Honored Contributor
Honored ContributorHi strykerg,
This is the fourth attempt to provide you with an answer, hopefully it might happen if I don't get any further interrupts by someone who doesn't understand the concept of return from interrupt herself ! Assume HEAD DESK ! Right first thing. ABANDON YOUR IDEAS OF THIS BEING SOFTWARE ! It isn't, you are trying to create a description of a digital circuit and NOT a program. This is done in a Hardware Description Language, not a programming language. You have an input called clk, the rest of the description shows you are trying to create a 22 bit counter (the variable counter of type reg). The first always block is being sensitized to action when it sees the postive edge of the clk. Each time it sees the this the value of counter is incremented by 2. However at start of day your counter does NOT have the value of 0 in it. In REAL electronics a register might come up as either '1' or '0' but it is entirely indeterminate. To represent this the counter value bits are at an unknown state. Adding 2 to an unknown state just has another unknown state. What you need to do is to make it go into a known state. This is usually done by applying what is called a reset. We can do this with your code (there is another way but I think it will be good to get you into a good methodology. Add this to the module description. module fpgatest ( ... input wire reset, ... and now we need to alter the alway blocks always (posedge clk) begin : counter_block if (reset) counter <= 22'b0000000000000000000000; else counter <= counter + 2; end end //counter_block What will happen is that when the reset signal goes high it will set the counter to all zeros. As this is Model Sim you can use the command force reset 1 for a few of the clocks and then force reset 0 Note you must have it high for at least 1 one clock. The value of counter will then have a known value for the counter and then it can start to count properly. What you have created is called a Synchronous Reset. You can make it Asynchronous if you wish but I'll let you research that. Now the second always block has a similar problem but I think you now know how to fix it using this the same technique but using a reset timed to the 'divided' clock you have generated so it has to be as long as one of your divided clocks. Note it's not great practice to create a second 'clock', ie bit 20 of counter to drive another counter, it needs to be treated as a clock within the FPGA fabric. Alternatively you can use the counter [20] as a gating mechanism, but again for now I'll leave you to research this, but it will mean that you can use the same reset as everything is run from the same clock. Good luck. HEY I managed to get this answer out to you :)
