1-Bit Full Adder Circuit Simulator

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Was du lernst

Add three bits (A, B, Cin) and produce sum + carry-out.
Read the full-adder truth table: 8 rows, sum = parity, carry-out = majority.
Recognise sum as XOR of all three inputs (parity function).
Recognise carry-out as the 3-input majority function.
Identify the full-adder as the universal cell for multi-bit arithmetic.

So funktioniert es

A full adder adds three single bits — A, B, and a carry-in (Cin) — and produces a 2-bit result: a sum bit (S) and a carry-out bit (Cout). The carry-in lets full-adders chain together to form multi-bit adders, where each higher bit feeds the next via the carry chain.

- Sum: S = A ⊕ B ⊕ Cin (high if an odd number of inputs are 1) - Carry-out: Cout = (A · B) + (Cin · (A ⊕ B)) (high if at least two of the three inputs are 1)

The carry-out logic is the majority function: out of three inputs, the carry is 1 whenever at least two of them are 1. This makes intuitive sense — adding three bits gives a value 0, 1, 2, or 3, and the high bit (carry-out) is 1 only for results ≥ 2.

With three binary inputs there are 2³ = 8 possible input combinations. The truth table has 8 rows; sum follows the parity pattern; carry-out follows the majority pattern.

A full adder is the workhorse of binary arithmetic. Every wide adder, subtractor, multiplier, and divisor inside a CPU is built from chained full-adders.

Wahrheitstabelle

Three inputs, two outputs. Sum is parity (XOR); Carry-out is majority (≥ 2 of 3 high).

Eingänge			Ausgang
A	B	Cin	Sum	Cout
0	0	0	0	0	0 + 0 + 0 = 0
0	0	1	1	0	0 + 0 + 1 = 1
0	1	0	1	0	0 + 1 + 0 = 1
0	1	1	0	1	0 + 1 + 1 = 2 (binary 10)
1	0	0	1	0	1 + 0 + 0 = 1
1	0	1	0	1	1 + 0 + 1 = 2
1	1	0	0	1	1 + 1 + 0 = 2
1	1	1	1	1	1 + 1 + 1 = 3 (binary 11)

Boolescher Ausdruck

S = A \oplus B \oplus C_{in}

Sum bit — XOR of all three inputs (the parity function for 3 bits).

C_{out} = AB + C_{in}(A \oplus B)

Carry-out — fires if both A and B are 1, OR if Cin is 1 and at least one of A/B is 1.

C_{out} = AB + AC_{in} + BC_{in}

Equivalent: the 3-input majority function. Carry happens when any pair of inputs is 1.

Schritt für Schritt ausprobieren

Stelle die Eingänge in der Einbettung oben ein, lies was passieren sollte und überprüfe es.

1

A = 0 B = 0 Cin = 0

Erwartet: S=0, Cout=0

Was du siehst: All zero — no addition. Default state.
2

A = 1 B = 0 Cin = 0

Erwartet: S=1, Cout=0

Was du siehst: Single bit set — sum is 1, no overflow.
3

A = 1 B = 1 Cin = 0

Erwartet: S=0, Cout=1

Was du siehst: Two bits set — sum=2 (binary 10). Cout fires; sum bit is 0.
4

A = 1 B = 1 Cin = 1

Erwartet: S=1, Cout=1

Was du siehst: All three bits set — sum=3 (binary 11). Both output bits high.

Verwendete Komponenten

Praxisanwendungen

CPU integer ALU. A 64-bit integer adder is 64 full-adders chained (with optional carry-lookahead acceleration). Every add, increment, decrement, and pointer arithmetic operation flows through this hardware.

Multipliers. Wallace-tree and Dadda-tree multipliers use arrays of full-adders to compress partial products into a final sum.

Floating-point arithmetic. The mantissa add/subtract path is a chain of full-adders after exponent alignment.

DSP MAC units. Multiply-accumulate (MAC) operations in DSP processors use a full-adder chain to accumulate products.

Counters. Incrementing a counter by 1 is essentially a full-adder chain with B fixed to 0...01 and Cin = 0.

Häufig gestellte Fragen

Why does a full adder have three inputs?

Two operand bits (A, B) plus the carry-in (Cin) from the previous bit position. The carry-in lets multi-bit additions propagate carries through the chain — bit 1's Cin is bit 0's Cout, etc.

Why is the sum just XOR of all three?

Sum is parity — high if an odd number of inputs are 1. Bits 0+0+0 = 0 (even), 0+0+1 = 1 (odd), 1+1+1 = 3 (odd). XOR over all inputs is exactly this parity calculation.

Why is the carry-out the majority function?

Adding three bits gives 0, 1, 2, or 3. The high bit (carry) is 1 only when the sum is ≥ 2 — i.e., when at least 2 of the 3 inputs are 1. That's the definition of majority.

Can a full adder be built from two half-adders?

Yes: the first half-adder adds A and B (gives sum1, carry1); the second half-adder adds sum1 and Cin (gives final S, carry2); the OR of carry1 and carry2 gives Cout. This decomposition matches the standard textbook diagram.

How many transistors is a full adder in CMOS?

About 28 transistors in static CMOS — 12 for sum (XOR-XOR), 16 for carry-out (majority). Optimized layouts using transmission gates can reach ~24 transistors. Carry-chain optimizations may share transistors across adjacent cells.