First slide

Logic gates

Question

The electronic circuit shown in the figure represents a

Easy

A

OR gate
B

AND gate
C

NOT gate
D

Universal gate

Solution

When A is connected to ground, base emitter junction is not forward based so the transistor will not conduct and current through $R_{0}$ is zero. Hence $V_{0} = 5 volt$ is when A = 0, Y = 1.
Again when A is connected to P, base-emitter junction is forward biased and collector-emitter junction is reverse biased. So the transistor conducts and current through $R_{0}$ is equal to $\frac{5}{R_{0}}$ . In this case $V_{0} = 5 - I_{0} . R_{0} = 5 - (\frac{5}{R_{0}}) R_{0} = 0$ . So when $A = 1, Y = 0$ . Hence the circuit represents a NOT gate

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