Page 43 - Computer_Science_F5

Computer Science circuit for the half adder is illustrated in From the truth table, you obtain the
Figure 1.34 which is simplified using the Boolean expression for sum (S) and
carry (C ) as follows:
XOR gate as shown in Figure 1.35.
2
A
ABC +
ABC
S =
ABC +
ABC +
A
A . B
1
1
1
1
0 FOR ONLINE READING ONLY
B
A A . B S = A . B + A . B S = ABC + 1 ABC + 1 ABC 1 + ABC 1
B B S = C 1 (AB AB+ ) C AB AB+ 1 ( + )
C = A . B S = C 1 (AB AB+ ) C AB AB+ 1 ( + )
Figure 1. 34: Half adder logic circuit S CA 1 (  ) B  CA 1 ( B )
A SC  (A B ) and C = AB AC+ + BC
S = A ⊕ B 1 2 1 1
B
A The Boolean expressions show that sum
C = A . B (S) and carry (C) are equivalent to XOR
A
B ( S  C  B ) and (C = 2 AB AC+ 1 + BC )
1
1
Figure 1. 35: Half adder logic circuit using gate. The logic circuit for the full adder
XOR and AND gates is illustrated in Figure 1.36.
Construction of full adder A S
B
A full adder adds three input bits. There C 1
are three inputs and two outputs on these
circuits. The two significant bits to be
added are represented by two inputs while
the third input represents the carry from the C 2
previous significant position. Table 1.13
shows the truth table for the full adder.
Figure 1.36: Full adder logic circuit using
Table 1.13: Full adder truth table three inputs XOR, OR and three
Input Output AND gates
A B C 1 Sum(S) C 2 The Boolean expression for a carry-out
0 0 0 0 (C ) can be simplified using XOR as
2
0 0 1 1 0 follows:
0 1 0 1 0 C = AB AC+ + BC
0 1 1 0 1 2 1 1
1 0 0 1 0 C = 2 AB AC+ 1 (B B+ ) BC A A+ 1 ( + )
1 0 1 0 1 C = AB AC B ABC+ + + AC B BC A+
1 1 0 0 1 2 1 1 1 1
1 1 1 1 1 C = 2 AB (1 C+ 1 + C 1 ) AC B BC A+ 1 + 1

34
for Advanced Secondary Schools

Computer Science Form 5.indd 34 23/07/2024 12:32