Verify the NAND and NOR gates as universal logic gates.
Logic trainer kit, NAND gates (IC 7400), NOR gates (IC 7402), wires.
A NOT produces complement of the input. It can have only one input, tie the inputs of a NAND gate together. Now it will work as a NOT gate. Its output is
Y = (A.A)’ Y = (A)’
The OR gate performs a logical addition commonly known as OR function. The output is high when any one of the inputs is high. The output is low level when both the inputs are low.
Y = ((A.B)’)’ Y = (A.B)
From DeMorgan’s theorems: (A.B)’ = A’ + B’ (A’.B’)’ = A’’ + B’’ = A + B So, give the inverted inputs to a NAND gate, obtain OR operation at output.
The output of a two input X-OR gate is shown by: Y = A’B + AB’. This can be achieved with the logic diagram shown in the below.
The NOR gate is a contraction of OR-NOT. The output is high when both inputs are low. The output is low when one or both inputs are high.
Gate No. | Inputs | Output |
1 | A, B | (AB)’ |
2 | A, (AB)’ | (A (AB)’)’ |
3 | (AB)’, B | (B (AB)’)’ |
4 | (A (AB)’)’, (B (AB)’)’ | A’B + AB’ |
Now the output from gate no. 4 is the overall output of the configuration.
Y = ((A (AB)’)’ (B (AB)’)’)’
= (A(AB)’)’’ + (B(AB)’)’’
= (A(AB)’) + (B(AB)’)
= (A(A’ + B)’) + (B(A’ + B’))
= (AA’ + AB’) + (BA’ + BB’)
= ( 0 + AB’ + BA’ + 0 )
= AB’ + BA’
Y = AB’ + A’B
X- NOR gate is actually X-OR gate followed by NOT gate. So give the output of X-OR gate to a NOT gate, overall output is that of an X-NOR gate.
Y = AB+ A’B’
The output is high when any one of the inputs is high. The output is low when both the inputs are low and both the inputs are high.
A NOR gate is an OR gate followed by NOT gate. So connect the output of OR gate to a NOT gate, overall output is that of a NOR gate.
Y = (A + B)’
NOR gate is actually a combination of two logic gates: OR gate followed by NOT gate. So its output is complement of the output of an OR gate. This gate can have minimum two inputs; output is always one. By using only NOR gates, we can realize all logic functions: AND, OR, NOT, X-OR, X-NOR, NAND. So this gate is also called universal gate.
A NOT produces complement of the input. It can have only one input, tie the inputs of a NOR gate together. Now it will work as a NOT gate. Its output is
Y = (A+A)’ Y = (A)’
A NOR produces complement of OR gate. So, if the output of a NOR gate is inverted, overall output will be that of an OR gate.
Y = ((A+B)’)’ Y = (A+B)
From DeMorgan’s theorems: (A+B)’ = A’B’ (A’+B’)’ = A’’B’’ = AB So, give the inverted inputs to a NOR gate, obtain AND operation at output.
The output of a two input X-NOR gate is shown by: Y = AB + A’B’. This can be achieved with the logic diagram shown in the left side.
Gate No. | Inputs | Output |
1 | A, B | (A + B)’ |
2 | A, (A + B)’ | (A + (A+B)’)’ |
3 | (A + B)’, B | (B + (A+B)’)’ |
4 | (A + (A + B)’)’, (B + (A+B)’)’ | AB + A’B’ |
Now the output from gate no. 4is the overall output of the configuration.
Y = ((A + (A+B)’)’ (B +( A+B)’)’)’
= (A+(A+B)’)’’.(B+(A+B)’)’’
= (A+(A+B)’).(B+(A+B)’)
= (A+A’B’).(B+A’B’)
= (A + A’).(A + B’).(B+A’)(B+B’)
= 1.(A+B’).(B+A’).1
= (A+B’).(B+A’)
= A.(B + A’) +B’.(B+A’)
= AB + AA’ +B’B+B’A’
= AB + B’A’
Y = AB+ A’B’
X-OR gate is actually X-NOR gate followed by NOT gate. So give the output of X-NOR gate to a NOT gate, overall output is that of an X-OR gate.
Y = A’B+ AB’
A NAND gate is an AND gate followed by NOT gate. So connect the output of AND gate to a NOT gate, overall output is that of a NAND gate.
Y = (AB)’
NAND & NOR are verified as universal gates successfully.
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