Overview

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. In pure silicon, free electrons and holes are equal in number. Now we add a little phosphorus, which gives extra free electrons and adds or removes no holes. So the holes should stay the same, right? But the numbers say they fall, from about ten to the power sixteen per cubic metre to only a few times ten to the power ten. So why should adding electrons remove holes? Today a concept check settles it, and we solve two problems from this bank. Let's begin.

0:34How to use this chapter

So, what is in this chapter? Part A has sixteen solved problems at JEE Main level. Okay? Then Part B has sixteen harder, multi-step problems. With me? Part C has twelve concept checks, the why questions, and then fourteen practice problems. Right? Now tell me, when should you read a solution? Straight away, or only after trying the problem yourself? Yes, exactly, only after trying it yourself! Attempt each part as a test, then read the solutions.

1:07The data every problem uses

Clear? Next, the course's data for the whole chapter. First, n i, n with a small i below, the intrinsic carrier concentration: free electrons, or holes, per unit volume in the pure crystal. It is one point five times ten to the power sixteen for silicon, and two point five times ten to the power nineteen for germanium, per cubic metre. Okay? Then mobility, the drift speed per unit field, written mu, the Greek letter, with a small e for electrons and a small h for holes, in square metres per volt per second. For silicon, mu e is zero point one three five, and mu h is zero point zero four eight. Right? For germanium, they are zero point three nine and zero point one nine. So who moves more easily, electrons or holes? Yes, exactly, it is the electrons that move more easily. In both materials, mu e is bigger than mu h, so electrons are more mobile than holes. Got it? The conductivity, sigma, a Greek letter, equals e, the electron charge, times bracket n e times mu e, plus n h times mu h. The resistivity, rho, also Greek, is one over sigma. With me?

2:28Part C: why do the holes fall?

Now one concept check from Part C, and our puzzle. In the figure, orange-edged boxes are conduction bands, blue-grey boxes valence bands, filled dots free electrons, and open circles holes. On the left, a pentavalent dopant makes n-type, with the dashed donor level, E D, capital E with a small D, about zero point zero one to zero point zero five electron volts below the conduction band. On the right, a trivalent dopant makes p-type, with the acceptor level, E A, just above the valence band. Under each, the double greater-than sign means much greater than: free electrons far outnumber holes in n-type, and holes far outnumber free electrons in p-type. Okay? In a pure crystal, each broken bond frees an electron and leaves a hole. So n e, n with a small e, the free electrons per unit volume, equals n h, the holes per unit volume, and both equal n i. Right? Now the key result. At a given temperature, n e times n h equals n i squared, pure or doped. Quick one: what is n i squared for silicon? One point five squared is two point two five, so n i squared is two point two five times ten to the power thirty two, per metre to the sixth. That is a Part A problem in one line! Clear? Now our puzzle. Donors raise n e, so n h must fall to keep the product fixed. But why, physically? What happens to the holes? Heat keeps making pairs, and recombination keeps removing them, as free electrons drop into holes. With many more free electrons around, the holes recombine faster, so their number drops. With me? Balance returns only when n e times n h is back to n i squared. So adding electrons does remove holes, and our puzzle is settled. Got it?

4:28Example 1: ne and nh in phosphorus-doped silicon

Now Example one, an easier one from Part A, to warm up. Silicon, with five times ten to the power twenty eight atoms per cubic metre, is doped with one phosphorus atom per ten to the power seven silicon atoms. Find n e and n h. Pause the video and try it first. Done? Okay, let's check. Phosphorus is pentavalent, so it is a donor, and the silicon becomes n-type. So how many donor atoms are in each cubic metre? N D, capital N with a small D, the number of donor atoms per unit volume, is five times ten to the twenty eight over ten to the seven, which gives five times ten to the twenty one per cubic metre. Right? Each donor gives one free electron, and that is far more than n i, so n e is about N D; the wavy equals sign means approximately. Okay? Now the holes. Phosphorus added none and removed none, so n h stays at n i, one point five times ten to the sixteen, right? No, it does not, and that is the trap from our puzzle. Since n e times n h equals n i squared, n h is two point two five times ten to the thirty two, over five times ten to the twenty one. That gives four point five times ten to the ten per cubic metre. Clear? So the holes fell from one point five times ten to the sixteen, to four point five times ten to the ten, just as our puzzle said. Very good, that problem is done!

6:06Example 2: donors and acceptors, ne, nh and conductivity

Now Example two, a harder one from Part B, to finish. Silicon is doped with ten to the power twenty two donors and four times ten to the power twenty one acceptors per cubic metre. Find n e, n h, and the conductivity. Pause the video and try it first. Done? Okay, let's check. Each donor gives one free electron, so n e is just ten to the twenty two, right? Not so fast, the acceptors are there too, and that is the trap. With both dopants, the larger one wins. Here capital N A, N with a small A, counts acceptor atoms per unit volume, and when N D is bigger than N A, n e is about N D minus N A, okay? So n e is ten to the twenty two, minus four times ten to the twenty one, which gives six times ten to the twenty one per cubic metre. Right? And n h is n i squared over n e: two point two five times ten to the thirty two, over six times ten to the twenty one. That gives three point seven five times ten to the ten per cubic metre. Got it? Now the conductivity. Of the two terms, n e times mu e, and n h times mu h, which one matters here? Yes, exactly, the electron term, since this is n-type, with far more electrons than holes. So sigma is about e times n e times mu e. With me? Six times ten to the twenty one, times one point six times ten to the minus nineteen, the electron charge, times zero point one three five, gives about one hundred and thirty siemens per metre; the capital S stands for siemens. Very good, that one is done too!

7:59Quick revision: ne nh = ni squared, ND minus NA, sigma

So, quick revision. Part A is JEE Main level, Part B is harder and multi-step, and Part C checks the ideas; attempt each as a test first. At a given temperature, n e times n h equals n i squared, pure or doped. So more of one carrier means fewer of the other, and with both dopants, if N D is bigger than N A, n e is about N D minus N A, the difference. In n-type material, sigma is about e times n e times mu e. Now try the practice problems on this page. All the best!