2-Bit Multiplier

Question

Problem: 2-Bit Multiplier. Solve this hands-on electronics task on EWskills.

Accepted Answer

Step 1: Long Multiplication Matrix

The circuit calculates multiplication just like long multiplication on paper. First, we generate the partial products by multiplying each bit, then we sum them by columns:

      A1  A0
   ×  B1  B0
   ---------
    A1B0 A0B0  (Row 1)
+ A1B1 A0B1    (Row 2, shifted left)
--------------------
P3   P2   P1   P0

Step 2: Column Breakdown

P0 Output: This bit is the first column. It has no other terms to add, so it is just (A0 . B0) using an AND gate.
P1 Output: This bit adds the two terms in the middle column ((A1 . B0) and (A0 . B1)). An XOR gate finds the sum for P1, and an AND gate saves the carry-out (Cout).
P2 and P3 Outputs: The last column contains the term (A1 . B1). We add the carry-out (Cout) to this term. An XOR gate gives the result for P2, and an AND gate outputs the final overflow bit for P3.

Boolean Equations:

P0 = A0 . B0
P1 = (A1 . B0) ^ (A0 . B1)
Cout = (A1 . B0) . (A0 . B1)
P2 = (A1 . B1) ^ C1
P3 = (A1 . B1) . C1

53. 2-Bit Multiplier