Doping: n-type and p-type

Transcript

0:00Introduction

Hello everyone! Here is a puzzle to start. Pure silicon has equal numbers of free electrons and holes. Now we mix in arsenic, one atom per million silicon atoms, and the free electrons go up more than three million times! So what happens to the holes? Do they go up, or down? Example one will tell us. This is doping, and it comes up often in JEE, so let's begin.

0:26Extrinsic semiconductors: n-type

So, here is what doping means. We add a small, controlled amount of impurity, about one atom per ten to the power six atoms, that is, one in a million, and get an extrinsic semiconductor. In the figure, the orange-edged boxes are conduction bands and the blue-grey boxes valence bands; filled dots are free electrons, open circles are holes. The left half is n-type, the right half p-type. Okay? Now we add a pentavalent atom, with five valence electrons: phosphorus, P, arsenic, A s, or antimony, S b. It replaces a silicon atom, and four of its electrons fill the four bonds. So what happens to the fifth electron? Yes, exactly, it has no bond, so it is held very weakly: only about zero point zero five electron volts in silicon, S i, and zero point zero one in germanium, G e; the wavy equals sign means approximately. That is the tiny gap under the orange box. With me? So at room temperature, this fifth electron is already free. The dopant gives away an electron, so we call it a donor. Clear? Now, n e, n with a small e below, is the number of free electrons per unit volume, and n h, with a small h, the number of holes. Capital N D, N with a subscript D, is the number of donor atoms per unit volume, and each gives one electron. So n e is about N D, much greater than n h; the double greater-than sign means much greater than. Electrons are the majority carriers, holes the minority. Right? Before it is freed, the fifth electron sits at the donor level, E D, E with a subscript D, the dashed line on the left. It lies just below E C, E with a subscript C, the bottom of the conduction band. Got it?

2:22p-type: trivalent acceptors

Now the right half. We add a trivalent atom, with three valence electrons: boron, B, aluminium, A l, gallium, G a, or indium, I n, on screen. Three electrons cannot fill four bonds, so one bond is short. How can that bond be completed? Yes, exactly, the dopant takes an electron from a neighbouring bond. Okay? That leaves a vacancy in the neighbour's bond, which is a hole. The dopant accepted an electron, so we call it an acceptor. With me? Capital N A, N with a subscript A, is the number of acceptor atoms per unit volume. Each acceptor makes one hole, so n h is about N A, much greater than n e. In p-type material, holes are the majority, electrons the minority. Clear? And the acceptor level, E A, E with a subscript A, is the dashed line on the right, just above E V, the top of the valence band. Right?

3:23One rule links electrons and holes

Now comes the most useful result. At a given temperature, n e times n h equals n i squared; the small two on top means squared. Here n i, with a small i, is the intrinsic carrier concentration, its value in the pure material. And it holds for intrinsic and doped material alike. Okay? So, back to our puzzle. Donors push n e way up, but n e times n h must stay n i squared. So what must happen to n h? It must fall, yes, exactly! Raise one carrier by doping, and the other falls. Example one will show us by how much.

4:05The crystal stays neutral

Right? Now a trap that catches many students. An n-type crystal has far more free electrons than holes. So the crystal must be negatively charged, right? No, it is not charged at all; that is the trap. Each free electron came from a neutral donor atom, which is left behind as a fixed positive ion, the plus sign on screen. So the crystal stays neutral. Okay? The same holds in p-type material. Each hole is balanced by a fixed negative acceptor ion, the minus sign on screen. Clear? With both dopants present, the larger one wins. If N D is greater than N A, as the greater-than sign shows, n e is about N D minus N A, the difference. Got it?

4:52Example 1: electron and hole density in As-doped silicon

Now Example one, our puzzle. Silicon has five times ten to the power twenty eight atoms per cubic metre, and n i is one point five times ten to the power sixteen per cubic metre, written m to the power minus three. It is doped with one part per million of arsenic. Find n e and n h. Pause the video and try it first. Done? Okay, let's check. Arsenic is pentavalent, so it is a donor, and the silicon becomes n-type. Right? One part per million is a factor of ten to the power minus six. So n e is about N D, five times ten to the power twenty eight, times ten to the power minus six, which is five times ten to the power twenty two per cubic metre. Okay? Now let us find the holes. Arsenic adds no holes, so n h stays at n i, one point five times ten to the power sixteen, right? No, that is exactly the trap, so be careful! Use the rule: n h equals n i squared over n e. One point five times ten to the power sixteen, squared, is two point two five times ten to the power thirty two. Divide by five times ten to the power twenty two, and n h is four point five times ten to the power nine per cubic metre. So the holes fell more than three million times. That solves our puzzle!

6:23Example 2: find ne and nh when holes are a million times ne

Now Example two, a harder one. Pure silicon is doped until its hole concentration is ten to the power six times its electron concentration. Find both concentrations. Pause the video and try it first, taking n i from Example one. Done? Okay, let's check. Holes are ten to the power six times the electrons, so they are the majority, and this is p-type. With me? Now here is a tempting shortcut. Keep the holes at n i, and just divide n i by a million to get the electrons. Will that work? No, it fails, because doping changes both carriers, and their product must stay n i squared. So use n e times n h equals n i squared, where n i squared is one point five times ten to the power sixteen, squared, which is two point two five times ten to the power thirty two. Okay? Now put in n h equal to ten to the power six times n e. Then ten to the power six times n e squared equals two point two five times ten to the power thirty two; the double arrow means which gives. So n e squared is two point two five times ten to the power twenty six. Right? Take the square root, and n e is one point five times ten to the power thirteen per cubic metre. Clear? And n h is ten to the power six times that, one point five times ten to the power nineteen per cubic metre. So the electrons fell a thousand times below n i, and the holes rose a thousand times above it; the shortcut fails. Got it?

8:01Quick revision: donors, acceptors and ne nh = ni squared

So, quick revision. n-type: donors like phosphorus, arsenic and antimony, and electrons are the majority. p-type: acceptors like boron, aluminium, gallium and indium, and holes are the majority. And n e times n h equals n i squared, even after doping, while the crystal stays neutral. Now try the practice problems on this page. All the best!