COMEDK 2023 Morning Shift | Semiconductor Devices and Logic Gates Question 9 | Physics

An LED is constructed from a $$p$$-$$n$$ junction diode using GaAsP. The energy gap is $$1.9 \mathrm{~eV}$$. The wavelength of the light emitted will be equal to

$$10.4 \times 10^{-26} \mathrm{~m}$$

$$654 \mathrm{~nm}$$

$$654 ~\mathop A\limits^o$$

$$654 \times 10^{-11} \mathrm{~m}$$

COMEDK 2023 Evening Shift

MCQ (Single Correct Answer)

-0

The energy gap between valance band and the conduction band for a given material is $$6 \mathrm{~eV}$$, then the material is :

A semiconductor

A metal

An insulator

A superconductor

COMEDK 2023 Evening Shift

MCQ (Single Correct Answer)

-0

In the given circuit the diode $$D_1$$ and $$D_2$$ have the forward resistance $$25 \Omega$$ and infinite backward resistance. When they are connected to the source as shown, the current passing through the $$175 \Omega$$ resistor is:

COMEDK 2023 Evening Shift Physics - Semiconductor Devices and Logic Gates Question 7 English

0.095 A

0.044 A

0.028 A

0.04 A

COMEDK 2023 Evening Shift

MCQ (Single Correct Answer)

-0

The reverse current in the semiconductor diode changes from $$20 \mu \mathrm{A}$$ to $$40 \mu \mathrm{A}$$ when the reverse potential is changed from $$10 \mathrm{~V}$$ to $$15 \mathrm{~V}$$, then the reverse resistance of the junction diode will be :

$$250 \mathrm{~k} \Omega$$

$$400 \Omega$$

$$400 \mathrm{~k} \Omega$$

$$250 \Omega$$

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