Question 5: Semiconductors and Logic Gates

Question

For the given logic circuit, which of the following inputs combination will make both LED-1 and LED-2 to glow?

Accepted Answer

The assertion is true.
In a reverse-biased diode, the current is very small and remains almost constant with increase in reverse voltage. This small current is called reverse saturation current. When the reverse voltage reaches a critical value called breakdown voltage, the current increases sharply.
So, Assertion A is correct.
Now consider the reason.
In reverse bias, the current below breakdown is due to minority charge carriers, not majority charge carriers. Majority carriers are pulled away from the junction in reverse bias and do not contribute to the small reverse current. Hence the statement in Reason R is false.
Therefore, the correct option is:
$$\boxed{	ext{Option C: A is true but R is false}}$$

Answer

$$ \mathrm{A}=0, \mathrm{~B}=1, \mathrm{C}=1 $$

Answer

$$ \mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=0 $$

Answer

$$ \mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=1 $$

Answer

$$ \mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0 $$

Semiconductors and Logic Gates

For the given logic circuit, which of the following inputs combination will make both LED-1 and LED-2 to glow?

Choose the correct answer: