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in Electronics by (61.1k points)

Draw circuit diagram for two input diode OR gate and AND gate. Also explain their working and give truth table.

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OR Gate:

An OR gate has two or more inputs with one output. The output Y is 1 when either input A or input B or both are 1s, that is; if any of the input is high, the output is high.

Since logic gates are based on such values for which only two values are possible 0 and 1, this type of Algebra is different. This type of Algebra is known as Boolean Algebra and the equations representing these relationships are called Boolean expressions.

For a two input OR gate if inputs are given as A and B respectively and output is F then Boolean expression is
Y = A + B
Here the + sign between A and B represents OR operation. Y = A + B means Y, A or B.
Where A = 0, 1 and B = 0, 1.
The truth table for this;

A B Y = A +B
0 0 0
1 0 1
0 1 1
1 1 1

It is clear from above table that out of the input values A and B if one or both are 1 then output Fig 1. When both A and B are zero then output F is also zero.

OR gate can also be explained by electrical switches connected in parallel. As shown in figure the switches A and B are connected parallel and are connected with a battery and a bulb Y.

If A and B’s closed state (ON state) is represented by 1 and open state (off state) is represented by 0 and the lighted and unlighted situation of the bulb is also 1 or 0 then for this circuit there are four possibilities.

(i) If A and B both are open then there is no current in Y, hence bulb is not lighted. In mathematical form A = 0, B = 0 then Y = A + B = 0.
(ii) If switch A is closed (A = 1) and B is open (B = 0) then battery will flow the current through switch A and light the bulb Y, hence Y = 1.
∴ A = 1, B = 0 then Y = A + B =1.
(iii) If A is open and B is closed then also bulb will be lighted.∴A = 0, B = 1, Y = A + B = 1.
(iv) If A and B both are closed (A = B = 1) then bulb will be lighted (Y = 1). ∴ A = B = 1, Y = A + B = 1.
All the above four possibilities are according to the truth table. Figure shows the symbol for OR gate.

(i) The value of 0 V or 5 V is provided to the inputs A and B. 0 V is represented by 0 and 5V represented by 1. R is the resistance at output Y and the other end of resistance R is earthed.
(ii) When A = B = 0 meaning there is no signal then no diode will be operated and no potential drop at R or Y = 0. This is equal to the first row of the truth table.
If input A is 5 V (1) or B = 0 V (0) then diode Di will be forward biased and behave as closed switch hence is in operated state. If D1 is an ideal diode then there is no potential drop.

So, 5 V potential drop will be on resistance R with earth or will get Y = 1 position. This will be similar to second row of truth table.
(iii) If A = 0 and B = 5 V (1) then in place of D1, D2 will be in state of operation. Still Y = 1. This is similar to third row.
(iv) If A and B both the diodes are (5 V) then both diodes will be in operation. This will be similar to fourth row of truth table.

AND Gate:

An AND gate has two or more inputs and one output. The output Y of AND gate is 1 only when input A and input B are both 1. The Boolean expression for this gate is represented as; Y = A . B
This represents AND operation. The above Boolean equation means, Y = A and B. The truth table for this:

A B Y = A.B
0 0 0
1 0 0
0 1 0
1 1 1

Hence, in AND gate when input of both is 1 then only output is 1.

The AND gate can be explained by electrical switches; (Figure). When both switches A and B are open (A = B = 0) then there is no current in circuit.

Hence, bulb Y is not lighted. This is equal to the first line of the truth table. If any of the two switches is open then also Y = 0. When both switches are closed (A = B = 1), then Y = 1, bulb is lighted. Figure shows AND sign.

In behaviour AND gate is also obtained by the help of diodes. Figure shows AND gate represented by diodes.

Here resistance R is connected with a battery of potential difference 5 V with the positive terminal and the negative end is earthed. Input terminals A and B are given 0 or 5 V level.

If A and B both are at zero potential (A = B = 0) then diodes Dand D2 both are forward biased. This represents the first line of truth table.

If A and B any of the two has value 5 V and the other has 0 V, then diode A is in operation mode and B is in non-operation mode. If diode A is ideal then there will be no potential drop. This represents the third line. If A = B = 5 V then Y – 1 (5 V) hence this represents the fourth line of truth table.

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