1. ⇒ (MHT CET 2023 13th May Morning Shift )

The following graph represents

MHT CET 2023 13th May Morning Shift Physics - Semiconductor Devices and Logic Gates Question 3 English

A. forward bias characteristics of a solar cell

B. reverse bias characteristics of a Zener diode

C. reverse bias characteristics of a photodiode

D. forward bias characteristics of a LED

Correct Option is (D)

With increasing voltage, the LED's current rises, leading to light emission due to charge carrier recombination.

2. ⇒ (MHT CET 2023 12th May Evening Shift )

Which one of the following statements is WRONG regarding LED?

A. LEDs are energy efficient

B. LEDs have long time if properly manufactured.

C. Brightness of light emitted by LED cannot be controlled.

D. Colours produced by LED do not fade out.

Correct Option is (C)

Currently no explanation available

3. ⇒ (MHT CET 2023 11th May Morning Shift )

Identify the correct circuit diagrams for the normal operation from the following.

MHT CET 2023 11th May Morning Shift Physics - Semiconductor Devices and Logic Gates Question 13 English

A. (A), (B)

B. (B), (C)

C. (A), (C)

D. (B), (D)

Correct Option is (D)

Option (B) is photodiode and option (D) is LED with reverse and forward biasing respectively. Therefore both will operate properly.

4. ⇒ (MHT CET 2023 10th May Morning Shift )

Identify the mismatch out of the following.

A. Zener diode - voltage regulator.

B. Photodiode-optocouplers.

C. Solar cell - Electrical energy into light.

D. Light emitting diode - optical communication.

Correct Option is (C)

Solar cell converts light energy into electrical energy.

5. ⇒ (MHT CET 2023 9th May Morning Shift )

A $5.0 V$ stabilized power supply is required to be designed using a $12 V$ DC power supply as input source. The maximum power rating of zener diode is $2.0 W$ . The minimum value of resistance $R_{s}$ in $Ω$ connected in series with zener diode will be

A. 16.5

B. 17.5

C. 18.5

D. 15.5

Correct Option is (B)

Using the series resistance formula for a zener diode

$\begin{aligned} R_{S} = \frac{(V_{S} - V_{Z})}{I_{Z max}} \\ I_{Z max} = \frac{P_{Z}}{V_{Z}} = \frac{2}{5} = 400 mA \\ ∴ & R_{S} = \frac{(12 - 5)}{400 \times 10^{- 3}} = \frac{7}{400} \times 10^{3} \\ = 17.5 Ω \end{aligned}$

⇒Semiconductor Devices