# Microelectronics Questions and Answers – Applications of a Diode

This set of Microelectronics Multiple Choice Questions & Answers (MCQs) focuses on “Applications of a Diode”.

1. The input is a sinusoid with a low frequency but 5V amplitude. If each diode has a threshold of 0.5V and R=2K, find the amplitude of Vout.

a) -4.5V
b) -3.33v
c) 2.5V
d) 0V

Explanation: The voltage drop across R is Vin – 0.5 and will reach the maximum voltage drop when the input voltage reaches its amplitude. Hence, the amplitude of the output voltage is -4.5V.

2. If the two diodes are identical, which on will turn on faster?

a) The left one
b) The right one
c) Both will start simultaneously
d) Cannot be determined

Explanation: The diode in the right side has a current limiting resistor. It would prevent the diode to turn on before a voltage greater than the threshold voltage is applied at the input. Hence, the left one will turn on faster.

3. The input is a sinusoid with a low frequency but 5V amplitude. If each diode has a threshold of 0.5V and R=2K, find the amplitude of Vout.

a) -0.5V
b) -4.5V
c) -3.33V
d) 0V

Explanation: The diode will turn on for the negative cycle and will sustain a voltage drop of -0.5V throughout the negative cycle. Hence, amplitude of the output voltage is -0.5V.

4. The input is a sinusoid with a low frequency but 5V amplitude. If each diode has a threshold of 0.5V and R=2K, find the amplitude of current through R.

a) 1mA
b) 3mA
c) 2.25mA
d) 1.5mA

Explanation: The amplitude of current will be (5-0.5)V/2K=2.25mA. This is the maximum current flowing through the diode with resistor R.

5. In the following circuit, what is Vout?

a) 0.5V
b) 0V
c) 1V
d) 0.7V

Explanation: The voltage drop across the branch with two diodes will be stuck to 0.5V due to the branch containing only one diode. They come in parallel and hence, the two diodes won’t turn on. Vout is always 0.

6. If the amplitude of Vin is VA, what is the output voltage in the circuit below?

a) VA
b) VA/e
c) VA/4
d) VA/8

Explanation: The capacitor gets charged to 0.5V. The output voltage will be stuck at VA since the capacitor doesn’t have a path to discharge in the negative cycle.

7. If a sinusoidal signal with 10V amplitude is applied to the following circuit, what happens to the current when the output reaches maximum? Assume that the threshold is 0.7V.

a) 0A
b) 2.9A
c) 9.3/e A
d) 9.3A

Explanation: The current becomes 0 when the capacitor voltage reaches its maximum. This is because the capacitor is completely filled with charge and it doesn’t have a path to discharge since the n side of the diode will block the flow of current.

8. If a sinusoidal signal with 10V amplitude is applied to the following circuit, what is the output of the following circuit?

a) A D.C. voltage
b) Output becomes 0
c) An A.C. output
d) A pulsating D.C.

Explanation: The discharging time constant is very large. The Capacitor cannot doesn’t discharge to 0 and hence, the output remains near to the maximum voltage of the capacitor while a very small amount of discharge happens. This leads to a pulsating D.C. output.

9. If a sinusoidal signal with 10V amplitude is applied to the following circuit, what is the output of the following circuit?

a) It is unstable
b) It’s a stable D.C. voltage
c) 0 V
d) 5V

Explanation: The discharging time constant is roughly 1ms. The capacitor discharges the entire accumulated charge via the resistor in the negative half and the output voltage becomes 0 again. The output voltage is very much unstable.

10. If a sinusoidal signal with 10V amplitude is applied to the following circuit, when is the output of the following circuit greater than 0?

a) The negative cycle
b) The positive cycle
c) The entire cycle
d) Always 0

Explanation: The capacitor will charge to maximum voltage when the amplitude of the sine wave becomes maximum. As the input decreases, the voltage across the capacitor being more than the input voltage starts reducing since the capacitor discharges via the resistor. Now we find that the discharging constant is very fast and hence, the capacitor discharges on its entirety. We observe an output during the positive cycle but it won’t be an exact sinusoidal.

11. In the following circuit, the threshold voltage of each ideal diode is 0.7V. What will be the maximum voltage during the positive cycle for the following circuit?

a) 5.7V
b) 5.55V
c) 2.565V
d) 1.565V

Explanation: The voltage drop on the p side of D1 should be 5.7V to turn it on. This voltage is maintained by D1 since there are no resistors on the same branch. Observe that the ideal diode approximation allows us to estimate the positive amplitude, it isn’t the exact value.

12. In the following circuit, the threshold voltage is 0.7V. What will be the minimum voltage during the negative cycle for the following circuit?

a) -4.565V
b) -3.565V
c) -2.565V
d) -2.7

Explanation: The voltage drop on the n side of D2 should be lesser than -2V by .7V. A voltage drop of -2.7V is seen at the output where the sinusoid gets clipped by the voltage source combination of a diode and another voltage source. The approximation allows us to approximate the large signal analysis and it should be noticed that this isn’t the exact answer.

13. In the following circuit, the threshold voltage of each ideal diode is 0.7V. What will be the minimum voltage during the positive cycle?

a) 6V
b) 4.5V
c) 0.7V
d) 9V

Explanation: The voltage across the resistors with the diodes is always Vin/3. Hence, the minimum voltage will be 0.7V. This is greater than the voltage drop across R1 by 0.7V.

14. In the following circuit, the threshold voltage of each ideal diode is 0.7V. What will be the minimum voltage during the negative cycle?

a) -1.4V
b) -9V
c) 0V
d) -5.2V

Explanation: The diode starts conducting when the negative voltage reaches about -1.4V while it stops when the voltage is -3.33V.It should be noticed that this approximation helps us in doing the analysis easily.

15. In the following circuit, V1=10sin(t). What will be the charging and the discharging constants?

a) High
b) Low
c) Very low
d) Moderate

Explanation: The frequency is simply 1 Hz. The charging and the discharging constants are quite high. The capacitor can only discharge through the resistor since the diode doesn’t allow any current from n to p side. The time constant is equal to the input frequency. The ripple amplitude becomes very low since the capacitor cannot discharge much before cycle changes.

Sanfoundry Global Education & Learning Series – Microelectronics.

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