Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Every silicon p-n junction has a built-in voltage across it, about zero point seven volt. Now, a cell also has a voltage across it, and a cell drives a current through a wire. So join the two ends of the junction with a wire. Do we get a free battery? Current for ever, from nothing at all? Something feels wrong, right? Today we do concept checks from this unit's problem bank, short why-questions on semiconductors and the junction, and our puzzle is one of them. JEE can ask questions like these, so let's begin.

0:38Check 1: on heating, does the resistance rise or fall?

So, check one, and first the formula we need. Conductivity, sigma, the Greek letter, equals e times, n e times mu e, plus n h times mu h. Here e is the charge of an electron, and n e and n h, the n with a small e or h below it, are the numbers of electrons and holes per unit volume. Mu e and mu h, the Greek letter mu, are their mobilities, the drift speed per unit field. Okay? Now the check. Heat a metal wire, and its resistance rises. Now heat a piece of pure silicon. Does its resistance rise or fall, and why? Let us find out from the middle picture, the semiconductor. Its gap, E g, the E with a small g, is small, under three electron volt, e V on screen. So heat lifts a few electrons across it, the black dots, and leaves holes behind, the open circles. In pure material, n e equals n h equals n i, the intrinsic carrier concentration, and n i is proportional to T to the power three over two, times e to the power minus E g over two k T, all in the exponent. Here T is the temperature in kelvin, k is the Boltzmann constant, this e is the exponential, and the sign like an open alpha means is proportional to. Right? This n i rises very steeply with temperature. Heating also increases the collisions, so the mobility falls a little, but the steep rise in the number of carriers outweighs it. So sigma rises, and R, the resistance, falls. Yes, on heating a semiconductor's resistance falls. With me? Now the metal, on the left. Its bands overlap, so E g is zero, and the number of free electrons is fixed. Heating cannot add carriers; it only increases the collisions, so the resistance rises. The insulator on the right has a big gap, and we skip it today. Clear?

2:45Check 2: one part in a million, a giant change

Now check two. Doping means adding a small, controlled amount of impurity, about one atom for every ten to the power six atoms, the ten with a small six above it. That is one part in a million. Okay? Now think about this. Just one atom in a million. Can that change the conductivity of silicon much? It changes it enormously, and here is the key fact. Pure silicon, S i on screen, has only about one free carrier for every ten to the power twelve atoms. That is a million million atoms, so the dopant atoms far outnumber these carriers. Right? In n-type silicon, the dopant is pentavalent, a donor. Its fifth electron is free at room temperature, so electrons are the majority carriers. So n e is about N D, the number of donors per unit volume, the capital N with a small D below it. The wavy equals sign means approximately. Got it? In p-type silicon, the dopant is trivalent, an acceptor, which creates a hole. So holes are the majority carriers, and n h is about N A, the number of acceptors per unit volume. Clear? So each dopant atom gives one carrier. Ten to the power twelve over ten to the power six gives ten to the power six. So the carriers multiply about a million times. Tiny doping, and a giant change in conductivity!

4:11Check 3: more electrons, so why fewer holes?

Now check three. Doping with donors increases n e, the electrons. But the holes, n h, fall; they do not stay the same. Donors add electrons, not holes, so why should the holes change at all? Let us start with pure silicon. Heat keeps making electron-hole pairs, and pairs keep disappearing by recombination, when an electron meets a hole. In equilibrium the two rates are equal. Okay? Now add donors. There are many more electrons around, so each hole meets an electron much sooner. So holes recombine faster, and their number falls. Right? This goes on until equilibrium is reached again, and that happens only when n e times n h has come back to n i squared. This holds at a given temperature, for pure and doped material alike. So more electrons means fewer holes. Got it?

5:07Example 1: why no net current in an unbiased p-n junction

Now example one, the easier one. Why is there no net current in an unbiased junction, although carriers keep crossing it? Pause the video and try it first. Done? Okay, let's check. In the figure, the p-side on the left has holes, the open circles, and the n-side on the right has free electrons, the black dots. Between the dashed lines is the depletion region. Carriers cross the junction all the time, so surely a current flows, correct? Careful, there are two currents here, not one. Majority carriers diffuse, holes from p to n and electrons from n to p, and both give a current from p to n, the diffusion current. With me? The carriers that cross recombine and leave fixed ions behind, the circles with minus and plus signs. These ions set up a field, the red arrow, capital E, pointing from n to p. Okay? This field sweeps the minority carriers across, the few electrons on the p-side and the few holes on the n-side. That is the drift current, from n to p, opposite to diffusion. Right? At equilibrium the two currents are exactly equal and opposite. So carriers keep crossing both ways, but the net current is zero. The graph below the junction is for example two. Very good!

6:30Example 2: why the barrier potential cannot drive a current

Now example two, a little harder, and it is our puzzle from the start. Our junction has a barrier potential across it. Why can it not drive a current through a wire joining its ends, like a cell? Pause the video and try it first. Done? Okay, let's check. In the graph below the junction, the potential is plotted along x, the distance across it. The n-side is higher than the p-side by V naught, the capital V with a small zero, the barrier potential, about zero point seven volt for silicon. But this barrier is an equilibrium condition; it is what balances diffusion and drift, as in example one. Okay? Now suppose a current did flow. The wire would warm up, for ever. So let me ask you this: where would that energy come from? From nowhere, and that is the whole point. Where the wire meets the p-side and the n-side, there are metal contacts. Their contact potentials exactly cancel the barrier, so the total voltage round the loop is zero. Right? There is no source of energy, so no current flows. The barrier looks like a cell, but it is not one, and that is the trap. Clear? So a current flows only if energy is supplied, for example by light. Light makes electron-hole pairs, and the junction field separates them, so we get a solar cell, with no external bias at all. That settles our puzzle!

7:58Quick revision: heating, doping, mass action law, junction

So, quick revision. On heating, a semiconductor's resistance falls, while a metal's rises. Doping of one part in a million multiplies the carriers about a million times, and n e times n h equals n i squared. In a junction, diffusion equals drift, and the barrier V naught alone drives no current. Now try the practice problems on this page. All the best!