Logic Gates Questions and Answers

Answer: A

The circuit acts as an AND gate.

The diodes' cathodes face the inputs and their anodes meet at the output, which is pulled up through R towards the positive supply. If either input is low, its diode conducts and holds the output low. Only when both inputs are high are both diodes reverse biased (or off), so the output rises to the high level.

The output is high only when all inputs are high, the AND function. An OR gate would have the diodes reversed (anodes at the inputs) with the resistor to ground. NAND and NOR need an inverting stage, which a diode network alone cannot provide.

Answer: B

The gate is the NAND gate, option (B).

The table gives output 0 only when both inputs are 1, and output 1 for every other input pair. That is the NOT of the AND function: \(Q = \overline{A \cdot B}\). The NAND symbol is the AND shape with a small circle (the inversion bubble) on the output, which is option (B).

An AND gate (A) would give 1 only for input 1, 1, the reverse of the table. An OR gate (C) would give 0 only for input 0, 0, and its output is 1 at 1, 1. A NOR gate (D) would give 1 only for input 0, 0. None of these matches the table, whose only 0 is at 1, 1.

Answer: C

NAND and NOR are considered standard logic gates.

NAND and NOR are among the basic gates available in every logic family, such as the 7400 (NAND) and 7402 (NOR) in TTL and the 4011 and 4001 in CMOS. They are also called universal gates, because any Boolean function can be built from NAND gates alone or from NOR gates alone, and in CMOS they are the simplest gates to build.

They are not special-purpose or complex gates; complex gates (such as AND-OR-INVERT) combine several functions in one stage. They are certainly not outdated, since they are the most widely used gates in digital design.

Answer: D

The circuit is a NAND gate.

Follow the signals through. The NOT gates give \(\bar{A}\) and \(\bar{B}\). The NOR gate then gives, by De Morgan's theorem,

\[ \overline{\bar{A} + \bar{B}} = A\cdot B \]

The final NAND gate takes \(AB\) and \(B\):

\[ Y = \overline{(AB)\cdot B} = \overline{AB} \]

since \(AB\cdot B = AB\). So \(Y\) is A NAND B.

The output depends on both inputs, so it is not a NOT gate, and it is the complement of AND, so it is neither AND nor OR: for \(A = B = 0\) it gives 1, which an AND or OR gate would not.

Answer: B

The output is 0.

An AND gate outputs 1 only when all its inputs are 1. With A = 1 and B = 0, one input is 0, so the output is 0 (\(1 \cdot 0 = 0\)).

1 would be the output of an OR gate, '10' is not a single-bit output, and a logic gate's output is always defined for valid inputs.

The AND gate behaves like multiplication of bits, so its output is 1 only for the input pair 1, 1.

Answer: C

The gate is the NAND gate.

NAND is a universal gate: NOT, AND and OR can each be built from NAND gates alone. NOT A is A NAND A, AND is a NAND followed by a NAND inverter, and OR follows from De Morgan's law as (NOT A) NAND (NOT B). So any Boolean function can be made from NAND gates. NOR is the other universal gate.

AND and OR cannot produce inversion, so they cannot make NOT on their own. XOR alone cannot build every function either, so it is not universal.

Answer: C

The universal gate is NAND.

A gate is called universal when every other logic function can be built from it alone. Tying the inputs of a NAND gate together gives NOT; a NAND followed by a NOT gives AND; and NOT-ing the inputs of a NAND gives OR (by De Morgan's theorem). From NOT, AND and OR any Boolean function follows. NOR is the other universal gate.

AND, OR and XOR cannot do this on their own: none of them can produce inversion (NOT) from its inputs, so they are not universal.

Answer: D

The gate is XOR (exclusive OR).

An XOR gate outputs 1 when its two inputs differ (0 and 1, or 1 and 0) and 0 when they are the same (both 0 or both 1). That is why it is used to compare bits and to form the sum bit in an adder.

OR gives 1 whenever at least one input is 1, including when both are 1. AND gives 1 only when both inputs are 1, which is when they are the same. NOT has a single input and simply inverts it.

Answer: C

The universal gate is NAND.

A universal gate can be used alone to build any Boolean function. With NAND, tying both inputs together gives NOT, a NAND followed by NOT gives AND, and NOT on each input followed by NAND gives OR (by De Morgan's law). Since AND, OR and NOT are complete, so is NAND. NOR is the other universal gate.

AND and OR cannot produce inversion on their own, and NOT has only one input, so none of them is universal.