Class 12 Exam > Class 12 Questions > In a p-n junction, as the diffusion process c...
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In a p-n junction, as the diffusion process continues the width of the depletion zone
- a)decreases
- b)increases
- c)remains the same
- d)oscillates
Correct answer is option 'B'. Can you explain this answer?
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In a p-n junction, as the diffusion process continues the width of the...
In reverse biasing, the positive terminal of the battery is connected to the n-type whereas the negative terminal is connected to the p-type junction. So the positive terminal tends to pull the electrons (near to the depletion layer) in n-type towards itself whereas the negative terminal pulls the holes towards itself which results in an increase in the width of the depletion layer.
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In a p-n junction, as the diffusion process continues the width of the...
Explanation:
A p-n junction is formed by joining a p-type semiconductor and an n-type semiconductor. In a p-type semiconductor, the majority carriers are holes and in an n-type semiconductor, the majority carriers are electrons.
When the p-type and n-type semiconductors are brought into contact, diffusion of majority carriers takes place. As a result, the region around the junction becomes depleted of majority carriers. This region is called the depletion region or the space charge region.
The width of the depletion region depends on the doping concentrations of the p-type and n-type semiconductors. The depletion region acts as a potential barrier to the flow of majority carriers.
As the diffusion process continues, the width of the depletion region increases. This is because the diffusion of majority carriers continues until the concentrations of the two types of majority carriers become equal in the depletion region.
Eventually, the width of the depletion region reaches a maximum value and becomes constant. At this point, the potential barrier becomes large enough to prevent any further diffusion of majority carriers.
Therefore, the correct option is (b) increases.
A p-n junction is formed by joining a p-type semiconductor and an n-type semiconductor. In a p-type semiconductor, the majority carriers are holes and in an n-type semiconductor, the majority carriers are electrons.
When the p-type and n-type semiconductors are brought into contact, diffusion of majority carriers takes place. As a result, the region around the junction becomes depleted of majority carriers. This region is called the depletion region or the space charge region.
The width of the depletion region depends on the doping concentrations of the p-type and n-type semiconductors. The depletion region acts as a potential barrier to the flow of majority carriers.
As the diffusion process continues, the width of the depletion region increases. This is because the diffusion of majority carriers continues until the concentrations of the two types of majority carriers become equal in the depletion region.
Eventually, the width of the depletion region reaches a maximum value and becomes constant. At this point, the potential barrier becomes large enough to prevent any further diffusion of majority carriers.
Therefore, the correct option is (b) increases.
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In a p-n junction, as the diffusion process continues the width of the...
If we increase the reverse bias and the widthof the depletion region does not change the electrons don't have enough energy to get to the P-side, therefore diffusion current dominates and that increases the width of the depletion region until we get an equilibrium
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In a p-n junction, as the diffusion process continues the width of the depletion zonea)decreasesb)increasesc)remains the samed)oscillatesCorrect answer is option 'B'. Can you explain this answer?
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