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f = ~s.a + s.b , thus is the function for a multiplexer. let a not gate with input x and output y. set a = 1 and b =0 to get a not gate. y = ~x.1 + x.0
For two input AND gate it is 7408for three input AND gate it is 7411
INVERTER gate
Gate current is denote by Ig
Short the inputs together. Logic: A High input, with the inputs shorted together, will be H+H at the input side of the NAND gate, therefore resulting in a low output. A Low input, with both inputs shorted together, is L+L for inputs, resulting in a High output. Also, a NOR gate can be used in exactly the same way.
f = ~s.a + s.b , thus is the function for a multiplexer. let a not gate with input x and output y. set a = 1 and b =0 to get a not gate. y = ~x.1 + x.0
You don't need two 4-to-1 multiplexers. You only need one 4-to-1 multiplexer, and something that functions as a 2-to-1, like a single 2-input OR gate with one input grounded.
many inputs one outputs
Its possible to design a 4-i/p xor gate using only 2-i/p nand gates Although the design turns out to be quite complex and comprises of 21 NAND gates : F = (A'B+AB')(C'D'+CD) + (A'B'+AB)(C'D+CD') Above given equation is the o/p equation for the circuit .
Assume that the two inputs of the XOR gate are I1 and I2, and that the two inputs of the multiplexer are M1 and M2 and the selection line is S: Connect I1 to M1 Connect Not(I1) to M2 Connect I2 to S
Gate diffusion input is defined as the new technique of low power digital combination of circuit design. This allows reduced power consumption and delayed propagation.
Gate diffusion input is defined as the new technique of low power digital combination of circuit design. This allows reduced power consumption and delayed propagation.
Connect the two inputs of the NAND gate together.
Tie unused input high through pullup resistor.
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))
To design an OR using 2:1 mux, we need to tie the "First" input to "Logic 1″ and the "Zeroth" input to the one of the input of the OR Gate. The other input of OR gate would be connected with the select line of the MUX. Now, the output of the MUX would be "1″ when any oth the two inputs would be "1″ otherwise it would be "0″ for all conditions.
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.