Adder MCQ Quiz - Objective Question with Answer for Adder - Download Free PDF

The correct answer is: 1) Each full adder waits for the carry from the previous full adder.
Explanation:
A serial binary adder processes binary addition one bit at a time, sequentially propagating the carry from one stage to the next. Here’s why Option 1 is correct:

Characteristic of Serial Adders:

Sequential Carry Propagation: Each full adder (FA) must wait for the carry-out from the previous FA before computing its sum and carry.

Example: To compute bit *i*, the adder needs the carry generated from bit i−1.

Operates Bit-by-Bit: Unlike parallel adders, serial adders do not process all bits simultaneously.

The correct answer is Sum and Carry.

Key Points

• A half-adder is a digital circuit used to add two single-bit binary numbers.
• The half-adder circuit has two inputs, typically designated as A and B.
• It produces two outputs:
  • Sum (S): Represents the sum of the two input bits.
  • Carry (C): Represents the carry-out bit, which is the overflow bit from the addition.
• The Sum (S) output is calculated using the XOR (exclusive OR) operation: S = A ⊕ B.
• The Carry (C) output is calculated using the AND operation: C = A ∧ B.
• These operations are fundamental in binary arithmetic and are used in more complex arithmetic circuits like full adders.

Additional Information

• Half-adders are crucial components in digital electronics and are used in the design of arithmetic logic units (ALUs).
• They are also foundational in the implementation of binary addition in computer processors and digital systems.
• While half-adders can add two single-bit numbers, they do not account for carry-in from previous additions, which is why full adders are used for multi-bit binary addition.
• Half-adders are typically implemented using logic gates like XOR and AND gates.

• Carry lookahead adders (CLAs) are indeed a method used in digital circuit design to speed up the process of adding binary numbers. Traditional ripple-carry adders have a delay associated with each stage, as the carry must propagate from one stage to the next, leading to longer overall addition times, especially for large numbers.
• CLAs minimize this delay by precomputing the carry signals for each stage based on the input values. This precomputation is done in parallel, allowing the carries to be generated simultaneously for all stages. As a result, the carry propagation time is greatly reduced, leading to faster addition operations.
• CLAs are often used in high-speed arithmetic circuits where fast addition is crucial, such as in processors and digital signal processing (DSP) applications. They offer improved performance compared to ripple-carry adders, especially for larger bit-width additions.
   

Additional Information 

CD40181:  It is a low-power four-bit parallel arithmetic logic unit (ALU) capable of providing 16 binary arithmetic operations.

CD40182: It is a high-speed look-ahead carry generator capable of anticipating a carry across four binary adders or groups of adders.

CD4527: It is a BCD rate multiplier (DRM) in a 16-lead DIP type package that provides an output pulse rate based upon the BCD input number.

CD4585: It is a 4-bit magnitude comparator designed for use in computer and logic applications that require the comparison of two 4-bit words.

The correct answer is Add the carry to the least significant bit (LSB)

Key Points

When you add binary numbers and there is a carry-out from the MSB, you incorporate this carry by adding it to the LSB to ensure the additional bit is accounted for in the result.

Example:
Adding the binary numbers 1011 (11 in decimal) and 1001 (9 in decimal):

Perform the initial addition:

   1011
+ 1001
 10100 (Binary result indicating carry-out)

Identify and add the carry to the LSB:

   10100 (Initial sum with carry-out)
    +     1 (Carry added to LSB)
   10101 (Corrected result)
Therefore, the sum is 10101 (21 in decimal). This method ensures that extra carry is properly added into the final result.

Based on this understanding, option 2 (Add the carry to the least significant bit (LSB)) is correctly chosen for this specific context and procedure.

The correct answer is Option 3.

Key Points

• In a full adder, the sum (S) is given by the boolean expression S = A ⊕ B ⊕ C_in, where ⊕ denotes the XOR operation.
• The carry out (C_out) is given by the boolean expression C_out = AB + BC_in + AC_in .
• Option 3 correctly lists these expressions, making it the correct answer.
• 

Formula:

The total time taken by the for ‘n’ bit ripple carry adder is

T_d = (n – 1) t_c + Maximum(t_c, t_s)

t_c = delay for carry through a single flip flop.

t_{s =} delay for sum

Data:

Each full adder takes 34 ns for computing

From this, Maximum(t_c, t_s) = t_s = 34 ns

T_d = 90 ns.

n = 8

Calculation:

T_d = (n – 1) t_c + Maximum(t_c, t_s)

90 = (8 – 1)t_c + 34

7tc = 56 ns.

∴tc = 8 ns.

Time taken by each full adder to find carry is 8 ns.

Hence, the correct answer is option 4.

Overflow is a condition that indicates that the result is too large or too small to fit in the original data type.

Here given 2’s complement representation means signed bit representation.

Two key points that must remember in a signed operation:

1. MSB is used to indicate the sign of the number.
2. For negative numbers, 2’s complement representation is used.

2’s complement overflow rules:

• If the sum of two positive numbers gives a negative output, then the sum is overflowed.
• If the sum of two negative numbers gives a positive output, then the sum is overflowed.
• In other cases, the sum is not overflowed.

Overflow for the signed numbers occurs when the carry-in into the MSB (most significant bit) is not equal to carry out. Now according to the rules,

If both A₇ and B₇ are 1 then S₇ must be 0. For this C₇ will be 1 and C₆ will be 0. Hence, it will give

C₇.C₆’

If both A₇ and B₇ are 0 then S₇ must be 1. For this C₇ will be 0 and C₆ will be 1. Hence, it will give

C₇’.C₆

When we combine these two, we get C₇.C₆’ + C₇’.C₆ (means C₇ XOR C₆) which is a condition to detect overflow. So, Option C is the answer.

Overflow is said to have occurred if:

Full adder:

• A one-bit full adder adds three one bit numbers usually these three operands are A, B and C_in.
• C_in is a bit carried in from the previous stage.

Example: Full adder CKT

Hence option 3 is the correct answer.

Note:

Formula:

The total time taken by the for ‘n’ bit ripple carry adder is

T_d = (n – 1) t_c + Maximum(t_c, t_s)

t_c = delay for carry through a single flip flop.

t_{s =} delay for sum

Data:

Each full adder takes 34 ns for computing

From this, Maximum(t_c, t_s) = t_s = 34 ns

T_d = 90 ns.

n = 8

Calculation:

T_d = (n – 1) t_c + Maximum(t_c, t_s)

90 = (8 – 1)t_c + 34

7t_n = 56 ns.

∴ t_n = 8 ns.

Time taken by each full adder to find carry is 8 ns.

Important Point:

If Maximum(t_c, t_s) = t_c then answer is 11.25 which is not in option

The correct answer is option 3.

1 Half Added and 16 full adders are needed

Explanation

Take 2 numbers for 17 bits, for the first 2 bits of each number we need one-half adder and for rest(16 bit of each number) we will have one carry of previous sum and two bits of each number so we need one full adder for each rest of bits, that is, 16 full adder.

Concept:

The truth table of a full-adder circuit is as shown:

So, the sum bit is simplified as

\(= \bar A\bar B{C_{in}} + \bar AB\overline {{C_{in}}} + A\bar B\overline {{C_{in}}} + AB{C_{in}}\)

⇒ (A ⊕ B) ⊕ C_in

The correct answer is: 1) Each full adder waits for the carry from the previous full adder.
Explanation:
A serial binary adder processes binary addition one bit at a time, sequentially propagating the carry from one stage to the next. Here’s why Option 1 is correct:

Characteristic of Serial Adders:

Sequential Carry Propagation: Each full adder (FA) must wait for the carry-out from the previous FA before computing its sum and carry.

Example: To compute bit *i*, the adder needs the carry generated from bit i−1.

Operates Bit-by-Bit: Unlike parallel adders, serial adders do not process all bits simultaneously.

Concept:

Longest latency for the sum to stabilize is the maximum time or delay a ripple carry adder take to add A and B.

Explanation:

Given A = 1 in decimal form,

In 2’s complement form, binary representation of A in binary form = 0000 0001.

In case of ripple carry adder, for sum we need 2 XOR gates.

For the longest latency, we must have C_in = 1 at every stage of the ripple carry adder.

Here, we have to find the value of B such that we get C_in= 1 at every stage and C_out = 1 at the last stage.

Maximum delay is when we take B = -1 in decimal (1111 1111 – in binary)

half Adder

Full Addere using 2 half adder and OR gate

Hence processing delay is 4.8, given all the bits are added by full adder, total processing delay will be 4 × 4.8 = 19.2

Also first bit could be added using half adder in that case processing delay will be 16.8

Concept:

Half adder circuit have two inputs and two outputs (sum and carry).

Sum (S) = A⊕B, Carry = A.B

Full Adder:

A digital circuit that performs addition is called a full adder. Hardware implements full adders using logic gates. Three one-bit binary values, two operands, and a carry bit are added using a complete adder. Two numbers are output by the adder: a sum and a carry bit. When compared to a half adder, which adds two binary digits, the term is used.

Explanation:

The adder known as a "full adder" adds three inputs and generates two outputs. A and B make up the first two inputs, and C-IN is the third input. The normal output is denoted as S, which represents SUM, while the output carry is designated as C-OUT.

The truth table of a full adder logic is:

The Sum output bit of a full adder is given by:

S = A ⊕ B ⊕ C

The carry output bit of a full adder is given by:

X1 = AB + BC + AC