Semiconductor Physics

Practice Questions:

Question 11: easy

A p–n photodiode is made of a material with a band gap of 2.0 eV. The minimum frequency of the radiation that can be absorbed by the material is nearly :

1. \[1\times 10_{14} Hz\]

2. \[20\times 10_{14} Hz\]

3. \[10\times 10_{14} Hz\]

4. \[5\times 10_{14} Hz\]

View Answer

Energy of EM wave E = hυ

⇒ 2 eV = 6.626 × 10^-34×υ

⇒ \[v= 5\times 10_{14} Hz\]

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Question 12: easy

Which one of the following statement is FALSE :

1. The resistance of intrinisic semiconductor decreases with increase of temperature.

2. Pure Si doped with trivalent impurities gives a p-type semiconductor.

3. Majority carriers in a n-type semiconductor are holes.

4. Minority carriers in a p-type semiconductor are electrons.

View Answer

Majority charge carrier in N-type semiconductor is electrons and in P-type semiconductor is holes

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Question 13: easy

Representation of N-type semiconductor is :

View Answer

In N type semiconductor number of electrons is more than number of holes

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Question 14: easy

Intrinsic semiconductor is electrically neutral. Extrinsic semiconductor having large number of current carriers would be :

1. Positively charged

2. Negatively charged

3. Positively charged or negatively charged depending upon the type of impurity that has been added

4. Electrically neutral

View Answer

An extrinsic semiconductor with a large number of current carriers is still electrically neutral overall because the total number of positive and negative charges remains balanced.

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Question 15: easy

When a PN junction diode is reverse biased :

1. Electrons and holes are attracted towards each other and move towards the depletion region

2. Electrons and holes move away from the junction depletion region

3. Height of the potential barrier decreases

4. No change in the current takes place

View Answer

Factual Question: When a PN junction diode is reverse biased potential barrier of depletion increases. Thus Electrons and holes move away from the junction depletion region.

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Question 16: easy

The dominant mechanism for motion of charge carriers in forward and reverse biased silicon P-N junction are :

1. Drift in forward bias, diffusion in reverse bias

2. Diffusion in forward bias, drift in reverse bias

3. Diffusion in both forward and reverse bias

4. Drift in both forward and reverse bias

View Answer

Theory: In forward biased mode diffusion current is more than drift current. Where as in reverse biased mode diffusion current decreased and become less than drift current.

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Question 17: easy

In the given circuit

The current through the battery is :

1. 0.5 A

2. 1 A

3. 1.5 A

4. 2 A

View Answer

Diode D2 and D3 are forward biased. Where as Diode D1 is reversed biased. No current will flow through D1.

Equivalent Resistance= 20/3 ohm

Current = Voltage / Resistance = 10/ (20/3) = 1.5 Ampere

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Question 18: easy

Phodiode, zenerdiode and solar cell is used in:

1. Forward Bias, Reverse Bias, Forward Bias

2. Reverse Bias, Reverse Bias, Unbiased

3. Reverse Bias, UnBias, Reverse Bias

4. Reverse Bias, Reverse Bias, Reverse Bias

View Answer

A Photodiode and Zener Diode operate in Reverse Bias, while a Solar Cell operates in an Unbiased condition (zero external voltage).

Photodiode: Works in reverse bias to detect light.
Zener Diode: Operates in reverse bias for voltage regulation.
Solar Cell: Generates power in an unbiased condition when exposed to light.

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Question 19: easy

Which of the following statement is not correct when a junction diode is in forward bias :

1. The width of depletion region decrease

2. Free electrons on n-side will move towards the junction

3. Holes on p-side move towards the junction

4. Electrons on n-side and holes on p-side will move away from junction

View Answer

In forward bias, the applied field opposes the barrier field, pushing major carriers (electrons on n-side and holes on p-side) towards the junction, thereby decreasing the depletion width. Thus, statement (4) is incorrect.

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Question 20: easy

The approximate ratio of resistances in the forward and reverse bias of the PN-junction diode is :

1. \(10^2 : 1\)

2. \(10^{-2} : 1\)

3. \(1 : 10^{-4}\)

4. \(1 : 10^4\)

View Answer

Forward resistance of a diode is very small (\(approx 10\Omega\)), while its reverse resistance is extremely high (\(approx 10^5 \Omega\)). The ratio \(R_f / R_r\) is of the order of \(10^{-4}\), which corresponds to \(1 : 10^4\).

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