I have the following code and I get an error. I am trying to use only one bit out of the four lines in the inputs A, B, and the output, Y. To be clear, I do not want to use the AND operator, I want to use the AND module that I made.
AND_Gate_Quad_SN74AHCT08N and0(.A({3'bzzz,clk_disable}), .B({3'bzzz,clk}), .Y({3'bzzz,clk_buffer}));
What would be a good way to accomplish this?
An output port cannot be connected to numeric literal constant.
There at least three options I can think of
Do a simple connection and live with a port size mismatch warning .y(clk_buffer)
Declare as a 4-bit wire [3:0] clk_buffer and only use clk_buffer[0].
Create a 3-bit dummy_wire and use the connection .y({dummy_wire,clk_buffer}).
Related
I've been using Verilog and SystemVerilog for many years, but I just came across something which seems weird.
I don't know if it's something I just never noticed or if there is something special about this I am missing.
Basically I have a module which has a 16-bit input, to which I connected an 8-bit signal. This seems like no big deal to me, in fact it's one of the things I love about Verilog/SystemVerilog, since unlike VHDL, you are allowed to do things like that.
I would expect the signal connected to the input port of my module to be right-aligned and zero-padded on the left, but instead it is being padded with Zs.
That causes my simulation to not work, because for example I use that signal to initialize a counter after substracting one from it, which leads XXXX.
Is this behaviour expected? Can I change something in my code to get the behaviour I expected?
This behavior seems reasonable. I ran a simulation on several simulators on EDA Playground, and I got a mix of results: some had the MSBs as 0, others had them as z.
The best approach is to explicitly drive the inputs with the value you desire. For example, if you want the MSBs to be 0, use something like:
module tb;
reg [7:0] data;
// ...
dut i1 ({ {8{1'b0}}, data });
endmodule
This creates a 16-bit expression which is connected to the module input port. The expression is a concatenation of an 8-bit constant and the 8-bit data signal.
If there is a size mismatch happening in your design or testbench, there should be a warning (if not an error) raised from your compiler/simulation tool. There is no default behavior defined to pad with 0 or undriven z, therefore it can be 0-padded or unknown depending on your tool, and you are on your own risk. The best practice is to watch carefully if there is such warning/error already reported, and try to resolve it during compilation/elaboration phase, or better to run a linting tool before compilation.
Hello I’m new to Verilog and I’ve been confused over this. In some examples the inputs and outputs are declared as wires or regs and on others they are just listed like this:
Module(input a, b, output c)
What is the default data type given to these values?
The default data type for a port in any direction is an unsigned 4-state scalar (1 binary digit). In Verilog, input and inout ports must be and default to the net type wire. output ports also default to wire but can also be specified as a variable reg. (SystemVerilog relaxes some of these rules by allowing input ports to also be declared as variables).
Now that you know what the implicit defaults are, please be a good engineer and never rely on them—be explicit.
Is it possible to use assign keyword with module instances?
Lets assume I have a module logarithm which works as it's intended. In some other module, I want to have: A = log(B) + log(C).
Is there any efficient way other than the following to do so?
wire [3:0] logB;
wire [3:0] logC;
Logarithm log(logB, B);
Logarithm log(logC, C);
assign A = logB + logC;
And is it known as a gate-level design or a data-flow one?
If it is not data-flow, would you please present a data-flow alternative to this code?
Well your assign keyword in the code is not working on module instances. Rather, it is working on the outputs of those module instances.
The way you are assigning to A should help you get the desired output provided you have set its width accordingly. And it is a data flow representation
There is one mistake in your given lines of code though. You can not use the same instance name log for both the instantiations of the module Logarithm. They have to be unique.
I got my code to work. I did not set my output to 3'b000 before running the case statements. it compiles and gives me desired outputs i think still verifying . I still have trouble with rtl compiler to take a shapshot of the circuit. Im sure every compiler is slightly diffrent so im not sure if anyone can help with this. I'm not sure why it hates everything. will re-post when done or if someones interested in helping thanks
First things that I see are that you need to use semicolon, not comma, to separate your statements.
Also, if you want to set the value of X during procedural statement (always block), then it needs to be a reg type. Signals declared as output are implicitly wire type, unless you declare it as output reg [3:0] X.
module exmaple(input a, b, input in[2:0], output d, e, output out[5:0])
I am new to Verilog and trying to understand what input in[2:0] means?
That isn't valid Verilog(IEEE-1364), it is SystemVerilog(IEEE-1800). SV allows ports to be declared as multi-dimensional arrays so in this case in is declared as an array of single bit wires.
Generally if you wanted a vector for a port you would use input [2:0] in which is valid in both Verilog and SystemVerilog. However if your port type cannot be a vector, such as integer or time then you would need to use this method.