Logic Gates and Boolean Algebra Revision Notes | A Level AQA Physics

1. Why Logic Gates Matter

Every decision a computer makes — from checking a password to controlling a traffic light — comes down to one question: is this signal on or off? Logic gates are the physical components that answer that question, and Boolean algebra is the language we use to describe what they do.

At A Level, you treat each gate as a building block: you need to know what it does, not how it is built internally from transistors. Think of a gate the way you think of a function in maths — you feed in inputs, a rule is applied, and you get an output.

2. The Language of Boolean Algebra

Boolean algebra uses three core operations, each with its own notation. The AQA specification requires you to recognise and use all three precisely.

The NOT operation flips the bit. It is the simplest gate — one input, one output, always the opposite value. In circuit diagrams it appears as a triangle with a small circle (bubble) at the output.

Picture a series circuit with two switches: the lamp only lights when switch A and switch B are both closed. That is exactly the AND rule. The dot notation ($A \cdot B$) mirrors multiplication — and notice that $1 \times 1 = 1$ while anything multiplied by 0 gives 0, which matches the gate's behaviour perfectly.

The plus sign in $A + B$ can look like ordinary addition, but remember: in Boolean algebra $1 + 1 = 1$, not 2. The OR gate asks 'is there any 1 present?' rather than counting how many.

3. Truth Tables for the Three Core Gates

A truth table is the definitive way to specify a gate's behaviour. For $n$ inputs there are $2^n$ rows — so a two-input gate has four rows, covering every possible combination of 0s and 1s.

Here are the truth tables for NOT, AND, and OR. Study the pattern in the output column for each — these patterns are what you need to recall instantly in an exam.

Notice the AND column: only the very last row gives a 1. The OR column: only the very first row gives a 0. These two patterns are the quickest way to check your work under exam conditions.

4. NAND, NOR, and XOR Gates

Beyond the three core gates, AQA expects you to handle NAND, NOR, and XOR as additional building blocks. Each one is defined by its truth table, and each has a Boolean expression derived from the core operations.

A NAND gate is an AND gate followed by a NOT: its output is $\overline{A \cdot B}$. The output is 0 only when both inputs are 1 — the exact opposite of AND. A NOR gate is an OR gate followed by a NOT: its output is $\overline{A + B}$. The output is 1 only when both inputs are 0.

An XOR gate (exclusive OR) outputs 1 when the inputs are different. Its Boolean expression is $A \oplus B$. When both inputs are the same (both 0 or both 1) the output is 0.

A useful memory hook: NAND and NOR are the 'inverse' gates — their outputs are the complements of AND and OR respectively. XOR is the 'difference detector' — it fires when the inputs disagree.

5. Writing Boolean Expressions for Combined Circuits

Real circuits chain gates together. The skill here is reading the circuit from left to right, writing the Boolean expression at the output of each gate in turn, and building up to the final output expression.

Think of it like substituting into a function: each gate's output becomes the input expression for the next gate. Work one gate at a time and you will not lose track.

Worked example: Writing the Boolean Expression for a Two-Gate Circuit

A circuit has inputs A and B. They feed into an AND gate whose output is labelled P. P and a third input C then feed into an OR gate to give the final output Q. Write the Boolean expression for Q.

1. Gate 1 is AND with inputs A and B. — Always start at the inputs and work towards the output.
2. $P = A \cdot B$ — The AND gate gives the product of its two inputs.
3. Gate 2 is OR with inputs P and C. — P is now treated as a single variable feeding the next gate.
4. $Q = P + C = (A \cdot B) + C$ — Replace P with its full expression. Brackets clarify the order of operations.
5. $Q = (A \cdot B) + C$ — Q = 1 whenever both A and B are 1, OR whenever C is 1 (or both conditions hold).

Answer: $Q = (A \cdot B) + C$

Once you have the Boolean expression, you can construct the full truth table by evaluating the expression for all eight possible combinations of A, B, and C (since there are three inputs, $2^3 = 8$ rows are needed).

6. Constructing a Full Truth Table for a Combined Circuit

With the expression $Q = (A \cdot B) + C$ in hand, the truth table is built column by column. First list all input combinations systematically, then evaluate any intermediate expressions (like $A \cdot B$), and finally compute the output.

Worked example: Completing the Truth Table for Q = (A · B) + C

Construct the full truth table for $Q = (A \cdot B) + C$ with inputs A, B, and C.

1. With 3 inputs there are $2^3 = 8$ rows. A: 0 0 0 0 1 1 1 1 B: 0 0 1 1 0 0 1 1 C: 0 1 0 1 0 1 0 1 — Cycle the rightmost input fastest (C alternates every row), then B, then A. This systematic approach guarantees no row is missed.
2. Row 1: 0·0=0, Row 2: 0·0=0, Row 3: 0·1=0, Row 4: 0·1=0, Row 5: 1·0=0, Row 6: 1·0=0, Row 7: 1·1=1, Row 8: 1·1=1 — AND is only 1 when both inputs are 1 — only rows 7 and 8 qualify here.
3. Row 1: 0+0=0, Row 2: 0+1=1, Row 3: 0+0=0, Row 4: 0+1=1, Row 5: 0+0=0, Row 6: 0+1=1, Row 7: 1+0=1, Row 8: 1+1=1 — OR is 1 whenever at least one of its inputs is 1. Apply this to each row using the intermediate column.
4. Q: 0, 1, 0, 1, 0, 1, 1, 1 — Q = 0 only in rows 1, 3, and 5 — where both A·B = 0 and C = 0.

Answer: Q outputs: 0, 1, 0, 1, 0, 1, 1, 1 for the eight input combinations in order.

7. Tracing Logic Levels Through a Circuit

Some exam questions give you a circuit diagram with specific input values and ask you to determine the output. The method is the same as building a Boolean expression, but you substitute numbers at each stage rather than letters.

Earlier we built the expression $Q = (A \cdot B) + C$ from a two-gate circuit. Now consider a slightly more complex circuit where a NOT gate is added: input A passes through a NOT gate to give $\overline{A}$, and then $\overline{A}$ and B feed into an AND gate to give output $P = \overline{A} \cdot B$.

Worked example: Tracing Logic Levels: NOT then AND

A circuit has input A passing through a NOT gate, then the NOT output and input B feed into an AND gate. Given A = 1 and B = 1, find the output P.

1. $P = \overline{A} \cdot B$ — The NOT gate acts on A first; its output then feeds the AND gate alongside B.
2. $\overline{A} = \overline{1} = 0$ — NOT inverts the input: 1 becomes 0.
3. $P = \overline{A} \cdot B = 0 \cdot 1 = 0$ — AND requires both inputs to be 1. Since $\overline{A} = 0$, the output is 0 regardless of B.
4. $P = 0$ — Even though B = 1, the NOT gate on A means the AND gate never sees two 1s simultaneously when A = 1.

Answer: $P = 0$

This pattern — a NOT gate feeding into an AND or OR gate — appears frequently in exam circuits. Always resolve the NOT first, then carry that value forward to the next gate.

8. Exam Technique and Common Mistakes

Logic gate questions are reliable mark-earners if you follow a disciplined method. The most common source of lost marks is not a gap in knowledge — it is a procedural slip under time pressure.

When an exam question asks you to 'describe' the behaviour of a gate, state the rule in terms of inputs and output without using 'because' — for example: 'The output is 1 only when both input A and input B are 1.' When asked to 'explain', link the Boolean expression to the truth table outcome: 'Because $A \cdot B = 1$ only when both A and B equal 1, the AND gate produces a high output only under that condition.'

Logic gates and Boolean algebra together form the foundation of all digital electronics at A Level: every gate is a building block described by a Boolean expression, every circuit is a combination of those blocks, and every combination can be fully specified by a truth table — which is exactly what the AQA specification requires you to demonstrate.