Topic: Analog Circuits
The voltage V in the circuit given below is :
- A. Less than 6 V
- B. More than 6 V
- C. Less than 12 V
- D. More than 12 V
Answer: B. More than 6 V
Explanation: The OPAMP operates as a comparator. The voltage at the inverting terminal $V_-$ is fixed by the voltage divider: $V_- = 12V \times \frac{10 k\Omega}{10 k\Omega + 10 k\Omega} = 6V$.
The non-inverting terminal $V_+$ is connected to $V_i$.
For the LED to turn ON, the output of the OPAMP must be positive, which requires $V_+ > V_-$.
Therefore, the LED will be ON when $V_i > 6V$.