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Digital Electronics

Number systems, Boolean algebra, K-maps, combinational and sequential circuits — the ECE exam and Viva foundation

4 modules8 lessons1 hr 22 minBeginner
Beginner
8 lessons in 4 modules
1h 22m of material
8 challenges
English audio
English captions
Certificate on completion

What you'll get

  • Verifiable certificate

  • XP on the leaderboard

Your certificate

TTieEduCertificate of Completionawarded toyour nameDigital ElectronicsFoundation Tier

Preview of the certificate this course issues. Yours will carry your name, score and a verifiable serial — signed, not editable.

What you will build

Traffic-light controller + mod-N counter (design doc)

Design two circuits end-to-end on paper (or in a free simulator like Digital or Logisim if you prefer): (1) a two-road traffic light as a Moore FSM with states, state table, state assignment, K-maps for next-state and output logic, and the final gate/flip-flop diagram; (2) a mod-N synchronous counter for an N of your choice between 5 and 10, with the terminal-count logic and a wrap-around proof. Submit a single PDF with both designs, the K-maps, and a short reflection on why you chose Moore over Mealy for the traffic light.

You will submit: One PDF: FSM state table + K-maps + circuit diagram for the traffic light; mod-N counter design with wrap logic; short reflection paragraph.

Binary, Octal and Hexadecimal

Positional bases, weight expansion and fast base-16/base-8 conversions.

What you will learn
  • Expand a binary number using positional weights
  • Convert between binary, octal and hexadecimal in both directions
  • Convert decimal to binary by repeated division

Every processor stores numbers as bits. Hex is just a compact way humans read four bits at a time — 0x3F is the byte 00111111.

Positional notation: a digit's value is digit x base^position, counting positions from 0 at the right. The decimal 345 = 3x100 + 4x10 + 5x1.

Binary (base 2): only 0 and 1. 1011_2 = 1x8 + 0x4 + 1x2 + 1x1 = 11_10.

Octal (base 8) and Hex (base 16): one octal digit = three binary bits; one hex digit = four bits. Convert by regrouping bits from the LSB, padding with leading zeros — never by going through decimal twice.

Decimal to any base: repeated division by the base; read remainders bottom-to-top (LSB first).

Hex grouping is four bits, not three

Octal groups of 3 and hex groups of 4 both start from the binary LSB. Pad the leftmost group with zeros.

Leading-zero padding mistakes

When converting 11011 to octal you must pad: 010 111 — not 11 011.

Convert 255 decimal to binary and hex
Express 255_10 in binary and in hexadecimal.
  1. 1

    Divide by 2 repeatedly

    255/2=127 r1, 127/2=63 r1, 63/2=31 r1, 31/2=15 r1, 15/2=7 r1, 7/2=3 r1, 3/2=1 r1, 1/2=0 r1.

  2. 2

    Read remainders LSB-first

    11111111.

  3. 3

    Group into nibbles for hex

    1111 1111 = FF in hex.

Solution

255_10 = 11111111_2 = FF_16

Convert 174 decimal

Find 174_10 in binary and hexadecimal.

Hint: Divide by 2 seven times; the result fits in one byte.

Common mistakes
  • Reading remainders top-to-bottom

    The first remainder is the least significant bit. Reading MSB-first gives a bit-reversed number.

  • Skipping the leading-zero pad

    1001 converted as octal without padding becomes 11 instead of 012.

Exam / interview angle

Convert A3_16 to decimal.

A3 = 10x16 + 3 = 163.

Source: Common university mid-term conversion question.

Lesson summary
  • Positional weight expansion is the foundation of every base conversion.
  • Binary-to-octal = 3-bit groups; binary-to-hex = 4-bit groups from the LSB.
  • Decimal-to-base: repeated division; remainders are LSB-first.

Semester exam prep

University mid-terms, finals and ECE Viva rounds

Units, important questions and previous-year papers — revisit before every exam.

1

Number Systems & Codes

Positional weight expansion; binary↔hex/octal by 3/4-bit grouping; decimal by repeated division. BCD = one nibble per digit. Gray changes one bit per step. ASCII: A=65, a=97, 0=48. Parity detects single-bit errors.

3 important questions
  1. Convert 174 to binary and hex.
  2. Why is Gray code used in rotary encoders?
  3. What does BCD(23) look like vs binary 23?
2

Boolean Algebra & K-maps

De Morgan: flip operators, invert literals. Absorption A+AB=A. SOP/POS. K-map Gray ordering; groups powers of two; wrap-around allowed; don’t-cares free.

3 important questions
  1. Simplify (A+B)·(A+B̄).
  2. Implement A+B with NAND only.
  3. How many groups cover Σm(0,1,2,5,7) of three variables?
3

Combinational Circuits

Half vs full adder equations; ripple-carry cascade. Subtractor: A+B̄+1. MUX: Y=S̄D0+SD1. Decoder lines = minterms + OR for SOP. Priority encoder for interrupts.

3 important questions
  1. Full adder Cout equation?
  2. How many select lines on a 32:1 MUX?
  3. Subtract with an adder — what changes on B and Cin?
4

Sequential Circuits & FSMs

Latch = level-sensitive; flip-flop = edge-triggered. D: Q+=D; JK: Q+=JQ̄+K̄Q; T: Q+=T⊕Q. Excitation tables drive counter/FSM design. Ripple vs synchronous counters. Moore = f(state); Mealy = f(state, inputs).

3 important questions
  1. JK flip-flop converted to D: what are J and K?
  2. Mod-6 counter terminal count?
  3. Moore vs Mealy — when is each preferred?

What you will be able to do

  • Convert between binary, octal, hexadecimal and decimal accurately
  • Encode and decode BCD, Gray and ASCII, and compute parity bits
  • Apply De Morgan's laws and absorption to simplify Boolean expressions
  • Minimize functions on 2-, 3- and 4-variable K-maps
  • Design and cascade adders, and build two’s-complement subtractors
  • Use MUX, decoders and priority encoders as standard building blocks
  • Work with D, JK and T flip-flops via characteristic and excitation tables
  • Design mod-N counters and sketch Moore/Mealy FSMs with state tables

About this course

University digital electronics courses often move too fast through number systems before landing students in K-maps and flip-flops they do not fully own. This course slows the foundation down to eight focused lessons, then accelerates once the basics are solid.

Codes and conversions come first because every later topic assumes them. Boolean algebra and K-maps are taught as a single minimization workflow — not as isolated tricks. Combinational circuits are presented as a fixed vocabulary (adders, MUX, decoders) that you recombine rather than re-derive. Sequential circuits close the course with the design recipe that answers "design a mod-6 counter" or "draw a traffic-light FSM" — the two questions that appear most often in exams and interviews.

Each lesson ends with a quiz graded on the server, and the course finishes with a 15-question timed exam from a 30-question bank. Passing earns a verifiable TieEdu certificate.

Who this course is for

  • ECE and EEE students in their first or second year
  • CS students wanting hardware fundamentals for interviews
  • Anyone preparing for university digital electronics mid-terms and finals
  • Aspirants facing Viva questions on gates, flip-flops and FSMs

What you should know before you start

Anything here is assumed, not taught. If it is not on the list, the course explains it.

  • Basic high-school algebra
  • Comfort with binary numbers (helpful but re-taught in Module 1)

Questions before you start

How much time does Digital Electronics take?

The course is 4 modules and 8 lessons — about 1 hr 22 min of material in total. Lessons unlock in order, so the time it takes you depends on how much of each lesson you actually read.

Do I get a certificate?

Yes. Finish every lesson and a certificate is issued automatically. Its serial is public: anyone can check it at https://tieedu.com/verify without an account, and a revoked certificate stops verifying.

Is it really free?

Yes — this course is free, and there is nothing to buy to start it.

How is my progress tracked?

On the server, lesson by lesson. Time is credited at real-time speed and capped, so a lesson cannot be marked complete faster than it can be read. That is why the percentage survives a new device.

How do ratings work?

Only learners who finished the course can rate it, and the average shown is calculated from those ratings alone. A course nobody has finished yet carries no rating rather than a default one.

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