Consider an eight-bit ripple-carry adder for computing the sum of A and B, where A and B are integers represented in 2’s complement form. If the decimal value of A is one, the decimal value of B that leads to the longest latency for the sum to stabilize is _________.
If we do 2's complement of 1 = 0000 0001, we get -1 = "1111 1111"
So, if B = -1, every carry bit is 1.
A half adder is implemented with XOR and AND gates. A full adder is implemented with two half adders and one OR gate. The propagation delay of an XOR gate is twice that of an AND/OR gate. The propagation delay of an AND/OR gate is 1.2 microseconds. A 4-bit ripple-carry binary adder is implemented by using four full adders. The total propagation time of this 4-bit binary adder in microseconds is ____________.
Here, each Full Adder is taking 4.8 microseconds. Given adder is a 4 Bit Ripple Carry Adder. So it takes 4*4.8 = 19.2 microseconds.
Consider the ALU shown below.
If the operands are in 2's complement representation, which of the following operations can be performed by suitably setting the control lines K and C0 only (+ and - denote addition and subtraction respectively)?
A + B, and A – B, but not A + 1
A + B, and A + 1, but not A – B
A + B, but not A – B or A + 1
A + B, and A – B, and A + 1
1) A+B when K=0 and C0 = 0. It is binary adder which performs addition of two binary numbers.
2) A - B = A+ B' + 1 when K=1 and C0 = 1 ;
Here XOR gates produce B' if K=1. Since 1⊕b= b'.
"1" in (A+B+1) is coming from C0.
Note: 2's complement of B is (B'+1). 3) A+1 when B=0, K=0, C0= 1.
The number of full and half-adders required to add 16-bit numbers is
8 half-adders, 8 full-adders
1 half-adder, 15 full-adders
16 half-adders, 0 full-adders
4 half-adders, 12 full-adders
But for rest of bits we need full address since carry from previous addition has to be included into the addition operation.
So, in total 1 half adder and 15 full adders are required.
An N-bit carry look ahead adder, where N is a multiple of 4, employs ICs 74181 (4 bit ALU) and 74182 (4 bit carry look ahead generator).
The minimum addition time using the best architecture for this adder is
proportional to N
proportional to log N
None of the above
Consider the circuit given below which has a four bit binary number b3b2b1b0 as input and a five bit binary number d4d3d2d1d0 as output. The circuit implements:
Binary of Hex conversion
Binary to BCD conversion
Binary to grey code conversion
Binary to radix-12 conversion
Whenever, b2 = b3 = 1, then only 0100, i.e., 4 is added to the given binary number. Lets write all possibilities for b.
Note that the last 4 combinations leads to b3 and b2 as 1. So, in these combinations only 0010 will be added.
1100 is 12
1101 is 13
1110 is 14
1111 is 15
in binary unsigned number system.
1100 + 0100 = 10000
1101 + 0100 = 10001, and so on.
This is conversion to radix 12.
more than two inputs
None of the above
● The first two inputs are A and B and the third input is an input carry as C-IN. When a full-adder logic is designed, you string eight of them together to create a byte-wide adder and cascade the carry bit from one adder to the next.
● Note: Here, answer they are given ambiguous because it takes minimum 3 input line
8 half adders, 8 full adders
1 half adders, 5 full adders
16 half adders, 0 full adders
4 half adders, 12 full adders
x NAND x
x NOR x
x NAND 1
X NOR 1
2. The full adder adds 3 one bit numbers, where two can be referred to as operands and one can be referred to as bit carried in. And produces 2-bit output, and these can be referred to as output carry and sum.
Half adder is used to add two numbers of the least significant bits, so one half adder is required. In order to add remaining 16 bits of two numbers , we require 16 full adders
One Half- Adder adds least significant bits of the two numbers. Three Full Adders
There is no carry in to the least significant bits. So one Half- Adder adds least significant bits of the two numbers.
Three Full Adders are used to add remaining bits along with carries.
(a)x = 1, y = 0 and C i (carry input) = 0
(b)x = 0, y = 1 and C i = 1 Compute the values of S(sum) and C o (carry output) for the above input values.
S = 1, C o = 0 and S = 0, C o = 1
S = 1, C o = 0 and S = 1, C o = 1
S = 1, C o = 1 and S = 1, C o = 0
S0 = 1, C o = 1 and S = 1, C o = 0
For the given x and y values, the correct option is A