Activity 2.2.6 — Majority Vote Circuit Design¶

Learning Objectives¶

By the end of this lesson, students will be able to:

Design a combinational logic circuit from a word problem specification
Create a truth table for a majority vote function (3 inputs, 2+ HIGH = output HIGH)
Simplify the Boolean expression using K-maps
Implement the circuit using logic gates
Extend the design to handle 5 inputs (3+ HIGH = output HIGH)

Vocabulary¶

Vocabulary (click to expand)

Term	Definition
Majority Function	A logic function where output is HIGH when more than half the inputs are HIGH
M-of-N Logic	Output is HIGH when at least M of N inputs are HIGH
Design from Specification	Creating a circuit starting from a written problem description

Part 1: Understanding Majority Vote Logic¶

The majority vote function is a classic combinational logic problem used in voting systems, error detection, and redundant systems.

Definition¶

For N inputs, the majority vote output is HIGH when more than N/2 inputs are HIGH.

For our case: - 3-input majority: Output HIGH when at least 2 of 3 inputs are HIGH - 5-input majority: Output HIGH when at least 3 of 5 inputs are HIGH

Key insight: "At least 2 out of 3" means: 2 HIGH, or 3 HIGH. Cases: (1,1,0), (1,0,1), (0,1,1), (1,1,1)

Part 2: 3-Input Majority Vote Design¶

Problem Statement¶

Design a circuit with 3 inputs (A, B, C) where the output Y is HIGH when at least 2 of the 3 inputs are HIGH.

Step 1: Create Truth Table¶

There are 8 possible input combinations (2^3 = 8):

A	B	C	Y (Majority)	At Least 2 HIGH?
0	0	0	0	No
0	0	1	0	No (only 1)
0	1	0	0	No (only 1)
0	1	1	1	YES (2)
1	0	0	0	No (only 1)
1	0	1	1	YES (2)
1	1	0	1	YES (2)
1	1	1	1	YES (3)

Step 2: Write Boolean Expression¶

The minterms where Y = 1: - m3 (A' B C): 011 - m5 (A B' C): 101 - m6 (A B C'): 110 - m7 (A B C): 111

Expression: Y = A'BC + AB'C + ABC' + ABC

Simplify: Group m5,m7: AC(B'+B) = AC Group m6,m7: AB(C'+C) = AB
Group m3,m7: BC(A'+A) = BC

Y = AC + AB + BC

Or factored: Y = (A·B) + (B·C) + (A·C)

Key insight: This is the classic majority vote expression: output is HIGH when any two inputs are HIGH.

Step 3: Simplify Using K-Map¶

3-variable K-map:

	BC=00	BC=01	BC=11	BC=10
A=0	0	0	1	0
A=1	0	1	1	1

Grouping: - Group of 2: m5,m7 → AC - Group of 2: m6,m7 → AB
- Group of 2: m3,m7 → BC

These three groups of 2 overlap at m7, which is fine.

Simplified Expression: Y = AB + BC + AC

This matches our algebraic result! ✓

Step 4: Draw Circuit¶

Circuit diagram:

A → AND →\
          OR → Y
B → AND →/  |
          ↗
C → AND →/  |
         ↗
(AB + BC + AC)

Using 3 AND gates (2-input) and 1 OR gate (3-input).

Step 5: Verify¶

Check each output condition: - A=1,B=1,C=0: AB=1 → Y=1 ✓ - A=1,B=0,C=1: AC=1 → Y=1 ✓ - A=0,B=1,C=1: BC=1 → Y=1 ✓ - A=1,B=1,C=1: All products =1 → Y=1 ✓

Part 3: Understanding the Expression¶

The expression Y = AB + BC + AC has important properties:

Alternative Forms¶

Sum of Products: Y = AB + BC + AC

Factored Form: Y = (A + B)(A + C)(B + C) — Try verifying this!

Interesting Property¶

Notice that AB + BC + AC = (A ⊕ B)·C + A·B [using XOR]

But the simple form AB + BC + AC is easiest to implement.

Implementation with NAND¶

Since NAND gates are universal and readily available, let's convert to NAND-only:

Y = AB + BC + AC = ((AB)' · (BC)' · (AC)')' [NAND of NANDs]

Circuit uses 4 NAND gates.

Part 4: Extending to 5-Input Majority Vote¶

Problem Statement¶

Design a circuit with 5 inputs where output Y is HIGH when at least 3 of the 5 inputs are HIGH.

Approach¶

With 5 inputs, we have 32 combinations (2^5 = 32). We'll need to identify all cases where output = 1:

Cases where Y = 1: - Exactly 3 inputs HIGH: C(5,3) = 10 combinations - Exactly 4 inputs HIGH: C(5,4) = 5 combinations - Exactly 5 inputs HIGH: C(5,5) = 1 combination

Total: 10 + 5 + 1 = 16 minterms

Simplification Strategy¶

Rather than write all 32 rows, we can use a systematic approach:

Identify all 3-input combinations (C(5,3) = 10)
Identify all 4-input combinations (C(5,4) = 5)
Identify all 5-input combination (1)

For 5 variables (A,B,C,D,E), each minterm represents a combination like ABC'D'E (where ' = complemented).

Alternative: Build from 3-Input Modules¶

A practical approach is to use multiple 3-input majority modules: - First stage: Two 3-input majority voters - Second stage: 3-input majority of the two stage outputs + remaining input

This is called cascading and is a common engineering approach.

Cascade Design¶

Inputs: A, B, C, D, E

Stage 1:
- Majority3_1: inputs A, B, C → M1
- Majority3_2: inputs D, E, (tie to ground or use one from first group) 

Actually, better approach:
Stage 1: Majority(A, B, C) → M1
         Majority(A, B, D) → M2  
         Majority(A, B, E) → M3
Stage 2: Majority(M1, M2, M3) → Final Output

This cascades the majority decision to ensure at least 3 of the 5 are needed.

Key insight: For complex functions, breaking into smaller modules that can be tested individually makes design and debugging much easier.

Part 5: Practical Applications¶

Real-World Uses of Majority Voting¶

Redundant Systems: In aerospace and nuclear systems, multiple sensors measure the same value. The majority vote determines the output, tolerating up to N-1 sensor failures.
Error Correction: Data transmission systems use majority voting to correct single-bit errors.
Fault-Tolerant Computing: Multiple processors vote on the correct output.
Election Systems: Democratic voting is essentially a majority function.

Example: Triple Modular Redundancy (TMR)¶

Three sensors measure temperature: - If 2 or 3 sensors agree, that's the accepted reading - One faulty sensor is tolerated

Circuit: Three sensors → 3-input majority voter → system decision

Practice Problem — Majority Vote¶

Problem 1: For a 4-input majority vote circuit (output HIGH when at least 3 of 4 inputs are HIGH), create the truth table for first 8 rows.

Show Solution

B	C	D	Y (Majority of 4)
0	0	0	0 (0 of 4)
0	0	1	0 (1 of 4)
0	1	0	0 (1 of 4)
0	1	1	1 (2 of 4 - not majority for 4)
1	0	0	0 (1 of 4)
1	0	1	1 (2 of 4 - not majority)
1	1	0	1 (2 of 4 - not majority)
1	1	1	1 (3 of 4 - YES majority!)

Note: For 4-input, "majority" means 3 or 4 HIGH (more than half, which is 2). So need at least 3 HIGH.

Problem 2: Implement Y = AB + BC + AC using only NAND gates. Draw the circuit.

Show Solution

Step 1: Original expression: Y = AB + BC + AC

Step 2: Write as NAND of NANDs: Y = ((AB)' · (BC)' · (AC)')'

Circuit: 1. NAND(A, B) → (AB)' 2. NAND(B, C) → (BC)' 3. NAND(A, C) → (AC)' 4. NAND(outputs of 1,2,3) → Y

Total: 4 NAND gates needed.

This is the NAND-only implementation!

Problem 3: The school cafeteria needs a voting system where 3 student representatives vote on a proposal (each has a switch). The proposal passes if at least 2 of 3 vote YES. Draw the complete design (truth table, simplified expression, circuit).

Show Solution

This is exactly the 3-input majority vote we designed in Part 2!

Truth Table: (From earlier) | A | B | C | Y | |---|---|---|---| | 0 | 0 | 0 | 0 | | 0 | 0 | 1 | 0 | | 0 | 1 | 0 | 0 | | 0 | 1 | 1 | 1 | | 1 | 0 | 0 | 0 | | 1 | 0 | 1 | 1 | | 1 | 1 | 0 | 1 | | 1 | 1 | 1 | 1 |

Simplified Expression: Y = AB + BC + AC

Circuit: - 3 input switches with pull-down resistors - 3 AND gates (for AB, BC, AC) - 1 OR gate (3-input) to combine - Output LED with resistor

Hardware: - Use 74HC08 (quad 2-input AND) + 74HC32 (quad 2-input OR) - Or use NAND-only: 74HC00 (quad 2-input NAND)

Summary¶

Majority vote: output HIGH when more than half inputs are HIGH
3-input majority: Y = AB + BC + AC (any two or all three)
K-map simplification shows three overlapping groups of 2
5-input majority can be designed directly or cascaded from 3-input modules
Real-world applications include redundant systems, error correction, fault tolerance

Key Reminders¶

"At least 2 of 3" means: exactly 2 OR exactly 3 HIGH
K-map grouping for majority: three overlapping groups of 2
Use factored form Y = (A+B)(A+C)(B+C) for POS implementation if needed
For larger inputs (5+), consider cascade design to simplify implementation

Assessment Rubric — Majority Vote Circuit¶

Criterion	4 — Exemplary	3 — Proficient	2 — Developing	1 — Beginning
Truth Table	Complete (8 rows for 3-input), all correct	Complete, minor errors	Incomplete	Missing/incorrect
Boolean Expression	Correct minterms or simplified form	Correct	Minor errors	Incorrect
Simplification	K-map grouped correctly, expression minimized	K-map correct	Attempted, errors	No attempt
Circuit	Clear, complete with gates labeled	Circuit shown	Partial circuit	No circuit
Extension (5-input)	Shows systematic approach or cascade	Approach identified	Mentioned	Not addressed

Custom activity — adapted from PLTW Digital Electronics