Question: The diagram shows the energy bands of a tunnel diode as the potential difference is increased. In this device high impurity atom density causes the occupied donor and unoccupied acceptor levels to spread into impurity bands which overlap respectively the n-type conduction- and the p-type valence bands. In all unbiased diodes, the depletion zone between the n-type and p-type bands constitutes a potential barrier (see Section 6.2) but in the tunnel diode it is so thin that significant tunnelling occurs. The current versus voltage plot shows that unlike a normal diode significant current begins to flow as soon as there is an applied voltage—before the bias voltage is Egap /e. It then decreases (so called negative resistance) before again increasing in the normal way. Explain this behavior.

Short Answer

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Answer

The tunneling effect of the diode has been explained.

Step by step solution

01

Given data 

Tunnel diode to be used.

02

Definition of Tunnel diode

Tunnel diode is a diode is a type of diode which is heavily diode at the junction.

The tunnel diode has negative coefficient of resistance.

As the voltage increases across the diode the current through the diode decreases.

03

Draw the diagram and explain the tunneling effect 

The given diagram is shown in Figure 1.

Tunnel diode is a diode is a type of diode which is heavily diode at the junction.

The tunnel diode has negative coefficient of resistance.

As the voltage increases across the diode, the current through the diode decreases.

As soon as the tunnel diode is forward biased, the current flows through the narrow junction barrier.

The electrons in the n-type region get aligned to the holes in the region, and a large amount of current flows through the diode.

The case when the voltage across the diode is increased the barrier potential starts to increase because of accumulation of the holes and the electrons at the junction this decrease the current through diode.

But when the voltage is increased further the current again starts to increase and diode is normal this is known as the tunneling effect.

Therefore, the tunneling effect of the diode has been explained.

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Most popular questions from this chapter

The bond length of theN2molecule is 0.11nm, and its effect the spring constant is 2.3×103N/m .

(a) From the size other energy jumps for rotation and vibration, determine whether either of these modes of energy storage should be active at 300K .

(b) According to the equipartition theorem, the heat capacity of a diatomic molecule storing energy in rotations but not vibrations should be52R(3 translational +2rotational degrees of freedom). If it is also storing energy in vibrations. it should be72R(adding 2 vibrational degrees). Nitrogen's molar heat capacity is 20.8J/mol.K at 300K. Does this agree with your findings in part (a)?

The accompanying diagrams represent the three lowest energy wave functions for three "atoms." As in all truly molecular states we consider, these states are shared among the atoms. At such large atomic separation, however, the energies are practically equal, so anelectron would be just as happy occupying any combination.

(a) Identify algebraic combinations of the states (for instance, 5+11/2+11/2 ) that would place the electron in each of the three atoms.

(b) Were the atoms closer together, the energies of states 1.11, and III would spread out and an electron would occupy the lowest energy one. Rank them in order of increasing energy as the atoms draw closer together. Explain your reasoning.

Carry out the integration indicated in equation (10.10)

Question: Referring to equations(10-2), lobe I of the hybrid states combines the spherically symmetric s state with the state that is oriented along thez-axis. and thus sticks out in the direction (see Exercises 28 and 33), If Figure is a true picture, then in a coordinate system rotated counterclockwise about they-axis by the tetrahedral angle, lobe II should become lobe. In the new frame. -values are unaffected. but what had been values in the 2x -plane become values in the -plane. according tox=x'cosα+z'sinα and z'cosα-x'sinα, whereα=109.5o iscos-1(-13) , or .

(a) Show that lobe II becomes lobe I. Note that since neither the 2s state nor the radial part of the p states is affected by a rotation. only the angular parts given in equations (10-1) need be considered.

(b) Show that if lobe II is instead rotated about thez-axis by simply shifting φby±1200 . it becomes lobes III and IV.

Exercise 29 outlines how energy may be extracted by transferring an electron from an atom that easily loses an electron from an atom that easily loses an electron to one with a large appetite for electrons , then allowing the two to approach , forming an ionic bond.

  1. Consider separately the cases of hydrogen bonding with fluorine and sodium bonding with fluorine in each case , how close must the ions approach to reach “break even” where the energy needed to transfer the electron between the separated atoms is balanced by the electrostatic potential energy of attraction? The ionization energy of hydrogen is 13.6 eV , that of sodium is 5.1 eV , and the electron affinity of fluorine is 3.40 Ev.
  2. Of HF and NaF , one is considered to be an ionic bond and the other a covalent bond . Which is which and Why?
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