AIEEE and Earlier JEE Main Questions: Semiconductors

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. One semiconductor sample has ten times more holes than free electrons. So which carrier gives most of its conductivity? The holes, surely? Hold that thought. Today we solve past A I triple E and JEE Main questions on semiconductors; one of them settles this puzzle. So, let's begin.

0:26Energy bands: four quick questions

In the figure, energy goes up along the arrow on the left. Each solid has a lower box, the valence band, and an upper box, the conduction band. The A I triple E paper of two thousand and four asked: is the band structure of solids due to Heisenberg's uncertainty principle, Pauli's exclusion principle, Bohr's correspondence principle, or Boltzmann's law? Yes, exactly, it is Pauli's exclusion principle: no two electrons can occupy the same state. So in a crystal, each sharp atomic level spreads out into many closely spaced levels, a band. Okay? Next, from the two thousand and two paper. Metals, superconductors, insulators or semiconductors: in which is the energy band gap the largest? The answer is insulators, that is correct. The band gap is written capital E with a small g below, E g, and the double arrows mark it. Under the pictures, E g is zero for a metal, where the bands overlap, below three electron volts, written e V, for a semiconductor, and above three for an insulator. With me? Now a question from two thousand and seven. Carbon, silicon and germanium, the C, S i and G e on screen, have four valence electrons each. Their gaps are five point four electron volts for diamond, the carbon in brackets, one point one for silicon, and zero point seven for germanium. At room temperature, where is the number of free electrons significant? It is significant in silicon and germanium, but small in carbon. Thermal energy at room temperature lifts electrons across the small gaps, but not the large one. In the middle picture, the dots are such electrons, and the small circles are the holes they leave. Got it? One more from two thousand and two. How does silicon behave at absolute zero, that is, zero kelvin? Yes, exactly, it behaves as an insulator; the capital K on screen means kelvin. At zero kelvin the valence band is full, and no electron has the thermal energy to cross the gap. Clear?

2:42Heating a metal and a semiconductor

Now let us bring in temperature. The two thousand and two paper asked: on heating, how does the specific resistance, the resistivity, of a conductor and of a semiconductor change? That is correct, it rises for the conductor and falls for the semiconductor. Rho, the Greek letter on screen, is the resistivity. Same heating, opposite results. Okay? The two thousand and three paper asked where this difference really comes from. Many students say more collisions, right? No, and that is the trap, because the collisions increase in both. In a metal the number of free electrons stays fixed, and heating only adds collisions, so the resistivity rises. With me? In a semiconductor, heating breaks bonds and creates many more carriers, which outweighs the extra collisions. So its resistivity falls. Right? So the answer is the variation of the number of charge carriers with temperature, the capital T on screen, not the collisions. Got it?

3:44Two carriers: current and conductivity

Now two formulas from the mobility lesson. The current, capital I, through an area capital A, is e, the electronic charge, times capital A, times the bracket, n e v e plus n h v h. Here n e and n h are the free electrons and holes per unit volume; the small letter below names the carrier. And v e and v h are their drift speeds. Okay? Mobility, mu, the Greek letter mu, is the drift speed per unit field. The conductivity, sigma, another Greek letter, equals e times the bracket, n e mu e plus n h mu h. And the resistivity rho is one over sigma. Right? Now the two thousand and six paper. The electron to hole concentration ratio is seven over five, and their current ratio is seven over four; capital I with a small e or a small h is the electron or the hole current. Find the ratio of the drift velocities, v e over v h. It is the same sample, so e and the area cancel, and the current ratio is n e v e over n h v h; the double arrow means this gives. So v e over v h is seven over four, times five over seven, which is five over four. Careful here, many students multiply by seven over five, but the concentration ratio must be flipped. Got it?

5:14Example 1: resistivity of an n-type semiconductor

Example one is from JEE Main twenty nineteen. In an n-type semiconductor, the electron density is ten to the power nineteen per cubic metre, the m to the power minus three on screen, and the mobility is one point six metre squared per volt per second. Find the resistivity, ignoring the holes. Pause the video and try it first. Done? Okay, let's check. With the holes ignored, only the electron term is left: sigma is n e, times e, times mu e, and rho is one over that. Right? Putting in the numbers, rho is one over the product of ten to the power nineteen, one point six times ten to the power minus nineteen, and one point six. With me? The powers of ten cancel, and one point six times one point six is two point five six. One over that is about zero point three nine; the wavy equals sign means approximately. So rho is about zero point four ohm metre, with the Greek capital omega for ohm. Easy, isn't it?

6:20Example 2: conductivity with both electrons and holes

Example two, the harder one, from JEE Main twenty seventeen, is our opening puzzle. There are five times ten to the power eighteen electrons per cubic metre, and five times ten to the power nineteen holes, ten times more. The mobilities are two point zero for electrons and zero point zero one for holes, in metre squared per volt per second. Find the conductivity. Pause the video and try it first. Done? Okay, let's check. There are ten times more holes, so we keep only the hole term, right? No, and that is exactly the trap. The electrons are two hundred times more mobile than the holes. So keep both terms: sigma equals e times the bracket, n e mu e plus n h mu h. Okay? First the electron term. Five times ten to the power eighteen, times two point zero, gives ten to the power nineteen. Right? Now the hole term. Five times ten to the power nineteen, times zero point zero one, gives five times ten to the power seventeen. That is twenty times smaller! So the electrons give most of the conductivity, and our puzzle is solved. With me? Add the two terms, and multiply by e, one point six times ten to the power minus nineteen coulomb. Got it? The bracket is one point zero five times ten to the power nineteen. The powers of ten cancel, and one point six times one point zero five is one point six eight. So sigma is one point six eight siemens per metre; capital S is siemens, and m to the power minus one means per metre. Clear?

8:09Quick revision: bands, heating, conductivity formula

So, quick revision. Bands come from Pauli's exclusion principle, insulators have the largest gap, and silicon at zero kelvin is an insulator. On heating, a metal's resistivity rises, but a semiconductor's falls, because it gets many more carriers. And sigma equals e times the bracket, n e mu e plus n h mu h, so compare both terms before dropping one. Now try the other questions on this page. All the best!