Half adder and full adder

WHY do we even need these?

WHAT we want: a circuit that, given input bits, outputs their arithmetic sum in binary.

The catch: $1 + 1 = 10_2$ — a two-bit answer. So any adder needs two outputs:

Sum (the current column's bit)
Carry (the bit that spills into the next column)

HOW we discover the logic: build the truth table, then read off the Boolean expression.

Half Adder

Deriving it from scratch

List every input combination and the arithmetic result:

$A$	$B$	$A+B$ (decimal)	Carry $C$	Sum $S$
0	0	0	0	0
0	1	1	0	1
1	0	1	0	1
1	1	2 = $10_2$	1	0

Why this step? The table is just "count the ones and write it in binary." No Boolean guessing yet — pure arithmetic truth.

Now read the columns:

$S$ is 1 exactly when $A$ and $B$ differ → that's XOR.
$C$ is 1 exactly when both are 1 → that's AND.

Full Adder

Deriving it from scratch

$A$	$B$	$C_{in}$	sum(dec)	$C_{out}$	$S$
0	0	0	0	0	0
0	0	1	1	0	1
0	1	0	1	0	1
0	1	1	2	1	0
1	0	0	1	0	1
1	0	1	2	1	0
1	1	0	2	1	0
1	1	1	3	1	1

Why this step? Again, just count the 1s among the three inputs and write the count in 2-bit binary $(C_{out}S)$ .

Reading Sum $S$ : $S=1$ when the number of 1s is odd (rows with count 1 or 3). "Odd number of 1s" is exactly the XOR of all three: $S = A \oplus B \oplus C_{in}$ Why? XOR flips output each time an input is 1; an odd count leaves it at 1.

Reading Carry $C_{out}$ : $C_{out}=1$ when at least two inputs are 1. Group the minterms: $C_{out} = AB + BC_{in} + AC_{in}$ This is the majority function — output the value held by the majority of the three inputs.

Why the second form is equivalent (steel-man the algebra): $(A\oplus B)C_{in} = (A\bar B + \bar A B)C_{in}$ . Add $AB$ . When $A=B=1$ , the $AB$ term already fires. When exactly one of $A,B$ is 1, then $A\oplus B=1$ so the second term fires iff $C_{in}=1$ . This reproduces "≥2 ones" exactly — matching $AB+AC_{in}+BC_{in}$ . ✓

Full adder from two half adders

Worked examples

Common mistakes

Recall Feynman: explain to a 12-year-old

Imagine you're adding 1 + 1 with only the buttons 0 and 1. The answer is "two," but you can't write "2" — you only have 0 and 1! So you write a 0 here and carry a little 1 over to the next spot, like carrying in normal addition. The half adder is a tiny machine that does this for two buttons. But sometimes a carry arrives from the right, so you're really adding three things. The full adder is the bigger machine that can handle all three. Line up a row of full adders and they can add huge numbers, one column at a time — that's how computers add!

Flashcards

What are the two outputs of any binary adder cell?

Sum (this column's bit) and Carry (bit passed to the next column).

Half adder Sum equation?

S = A \oplus B

(XOR).

Half adder Carry equation?

C = A \cdot B

(AND).

Why is it called a "half" adder?

It adds only two bits and cannot accept a carry-in from a previous stage.

How many inputs does a full adder have?

Three:

A

B

, and

C_{in}

Full adder Sum equation?

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

(odd/parity of inputs).

Full adder Carry-out equation?

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

(majority function).

Full adder from two half adders — final carry?

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

, i.e. OR of the two half-adder carries.

Common mistake: is half-adder carry OR?

No — it's AND; carry only when BOTH bits are 1.

What multi-bit adder do you get by chaining full adders?

A ripple-carry adder (each

C_{out}

feeds next

C_{in}

For

A=1,B=1,C_{in}=1

, what is

(C_{out},S)

(1,1) = 3

Connections

XOR gate — the parity behavior powering the Sum output
AND gate and OR gate — build the carry logic
Ripple-carry adder — full adders chained
Carry-lookahead adder — faster carry, avoids ripple delay
Truth tables and minterms — the derivation method used here
Two's complement subtraction — adders reused for subtraction
Combinational circuits — the parent family (no memory, output = f(inputs))

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab hum do numbers add karte hain to har column me ek digit likhte hain aur kabhi-kabhi ek carry aage bhej dete hain. Binary me bhi bilkul yahi hota hai, bas digits sirf 0 aur 1 hote hain. Problem yeh hai ki $1+1 = 10_2$ , yani do-bit ka answer. Isliye kisi bhi adder ke do outputs chahiye — ek Sum (is column ka bit) aur ek Carry (agle column me jaane wala bit).

Half adder sirf do bits ( $A$ aur $B$ ) add karta hai. Truth table banao, dekho: Sum tab 1 hota hai jab dono bits alag hain — yeh to XOR hai. Aur Carry tab 1 jab dono 1 hain — yeh AND hai. Bas: $S=A\oplus B$ , $C=A\cdot B$ . Ise "half" kehte hain kyunki yeh pichle stage ka carry-in accept nahi kar sakta.

Full adder teen bits add karta hai — $A$ , $B$ , aur $C_{in}$ . Sum tab 1 hota hai jab 1's ki ginti odd ho — matlab teeno ka XOR: $S=A\oplus B\oplus C_{in}$ . Aur carry-out tab 1 jab kam se kam do inputs 1 hon — ise majority function kehte hain: $C_{out}=AB+AC_{in}+BC_{in}$ .

Sabse important baat: Boolean me $+$ ka matlab OR hota hai, arithmetic addition nahi. Isliye Sum kabhi bhi $A+B$ (OR) nahi hoga — woh hamesha XOR hai. Aur agar bahut saare full adders ko chain kar do (har ka $C_{out}$ agle ka $C_{in}$ ), to bade numbers add ho jaate hain — isi ko ripple-carry adder kehte hain, aur yahi CPU ke andar addition ka core hai.

WHY do we even need these?

Half Adder

Deriving it from scratch

Full Adder

Deriving it from scratch

Full adder from two half adders

Worked examples

Common mistakes

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Combinational Circuits

Connections