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A/L ICT syllabus
Grade 1226 periodsLogic Gates

Fundamentals of Digital Circuits

Uses logic gates to design basic digital circuits and devices

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    Exam objectives

    • Name and draw the symbols of NOT, AND, OR, XOR, NAND, NOR and XNOR gates
    • Construct truth tables for basic and combinational gates, up to three inputs
    • Explain why NAND/NOR are universal gates and build a given gate using only NAND/NOR

    Important areas to cover

    • Drawing gate symbols and truth tables from a given expression is asked almost every year
    • "Build gate X using only NAND/NOR" is a recurring, higher-mark question

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    Exam objectives

    • State and apply Boolean postulates/laws: commutative, associative, distributive, identity, redundancy, De Morgan's
    • Convert a truth table to SOP/POS form and transform between SOP and POS
    • Simplify Boolean expressions using theorems and using a Karnaugh map

    Important areas to cover

    • Karnaugh map simplification (up to 3-4 variables) is one of the highest-value recurring essay questions in this unit — practise until it is routine
    • De Morgan's law application is a frequent MCQ/short-answer trap question

    Notes

    Why simplify at all

    Two circuits with the same truth table behave identically, so the smaller one always wins. Every gate removed is a gate that costs money, draws power, takes board space and adds propagation delay. Simplification is not tidying up; it is the design work.

    The laws worth knowing by heart

    - Identity: A + 0 = A and A · 1 = A - Null: A + 1 = 1 and A · 0 = 0 - Idempotent: A + A = A and A · A = A - Complement: A + A' = 1 and A · A' = 0 - Double negation: (A')' = A - Absorption: A + AB = A and A(A + B) = A - Distributive: A(B + C) = AB + AC, and also A + BC = (A + B)(A + C) - De Morgan: (A · B)' = A' + B' and (A + B)' = A' · B'

    Two of those catch people out. The second distributive law has no equivalent in ordinary algebra, so it does not feel right and gets skipped. And De Morgan is the one that appears most often, because it is how any expression is converted to NAND-only or NOR-only form — which is asked because real circuits are built from universal gates.

    Sum of products and product of sums

    From a truth table, SOP is built from the rows where the output is 1: one product term per row, each variable negated where it is 0, all added together. POS is built from the rows where the output is 0: one sum term per row, each variable negated where it is 1, all multiplied together. Both describe the same function.

    Reading a Karnaugh map

    Write the rows and columns in Gray code — 00, 01, 11, 10 — never in counting order. The whole method depends on it, because neighbouring cells must differ in exactly one variable: 01 to 11 changes one bit, but 01 to 10 changes two.

    Then group the 1s by these rules:

    • Groups must be rectangular and of size 1, 2, 4, 8 or 16. Never 3, never 6
    • Bigger is better: each doubling of a group removes one more variable from its term
    • Groups may overlap, and usually should
    • The edges wrap around. The left column is adjacent to the right, the top row to the bottom, and the four corners form one group of four
    • Every 1 must be inside at least one group
    • Use as few groups as you can

    For each group, write down the variables that stay the same right across it, negated where they are 0, and drop the variables that change. Add the terms together.

    Don't-care conditions

    An X marks a combination that cannot occur or whose output does not matter — an invalid BCD code, for instance. Treat each X as a 1 wherever that makes a group bigger, and as a 0 wherever it does not. You are never obliged to cover an X.

    The trap in the exam

    Stopping at the first set of groups that happens to cover every 1. The question asks for the simplest expression, not for a correct one, and a correct-but-larger answer loses marks.

    Find the forced groups first. If a cell can be reached by only one possible group, that group has to be in the answer. Put those down, and the choices left over shrink sharply.

    Try it — group the map and read off the answer

    Click a cell to cycle it through 0, 1 and X (don't care). The simplest sum-of-products is worked out as you go, and every group it used is outlined on the map.

    The four corner cells form one group of four, because the left edge is adjacent to the right edge and the top to the bottom. Miss it and you write four separate terms instead of one.

    Rows are AB, columns are CD, both in Gray code — so neighbouring cells differ by exactly one variable.
    AB\CD00011110
    00
    01
    11
    10

    Before simplifying

    F = A'B'C'D' + A'B'CD' + AB'C'D' + AB'CD'

    4 minterms

    Simplified

    F = B'D'

    One group, and it is forced.

    A "forced" group is one covering a cell no other group can reach, so it must be in the answer. Find those first in the exam — the rest of the grouping then has far fewer choices left in it.

  3. Join this class

    Exam objectives

    • Derive a logic expression and truth table from a stated real-world requirement, up to three inputs
    • Draw the resulting digital circuit using logic gates

    Important areas to cover

    • "Design a circuit for scenario X" (e.g. an alarm or voting system) is a classic essay-length question that combines 4.1-4.3

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  4. Join this class

    Exam objectives

    • Derive the truth table and logic expression for a half adder and a full adder
    • Explain how a flip-flop uses a feedback loop to store one bit

    Important areas to cover

    • Half adder / full adder truth table and circuit diagram is a very frequently repeated essay component
    • A flip-flop's role in memory is a common question linking back to Unit 2's registers

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