Comparators

1. What are we actually building?

WHAT is the simplest one? A 1-bit comparator. Compare a single bit $A$ against a single bit $B$ .

HOW do we start? Write the truth table — 2 inputs → 4 rows.

$A$	$B$	$A>B$	$A=B$	$A<B$
0	0	0	1	0
0	1	0	0	1
1	0	1	0	0
1	1	0	1	0

2. Deriving the 1-bit logic from scratch

WHY these equations? Read the truth table column by column and translate each "1" into a product term.

$A>B$ : the only row where this is 1 is $A=1, B=0$ . That is the product term $A\bar B$ . $G = A\bar B$

$A<B$ : the only 1 is at $A=0, B=1$ , i.e. $\bar A B$ . $L = \bar A B$

$A=B$ : 1 when both are 0 or both are 1. That is exactly the XNOR: $E = \bar A\bar B + AB = \overline{A\oplus B}$

3. Building a multi-bit comparator (the real trick)

WHY not just a giant truth table? For $n$ -bit numbers you'd need $2^{2n}$ rows — for 4-bit that's 256 rows. Impossible to hand-design. We need a structured, first-principles method: think like a human comparing numbers.

HOW to formalise for bits. Let $A = A_{n-1}\dots A_0$ , $B=B_{n-1}\dots B_0$ .

Define per-bit equality $x_i = \overline{A_i\oplus B_i}$ (1 when bit $i$ matches).

$A>B$ happens if, scanning from the top, the highest differing bit has $A_i=1, B_i=0$ , and all higher bits were equal:

G = A_{n-1}\bar B_{n-1} + x_{n-1}A_{n-2}\bar B_{n-2} + x_{n-1}x_{n-2}A_{n-3}\bar B_{n-3} + \dots

WHY the $x$ chain? The term $x_{n-1}x_{n-2}$ means "all bits above position $n-3$ were equal, so this is the first place they differ" — exactly the human rule.

Symmetrically:

L = \bar A_{n-1}B_{n-1} + x_{n-1}\bar A_{n-2}B_{n-2} + \dots

And equality is "every bit matches": $E = x_{n-1}\,x_{n-2}\cdots x_1\,x_0$

4. Worked example — 2-bit comparator from scratch

Compare $A=A_1A_0$ with $B=B_1B_0$ .

Step 1 — per-bit equals. $x_1=\overline{A_1\oplus B_1}$ , $x_0=\overline{A_0\oplus B_0}$ . Why this step? These are the reusable "match" signals the human method needs.

Step 2 — Equal. Both bits match: $E = x_1 x_0$ Why? Numbers are equal only if MSB and LSB agree.

Step 3 — Greater. MSB decides first; if MSBs tie, LSB decides: $G = A_1\bar B_1 + x_1\,A_0\bar B_0$ Why? First term: MSB of $A$ wins outright. Second: MSBs equal ( $x_1$ ), so LSB $A_0\bar B_0$ breaks the tie.

Step 4 — Less. Mirror image: $L = \bar A_1 B_1 + x_1\,\bar A_0 B_0$

Check with $A=10_2=2$ , $B=01_2=1$ : $A_1\bar B_1 = 1\cdot 1 =1$ → $G=1$ . ✔ (2 > 1)

Check with $A=01$ , $B=01$ : $x_1=1,x_0=1$ → $E=1$ . ✔

5. Worked example — the IC 7485 (4-bit) cascade

WHY care? Real chips like the 7485 take cascade inputs $(A>B)_{in},(A=B)_{in},(A<B)_{in}$ so you can chain them for wider numbers.

Rule when tie occurs internally: if the 4 local bits are all equal, the chip passes through the cascade input.

(A>B)_{out} = G_{local} + E_{local}\cdot(A>B)_{in}

Why this step? $E_{local}$ means "these 4 bits didn't decide anything, so trust the less-significant chip below" — again the human first-differing-digit rule, now across whole chips.

Cascade wiring for 8-bit: feed the low nibble's outputs into the high nibble's cascade inputs. The low chip's ground-level inputs are tied $(A=B)_{in}=1$ (so if everything equals, output equals).

6. Common mistakes

7. Active recall

Recall Test yourself (hide answers)

Which gate implements 1-bit equality? → XNOR
Which bit decides magnitude first? → the MSB
Why guard lower-bit terms with $x_i$ ? → so they only act when higher bits tied
What does $E_{local}$ do in a cascaded 7485? → passes cascade input through
Sum identity that must always hold? → $G+E+L=1$

Recall Feynman: explain to a 12-year-old

Imagine two kids showing number cards, biggest first. They flip the front card: if one is bigger, that kid wins the whole game — no need to look further. If the front cards match, they move to the next card, and so on. If every card matches, it's a tie. A comparator is a tiny electric machine that plays this exact card game with 1s and 0s, instantly.

8. Flashcards

What three outputs does a comparator produce?

A>B

(G),

A=B

(E),

A<B

(L), mutually exclusive.

1-bit equality equation?

E=\overline{A\oplus B}

(XNOR).

1-bit greater-than equation?

G=A\bar B

1-bit less-than equation?

L=\bar A B

Why compare MSB first?

Magnitude is dominated by the most significant bit; the first differing bit decides.

Purpose of the

x_i=\overline{A_i\oplus B_i}

signals?

Per-bit equality flags used to guard lower-bit decision terms.

2-bit greater-than expression?

G=A_1\bar B_1 + x_1 A_0\bar B_0

n-bit equality expression?

E=x_{n-1}x_{n-2}\cdots x_0

(all bits match).

What identity always holds among comparator outputs?

G+E+L=1

Role of cascade inputs in the 7485 IC?

Chain multiple comparators; used when local bits are all equal to break the decision at less-significant stage.

7485 cascaded output formula for A>B?

(A>B)_{out}=G_{local}+E_{local}(A>B)_{in}

Common XOR-vs-XNOR trap?

XOR detects difference (inequality); equality needs XNOR.

9. Connections

XOR and XNOR Gates — the equality primitive
Truth Tables and Minterms — how we read out $G,E,L$
Binary Subtraction — subtracting and checking the sign bit is an alternate comparison method
Multiplexers — another decision-making combinational block
Ripple Carry Adder — contrast: adders go LSB-first, comparators go MSB-first
7485 IC — the standard cascadable comparator chip

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Comparator ka kaam simple hai: do binary numbers $A$ aur $B$ ko compare karo aur batao kaun bada hai — teen output milte hain: $A>B$ , $A=B$ , aur $A<B$ . Ek time pe sirf ek hi output high hota hai, isliye $G+E+L=1$ hamesha true rahta hai. Sabse chhota comparator 1-bit ka hota hai: $G=A\bar B$ , $L=\bar A B$ , aur equality ke liye XNOR use hota hai kyunki XNOR tabhi 1 deta hai jab dono bits same ho.

Sabse important intuition: hum numbers ko MSB se compare karte hain, LSB se nahi. Jaise aap 3746 aur 3752 compare karte ho — aage se digit dekhte ho, jahan pehli baar difference milta hai wahi decide kar deta hai, baaki digits matter nahi karte. Hardware bhi bilkul yehi karta hai. Isliye har lower-bit term ko $x_i=\overline{A_i\oplus B_i}$ (matching flag) se "guard" karte hain — matlab lower bit tabhi bolega jab upar ke saare bits equal ho.

Common galti: log equality ke liye XOR laga dete hain, par XOR to "difference" batata hai, ulta jawab. Equality ke liye hamesha XNOR. Doosri galti: LSB se compare karna — galat, kyunki $1000 > 0111$ hota hai even though LSBs mein $B$ bade dikhte hain.

Real chips jaise 7485 mein cascade inputs hote hain taaki chhote comparators ko jodkar bade numbers compare kar sako. Rule: agar local 4 bits sab equal ho ( $E_{local}=1$ ), to niche wale (less-significant) chip ka jawab pass kar do — same MSB-first logic, bas ab poore chips ke level pe.