Fpga lut to asic gates code#

(when output(0) and output(1) depends on different values of input) Total LUTs required = LUT + LUT = 2* = 2* = 6, where LIF is least integer function or Ceiling function. Here we got 1 as our final quotient after two times division by 4, so logic levels count will be 2.Ĭonsider a 8 bit input sequential circuit, where output will be decided by the status or value of “sel” signal. Total LUTs required = (8/4) + LIF(8/16) = 2 + 1 = 3, where LIF is least integer function or Ceiling function.

Based on the current value of “sel” signal, the corresponding value of input will be taken out as output.īelow mentioned diagram depicts the behaviour of the above code, and here one can clearly see that 3 LUTs and 1 flip-flop are needed, and 2 logic levels are present. Here, we will look into some examples and compare the LUTs count calculated from the above mentioned formulae and the value estimated by Vivado tool.Ĭonsider a 8 bit input sequential circuit, where output will be decided by the status or value of “sel” signal. Thus, the number dividing N will vary with the type of LUT. Some have 4 inputs whereas others have 6 inputs. LUT inputs may vary as different input LUTs are present on the FPGA. If it isn’t the case then we will divide the integer N by 6. In this above example of 128:1 MUX, we have considered that two inputs among the six, are already occupied by select lines. N will only include those bits of input on which the output depends. If N is not divisible by 4 then we will take smallest integer value of (N/4). Then we simply add all the LUTs required in all the logic levels. This process continues until we get 1 as our final number. This will be implemented in different logic levels levels.Ĭonsider a input of N bits, now one LUT can accommodate 4 bits therefore N/4 LUTs can carry N bits, now N/4 output will be there in first logic level (where all the outputs are evaluated simultaneously) and will travel to the next stage or logic level where we are gonna require (N/4)/4 (=N/16) LUTs to process N/4 inputs. We have seen if the MUX input is 4 bit then it can be implemented using a single LUT6, but what happens if input contains more than 4 bits. Then the Schematic would look like as depicted by below illustration. A single LUT6 contains 6 input pins and a output pin, which can be used to implement a 4:1 MUX (where 4 inputs + 2 select lines = 6 inputs).Ĭonsider an example of MUX consisting of 4 bit input and 2 bit sel line and is implemented in HDL as, Let’s step forth with an easy example which depicts the schematic of a MUX obtained from Vivado tool.

Fpga lut to asic gates code#

The incoming sections are going to provide a brief understanding of manually estimating the number of LUTs required to implement a piece of HDL code on hardware. The results are already stored, which gets loaded as the FPGA is powered up. An LUT carries a customized truth table for every possible input. It simply generates output based on the input combination. The LUTs mainly define the behaviour of the combinational logic designed with a VHDL or Verilog code. These CLBs along with programmable routing can form a complex web of combinational and sequential circuits. The CLBs (Configurable LogicBlocks), which are present as a part of programmable resources, contains flip-flops, Look-Up tables (LUTs) and multiplexers. Every FPGA has fixed number of programmable logic, I/O banks and memory elements. The number of resources used and their allocation will vary with the logic supporting the design.

To implement a specific design on a FPGA, a number of strategies can be followed while writing an HDL code.