The Photodiode

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle to start. For a big current through an ordinary diode, we use forward bias, right? But one diode is connected the other way round, on purpose, in reverse bias, where only a tiny current flows. It is the photodiode, a diode that detects light. So why choose the direction in which almost nothing flows? By the end of this lesson, you will answer that with numbers. Questions on the photodiode do appear in JEE, so let's begin.

0:34What is a photodiode?

So, what is a photodiode? It is a p-n junction with a transparent window, so that light can reach the junction. On the right you see two graphs: the dark blue one, labelled Photodiode, is ours, and the orange one is the solar cell, which has its own lesson. Okay? And it is operated in reverse bias, exactly our puzzle from the start, so we will come back to it. Right? Now light falls on the junction, and each photon brings a small packet of energy. Let me ask you this: can every photon, of any colour, make an electron-hole pair? No, it cannot, because only a photon whose energy, h nu, is more than the band gap can do it. Here h is Planck's constant, and nu, the Greek letter written like a v, is the frequency. E g, the capital E with a small g below, is the band gap, and the sign between them means greater than. Such a photon makes an electron-hole pair in or near the depletion region. Clear? Now the junction field acts. It separates the electron and the hole, pushing them in opposite directions, and that flow of charge is the photocurrent. With me? And the more intense the light, the more pairs, so the photocurrent is proportional to the intensity; the sign like an open alpha means proportional to. On the dark blue graph, the current, capital I, goes up, and the voltage, capital V, goes across. Its curves are for intensities I one, I two and I three, each I with a small number below it, and the line just below the curves says I three is the brightest. Brighter light pushes the flat reverse part further down, so more reverse current flows.

2:14Which light can it detect?

Got it? Now, which light can it detect? The photon needs h nu greater than E g, and its energy, E, is h c over lambda, with c the speed of light and lambda, the Greek letter, the wavelength. So, if the wavelength gets longer, does the photon energy go up or down? It goes down, yes, exactly, because lambda sits in the denominator. So lambda must be less than or equal to h c over E g, and that sign means less than or equal to. So h c over E g is the longest wavelength it can detect. Okay? Here is the shortcut from our lesson on energy bands. Lambda max, the lambda with a small max below it, in nanometres, written n m, is about twelve hundred and forty over E g in electron volts, written e V, and the wavy equals sign means approximately. Right? So light of too long a wavelength gives no photocurrent. Now, if we make it ten times brighter, will the photodiode pick it up? No, it will not, and this is the trap. Brighter light means more photons, but each one still carries too little energy, so no pairs and no photocurrent. The intensity decides how many photons arrive; the wavelength decides whether they make pairs at all. Clear?

3:34Why reverse bias?

Now, back to our puzzle. On the left of the dark blue graph you see the words reverse bias; that is where the photodiode works. So why not forward bias, where the current is big? Here is the key. Light produces equal numbers of extra electrons and extra holes, the same number of each. Okay? Now, the majority carriers are already present in huge numbers, so these extra ones are a drop in the ocean. Their fractional change is negligible. Right? But the minority carriers are very few to begin with, so the same extra number is enormous for them. Their fractional change is huge. With me? And the reverse current is carried by these minority carriers, so it changes measurably with light, while the forward current does not. That is why the photodiode is used in reverse bias, and that solves our puzzle. One more term to learn. Even with no light, a small reverse current flows, called the dark current. When light falls, the photocurrent adds on top of it. Clear?

4:40Example 1: can a 1.4 eV photodiode detect 1000 nm, 800 nm?

Alright, example one, an easy one. This photodiode is made from a semiconductor of band gap one point four electron volts. Can it detect light of one thousand nanometres, part small a, or eight hundred nanometres, part small b? Pause the video and try it first. Done? Okay, let's check. First, a quick question for you. Which photon carries more energy, the one at one thousand nanometres or the one at eight hundred nanometres? The eight hundred nanometre photon, since a shorter wavelength means more energy. Now part small a: twelve hundred and forty over one thousand gives one point two four electron volts. That is less than one point four, as the less than sign shows, so no pair is made, and this light is not detected. Okay? Now part small b: twelve hundred and forty over eight hundred gives one point five five electron volts. That is more than one point four, so yes, this light is detected. Right? And a faster check uses lambda max. Twelve hundred and forty over one point four gives about eight hundred and eighty six nanometres. Eight hundred is below it, so it is detected; one thousand is above it, so it is not. Got it?

6:01Example 2: fractional change in electrons and holes

Now a harder one, and it settles our puzzle with numbers. An n-type semiconductor has n, its electrons per cubic metre, equal to ten to the power twenty two, and p, its holes per cubic metre, equal to ten to the power ten; m to the power minus three means per cubic metre. Light creates ten to the power fifteen extra electrons and as many extra holes per cubic metre. Find the fractional change in each, and what it shows about the photodiode. Pause the video and try it first. Done? Okay, let's check. First, in this n-type material, which carriers are the majority, the electrons or the holes? The electrons, yes, exactly, because n-type means electrons are the majority. Their fractional change is ten to the power fifteen over ten to the power twenty two, which gives ten to the power minus seven. That is one part in ten million, practically nothing. Okay? Now the holes, the minority carriers. Ten to the power fifteen over ten to the power ten gives ten to the power five. So the extra holes are a hundred thousand times the holes already there! The same extra number for both, yet a huge difference. Right? So only the current carried by the minority carriers, which is the reverse current, changes measurably with light. The forward current does not change measurably. With me? And that is exactly why the photodiode is used in reverse bias. This is the answer to our puzzle, in numbers. See how neat that is?

7:40Quick revision: reverse bias and lambda max = 1240/Eg

So, quick revision. The photodiode is a p-n junction with a window, used in reverse bias. Light with h nu greater than E g makes pairs, and the photocurrent is proportional to the intensity. It detects only wavelengths up to h c over E g, about twelve hundred and forty over E g in electron volts, giving nanometres. And reverse bias works because the minority carriers change by a huge fraction. Now go and try the practice problems on this page. All the best!