---
title: "Seven-segment \"a\" segment using NOR gates"
subject: "Digital Logic"
chapter: 4
key: "semester_3/digital_logic:a-segment-nor-gates"
source: https://www.bctengineer.com/q/semester_3/digital_logic/a-segment-nor-gates
site: BCT Engineering
---

# Seven-segment "a" segment using NOR gates

Digital Logic · Chapter 4: Data Processing Circuits

- Asked in: 2080 Baisakh, 4+3 marks
- Video: [Seven Segment Display Decoder (Part 2)](https://www.youtube.com/watch?v=_qGFpgpqf7s) — Neso Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/digital_logic/a-segment-nor-gates
- Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-4

## Question

Draw the simplest logic circuit for "a" segment of the BCD-to-seven segment display decoder and realize the simplest logic expression using only NOR gates.

## Short answer

Simplest SOP: a = A + C + BD + B'D' = A + C + (B XNOR D). NOR-only realization uses POS: a = (A+B+C+D')(B'+C+D) with five NOR gates: two inverters, one 4-input, one 3-input, and one 2-input NOR.

## Answer

Segment a of a BCD-to-seven-segment decoder is ON for digits 0, 2, 3, 5, 6, 7, 8, 9 and OFF only for 1 and 4. Inputs 10–15 are don't-cares.

$$
a = \Sigma m(0,2,3,5,6,7,8,9) + d(10\text{–}15)
$$

#### Simplest SOP and its circuit

| AB \ CD | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 00 | 1 | 0 | 1 | 1 |
| 01 | 0 | 1 | 1 | 1 |
| 11 | X | X | X | X |
| 10 | 1 | 1 | X | X |

Groups: rows 11, 10 give $A$; columns 11, 10 give $C$; cells 5, 7, 13, 15 give $BD$; corners 0, 2, 8, 10 give $B'D'$.

$$
a = A + C + BD + B'D' = A + C + (B \odot D)
$$

```text
 B --+-[AND]--------+
 D --+              |
 B'--+-[AND]--------+--[OR 4]--- a
 D'--+              |
 A -----------------+
 C -----------------+
```

#### Realization using only NOR gates

NOR-NOR logic directly implements a **product-of-sums**. Take the 0s of the map (cells 1 and 4) with don't-cares:

- Cell 1 ($0001$): cannot combine with 9 (9 is a 1), so the maxterm is $(A + B + C + D')$.
- Cell 4 ($0100$) combines with don't-care 12: $(B' + C + D)$.

$$
a = (A + B + C + D')(B' + C + D)
$$

Using De Morgan's theorem:

$$
a = \overline{\overline{(A + B + C + D')} + \overline{(B' + C + D)}}
$$

So: two first-level NORs form the complemented sums, a second-level NOR combines them, and the complemented inputs $B'$ and $D'$ are made with NOR gates used as inverters (both inputs tied together).

```text
 B --[NOR]-- B'      D --[NOR]-- D'
 A  --+
 B  --+--[NOR 4]--N1--+
 C  --+               |
 D' --+               +--[NOR 2]--- a
 B' --+               |
 C  --+--[NOR 3]--N2--+
 D  --+
```

Total: 2 NOR inverters + one 4-input NOR + one 3-input NOR + one 2-input NOR = 5 NOR gates.

**Check:** digit 1 ($0001$): $N_1 = \overline{0+0+0+0} = 1$, so $a = 0$. Digit 4 ($0100$): $N_2 = 1$, so $a = 0$. Digit 9 ($1001$): $N_1 = 0$, $N_2 = \overline{1+0+1} = 0$, so $a = 1$. Correct.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For which BCD digits is segment a OFF?
   A. 0 and 7
   B. 1 and 4
   C. 2 and 5
   D. 8 and 9
   Answer: B. Segment a is lit for 0, 2, 3, 5, 6, 7, 8, 9 and off only for digits 1 and 4. (Key point: Segment a truth set)

2. Why do we take the 0s of the K-map to build the NOR-only circuit?
   A. NOR-NOR directly implements a sum-of-products
   B. NOR-NOR directly implements a product-of-sums
   C. Zeros need fewer gates than ones in every map
   D. Don't-cares can only be grouped with 0s
   Answer: B. A NOR-NOR network implements a product-of-sums, so the function is written as a product of maxterms taken from the 0 cells. (Key point: NOR-NOR gives POS)

3. How is the 0 at cell 4 (0100) grouped to get the simplest maxterm?
   A. Alone, giving (A + B' + C + D)
   B. With 9, giving (A + B' + C)
   C. With don't-care 12, giving (B' + C + D)
   D. With cell 1, giving (A + B + C)
   Answer: C. Cell 4 pairs with don't-care 12 (1100), which removes A and gives the maxterm (B' + C + D). (Key point: Using don't-cares for maxterms)

4. The NOR-only circuit for a = (A+B+C+D')(B'+C+D) uses how many NOR gates in total?
   A. 3
   B. 4
   C. 6
   D. 5
   Answer: D. Two NOR inverters (for B' and D'), one 4-input NOR, one 3-input NOR and one 2-input NOR give 5 gates. (Key point: NOR gate count)

## Related in Digital Logic

- [BCD-to-seven-segment decoder, segments a and e](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Abcd-seven-segment-a-e) — Digital Logic, chapter 4
- [DEMUX vs decoder; seven-segment "b" segment](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Ademux-decoder-b-segment) — Digital Logic, chapter 4
- [Seven-segment decoder "b" segment](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Ab-segment-seven-segment) — Digital Logic, chapter 4, asked 2×
- [BCD-to-7-segment decoder; frequency counter](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Abcd-7segment-frequency-counter) — Digital Logic, chapter 11
- [2-input XOR using NOR gates only](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Axor-using-nor-only) — Digital Logic, chapter 2, asked 2×
- [Seven-segment "e" segment](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Ae-segment-seven-segment) — Digital Logic, chapter 4

## Related in other subjects

- [BCD counter 0-9999 on seven segment VHDL](https://www.bctengineer.com/question.md?key=semester_6%2Fembedded_system%3Avhdl-bcd-counter-seven-segment) — Embedded System, chapter 9
- [Display 99 to 00 on seven segment](https://www.bctengineer.com/question.md?key=semester_6%2Fembedded_system%3A8051-seven-segment-99-00) — Embedded System, chapter 9
- [Seven segment display and clock counter](https://www.bctengineer.com/question.md?key=semester_6%2Fembedded_system%3Aseven-segment-clock-design) — Embedded System, chapter 9
- [Display 0 to 9 on seven segment](https://www.bctengineer.com/question.md?key=semester_6%2Fembedded_system%3A8051-seven-segment-0-9) — Embedded System, chapter 9, asked 2×