---
title: "Seven-segment \"e\" segment"
subject: "Digital Logic"
chapter: 4
key: "semester_3/digital_logic:e-segment-seven-segment"
source: https://www.bctengineer.com/q/semester_3/digital_logic/e-segment-seven-segment
site: BCT Engineering
---

# Seven-segment "e" segment

Digital Logic · Chapter 4: Data Processing Circuits

- Asked in: 2078 Bhadra, 6 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/e-segment-seven-segment
- Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-4

## Question

Find out the simplest logic circuit as far as possible for the 'e' segment of the seven segment display decoder.

## Short answer

Segment e is ON for digits 0,2,6,8. The minimal SOP expression is e = D'(B'+C) = B'D' + CD', implemented with one OR, one AND and an inverter on D. Don't cares 10–15 are used.

## Answer

For a BCD-to-seven-segment decoder the input is a BCD digit $ABCD$ ($A$ = MSB). Codes 1010–1111 never occur, so they are **don't cares**. Segment **e** is the lower-left segment (common-cathode display: 1 = ON).

```text
   --a--
  f     b
   --g--
  e     c
   --d--
```

#### Truth table for e

Segment e is ON only for digits 0, 2, 6 and 8.

| Digit | A B C D | e |
|---|---|---|
| 0 | 0000 | 1 |
| 1 | 0001 | 0 |
| 2 | 0010 | 1 |
| 3 | 0011 | 0 |
| 4 | 0100 | 0 |
| 5 | 0101 | 0 |
| 6 | 0110 | 1 |
| 7 | 0111 | 0 |
| 8 | 1000 | 1 |
| 9 | 1001 | 0 |
| 10–15 | 1010–1111 | X |

$$
e = \Sigma m(0,2,6,8) + d(10,11,12,13,14,15)
$$

#### K-map

```text
        CD=00 01  11  10
AB=00     1   0   0   1
AB=01     0   0   0   1
AB=11     X   X   X   X
AB=10     1   0   X   X
```

- **Quad 1:** the four corners $m_0, m_2, m_8, m_{10}$ give $B'D'$.
- **Quad 2:** column $CD = 10$: $m_2, m_6, m_{14}, m_{10}$ give $CD'$.

$$
e = B'D' + CD' = D'(B' + C)
$$

#### Simplest circuit

The factored form needs only one OR, one AND and two inverters:

```text
B --[NOT]-- B' --+
                 +--[OR]--+
C ---------------+        +--[AND]--> e
D --[NOT]-- D' -----------+
```

Check: digit 2 (0010): $D' = 1$, $B' + C = 1$, so $e = 1$; digit 4 (0100): $B' = 0$, $C = 0$, so $e = 0$. Both correct.

## Check yourself

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

1. Why can minterms 10-15 be used as don't cares in the e-segment K-map?
   A. They always turn segment e off
   B. They always turn segment e on
   C. BCD codes 1010-1111 never occur as inputs
   D. They are the same as digits 0-5
   Answer: C. In a BCD decoder those input codes are invalid, so their outputs may be taken as 0 or 1 to help grouping. (Key point: Don't cares)

2. For which digits is segment e ON (1)?
   A. 1, 3, 4 and 7
   B. 0, 2, 6 and 8
   C. 0, 2, 3 and 5
   D. 2, 4, 6 and 8
   Answer: B. Segment e is the lower-left bar, used only by 0, 2, 6 and 8 among the digits 0-9. (Key point: Segment e digits)

3. Which quad of corners m0, m2, m8, m10 gives which term?
   A. C'D'
   B. BD'
   C. A'D
   D. B'D'
   Answer: D. In those corner cells B = 0 and D = 0 stay constant while A and C change, so the term is B'D'. (Key point: K-map groups)

4. What is the simplified expression for segment e?
   A. e = B'D' + CD' = D'(B' + C)
   B. e = B'D + CD'
   C. e = BD' + C'D
   D. e = A'D' + BC
   Answer: A. Quad 1 gives B'D' and the column CD=10 gives CD'; factoring D' gives D'(B' + C). (Key point: Final simplification)

## Related in Digital Logic

- [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-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
- [Seven-segment "a" segment using NOR gates](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Aa-segment-nor-gates) — Digital Logic, chapter 4
- [Seven-segment "b" and "d" segments](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Ab-d-segments) — Digital Logic, chapter 4
- [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

## Related in other subjects

- [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×
- [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