you will need 2 two input AND gates to do this. connect the output of the first to one input of the second. you now have a three input AND gate.
just remember when calculating timing that 2 inputs of the 3 have twice the gate delay of the remaining input, thus the output will have skew and possibly glitches. if timing is critical or glitching can't be tolerated it may be best to use an actual three input AND instead of kludging one.
Use 2 AND gates to input the 4 i/p signals and then use the outputs of these 2 AND gates in the 3rd AND gate to get the desired output.
with a circuit called a parity tree
NOR gate = not(A or B) = A nor BAND gate = A and BAND gate = not(not A or not B)AND gate = not(not(A or A) or not(B or B))AND gate = (A nor A) nor (B nor B)Therefore using 2 input NORs to make a 2 input AND you need three NORs. If you wanted something different (e.g. a 5 input AND) the above proof can be modified appropriately to get your answer.
Use 4 NOR gates. For the 1st NOR gate, inputs should be x' and y For the 2nd NOR gate,inputs should be y' and x The outputs of NOR 1 and NOR 2 are taken as inputs of NOR gate 3 The output of NOR 3 is the complemented form of the output required, so, just complement the output of NOR gate 3 with another NOR gate and Viola!, you have your HALF ADDER OUTPUT PS:I have used a double rail logic, where both x:x' and y:y' are available
It's a "quad, 2 input nor gate". To understand the significance of a "nor" gate, you need to understand a little about digital logic. An "or" gate takes 2 or more digital inputs and if either is "on", the output will be on. (asserted high). A "nor" gate inverts the output of the "or" gate, meaning that when either of the outputs are "on", the output will be "off" (asserted low). The two input part of the description just indicates that it only accepts two inputs. So, simply stated: If either (or both) input(s) of a quad, 2 input nor gate is (are) asserted high, the output will be low. If both inputs are off (low), the output will be high.
All inputs hae to be low i.e 0.
You can't make XOR out of NOT alone. Mathematically, NOT takes only a single argument, and its gate, an inverter, takes a single input. There's no way to combine two inputs giving a single output with one input gates. You need some two input gates to do the job. They can be AND, OR, NAND, NOR, or some combination, but you need something. That said, they don't have to be IC gates; you can combine two inputs with diodes to make an OR gate, so you could make XOR with only inverters and diodes, i.e. no other gate symbols on your schematic, but it would mot be making XOR out of NOT.
This is made by joining the inputs of a NOR gate. As a NOR gate is equivalent to an OR gate leading to NOT gate, this automatically sees to the "OR" part of the NOR gate, eliminating it from consideration and leaving only the NOT part. Truth Table Input A Output Q 0 1 1 0
A 2 input NAND gate requires 4 NOR gates.A 3 input NAND gate requires 5 NOR gates.A 4 input NAND gate requires 6 NOR gates.etc.
an 2 input AND gate can be realize using 3 NOR gates.Let ,A and B are the input and x be the output.x=A.B= NOR(NOR(A) NOR(B))
NOR gate = not(A or B) = A nor BAND gate = A and BAND gate = not(not A or not B)AND gate = not(not(A or A) or not(B or B))AND gate = (A nor A) nor (B nor B)Therefore using 2 input NORs to make a 2 input AND you need three NORs. If you wanted something different (e.g. a 5 input AND) the above proof can be modified appropriately to get your answer.
Use 4 NOR gates. For the 1st NOR gate, inputs should be x' and y For the 2nd NOR gate,inputs should be y' and x The outputs of NOR 1 and NOR 2 are taken as inputs of NOR gate 3 The output of NOR 3 is the complemented form of the output required, so, just complement the output of NOR gate 3 with another NOR gate and Viola!, you have your HALF ADDER OUTPUT PS:I have used a double rail logic, where both x:x' and y:y' are available
By using 5 NOR gates, we can implements half-subtractor. The inputs for 1st NOR gate are A and B, for 2nd NOR gate inputs are the output of 1st NOR gate and A input, for 3rd NOR gate inputs are the output of 1st NOR gate and B input, for 4th NOR gate the inputs are gates 2 and 3, and for last gate input is the output of the 4th gate.
It's a "quad, 2 input nor gate". To understand the significance of a "nor" gate, you need to understand a little about digital logic. An "or" gate takes 2 or more digital inputs and if either is "on", the output will be on. (asserted high). A "nor" gate inverts the output of the "or" gate, meaning that when either of the outputs are "on", the output will be "off" (asserted low). The two input part of the description just indicates that it only accepts two inputs. So, simply stated: If either (or both) input(s) of a quad, 2 input nor gate is (are) asserted high, the output will be low. If both inputs are off (low), the output will be high.
It's a "quad, 2 input nor gate". To understand the significance of a "nor" gate, you need to understand a little about digital logic. An "or" gate takes 2 or more digital inputs and if either is "on", the output will be on. (asserted high). A "nor" gate inverts the output of the "or" gate, meaning that when either of the outputs are "on", the output will be "off" (asserted low). The two input part of the description just indicates that it only accepts two inputs. So, simply stated: If either (or both) input(s) of a quad, 2 input nor gate is (are) asserted high, the output will be low. If both inputs are off (low), the output will be high.
Well, if you mean to make the truths of an AND gate similiar to the ones of OR gates, then you can do the following: 1. Put an inverter at every input of the AND gate. This will make it act like an NOR gate 2. Put an inverter at the output of the gate. This will invert the truths and turn the NOR gate into an OR gate
A nor gate provides an output of 0 when any input is 1.Nor gate provides the opposite of or gate. An or gate provides a 1 or true output when any of the inputs is 1 or true. Therefore the opposite output would be provided by a nor gate.
IC 7402 is different from the other type of IC like 7404,7408,7432 and 7400 when it is connected to have an output desired... if you noticed all i mentioned ic is connected from left to the right. input pin 1 and 2, output pin 3 input pin 4 and 5, output pin 6 input pin 13 and 12, output pin 11 input pin 10 and 9, output pin 8 while in nor gate, to have the desired output it must be connected from right to left... input pin 2 and 3, output pin 1 input pin 5 and 6, output pin 4 input pin 8 and 9, output pin 10 input pin 11 and 12, output pin 13
All inputs hae to be low i.e 0.