intf.aclk = 0;
end
initial begin
intf.areset_n = 1;
#2000 intf.areset_n = 0;
#2000 intf.areset_n = 1;
end
This is the mem module:
module mem
(
test_intf intf
);
always #(posedge intf.clk or negedge intf.reset_n)
begin
//end
end
I checked that the clk_reset interface ports are working (all outputs).
But, the problem is that the output clock signals from clk_reset can't reach the mem module by interface. So, intf.clk signal is 'x' in the mem module.
How do I pass some interface's signals into another interface (module)?
Your simulator should have given you a compile error. The interface is missing the declaration of the clk signal. Add it like this:
logic clk;
Since you want to use clk inside mem, you need to add clk to the CLK_RESET modport list:
modport CLK_RESET (output aclk, bclk, clk, areset_n);
Finally, you need to generate the clk signal so that it is not X. Here are the required changes:
interface test_intf;
logic reset_n;
logic aclk;
logic bclk;
logic clk;
logic areset_n;
logic wr_in;
logic rd_en;
logic [ 9:0] wr_addr;
logic [23:0] wr_data;
logic [ 9:0] rd_addr;
logic [23:0] rd_data;
modport CLK_RESET (output aclk, bclk, clk, areset_n);
modport MEM (output rd_data, input clk, reset_n, wr_in, rd_en, wr_addr, wr_data, rd_addr);
endinterface
module clk_reset(
test_intf intf
);
always #(5) intf.aclk = ~intf.aclk;
initial begin
intf.aclk = 0;
end
always #(5) intf.clk = ~intf.clk;
initial begin
intf.clk = 0;
end
initial begin
intf.areset_n = 1;
#2000 intf.areset_n = 0;
#2000 intf.areset_n = 1;
end
endmodule
Related
I have this 4 bit ring counter that I'm trying to make, and I feel like I'm so close, but I can't figure out how to make one input depend on the previous state's output. Here's what I have:
`default_nettype none
// Empty top module
module top (
// I/O ports
input logic hz100, reset,
input logic [20:0] pb,
output logic [7:0] left, right
);
// Your code goes here...
q[3:0];
assign q[3:0] = right[3:0];
hc74_set setFF(.c(pb[0]), .d(pb[1]), .q(right[0]), .sn(pb[16]));
hc74_reset resetFF1(.c(pb[0]), .d(pb[1]), .q0(right[1]), .rn(pb[16]));
hc74_reset resetFF2(.c(pb[0]), .d(pb[1]), .q1(right[2]), .rn(pb[16]));
hc74_reset resetFF3(.c(pb[0]), .d(pb[1]), .q2(right[3]), .rn(pb[16]));
endmodule
// Add more modules down here...
// This is a single D flip-flop with an active-low asynchronous set (preset).
// It has no asynchronous reset because the simulator does not allow it.
// Other than the lack of a reset, it is half of a 74HC74 chip.
module hc74_set(input logic d, c, sn,
output logic q, qn);
assign qn = ~q;
always_ff #(posedge c, negedge sn)
if (sn == 1'b0)
q <= 1'b1;
else
q <= d;
endmodule
// This is a single D flip-flop with an active-low asynchronous reset (clear).
// It has no asynchronous set because the simulator does not allow it.
// Other than the lack of a set, it is half of a 74HC74 chip.
module hc74_reset(input logic d, c, rn,
output logic q, qn);
assign qn = ~q;
always_ff #(posedge c, negedge rn)
if (rn == 1'b0)
q <= 1'b0;
else
q <= d;
endmodule
This is on an FPGA simulator, which is why there are a few things like pb (these are push buttons) and left, right outputs which are sets of 8 LEDs each.
Let's first make sure we are on the same page
Based on wikipedia description of a ring counter
This could be implemented as follows:
module top (
// I/O ports
input logic reset_n,
input logic clk,
output logic [3:0] ring
);
// Your code goes here...
always #(posedge clk or negedge reset_n) begin
if(~reset_n) begin
ring = 4'b0001;
end
else begin
ring[0] <= ring[3];
ring[1] <= ring[0];
ring[2] <= ring[1];
ring[3] <= ring[2];
end
end
endmodule
The output ring is a 4-bit one hot vector, reset_n = 0 makes ring = 0001 every clock with reset_n = 1 rolls the ring to the right, [0001, 0010, 0100, 1000, 0001, ...].
But you want to use instances of the flops you defined. Notice that in an assignment a <= b, a is the output of the flop (q port), and b is the input of the flop (d port).
module top (
// I/O ports
input logic reset_n,
input logic clk,
output logic [3:0] ring
);
// Your code goes here...
hc74_set setFF(.c(clk), .d(ring[3]), .q(ring[0]), .sn(reset_n));
hc74_reset resetFF1(.c(clk), .d(ring[0]), .q0(ring[1]), .rn(reset_n));
hc74_reset resetFF2(.c(clk), .d(ring[1]), .q1(ring[2]), .rn(reset_n));
hc74_reset resetFF3(.c(clk), .d(ring[2]), .q2(ring[3]), .rn(reset_n));
endmodule
You have to connect the ports accordingly, I just used clk for the clock and reset_n for the negated reset signal.
I have an system verilog interface
interface add_sub_if(
input bit clk,
input [7:0] a,
input [7:0] b,
input doAdd,
input [8:0] result
);
clocking dut_cb #(posedge clk);
output a;
output b;
output doAdd;
input result;
endclocking // cb
modport dut(clocking dut_cb);
endinterface: add_sub_if
And i have a SV module which uses this interface
module dummy(add_sub_if.dut _if);
....
endmodule: dummy
What is the ideal way to hook up this in my TB?
If i instantiate the interface, i need to create wires.
If i use bind then also i need to do a port mapping of the individual signals, that beats the convenience of having an interface.
Another add on question is, how to assign one such interface to another interface?
Thanks in advance,
Rajdeep
You can find a simple example of an interface definition and usage in the IEEE Std 1800-2012 Section 3.5 (Interfaces). It shows how to define the interface and hook it up to the design (the way you have already done it). It also shows how the interface can be instantiated (and connected) inside a top level module/wrapper (I copied the code below directly from the spec for your convenience):
interface simple_bus(input logic clk); // Define the interface
logic req, gnt;
logic [7:0] addr, data;
logic [1:0] mode;
logic start, rdy;
endinterface: simple_bus
module memMod(simple_bus a); // simple_bus interface port
logic avail;
// When memMod is instantiated in module top, a.req is the req
// signal in the sb_intf instance of the 'simple_bus' interface
always #(posedge clk) a.gnt <= a.req & avail;
endmodule
module cpuMod(simple_bus b); // simple_bus interface port
...
endmodule
module top;
logic clk = 0;
simple_bus sb_intf(.clk(clk)); // Instantiate the interface
memMod mem(.a(sb_intf)); // Connect interface to module instance
cpuMod cpu(.b(sb_intf)); // Connect interface to module instance
endmodule
Once you have the interface hooked up, then you can have drive/sample all the signals from a testcase program (just remember that you have to pass the interface to it). In this case, it would be be something like:
program testcase(simple_bus tb_if);
initial begin
tb_if.mode <= 0;
repeat(3) #20 tb_if.req <= 1'b1;
[...]
$finish;
end
endprogram
For a real-world example, you can check the source code of a UVM testbench that is available on my GitHub page. The interface hook-up is done in the xge_test_top.sv file.
Here is fsm dut with testbench.
This fsm dut performs state transition 1 - 0 - 1 - 0 in sequence.
Test bench verifies dut is properly working or not.
verilog module code(dut) :
module melay_fsm(o,clk,rst,i);
output o;
input i,clk,rst;
reg o;
reg [1:0]state;
// [1:0]state;
always#(posedge clk,posedge rst)
begin
if(rst)
begin
state <=2'b00;
end
else
begin
case(state)
2'b00:
begin
if(i)
state<=2'b01;
else
state<=2'b00;
end
2'b01:
begin
if(!i)
state<=2'b10;
else
state<=2'b01;
end
2'b10:
begin
if(i)
state<=2'b11;
else
state<=2'b00;
end
2'b11:
begin
if(!i)
state<=2'b00;
else
state<=2'b01;
end
endcase
end
end
always#(posedge clk,negedge rst)
begin
if(rst)
o<=1'b0;
else if(state==2'b11 && i==0)
o<=1'b1;
else
o<=1'b0;
end
endmodule
System verilog module code(testbench) :
interface melay_intf(input bit clk);
logic o,rst,i;
wire clk;
clocking c1#(posedge clk);
input o;
output i,rst;
endclocking
modport tes(clocking c1);
endinterface
module top;
bit clk;
always
#1 clk = ~clk;
melay_intf i1(clk);
melay_fsm a1(.o(i1.o),.clk(i1.clk),.rst(i1.rst),.i(i1.i));
melay_tes(i1);
endmodule
program melay_tes(melay_intf i1);
initial
#100 $finish;
initial
begin
i1.rst <= 0;
#4 i1.rst <= 1;
#4 i1.rst <= 0;
i1.i = 1;
#2 i1.i = 0;
#2 i1.i = 1;
#2 i1.i = 0;
#2 i1.i = 1;
#2 i1.i = 0;
repeat(10)
begin
i1.i = 1;
#2 i1.i = $urandom_range(0,1);
end
end
initial
$monitor("output = %d clk = %d rst = %d i = %d",i1.o,i1.clk,i1.rst,i1.i);
initial
begin
$dumpfile("mem.vcd");
$dumpvars();
end
endprogram
Important thing to note down here is connection of signals in top module.
melay_fsm a1(.o(i1.o),.clk(i1.clk),.rst(i1.rst),.i(i1.i));
Please, observe properly how I bind interface with testbench and dut.
Please, observe following things.
I defines interface with all the dut's signals.
I took instance(i1) of interface(melay_intf) in top module.
I took instance (a1) of dut (melay_fsm) in top module.
Now observes melay_fsm a1(.o(i1.o),.clk(i1.clk),.rst(i1.rst),.i(i1.i))
All the dut's signals are connected with interface.
I passed instance of interface(i1) in testbench. melay_tes(i1)
So, testbench can access interface signals and interface signals are connected to dut's signals.
Now, you can access dut's signals in your test bench with help of interface.
I think now you can understand proper flow.
Please ask the question if you have any doubt.
You can bind interface in system verilog module.
Here, I provides sample code with help of that you can understood how to bind interface in system verilog module and with dut.
Here I provides verilog module and system verilog module. Main part of code is interface from which verilog and system verilog module are connected.
verilog module code(dut) :
module dff(qn,d,clk,reset);
output qn;
input d,clk,reset;
reg qn;
always#(posedge clk,negedge reset)
begin
if (!reset)
begin
qn=1'bx;
end
else if (d==0)
begin
qn=0;
end
else if (d==1)
begin
qn=1;
end
end
endmodule
System verilog module code(testbench) :
interface melay_intf(input bit clk);
logic o,clk,rst,i;
clocking c1#(posedge clk);
input o;
output i,rst;
endclocking
endinterface
module top;
bit clk;
always
#1 clk = ~clk;
melay_intf i1(clk);
dff d1(.o(i1.o),.clk(i1.clk),.rst(i1.rst),.i(i1.i));
melay_tes(i1.tes);
endmodule
program melay_tes(melay_intf i1);
initial
#100 $finish;
initial
begin
i1.rst <= 0;
#4 i1.rst <= 1;
#4 i1.rst <= 0;
i1.i = 1;
#2 i1.i = 0;
#2 i1.i = 1;
#2 i1.i = 0;
#2 i1.i = 1;
#2 i1.i = 0;
repeat(10)
begin
i1.i = 1;
#2 i1.i = $urandom_range(0,1);
end
end
initial
$monitor("output = %d clk = %d rst = %d i = %d",i1.o,i1.clk,i1.rst,i1.i);
initial
begin
$dumpfile("mem.vcd");
$dumpvars();
end
endprogram
Here important part is interface and in it I used clocking block for synchronization purpose. Here clocking c1#(posedge clk); so all signals which are mention inside the clocking block which are i,o,rst.All this signal change its value at every posedge of clk signal.
Here dff d1(.o(i1.o),.clk(i1.clk),.rst(i1.rst),.i(i1.i)); Important thing that you find in top module I made connection between verilog signals and system verilog signals.
You can find verilog module name is "dff". I took the instance of dff verilog module and made the connection. Here i1.o,i1.clk,i1.rst,i1.i is system verilog signals which are connected to o,clk,rst,i signals of verilog module of with dot convention.
i have written a code in system verilog for interface. but it it giving me the error at clk. the error is Undefined variable clk....
code is
error at always(posedge clk)
interface simple_bus(input logic clk);
// Define the interface
logic req, gnt;
logic [7:0] addr, data;
logic [1:0] mode;
logic start, rdy;
endinterface: simple_bus
module memMod(simple_bus a);
// simple_bus interface port logic avail;
//logic clk;
always #(posedge clk)
a.gnt <= a.req & avail;
endmodule
when using clock in always block it is giving the error "Undefined variable: clk"
clk is input to the given interface simple_bus, you need to hierarchically access it. Such as a.clk.
So your module code would be:
module memMod(simple_bus a);
always #(posedge a.clk)
a.gnt <= a.req & avail;
endmodule
Edit:
I have tried your code at my end and it's working. PFB exa code.
interface simple_bus(input logic clk);
// Define the interface
logic req, gnt;
logic [7:0] addr, data;
logic [1:0] mode;
logic start, rdy;
endinterface: simple_bus
module memMod(simple_bus a);
// simple_bus interface port logic avail;
//logic clk;
always #(posedge a.clk) a.gnt <= a.req;
endmodule
module main();
logic clk;
simple_bus sb(clk);
memMod m(sb);
initial repeat(10) clk = #5 ~clk;
endmodule
I am trying to make a block for an 8-bit multiplier, and the testbench is giving me a result that basically says that I don't know what I'm doing with my wires and regs. To make this easier to answer, I'm going to display my code, and then the parts that I think are important:
module multiplier_result(
input ADD_cmd,
input LOAD_cmd,
input SHIFT_cmd,
input reset,
input [7:0] B_in,
input [7:0] Add_out,
input cout,
output wire [7:0] RB,
output wire [15:0] RC,
output wire [8:0] temp_reg,
output wire LSB
);
wire [8:0] from_mux;
reg[16:0] balreg;
reg tempadd;
//assign the outputs. all combinational
assign RB = balreg[15:8];
assign RC = balreg[15:0];
assign LSB = balreg[0];
assign temp_reg = balreg[16:8];
mux_9 mux(
.sel(~ADD_cmd),
.Add_out(Add_out),
.cout(cout),
.mux_out(from_mux),
.temp_reg(temp_reg)
);
always # (*) begin
if(reset) begin
balreg[16:0] = 17'd0;
tempadd = 1'b0;
end
else
begin
if(LOAD_cmd)
begin
balreg[16:8] = 9'b000000000;
balreg[7:0] = B_in;
end
if(SHIFT_cmd)
begin
balreg[16:8] = from_mux;
balreg = balreg >> 1;
end
end
end
endmodule
Now, here is what's troubling me:
Here I'm assigning wires to different bits of the balreg register (in black). What is going on in my head (please excuse my paint skills):
But for some reason, LSB gets what it's supposed to, while RB and RC get high impedance. Here is the simulate result, followed by the code I used (just a simple test case)
module multiplier_result_tb(
);
reg ADD_cmd;
reg LOAD_cmd;
reg SHIFT_cmd;
reg reset;
reg [7:0] B_in;
reg [8:0] Add_out;
wire [7:0] RB;
wire [15:0] RC;
wire [8:0] temp_reg; //size 9
wire LSB;
multiplier_result dut(ADD_cmd,LOAD_cmd,SHIFT_cmd,reset,B_in,Add_out,RB,RC,temp_reg,LSB);
initial begin
LOAD_cmd = 0;
#10;
LOAD_cmd = 1;
reset = 0;
B_in = 8'b00001010;
Add_out = 9'd0;
ADD_cmd = 0;
SHIFT_cmd = 0;
end
endmodule
I'm not following these results at all. The balreg register is all set up, so the RB and RC wires MUST be defined, but according to the simulation, they are high impedance.
The only conclusion that I get at, is that I really don't know what's going on with the types (the model I had in my had worked for me so far).
Any help, ideas, tips are much appreciated.
You only connected 10 of the 11 ports of the dut. Didn't you get a warning? You are making connections by position, not by name. You connected RB to input cout. You need to drive cout in your testbench.
Another way to make connections is by name. This is more verbose, but it can make your code clearer:
multiplier_result dut (
// Inputs:
.ADD_cmd (ADD_cmd),
.Add_out (Add_out),
.B_in (B_in),
.LOAD_cmd (LOAD_cmd),
.SHIFT_cmd (SHIFT_cmd),
.cout (cout),
.reset (reset),
// Outputs:
.LSB (LSB),
.RB (RB),
.RC (RC),
.temp_reg (temp_reg)
);
I am making a parallel to serial converter using ring counter in verilog. The ring counter is working fine but the Parallel to serial converter is not working properly and I am getting x undefined result. I am providing the code kindly help me finding the problem.
TOP
module PtoSTOP;
reg clk,rst;
wire [3:0] myout;
wire out;
Ring a(clk,rst,myout);
parToser x(myout,clk,rst,out);
initial begin
clk=1;
rst=1;
#1 rst=0;
end
always
#2 clk=~clk;
endmodule
Parallel TO Serial Converter
module parToser(myout,clk,rst,out);
input clk,rst;
input [3:0] myout;
output reg out;
reg [2:0]i;
always #(posedge clk or posedge rst) begin
if(rst) begin
out <= 0;
i <= 0;
end
else begin
out <= myout[i];
i <= i+1;
end
end
endmodule
RingCounter
module Ring(clk,rst,myout);
input clk,rst;
output reg [3:0]myout;
always #(posedge clk or posedge rst) begin
if(rst)
myout<=1;
else
myout<=myout<<1;
end
endmodule
I think the main issue you are seeing is part of parToser.
You have reg [2:0]i; which you increment and use to address input [3:0] myout; but i can hold values 0 to 7, half of which is outside the address range of [3:0] myout. You should be seeing a simulation error about out of range addressing.
Also you have included a few flip-flops with a reset condition but not added the reset to the sensitivity list in 'parToser' & 'Ring':
always #(posedge clk)
Should be:
always #(posedge clk or posedge rst)
With out this trigger your out, i and myout variables will be x, as they have not been set to a known condition.
NB: parToser i = i+1; should be i <= i+1;