The width of the space charge region at a p-n junction is reduced by applying a voltage bias over the p-n junction. What is the correct statement on both the bias voltage and the p-n junction?
The voltage is a forward bias and the diffusion of the majority charge carriers becomes more dominant.
The voltage is a reverse bias and the diffusion of the majority charge carriers becomes more dominant.
The voltage is a forward bias and the drift of the minority charge carriers becomes more dominant.
The voltage is a reverse bias and the drift of the minority charge carriers becomes more dominant.
A boat beguins on the best side of a river and heads straight east across the river with a speed of 1.9 Ft/s (relative to the water). the river water flows NORTH at the speed of 2.^ Ft/s (relative to the shore). The resultant veliocty of the boat ( relative to the shore) is approximately ___ Ft/s at __ degrees (-CCW from east)
To determine the correct statement, we need to understand the behavior of a p-n junction under different bias voltages.
A p-n junction is formed by joining a p-type semiconductor (which has excess holes as majority carriers) and an n-type semiconductor (which has excess electrons as majority carriers). At the interface of the p and n regions, a region called the space charge region forms. This region contains immobile, ionized dopants that create an electric field that opposes the flow of majority carriers across the junction.
When a voltage bias is applied, two types of biases can occur: forward bias and reverse bias.
In a forward bias, the positive terminal of the voltage source is connected to the p-side of the junction, and the negative terminal is connected to the n-side. This bias causes the majority charge carriers (holes in the p-region and electrons in the n-region) to move towards the junction and combine, reducing the width of the space charge region. The flow of majority carriers across the junction dominates, leading to a higher current flow.
On the other hand, in a reverse bias, the positive terminal of the voltage source is connected to the n-side of the junction, and the negative terminal is connected to the p-side. This bias widens the space charge region by increasing the electric field across the junction. The majority carriers are pushed deeper into their respective regions, and only a small leakage current due to very few minority carriers (electrons in the p-region and holes in the n-region) crossing the junction occurs. Here, minority carriers' behavior dominates because of the opposing electric field.
Now, let's analyze the answer choices:
1. The voltage is a forward bias and the diffusion of the majority charge carriers becomes more dominant.
- This statement is correct because a forward bias (positive terminal on the p-side) reduces the width of the space charge region due to the diffusion of majority charge carriers (holes and electrons) dominating.
2. The voltage is a reverse bias and the diffusion of the majority charge carriers becomes more dominant.
- This statement is incorrect because in a reverse bias, the diffusion of majority carriers is not dominant. The minority carriers' behavior, namely drift under the electric field, becomes more dominant.
3. The voltage is a forward bias and the drift of the minority charge carriers becomes more dominant.
- This statement is incorrect because in a forward bias, diffusion of majority carriers is more dominant, not the drift of minority carriers.
4. The voltage is a reverse bias and the drift of the minority charge carriers becomes more dominant.
- This statement is correct because in a reverse bias, the drift of minority charge carriers (electrons in the p-region and holes in the n-region) becomes more dominant due to the widened space charge region and the increased electric field across the junction.
So, the correct statement is: The voltage is a reverse bias, and the drift of the minority charge carriers becomes more dominant.