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
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
You can use a second gate on the same IC to invert the output (most OR-gate ICs I know of have 4 gates per chip). Do this by tying the output of the first gate to both inputs of the spare gate.
No. OR is not functionally complete, so you can not use it to derive any other logical expression. The reason for this is because you can only construct the following expressions out of only OR gates: A OR B A OR A Because of the Idempotency theorem, A OR A simply reduces to A, so we are left with A OR B, which we can not use to derive any other logical circuits. At the very least, we would also need a NOT gate. This is why NOR and NAND are functionally complete: you can derive a NOT gate by using A NAND A or A NOR A.
No. OR is not functionally complete, so you can not use it to derive any other logical expression. The reason for this is because you can only construct the following expressions out of only OR gates: A OR B A OR A Because of the Idempotency theorem, A OR A simply reduces to A, so we are left with A OR B, which we can not use to derive any other logical circuits. At the very least, we would also need a NOT gate. This is why NOR and NAND are functionally complete: you can derive a NOT gate by using A NAND A or A NOR A.
These circuits use nMOS for implementation of a whole gate + one pMOS which is connected between positive supply and nMOS.
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.
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.
baby just get artix and valencia then use final strike on the gate keeper and then use holy on the gate keeper then use throw on the gate keeper then after the gate keeper,s turn use rappid attack on the gate keeper then use attack on the gate keeper then use throw on the gate keeper then keep on reapeeting step 2
NAND gates are universal gates and can be used to construct any of the logic gates (AND, OR, NOT, NOR, XOR, XNOR). The easiest way to figure this out is to use basic Boolean Laws. For instance, to create a NOT gate (A'), tie one of the NAND gate's input to logic high: (A+1)' = A'. To create an AND gate (AxB), use two NANDs in series, with the second one configured as an inverter: (AxB) = ((AxB)')'
He went through the attic AND GATE then. The AND GATE could be used as a physical gate.
Please open the gate. I requested the security to open the gate.
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