Verilog Code For Serial Adder Subtractor Cinnamon

Verilog Code For Serial Adder Subtractor Cinnamon

Verilog Code For Serial Adder Subtractor Cinnamon 4,2/5 2787 votes

I'm trying to implement a serial adder/subtractor in VHDL, I've done it the ripple carry way before but now I'm supposed to implement the same functionality by just using one full adder cell instead of N-amount of cells so I have to shift the bits from the vectors in to the full adder/subtractor and store the result in another vector which I just shift the index for as well. The logic behind it is very easily understood, you just have a counter for the index and so on. But I obviously encounter problems since I'm probably still thinking a bit too much software programming I guess. The problem I have is that the counter increases to 1 right away when it enters the process so that when I try to add the vectors a = 0101 0101 and b = 1010 1010 I get y = 1111 111X and the carrys = 0000 000X. I've tried to start storing the result at index_counter - 1 since the index jumps to 1 on the first clock cycle, but then I get a fatal error in the simulation. I've been trying to solve this problem for a few hours now and can't seem to figure out how to do it, so could you please take a look and see what I can do to fix it? Would be very much appreciated!

I do have signals for saturation and overflow that I am going to implement later on but they aren't really used at the moment so don't worry about them. The design shouldn't update the final result before the addition/subtraction is finished if you're wondering why I have done it the way I have.

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Assign sum = a + b + ci; endmodule. Verilog code for an unsigned 8-bit adder with carry out. Share the post. Verilog code for adders/subtractors. Designs are usually made up of combinatorial logic and macros (for example, flip-flops, adders, subtractors, counters, FSMs, RAMs). Following is the Verilog code for an 8-bit shift-left register with a positive-edge clock, serial in and serial out.

I haven't tried to debug this, but I think it is a timing problem: At the very first rising clock edge, you define both the inputs to your adder and register the adder's outputs (which are X at this point, because its inputs weren't defined before). With the idea of storing the result at index_counter - 1 you are on the right track however you need to take care that you never try to store the result in a non-existing register such as carry(-1) or temp_sum(-1).

4 bit adder subtractor calculator

You can use if index_counter > 0 for that for example. Similarly, you then need to allow for one more cycle at the end of the calculation or assign the last result bit directly to your y output or you'll miss the last result bit. Hope that helps!

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