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Reverse bias applied to a junction diode [2003]
  • a)
    increases the minority carrier current
  • b)
    lowers the potential barrier
  • c)
    raises the potential barrier
  • d)
    increases the majority carrier current
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Reverse bias applied to a junction diode [2003]a)increases the minorit...
In reverse biasing, the conduction across the
p-n junction does not take place due to
majority carriers but takes place due to
minority carriers if the voltage of external
battery is large. The size of the depletion
region increases thereby increasing the
potential barrier.
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Most Upvoted Answer
Reverse bias applied to a junction diode [2003]a)increases the minorit...
Explanation:

When a p-n junction diode is reverse biased, the positive terminal of the battery is connected to the n-type material and the negative terminal is connected to the p-type material. This causes the potential difference across the junction to increase.

As a result of this reverse bias, the following changes take place:

1. Widening of Depletion Region: The reverse bias voltage attracts the majority carriers away from the junction, leaving behind more negative ions in the p-region and more positive ions in the n-region. This increases the width of the depletion region.

2. Decrease in Majority Carrier Current: As the width of the depletion region increases, the number of majority carriers (electrons in n-region and holes in p-region) available for conduction decreases. This leads to a decrease in majority carrier current.

3. Increase in Minority Carrier Current: The minority carriers (holes in n-region and electrons in p-region) which are not involved in the conduction process, move across the junction due to the increased electric field. This leads to an increase in minority carrier current.

4. Increase in Potential Barrier: The potential barrier is the energy required for the majority carriers to cross the depletion region. As the width of the depletion region increases, the potential barrier also increases. Hence, option 'C' is the correct answer.

Note: When a p-n junction diode is forward biased, the positive terminal of the battery is connected to the p-type material and the negative terminal is connected to the n-type material. This causes the potential difference across the junction to decrease, resulting in a decrease in the width of the depletion region and an increase in majority carrier current.
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Read the following text and answer the following questions on the basis of the same:Light Emitting Diode:It is a heavily doped p-n junction which under forward bias emits spontaneous radiation. The diode is encapsulated with a transparent cover so that emitted light can come out. When the diode is forward biased, electrons are sent from n → p (where they are minority carriers) and holes are sent from p → n (where they are minority carriers). At the junction boundary, the concentration of minority carriers increases as compared to the equilibrium concentration (i.e., when there is no bias).Thus at the junction boundary on either side of the junction, excess minority carriers are there which recombine with majority carriers near the junction. On recombination, the energy is released in the form of photons. Photons with energy equal to or slightly less than the band gap are emitted. When the forward current of the diode is small, the intensity of light emitted is small. As the forward current increases, intensity of light increases and reaches a maximum. Further increase in the forward current results in decrease of light intensity. LED's are biased such that the light emitting efficiency is maximum. The V-I characteristics of a LED is similar to that of a Si junction diode. But, the threshold voltages are much higher and slightly different for each colour. The reverse breakdown voltages of LED's are very low, typically around 5 V. So care should be taken that high reverse voltages do not appear across them. LED's that can emit red, yellow, orange, green and blue light are commercially available.Q. During recombination at the junction, emitted photons have

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Reverse bias applied to a junction diode [2003]a)increases the minority carrier currentb)lowers the potential barrierc)raises the potential barrierd)increases the majority carrier currentCorrect answer is option 'C'. Can you explain this answer?
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