Full adder using a 2×4 decoder

A full adder adds three bits A, B and carry-in $C_{in}$ and gives sum S and carry-out $C_o$.

Truth table

$$S = \Sigma m(1,2,4,7), \qquad C_o = \Sigma m(3,5,6,7)$$

Realization with logic gates

From K-maps / algebra:

$$\begin{aligned} S &= A \oplus B \oplus C_{in} \\ C_o &= AB + BC_{in} + AC_{in} = AB + C_{in}(A \oplus B) \end{aligned}$$

A, B ────────[XOR]── P = A⊕B
P, Cin ──────[XOR]──────────────── S

P, Cin ──────[AND]── Cin·P ──┐
                             ├─[OR]── Co
A, B ────────[AND]── AB ─────┘

(Two XOR, two AND and one OR gate.)

Realization with a 2×4 decoder

A 2×4 decoder has only two inputs, so it cannot generate all 8 minterms. Feed A and B to the decoder; its outputs are the minterms of A, B:

$$Y_0 = \overline{A}\,\overline{B},\quad Y_1 = \overline{A}B,\quad Y_2 = A\overline{B},\quad Y_3 = AB$$

Write S and $C_o$ for each AB combination in terms of $C_{in}$:

$$\begin{aligned} S &= C_{in}(Y_0+Y_3) + \overline{C_{in}}(Y_1+Y_2) = C_{in} \oplus (Y_1+Y_2) \\ C_o &= Y_3 + C_{in}(Y_1+Y_2) \end{aligned}$$

($Y_1 + Y_2 = A \oplus B$, so these are the same equations as the gate realization.)

          +----------+
 A ──────>| I1    Y0 |── (not used)
 B ──────>| I0    Y1 |── Y1
          |  2x4  Y2 |── Y2
          |  DEC  Y3 |── Y3
          +----------+

Y1, Y2 ──────[OR]── P (= A⊕B)
P, Cin ──────[XOR]─────────────── S

P, Cin ──────[AND]── Cin·P ──┐
                             ├─[OR]── Co
Y3 ──────────────────────────┘

Gates needed besides the decoder: two OR gates, one XOR gate and one AND gate.

Check: A=1, B=1, $C_{in}$=0: $Y_3 = 1$, $P = 0$, so $S = 0 \oplus 0 = 0$ and $C_o = Y_3 = 1$; correct ($1+1 = 10_2$).

(With a 3×8 decoder, the usual method $S = Y_1+Y_2+Y_4+Y_7$, $C_o = Y_3+Y_5+Y_6+Y_7$ would need no XOR gate.)