design circuit with inputs A, B ,C and D the output of this circuit is highly only when the majority of inputs are high
then show to implement circuit using all NAND - gates
Answers
Answer:
A majority” is not something typically handled by logic, but as there’s only three, a majority is when two are in the required condition. All three A, B & C fed into NOT gates. Outputs from NOT gates into ANDs for A&B, A&C, B&C. All three AND outputs into an OR (assuming three-input OR, or two go into one, then the output of that to another OR along with the third). Could reduce it all to NAND gates if you’re at all bothered about doing so.
Explanation:
Can you design a logic circuit whose output is high only when a majority of inputs A, B, and C are low?
“A majority” is not something typically handled by logic, but as there’s only three, a majority is when two are in the required condition. All three A, B & C fed into NOT gates. Outputs from NOT gates into ANDs for A&B, A&C, B&C. All three AND outputs into an OR (assuming three-input OR, or two go into one, then the output of that to another OR along with the third). Could reduce it all to NAND gates if you’re at all bothered about doing so.
How do you design a logic circuit that has three inputs, A, B, and C, and whose output will be high only when majority of the inputs are high?
How could I design a logic circuit that counts the number of 1s in 3 inputs (A,B, C) and outputs a two-bit binary number representing that count of 1s?
How can I design a combinational circuit with three inputs, x, y, and z, and three outputs A, B, and C. When the binary input is 0, 1, 2, or 3, the binary output is four greater than the input?
A logic gate have three inputs. The output will show "1" if and only if any two of those three inputs are "1". Is there a possible circuit and equation for that?
Which logic gate has output high if and only if all inputs are low?
As three inputs are low l
Let
A=0,
B=0,
C=0
In the below photo u can see that output is high
First of all do the and operation for A and b
Then for AB and C which gives A.B.C then apply inverter to make it high that's why I used nand gate.
U can also use OR gate but remember to apply inverter(not) at the end
Sure
(~A & ~B & C) | (A & ~B & ~C) | (~A & B & ~C) | (~A & ~B & ~C).
Which is what you already stated in the question. Working out the actual gates is your homework.
Well, design one that is high when a majority are high and invert the output.
And that one is the carry logic for a full adder, so it has already been designed.
Why should a circuit have low output impedance and high input impedance?
This concept of - High input impedance and low output impedance is mostly used in amplifier circuits.
Imagine you have an amplifier circuit that can amplify very small magnitudes of currents and can provide enough amplified output Current (let’s say to drive a Speaker at the output)
Now ,since the input signal is a small magnitude current signal ,it might have been generated by a Low power Voltage source (Can be a voltage produced by a transducer based microphone).
Remember: Here our aim is to utilize the minimum possible input current for amplification.
Let’s say the amplifier input stage is con
How could I design a logic circuit that counts the number of 1s in 3 inputs (A,B, C) and outputs a two-bit binary number representing that count of 1s?
Let's say for inputs [math]A, B, C[/math] outputs are [math]Y_1[/math] and [math]Y_0[/math]
So truth table for [math]Y_1, Y_0[/math] are as follows
So, let's simplify [math]Y_1[/math] using K-map
[math]Y_1=AB+BC+CA[/math]
For [math]Y_0[/math]
[math]Y_0=A'B'C+A'BC'+AB'C'+ABC[/math]
[math]=C'(A\oplus B)+C(\overline A\oplus B)[/math]
[math]=A\oplus B\oplus C[/math]
In conclusion it's just the output of FULL ADDER.
Cheers.
How can I design a combinational circuit with three inputs, x, y, and z, and three outputs A, B, and C. When the binary input is 0, 1, 2, or 3, the binary output is four greater than the input?
This question is not giving any information regarding the outputs of the inputs 4,5, 6 and 7. So, these four inputs can be taken as don't care terms. I will implement this using K map and truth table.
The truth table is shown below. You can see that when the input is 0, we are adding 4 to it and the output will be 4 in binary. Similarly all the outputs can be obtained.
Now, you can draw the K maps for outputs A, B and C. I am not showing the K maps. If you are unable to draw the K maps, you can visit here and check additional question number 1. There are some other questions from Morris Mano's book, which may be helpful for you.
SOLUTION OF COMBINATIONAL CIRCUIT MORRIS MANO(2ND EDITION)
The expressions for the outputs will be as follows:
A=1
B=y
C=z
Thank you
Happy reading:)