It depends on the OR gate. Typically, there are two, three or four inputs, but sometimes there are more.
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
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
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.
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
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.
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
-- Take two 2-input AND gates.-- Attach 'y' to one input of one gate, and jumper that point toone input of the other gate.-- Attach 'x' to the free input of one gate, and attach 'z' to thefree input of the other gate.-- Attach the outputs of both gates to the inputs of a single OR gate.-- The output of the OR gate is the Boolean function ( XY + YZ ).=======================================Better implementation (faster, cheaper, easier to build, less hardware to fail):-- Attach 'x' and 'z' to the inputs of a single OR gate.-- Attach 'y' and the output of the OR gate to the two inputs of a single AND gate.-- The output of the AND gate is the same Boolean function.
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.
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