0
Want this question answered?
Be notified when an answer is posted
Add your answer:
Earn +20 pts
Binary adder-subtractor combine circuit?
Switch Mode (SM) is a control input to the circuit to switch between addition or subtraction operations. Adder When SM = 0 the circuit is equivalent to Binary Adder. A + B (bit ) XOR 0 = A + B (bit) Subtraction When SM = 1 the circuit is equivalent to Binary subtraction. A + B (bit ) XOR 1 = A + invert(B (bit)) 'B' input becomes inverted in this case.
What is meant by adder circuit?
truth table a b out carry 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 it may be noted that out can be achieved with XOR gate and carry with an AND gate for a two bit adder you need twice the hardware if you can add in decimal with pencil and paper perhaps binary will be easier to understand
How using two-input logic gates of your choice design a circuit to determine if two 3-bit numbers are the same or different?
Three 2-input XOR gates and one 3-input NOR gate will do the work. Connect each output of each XOR gate to one input of the 3-input NOR gate and apply the two 3-bit words to the inputs of the XOR gates. If X (X2X1X0) and Y(Y2Y1Y0) are two 3-bit words, X2 and Y2 will connect to one XOR gate, X1 and Y1 to the next XOR gate and X0 and Y0 to the last XOR gate. You could see the result of the operation on a LED connected to the output of the NOR gate. Other implementations are also possible of course. The solution above is absolutely correct, but includes a 3 input gate. If the task is to use only two input gates, then a small change will be needed. Take the outputs from any two XOR gates into a 2 input OR gate. Then take the output of the OR gate and the output of the third XOR gate into a 2 input NOR gate. The operation remains identical to the first solution but adheres to the brief of using gates with 2 inputs. In the real world, there is probably no reason to impose such a limitation on a design so the first solution would normally be the preferred route to take.
How many logic gates are needed to build the 4 bit carry-lookahead adder?
10 ANd and 4 Or gate..
Why half adder is so called?
A full adder has a sum bit and a carry bit. A half adder just has a sum bit.
What is the difference between 64 bit carry select adder and carry select adder?
Carry select adder is used to select the carry during addition of two numbers. If those numbers are of 64 bits, then we call it as a 64 bit carry select adder.
Operation of an adder capable of adding 2 one bit binary digits?
The 2-bit adder has two input lines and two output lines.The two input lines are the two bits to be added, call them "Bit-A" and "Bit-B".The two output lines are "sum" and "carry".Here are the states of the "sum" and "carry" outputs, for every combination of inputs "Bit-A" and "Bit-B":Bit-A . . . Bit-B . . . Sum . . . Carry0 . . . . . . . . 0 . . . . . . 0 . . . . . . 00 . . . . . . . . 1 . . . . . . 1 . . . . . . 01 . . . . . . . . 0 . . . . . . 1 . . . . . . 01 . . . . . . . . 1 . . . . . . 0 . . . . . . 1The Boolean functions for the two output lines are:SUM = [ (Bit-A) XOR (Bit-B) ]CARRY = [ (Bit-A) AND (Bit-B) ](Actually, I think the "adder" only has one output line, called "sum"; if it also has the "carry" output,then it's called a "full adder". To me, a full adder is a snake plus a dead frog.)
How many xor gates required for 4 bit parity checker?
3*xor- two input
What is VHDL program for half adder in behavioral model?
PROGRAM:Library ieee;use ieee.std_logic_1164.all;entity ha1 isport(a,b:in bit;s,c:out bit);end ha1;architecture ha1 of ha1 isbegins
What are applications of half adder circuit?
Full adder circuit:Full adder reduces circuit complexibility. It can be used to construct a ripple carry counter to add an n-bit number. Thus it is used in the ALU also. It is used in Processor chip like Snapdragon, Exynous or Intel pentium for CPU part . Which consists of ALU (Arithmetic Block unit) . This Block is used to make operations like Add, subtract, Multiply etcA full adder adds binary numbers and accounts for values carried in as well as out. A one-bit full adder adds three one-bit numbers, often written as A, B, and Cin; A and B are the operands, and Cin is a bit carried in from the previous less significant stage.The full adder is usually a component in a cascade of adders, which add 8, 16, 32, etc. bit binary numbers.
Draw the logic circuit for a haly adder using nand gates only?
____ ____ c ----->|xor |------------> s a ->|xor |-+------------>|____| b ->|____| | _____ _____ +--->|nand |------>|nand |--> c c ----->|_____| +-->|_____| _____ | a ----->|nand |--+ b ----->|_____|
Convert the gray code 10101111 to binary?
To convert Gray code to binary code you must be familiar with the logical XOR operator. XOR outputs a 1 bit if either of two input bits is 1, but not both. The truth table for XOR, for all possible inputs p and q, is as follows:p q output0 0 00 1 11 0 11 1 0The algorithm to convert from Gray code to binary code is as follows:Step 1: Fix the most-significant bit, the MSB, which is always the same for both codes. If there are no more bits, we're done, otherwise proceed to step 2.Step 2: XOR the most recently fixed binary bit with the next available Gray bit. Fix the result as the next binary bit.Step 3: If there is at least one more Gray bit available, go to step 2. Otherwise we're done.Therefore, to convert 10101111 from Gray to binary, we proceed as follows:Gray = 10101111Fix MSB = 11 XOR 0 = 11 XOR 1 = 00 XOR 0 = 00 XOR 1 = 11 XOR 1 = 00 XOR 1 = 11 XOR 1 = 1Thus: Binary = 11010101Note that we carry the fixed bit (the bold bit) onto the next line as the l-value (left operand) of XOR. The r-value (right operand) of XOR is always the next available Gray bit after the MSB. Reading the fixed bits from top to bottom reveals the binary code.We can also write this as follows:Gray = 10101111Binary = 1 XOR 0 = 1 XOR 1 = 0 XOR 0 = 0 XOR 1 = 1 XOR 1 = 0 XOR 1 = 1 XOR 1 = 1Reading the fixed (bold) bits left to right reveals the binary code.
