IIT-JEE Questions: Semiconductors

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle to start. Cool a piece of copper and a piece of germanium from room temperature down to eighty kelvin. The resistance of one goes down, but that of the other goes up. Which one goes up, and why? Keep your guess in mind. Today we solve old IITJEE questions on semiconductors. Questions like these come up often. So, let's begin.

0:28Idea 1: the band gap and light

So, idea one is the energy gap. Energy goes up along the arrow on the left. In each picture, the upper box is the conduction band and the lower box the valence band. The gap between them is written capital E with a small g below, E g, and the double arrow marks it. Okay? In a metal, on the left, the bands overlap, so E g is zero. The semiconductor in the middle has a gap below three electron volts, written e V: silicon has one point one, germanium zero point seven. Its dots are electrons that crossed the gap, and the small circles are the holes they left. The insulator on the right has a gap above three, like diamond at five point four. Right? Now shine light on a semiconductor. Which photons can lift an electron from the valence band across the gap? Yes, exactly, only a photon with enough energy can do it. Its energy is h nu: h is Planck's constant, and nu, the Greek letter like a v, is the frequency. So h nu must be greater than or equal to E g, the gap. Clear? Now use the wavelength, lambda, another Greek letter. Longer wavelength means less energy. So lambda must be less than or equal to lambda max, the maximum wavelength, which is h c over E g, where c is the speed of light. With me? Now a handy shortcut. Lambda max in nanometres, written n m, equals twelve hundred and forty, over E g in electron volts. Got it?

2:03Idea 2: electrons, holes and temperature

Now idea two: who carries the current? When a bond breaks in pure silicon or germanium, an electron goes up and leaves a hole behind. The hole acts as a carrier of charge plus e, where e is the size of the electron's charge. Okay? In an intrinsic, that is pure, semiconductor, each broken bond gives one electron and one hole. So n e, the free electrons per unit volume, equals n h, the holes per unit volume, and both equal n i, the intrinsic carrier concentration; the small e and h below the n name the carrier, and the small i stands for intrinsic. Clear? Now let me ask you this: we heat a semiconductor. Does its resistance go up, or come down? Yes, exactly, it comes down, because heating breaks many more bonds. So n i rises steeply and the resistance, capital R, falls: a negative temperature coefficient. Got it? But a metal like copper is different. Its free electrons are fixed in number, and heating only makes them collide more often with the ions, so its resistance rises. With me? So, back to our puzzle, from the nineteen eighty eight paper. Copper and germanium are cooled to eighty kelvin; the capital K means kelvin. What happens to each resistance? The resistance of copper falls, as its electrons collide less often. That of germanium rises, because at low temperature far fewer bonds are broken, so there are far fewer carriers. So germanium goes up, and our puzzle is solved.

3:44Idea 3: doping, n-type and p-type

Right? Now idea three: doping, adding a tiny, controlled amount of impurity. On the left, we add a pentavalent atom, with five valence electrons: phosphorus, arsenic or antimony, symbols in the brackets. Four electrons fill the bonds, and the fifth sits on the dashed line marked E D, with a small D below, the donor level. The wavy equals sign means about: this level is just zero point zero one to zero point zero five electron volts below the conduction band, so that electron is easily freed. Okay? So the atom is a donor, and electrons are the majority carriers. Below the picture, n e is much greater than n h; the double angle sign means much greater than. The false statement in the nineteen ninety seven paper said the opposite, that holes are the majority carriers in n-type. Right? Next, a question from the nineteen eighty eight paper. To make pure silicon p-type, should we dope it with phosphorus, boron, antimony or aluminium? The answer is boron and aluminium, because p-type needs trivalent atoms, with three valence electrons. The brackets hold the symbols for boron, aluminium, gallium and indium. Each is one bond short, so it accepts an electron from a neighbour and leaves a hole. Clear? So in p-type material, holes are the majority carriers, and n h is much greater than n e. The dashed line on the right, E A with a small A below, is the acceptor level, just above the valence band. Got it? One rule ties it all together. In equilibrium, at a fixed temperature, n e times n h equals n i squared, with the small two written high, for pure and doped material alike. So raising one carrier lowers the other.

5:35Example 1: band gap from 2480 nm cutoff wavelength

Okay? Example one is from the nineteen ninety seven paper. The electrical conductivity of a semiconductor increases when radiation of wavelength shorter than twenty four hundred and eighty nanometres falls on it. Find its band gap. Pause the video and try it first. Done? Okay, let's check. Only wavelengths below twenty four hundred and eighty nanometres can lift electrons across the gap, so lambda max is twenty four hundred and eighty nanometres. Right? Now, one student puts twenty four hundred and eighty over twelve hundred and forty, and gets two electron volts. Is that right? No, that answer is upside down, and that is the trap. E g in electron volts is twelve hundred and forty, over lambda max in nanometres, since a longer wavelength means a smaller gap. Okay? So E g is twelve hundred and forty over twenty four hundred and eighty, exactly one half, which is zero point five electron volts. That is below three electron volts, so it is a semiconductor. Clear?

6:44Example 2: resistivity vs temperature statement question

Now the harder one, Example two, from the nineteen ninety three paper. Statement Y says the resistivity of a semiconductor decreases with increase of temperature. Statement Z says that in a conducting solid, the rate of collisions between free electrons and ions increases with increase of temperature. We have to comment on them. Pause the video and try it first. Done? Okay, let's check. Statement Y is true, from idea two: heating raises n i steeply, so the resistivity of a semiconductor falls. Right? And statement Z is also true: more collisions on heating is exactly why the resistivity of a metal rises. With me? So both statements are true. Then Z must be the reason for statement Y, right? No, and this is exactly the trap. More collisions on their own would make the resistivity go up, not down. So Z cannot explain a resistivity that falls. Okay? In a semiconductor the collisions do increase, but heating creates so many more carriers that the rise in n i outweighs them. That is the real reason. Clear? So both are true, but Z is not the reason for statement Y, and that is the answer. It is the same physics as our copper and germanium puzzle. Got it?

8:05Quick revision: band gap, temperature and doping

So, quick revision. Lambda max in nanometres is twelve hundred and forty, over E g in electron volts, so a longer wavelength means a smaller gap. On heating, a semiconductor's resistance falls because n i rises, while a metal's rises. And n-type means donors with electrons in the majority; p-type means acceptors with holes in the majority. Well done, you have just cracked four old IITJEE questions. Now try the other questions on this page. All the best!