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Chapter 18: Problem 43

Briefly describe electron and hole motions in a \(p-n\) junction for forward and reverse biases; then explain how these lead to rectification.

Short Answer

Expert verified
Question: Explain how the movements of electrons and holes in a p-n junction under forward and reverse biases result in rectification. Answer: Under forward bias, electrons move from the n-region to the p-region while holes move from the p-region to the n-region, allowing current to flow and representing the positive half of the AC signal. In reverse bias, increased barrier prevents electron and hole movements, disallowing current flow and representing the negative half of the AC signal. This selective allowance and prohibition of the current in a p-n junction diode result in the rectification process, converting AC signals to DC.

Step by step solution

01

Understanding Forward Bias

In a forward bias arrangement, the positive terminal of the battery is connected to the p-type and the negative terminal to the n-type. This bias reduces the potential barrier, enabling current to flow. Electrons from the n-region now have enough energy to cross the junction and travel into the p-region. They then combine with holes in the p-region.
02

Understanding Reverse Bias

For a reverse bias, the connections are flipped. The positive terminal is connected to the n-type and the negative terminal to the p-type. Here, the potential barrier increases instead, hindering electrons and holes' movement across the junction. Electrons in the n-region are drawn towards the positive terminal of the battery, and holes in the p-region are attracted to the negative terminal.
03

Analyzing Electron and Hole Motions

Under forward biasing, electrons move from the n-region to the p-region and holes move from the p-region to the n-region. In the reverse bias, their movement is prohibited due to the increased barrier.
04

Understanding Rectification Process

Rectification is the conversion of alternating current (AC) to direct current (DC). This process is the basic principle of a diode. In a p-n junction diode under forward bias, current can pass through, representing the positive half of the AC signal. On the contrary, when it's reverse biased, almost no current flows, representing the negative half of the AC signal. This allows the diode to convert the AC signal into a form of DC. In essence, the electron and hole movements under various biases in a p-n junction lead to rectification by allowing or disallowing current.

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Most popular questions from this chapter

For \(\mathrm{CaO}\), the ionic radii for \(\mathrm{Ca}^{2+}\) and \(\mathrm{O}^{2-}\) ions are \(0.100\) and \(0.140 \mathrm{~nm}\), respectively. If an externally applied electric field produces a \(5 \%\) expansion of the lattice, compute the dipole moment for each \(\mathrm{Ca}^{2+}-\mathrm{O}^{2-}\) pair. Assume that this material is completely unpolarized in the absence of an electric field.

A parallel-plate capacitor using a dielectric material having an \(\epsilon_{r}\) of \(2.2\) has a plate spacing of \(2 \mathrm{~mm}(0.08\) in.). If another material having a dielectric constant of \(3.7\) is used and the capacitance is to be unchanged, what must the new spacing be between the plates?

In your own words, explain the mechanism by which charge-storing capacity is increased by the insertion of a dielectric material within the plates of a capacitor.

(a) The room-temperature electrical conductivity of a silicon specimen is \(500(\Omega \cdot \mathrm{m})^{-1}\). The hole concentration is known to be \(2.0 \times\) \(10^{22} \mathrm{~m}^{-3}\). Using the electron and hole mobilities for silicon in Table \(18.3\), compute the electron concentration. (b) On the basis of the result in part (a), is the specimen intrinsic, \(n\)-type extrinsic, or \(p\)-type extrinsic? Why?

Compare the temperature dependence of the conductivity for metals and intrinsic semiconductors. Briefly explain the difference in behavior.
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