To realize basic logic gates using universal gates

Aim

To realize basic logic gates using universal gates.

Apparatus Required:

Digital trainer kit, IC 7400, IC 7402, Connecting wires.

Theory

A universal gate is a gate which can implement any Boolean function without need to use any other gate type. The NAND and NOR gates are universal gates. In practice, this is advantageous since NAND and NOR gates are economical and easier to fabricate.

NAND gate

The NAND gate produces a LOW output when all of the inputs are HIGH. The abbreviation for this gate is NAND & the operation is same as AND followed with NOT. When any of theinputs are LOW, the output is HIGH. The standard symbol for an NAND gate is shown in figure below along with the associated TruthTable.

NOR gate

The NOR gate produces a HIGH output when all of the inputs are LOW. The abbreviation for this gate is NOR & the operation is same as OR followed with NOT. When any of the inputs are HIGH, the output is LOW. The standard symbol for an NOR gate is shown in figure below along with the associated Truth Table.

NAND Gate as a Universal Gate

To prove that any Boolean function can be implemented using only NAND gates, we will show that the AND, OR, and NOT operations can be performed using only these gates.

Implementing an Inverter Using only NAND Gate

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)’

Implementing AND Using only NAND Gates

A NAND produces complement of AND gate. So, if the output of a NAND gate is inverted, overall output will be that of an AND gate.

Y = ((A.B)’)’

=> Y =(A.B)

Implementing OR Using only NAND Gates

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

NOR Gate as a Universal Gate

To prove that any Boolean function can be implemented using only NOR gates, we will show that the AND, OR, and NOT operations can be performed using only these gates.

Implementing an Inverter Using only NOR 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)’

Implementing OR Using only NOR Gates

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)

Implementing AND Using only NOR Gates

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.

Procedure:-

1. Place the breadboard gently on the observationtable.
2. Fix the IC which is under observation between the half shadow lineof breadboard, so there is no shortage ofvoltage.
3. Connect the wire to the main voltage source (Vcc) whose other end is connected to last pin of the IC (14 place from thenotch).
4. Connect the ground of IC (7th place from the notch) to the ground terminal provided on the digital labkit.
5. Give the input at any one of the gate of the ICs by using connecting wires. (Inaccordance to ICprovided).
6. Connect output pins to the led on digital labkit.
7. Switch on the powersupply.
8. If led glows then output is true, if it doesn’t glow output is false, which is numerically denoted as 1 and 0respectively.
9. Verify the truthtable.

Precautions:

1. All ICs should be checked before starting theexperiment.
2. All the connection should betight
3. Always connect ground first and then connectVcc
4. Suitable type wire should be used for different types ofcircuit.
5. The kit should be off before changing theconnections.
6. After the completion of experiment, switch off the supply of theapparatus

Result:Thus the basic logic gates using universal gates are realized and the truth table is verified.

Questions:

Q1:What are universal gates?

Ans:NAND and NOR gates are called universal gates as any type of logic gates or logic Functions can be implemented by these gates.

Q2:When the output of a NOR gate is high?

Ans:If all the inputs arelow

Q3:State De-Morgan’stheorem.

Ans:(x+y)’ =x’y’

(xy)’ = x’ + y’