Question

Design a 4-bit AND/OR unit that accepts two 4-bits numbers, A and B, and either AND them or OR them depending on a control signal A/O (when the control signal is 0 perform AND, otherwise if the signal is 1 then perform OR – bitwise logical operation). The result will be stored in C.

Answer 1

As per your requirement the below one is solution please follow it

Let’s first talk about a 1 bit circuit:

We will first prepare the truth table for the given scenario. When C is 0, we will perform AND operation between A and B, when C is 1, we will perform OR operation on A and B.

Output 0 0 0 0 0 0 0 0 A+B 1 0 0 A*B 0 0 A+B 1 0 A*B 1 A+B 1

Now to prepare the circuit on basis of this, we will follow a simple method for designing circuit, which is called the Sum-Of-Products, or SOP, form. So, a Sum-Of-Products Boolean expression is basically a set of Boolean terms added together, each term being a product of Boolean variables. An example would be: ABC + BCD + DEF, the sum of products “ABC,” “BCD,” and “DEF.”

Now for above truth table, how to create the SOP expression

For the above truth table, for each row, if output is 0, we ignore the row. As this will not play any role in the output. For the rows, which outputs a 1, we will try to negate the individual bits(if required) to prepare a product term of A, B, C which results 1 for that particular row.

SOP Expression 0 0 0 N/A 0 0 N/A 0 0 N/A 0 ABC 0 0 N/A 0 ABC 0 ABC ABC

So overall SOP Expression would be sum of individual SOP expressions, where applicable.

Overall SOP expression = $\bar{A}BC$ + $A\bar{B}C$ + $AB\bar{C}$ +

Now to simplify above Expression, so that we can design the circuit:

SOP expression = $\bar{A}BC$ + $A\bar{B}C$ + $AB\bar{C}$ +

                                = $BC (\bar{A} + A) + A\bar{B}C + AB\bar{C}$ = $BC (1) + A\bar{B}C + AB\bar{C}$
=$BC + A\bar{B}C + AB\bar{C}$

Using principle $(A + \bar{A} = 1)$ and

                                = $BC + A\bar{B}C + AB\bar{C}$= $BC + AB\bar{C} + A\bar{B}C$

                                = $B(C + A\bar{C}) + A\bar{B}C$

Using principle $(A + B\bar{A} = A+B)$

                                = $B(C + A) + A\bar{B}C$ = $BC + AB + A\bar{B}C$

                                =$AB + C (B+ A\bar{B})$

Using principle $(A + B\bar{A} = A+B)$

                                =

                                =

Now this looks simplified:

For 1 bit, result will be

Now the circuit can be designed as(For 1 bit):

Now the circuit can be designed as(For 1 bit) AB AND AB+BC OR AB+BC+AC AND OR AC AND Similarly, In case you want it for 4bits of input A and B with 1 bit for control bit C, Circuit can be designed in following way: Al AND A3 91-83 81 81°C 1st output 81t One bit control signal(C) A1 A2 4282 AND A2 82 82C 82 82*C A) 83 C 3rd Output sit OR AND 4484 A4 AND 494 84 CC 04°C 4th output est AND OR