Adders

QUARTER ADDER

A quarter adder is a circuit that can add two binary digits but will not produce a carry. This circuit will produce the following results:

0 plus 0 = 0

0 plus 1 = 1

1 plus 0 = 1

1 plus 1 = 0 (no carry)

You will notice that the output produced is the same as the output for the Truth Table of an X-OR. Therefore, an X-OR gate can be used as a quarter adder.

The combination of gates in the figure below will also produce the desired results. When A and B are both LOW (0), the output of each AND gate is LOW (0); therefore, the output of the OR gate is LOW (0). When A is HIGH and B is LOW, then B is HIGH and AND gate 1 produces a HIGH output, resulting in a sum of 1 at gate 3. With A LOW and B HIGH, gate 2 output is HIGH, and the sum is 1. When both A and B are HIGH, neither AND gate has an output, and the output of gate 3 is LOW (0); no carry is produced.

Quarter adder.

A half adder is designed to combine two binary digits and produce a carry.

The figure below shows two ways of constructing a half adder. An AND gate is added in parallel to the quarter adder to generate the carry. The SUM column of the Truth Table represents the output of the quarter adder, and the CARRY column represents the output of the AND gate.

Half adders and Truth Table.

The full adder becomes necessary when a carry input must be added to the two binary digits to obtain the correct sum. A half adder has no input for carries from previous circuits.

One method of constructing a full adder is to use two half adders and an OR gate as shown in the figure below. The inputs A and B are applied to gates 1 and 2. These make up one half adder. The sum output of this half adder and the carry-from a previous circuit become the inputs to the second half adder. The carry from each half adder is applied to gate 5 to produce the carry-out for the circuit.

Full adder and Truth Table.

First, let�s add 1 and 0. When either A or B is HIGH, gate 1 has an output. This output is applied to gates 3 and 4. Since the carry-in is 0, only gate 3 will produce an output. The sum of 1 2 and 0 is 1 2.

Now let�s add 1 2 and 1 2. If A and B are both HIGH, the output of gate 1 is LOW. When the carry-in is 0 (LOW), the output of gate 3 is LOW. Gate 2 produces an output that is applied to gate 5, which produces the carry-out. The sum of 1 2 and 1 2 is 10 2, just as it was for the half adder.

When A and B are both LOW and the carry-in is 1, only gate 3 has an output and produces a sum of 1 2 with no carry-out.

Now, let�s add A or B and a carry-in. For example, let�s assume that A is HIGH and B is LOW. With these conditions, gate 1 will have an output. This output and the carry-in applied to gates 3 and 4 will produce a sum out of 0 and a carry of 1. This carry from gate 4 will cause gate 5 to produce a carry-out. The sum of A and a carry (1 2 plus 1 2) is 10 2.

When A, B, and the carry-in are all HIGH, a sum of 1 and a carry-out are produced. First, consider A and B. When both are HIGH, the output of gate 1 is LOW, and the output of gate 2 is HIGH, giving us a carry-out at gate 5. The carry-in produces a 1 output at gate 3, giving us a sum of 1. The output of the full adder is 1 1 2. The sum of 1 2 plus 1 2 plus 1 2 is 1 1 2.

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