NOR gate means not an OR gate which means output of this gate is just reverse of that of a similar OR gate. We know that output an OR gate is 0 only when all inputs of OR gate is 0. But in the case of NOR gate the output is 1 only when all inputs are 0. In all other cases, that is for all other combinations of inputs the output is 0. Hence, the truth table of a NOR gate is shown below,
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It is just reverse of the AND gate truth table which is as follows, Like OR gate, NOR gate may also have more than two inputs. A NOR gate is also referred as universal gate. Because all binary operations can be realized by only using NOR gates. As we know that there are only three basic operations AND, OR and NOT. Also we know that all complex binary operations can be realized by using these three basic operations. If we can prove that AND, OR and NOT operations can be realized by using only NOR gate, then we can easily say that all binary operations can be realized by using only NOR gates.
NOT Gate by Using NOR GatesIf inputs of two inputs OR gate are 1, then output is 1 and when both inputs of two inputs OR gate is 0, the output is 0. As the NOR gate is reverse of OR gate when both inputs of NOR gate are 1, the output will be 0 and when both inputs of two inputs NOR gate 0, the output will be 1. Now if we short circuit both inputs of a NOR gate, then we will get, Where, X is either 0 or 1. So when X = 1, the output of the above gate is 1. This is nothing but a NOT operation. So we have seen a NOT gate can easily be realized by using NOR gate only.
Realizing OR Gate Using NOR GateNOR gate is an inverted OR gate, so inverted NOR gate will be OR gate. A NOT gate followed by a NOR gate will be equivalent to an OR gate. Here we will use the NOT gate which is realized from another NOR gate. The logic circuit of this OR gate will be as shown below,
AND Gate from NOR GateThe NOR operation of two input NOR gate is, Now, according to D.Morgan’s theorem, Now, if we put inverted A instead of A and inverted B instead of B in the above equation, we will get, This is an AND operation. So from above equation it is clear that if inputs of a NOR gate are inverted then the output will be AND equivalent. So, if we connect NOT gate at both inputs of a NOR gate, we will get an AND gate equivalent. As we are only allowed to use NOR gate, we will use here the NOT gates which are realized from NOR gate. The logical equivalent circuit will be as shown below, Now we have proved that all three basic logical gates, NOT, OR and AND gates can be realized by using only NOR gates. So any even much complex binary operation can be realized by using only NOR gate hence it is quite justified to cal a NOR gate as universal logic gate.
Realizing NOR Gate
Resistor Transistor Logical GateA NOR gate can be realized by using two bipolar junction transistors. The basic circuit is as shown. This circuit is made by two parallel connected transistors.
In the circuit when both A and B are given +5 V, base of the both transistors get quite a high potential to make the transistors ‘ON’. As both transistors T1 and T2 are in ON condition, supply voltage at terminal C gets path to the ground through resistor R. Entire supply voltage ideally drops across resistor R and the output terminal of the circuit will not get any voltage and hence it will be at logical 0 state. Practically, a transistor cannot be ideally short circuited in ON condition. There will be a voltage drop between emitter and collector of the transistor even in ON condition. So practically entire supply voltage will not drop across resistor R instead a small portion of it (0.6 V) is dropped across transistor (saturated). That is, 5 - 0.6 = 4.4 V. Hence, voltage appears at output terminal X is not practically zero instead it will be 0.6 V which is considered as logical 0. Now, if either of inputs A and B is given with +5 V, the only corresponding transistor will be in ON condition. But in this case also supply voltage will get path to the ground through R and ON transistor and similarly the output will be in logical 0 states which is 0.6 V. Now if both of the inputs A and B are given with 0V or grounded, both transistors will be in OFF condition as in this case the base of both transistors does not have enough potential to make the transistors ON. As the supply voltage will not get nay path to the ground, the supply voltage will appear at output X, hence output is in logical 1 or high state. Hence, in the circuit, the output is 1, only when both of the inputs are 0 and in all other conditions, the output is 0.