Formation of the Junction

Transcript

0:00Introduction

Hello everyone, let us start with a small puzzle. Take one crystal of silicon, p-type on the left, full of holes, and n-type on the right, full of free electrons. Surely the holes and electrons rush across and wipe each other out, right? But they do not. The action stops in a thin layer near the junction, about a micrometre thick at most. So what stops them? Let us find out. This junction is the heart of every diode, and it comes up often in JEE questions. So, let's begin.

0:33Diffusion across the junction

Here is our junction. On the left is the p-side, where the small open circles are holes, the majority carriers there. On the right is the n-side, where the black dots are free electrons, the majority there. We will come to the middle strip, between the dashed lines, in a minute. Okay? Lots of holes on the left, and hardly any on the right. So which way will they move? Yes, exactly, the holes move from p into n, from crowded to empty, just because of the difference in concentration. That is diffusion. And the electrons diffuse the other way, from n into p. The little arrow on screen means goes to. With me? Now, a question many students get wrong. The electrons move from n to p. So which way is their current? Yes, the current is from p to n, because electrons are negative, so their current is opposite to their motion. So both carriers give a current from p to n, the diffusion current. Got it?

1:33The depletion region

Now, a hole that crosses into the n-side is surrounded by free electrons. Soon it meets one, and the two recombine. The same happens to an electron that crosses into the p-side. Right? So the mobile carriers near the junction vanish. What is left behind there? Fixed ions are left behind, yes, absolutely. On the p-side, the acceptor atoms become uncovered negative ions, the circles with a minus sign. On the n-side, the donor atoms become uncovered positive ions, the circles with a plus sign. They are locked in the crystal and cannot move. Clear? So this layer between the dashed lines has no mobile carriers. That is why we call it the depletion region. It is very thin, about zero point one to one micrometre. The wavy equals sign means approximately, and mu, the Greek letter before the m, means micro, a millionth. With me?

2:31Drift current and equilibrium

Now we have positive ions on the n-side, and negative ions on the p-side. Which way does their electric field point? It points from positive to negative, so from n to p, the red arrow labelled capital E, for the field. Notice that it pushes the diffusing carriers back. Okay? Now, the few electrons on the p-side and the few holes on the n-side are minority carriers. If one wanders into the depletion region, the field sweeps it straight across. That is the drift current, from n to p, opposite to diffusion. Right? So more diffusion uncovers more ions, and the field grows stronger. Does the diffusion go on for ever? No, it settles down, and that answers our puzzle: the growing field stops the carriers. At equilibrium, the diffusion and drift currents are equal and opposite, so the net current is zero. Now here is a trap. Many students say no current flows at all. That is wrong. Both currents flow, and they simply cancel. Got it?

3:36The barrier potential

The field points from higher potential to lower potential, and ours points from n to p. So which side is at the higher potential? The n-side is higher, that is correct. In the graph below, the potential rises across the depletion region by V naught, a capital V with a small zero below it. This is the barrier potential. Clear? It is about zero point seven volt for silicon, S i on screen, and about zero point three volt for germanium, G e on screen. Right? And for a depletion region of width d, the average field is E equals V naught over d, the potential step over the width. With me?

4:15Equal charges on the two sides

Next, the uncovered charges on the two sides are equal, so the region as a whole is neutral. N A, a capital N with a small A below it, is the number of acceptors per unit volume, and x p, an x with a small p below it, is how far the region goes into the p-side. N D and x n mean the same for the n-side. So N A times x p equals N D times x n. Okay? Now suppose the p-side is lightly doped, so N A is small. Which part of the region will be wider? The p part is wider, yes, exactly. With fewer ions per unit volume, the region must reach deeper to uncover the same charge. So remember this trick: the depletion region leans into the lightly doped side. Got it? And with heavier doping on both sides, fewer atomic layers give that charge, so the region becomes thinner. The double arrow means implies.

5:10Example 1: average field across a 0.5 micron region at 0.7 V

Clear? Now, example one, an easy one. We have a silicon junction with a barrier potential of zero point seven volt, and a depletion region zero point five micrometre wide. Find the average electric field in it. Pause the video and try it first. Have you tried it? Okay, let us check. V naught over d is zero point seven over zero point five, which gives one point four volts per metre. Is that the answer? No, it is not, and that is exactly the trap. The width is in micrometres. Micro means ten to the power minus six, so zero point five micrometre is five times ten to the power minus seven metre, with the minus seven written as a small power. Okay? So E equals zero point seven over five times ten to the power minus seven. That gives one point four times ten to the power six volts per metre, where metre to the power minus one means per metre. Over a million volts per metre, from less than one volt. Amazing, right?

6:14Example 2: split a 1.01 micron width between p and n sides

Now example two, a harder one. The p-side is doped with ten to the power twenty two acceptors, and the n-side with ten to the power twenty four donors, per cubic metre. The depletion region is one point zero one micrometre wide. How far does it extend into each side? Pause the video and try it first. Have you got your answer? Okay, let us check. Before calculating anything, which side gets the wider part? The p-side gets the wider part, yes, exactly, because it is lightly doped. Equal charges give ten to the power twenty two times x p equals ten to the power twenty four times x n. So x p equals one hundred times x n. Clear? The two parts make the whole width. So x p plus x n, which is one hundred and one times x n, equals one point zero one micrometre. Right? So x n is zero point zero one micrometre, and x p is one point zero zero micrometre. Almost the whole region lies in the lightly doped side, just as we guessed.

7:21Quick revision: equilibrium, V0, E = V0/d, NA xp = ND xn

So, quick revision. Diffusion uncovers fixed ions, and that layer with no mobile carriers is the depletion region. The field points from n to p, and at equilibrium diffusion equals drift, so the net current is zero. V naught is about zero point seven volt for silicon and zero point three volt for germanium, E equals V naught over d, and N A times x p equals N D times x n. Now go and try the practice problems on this page, and all the best!