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Electronic Devices - Logic Gates

Grade 12ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Digital signals are discrete signals having only two possible levels, represented as 'High' (Logic 11) or 'Low' (Logic 00).

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A Logic Gate is a digital circuit that follows a logical relationship between the input and output voltages.

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The OR Gate: The output YY is 11 if either input AA or input BB (or both) are 11. It follows the Boolean expression Y=A+BY = A + B.

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The AND Gate: The output YY is 11 only if both inputs AA and BB are 11. It follows the Boolean expression Y=A⋅BY = A \cdot B.

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The NOT Gate: It is an inverter that produces an output YY which is the complement of the input AA. It follows the Boolean expression Y=AˉY = \bar{A}.

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The NAND Gate: It is a combination of an AND gate followed by a NOT gate. The output is 00 only when all inputs are 11. Expression: Y=A⋅B‾Y = \overline{A \cdot B}.

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The NOR Gate: It is a combination of an OR gate followed by a NOT gate. The output is 11 only when all inputs are 00. Expression: Y=A+B‾Y = \overline{A + B}.

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Universal Gates: NAND and NOR gates are called universal gates because any logic gate (AND, OR, NOT) can be realized using only NAND or only NOR gates.

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De Morgan's First Theorem: The complement of a product is equal to the sum of the complements: A⋅B‾=Aˉ+Bˉ\overline{A \cdot B} = \bar{A} + \bar{B}.

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De Morgan's Second Theorem: The complement of a sum is equal to the product of the complements: A+B‾=Aˉ⋅Bˉ\overline{A + B} = \bar{A} \cdot \bar{B}.

📐Formulae

Y=A+B (OR Gate)Y = A + B \text{ (OR Gate)}

Y=A⋅B (AND Gate)Y = A \cdot B \text{ (AND Gate)}

Y=Aˉ (NOT Gate)Y = \bar{A} \text{ (NOT Gate)}

Y=A⋅B‾ (NAND Gate)Y = \overline{A \cdot B} \text{ (NAND Gate)}

Y=A+B‾ (NOR Gate)Y = \overline{A + B} \text{ (NOR Gate)}

Y=A⊕B=ABˉ+AˉB (XOR Gate)Y = A \oplus B = A\bar{B} + \bar{A}B \text{ (XOR Gate)}

A⋅B‾=Aˉ+Bˉ (De Morgan’s I)\overline{A \cdot B} = \bar{A} + \bar{B} \text{ (De Morgan's I)}

A+B‾=Aˉ⋅Bˉ (De Morgan’s II)\overline{A + B} = \bar{A} \cdot \bar{B} \text{ (De Morgan's II)}

💡Examples

Problem 1:

Construct the truth table for the logic circuit where a NAND gate's output is fed into both inputs of another NAND gate.

Solution:

Let the first NAND gate have inputs AA and BB. Its output is Y1=A⋅B‾Y_1 = \overline{A \cdot B}. This output Y1Y_1 is fed into both inputs of a second NAND gate. The final output YY is Y1⋅Y1‾=Y1ˉ=(A⋅B‾)‾=A⋅B\overline{Y_1 \cdot Y_1} = \bar{Y_1} = \overline{(\overline{A \cdot B})} = A \cdot B.

Explanation:

By connecting the two inputs of a NAND gate together, it functions as a NOT gate. Therefore, a NAND followed by a NOT (acting NAND) results in an AND operation.

Problem 2:

Identify the logic gate that corresponds to the following truth table: Input A=0,B=0⇒Y=1A=0, B=0 \Rightarrow Y=1; A=0,B=1⇒Y=0A=0, B=1 \Rightarrow Y=0; A=1,B=0⇒Y=0A=1, B=0 \Rightarrow Y=0; A=1,B=1⇒Y=0A=1, B=1 \Rightarrow Y=0.

Solution:

The output is 11 only when both inputs are 00. This matches the Boolean expression Y=A+B‾Y = \overline{A + B}, which is the NOR gate.

Explanation:

In an OR gate, the outputs would be 0,1,1,10, 1, 1, 1. Inverting these results (NOT-OR) gives 1,0,0,01, 0, 0, 0, which matches the given table.

Problem 3:

Using De Morgan's theorem, simplify the Boolean expression Y=Aˉ+Bˉ‾Y = \overline{\bar{A} + \bar{B}}.

Solution:

According to De Morgan's second theorem, X+Y‾=Xˉ⋅Yˉ\overline{X + Y} = \bar{X} \cdot \bar{Y}. Let X=AˉX = \bar{A} and Y=BˉY = \bar{B}. Then, Y=Aˉ‾⋅Bˉ‾Y = \overline{\bar{A}} \cdot \overline{\bar{B}}. Since double inversion cancels out, Aˉ‾=A\overline{\bar{A}} = A and Bˉ‾=B\overline{\bar{B}} = B. Thus, Y=A⋅BY = A \cdot B.

Explanation:

This shows that a NOR gate with inverted inputs behaves exactly like an AND gate.