Vhdl Program For Parity Generator

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I have a little problem coding a 4-bit odd parity generator with only NAND as logical operator in VHDL. I have the entity done, like you see in code but struggle doing the architecture with only NAND as logical operator. Could you help me doing this? I know how to do this with xor gates but nand is a bit harder for me. Thanks for replies Deadbrain Code: library ieee; use ieee.std_logic_1164.all; entity parity_generator is port ( A: in bit; B: in bit; C: in bit; D: in bit; P: out bit); end parity_generator; architecture parity_gen of parity_generator is begin --parity_gen p.

Your code is VHDL not Verilog, remove the tag? Also doesn't relate to the Windows CE tag. Show all your work for your homework problem and tell us specifically where you are stuck. A hint, a two input XOR gate can be constructed with 4 or 5 two input NAND gates, 4 bit odd parity requires 3 two input XOR gates and an inverter (another NAND gate). Without the parameters of your assignment it seems likely instantiating gates would be required, a single expression requires'xor' operator overload. – Mar 28 '18 at 21:02.

Design 8 bit Ripple Carry Adder using VHDL Coding and Verify using Test Bench Given below code will generate 8 bit output as sum and 1 bit carry as cout. It also takes two 8.

So today we will see an application of XOR Gate, which is parity generation. A parity bit is used for the purpose of detecting errors during the transmission of binary information.

A parity bit is an extra bit included with the binary message to make the number of ones either even or odd. An error is detected if the checked parity doesn’t correspond with the one transmitted. The circuit that generates the parity bit is called parity generator. So with this explanation, let us design an even parity generator. Here is the truth table of an even parity generator. INPUTS OUTPUT x y z Parity Bit 0 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 0 1 1 1 1 As seen from the truth table, in this even parity generator, if the number of 1’s in the input are odd, the output is 1 making the total numbers of ‘1’ be even.

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If the number of 1’s in input is even, the output is 0 since the number of input ‘1’ is already even. By closely observing the truth table, it can be understood that the output is a mere XOR of input bits. The three bits in the message, together with the parity bit are transmitted to their destination, where they are applied to a parity checker circuit to check for possible errors.