Logic Gates Questions and Answers

Answer: A

Correct answer: competitive with TTL

Early CMOS was markedly slower than TTL, but successive process improvements have brought its switching speed to roughly the same level, which is why CMOS has taken over while keeping its advantage in power.

Answer: B

Correct answer: 1 ⊕ 1 ⊕ 0 = 1

XOR gives 1 when an odd number of its arguments is 1. Here two of the three are 1, an even number, so the result is 0, not 1. Checking the others: \(1\oplus 0 = 1\) (one 1), \(1\oplus 1\oplus 1 = 1\) (three 1s), and \(1\oplus 1 = 0\) (two 1s).

Answer: D

Correct answer: 16

A truth table needs one row per input combination, and four inputs give \( 2^4 = 16 \).

Answer: B

Correct answer: True

The 7408, like the rest of the 14-pin family, takes V_CC on pin 14 and ground on pin 7, and that single supply serves all four gates in the package.

Answer: C

Correct answer: Whenever a 1 is present at an input

NOR is the complement of OR. OR is 1 as soon as any input is 1, so NOR is 0 in that case, and it is 1 only when every input is 0.

Answer: D

Correct answer: X NOR 1

Work each one out. \( X\,\text{NAND}\,X=(X\cdot X)'=X' \); \( X\,\text{NOR}\,X=(X+X)'=X' \); \( X\,\text{NAND}\,1=(X\cdot 1)'=X' \). But \( X\,\text{NOR}\,1=(X+1)'=1'=0 \), which is a constant, not \(X'\).

Answer: B

Correct answer: high speed

ECL keeps its transistors out of saturation, so they switch very fast; the price is a large standing current and therefore high power dissipation. Speed is what it is chosen for, and power is what it costs.

Answer: A

Correct answer: OR gate

OR gives 1 when at least one input is 1, which is the condition described. A NOR gate would give 0 in that case, and an AND gate needs every input HIGH.

Answer: B

Correct answer: 1 + 1 + 0 + 0 = 1

In Boolean algebra the plus sign is OR, not arithmetic addition, and OR gives 1 if any operand is 1. So the sum of several 1s is still 1, and the result is a single bit rather than 01 or 00.

Answer: D

Correct answer: NOR gate

NOR and NAND are the two universal gates: either alone can build NOT, AND and OR, and therefore any Boolean function. AND, OR and XOR cannot produce a complement by themselves.