A four input and gate is a logic gate with four inputs. The output is true only when all four inputs are true.
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.
this shows you everything you need about them Pin Number Description 1 A Input Gate 1 2 B Input Gate 1 3 Y Output Gate 1 4 A Input Gate 2 5 B Input Gate 2 6 Y Output Gate 2 7 Ground 8 Y Output Gate 3 9 B Input Gate 3 10 A Input Gate 3 11 Y Output Gate 4 12 B Input Gate 4 13 A Input Gate 4 14 Positive Supply
"not" gate
one terminal of the ex-or gate should be connected to a high input(1 or 5V). the other terminal is used as the input terminal,since not gate has only one input.when we give a high to this input, now the inputs of exor gate are 1 & 1.so output is 0.when we give a low to the input , then the inputs of ex-or gate are 0 and 1, output is 1. Hence it works as a controlled inverter.
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.
It depends on the OR gate. Typically, there are two, three or four inputs, but sometimes there are more.
for a two input gate to represent as an n-input gate excatly n-1 two input gates are required. this implies that for a two input OR gate to represent a four input OR gate exactly three two input OR gates are required let F is =a+b+c+d =(((a+b)+c)+d) =((a+b)+(c+d)) in both the above cases + is used three times so three two input OR gates make a four input OR gates. This discussion doesnot hold good for NAND gates an example can illlustrate the reson:- take F=(a.b.c.d)'=a'+b'+c'+d' --------------------------->(1) (this is obtained by a four input NAND gate) let us take this in the manner we did it for an OR gate and we will then verify the result. =((a.b)'(c.d)')' =((a'+b').(c'+d'))' =(a'+b')'+(c'+d')' =ab+cd <------------------------(2) (1)is not equal to (2) so we can say that a NAND gate cannot be replaced in the manner as OR gate is replaced
To produce a 3-input OR gate when only 2-input OR gates are available: Use 3 OR gates Inputs to Gate A are input 1 and input 2 Input to Gate B is input 3 (if 2 inputs are necessary, include input...
For two input AND gate it is 7408for three input AND gate it is 7411
You would connect the output of the first AND gate to one input of the second AND gate. You are left with 2 inputs on the first AND gate and 1 input on the second AND gate. The final output is from the second AND gate.
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.
this shows you everything you need about them Pin Number Description 1 A Input Gate 1 2 B Input Gate 1 3 Y Output Gate 1 4 A Input Gate 2 5 B Input Gate 2 6 Y Output Gate 2 7 Ground 8 Y Output Gate 3 9 B Input Gate 3 10 A Input Gate 3 11 Y Output Gate 4 12 B Input Gate 4 13 A Input Gate 4 14 Positive Supply
output is feedback in input
A not gate is a logical gate which inverts a digital signal. If the input to a not gate is 1, then the output will be 0. If the input is 0, then the output will be 1.
NOT Gate
INVERTER gate
inverter gate