Overview

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. This chapter gives you fifteen problems at a time, and just twenty five minutes. Work out how long each problem gets, and keep that number ready. We will check it soon. In this video, we pack the tools for the whole unit on electronic devices, then solve two problems from the sets. Let's begin.

0:24How this chapter works

So, here is how this chapter works. There are three timed sets of fifteen original problems each, in the style of recent JEE Main papers. Okay? The problems mix all the chapters of the unit: semiconductors, the junction diode, rectifiers, special diodes and logic gates. So first, spot the chapter. Right? Give yourself twenty five minutes per set, then check the solutions. Now our puzzle: twenty five minutes is one thousand five hundred seconds, the small s, and the division sign says divide by fifteen. That is just one hundred seconds per problem, so keep the tools ready in your head. Clear? Where a problem needs them, use these mobilities. Mu, the Greek letter like a u with a tail, is the mobility, with a small e for electrons and a small h for holes. Silicon, S i, has zero point one three five and zero point zero four eight; germanium, G e, has zero point three nine and zero point one nine, all in square metres per volt per second. With me?

1:27Tools 1: light, carriers and doping

Now, let us pack the first set of tools: light, carriers and doping. Light can make an electron-hole pair only if h nu is greater than or equal to E g, the band gap, E with a small g below. Here h is Planck's constant and nu, the Greek letter like a v, is the frequency. The shortcut: the wavelength lambda, the Greek letter like an upside down y, in nanometres, n m, equals twelve hundred and forty over the energy in electron volts, written e V, okay? Next, n e and n h, with a small e and a small h below, are the numbers of electrons and holes per unit volume. In a pure, intrinsic semiconductor, both equal n i, the intrinsic carrier concentration. Right? Pentavalent donors, with five valence electrons, make n-type material, where electrons are the majority. Clear? Trivalent acceptors, with three, make p-type material, where holes are the majority. Now, doping pushes one kind of carrier far up. What happens to the other kind? It falls, and many students miss this; donors raise n e, so n h comes down. In equilibrium, n e times n h equals n i squared, so if one goes up, the other comes down. And the crystal stays neutral either way.

2:46Tools 2: the junction diode and rectifiers

Got it? Now, the second set of tools, for the junction diode and rectifiers. Every p-n junction has a barrier potential, V naught, the V with a small zero below: about zero point seven volts for silicon, and zero point three volts for germanium. With me? Forward bias lowers the barrier to V naught minus the applied voltage V, and the depletion layer gets thinner. Reverse bias raises it to V naught plus V, and the layer gets wider. Okay? Now the trap: before using zero point seven volts or any formula, first decide whether the diode conducts at all. If it does, the current, capital I, is V minus V k, over capital R plus r f, where capital R is the resistor in series; V k, small k below, is the knee voltage, and r f, small f below, the forward resistance. Clear? For an ideal half-wave rectifier, the dc output, V d c, is V m over pi, with V m the peak voltage and pi the Greek letter; its output frequency f equals the input frequency. Now, a full-wave rectifier uses both halves of each cycle. What is its output frequency? Yes, exactly, it is two f, twice the input frequency, and many students write just f here. Its dc output, two times V m over pi, is also double. Got it?

4:08Tools 3: special diodes and logic gates

Now, the third set of tools, for special diodes and logic gates. In a regulator, the Zener diode is reverse biased, across the load. In breakdown, the output stays at V Z, the V with a capital Z below, the Zener voltage. Right? Then capital I, the current through the series resistor R S, with a small S below, is V in minus V Z, over R S, where V in, with small i n below, is the input voltage. The load current, I L, is V Z over R L, the load resistance, and the Zener current, I Z, is capital I minus I L, the difference. Okay? The LED works in forward bias, the photodiode in reverse bias, and the solar cell with no external bias at all. Clear? For gates, capital A and capital B are the inputs. NAND is A dot B with a bar on top, giving zero only for the inputs one and one, in brackets. NOR is A plus B with a bar on top, giving one only for the inputs zero and zero. With me? And NAND and NOR are universal gates: any logic circuit can be built from NAND gates alone, or NOR gates alone. Got it?

5:18Example 1: can 775 nm light make pairs in GaAs?

Alright, Example one, from Set one. Can light of wavelength seven hundred and seventy five nanometres create electron-hole pairs in gallium arsenide, with a band gap E g of one point four three electron volts? Pause the video and try it first. Done? Okay, let's check. The photon energy, capital E, in electron volts, is twelve hundred and forty over the wavelength in nanometres. Twelve hundred and forty over seven hundred and seventy five gives exactly one point six electron volts. Right? Now, one point six is not equal to the gap, one point four three. So does the light fail, since its energy does not match the gap exactly? No, it does not fail; the rule is h nu greater than or equal to E g, not exactly equal. One point six is more than one point four three, as the greater-than sign shows, so yes, pairs are created, well inside our hundred seconds. Clear?

6:16Example 2: Zener current, 15 V input, 1 kilo-ohm load

Now Example two, a harder one, also from Set one. In a Zener regulator, V in is fifteen volts, R S is two hundred and fifty ohms, V Z is ten volts, and R L is one kilo-ohm, where omega means ohm and the small k means kilo. Find the Zener current. Pause the video and try it first. Done? Okay, let's check. In the circuit on the right of the figure, the Zener sits across the load. Before any formula, what must we check? Yes, exactly, whether the diode breaks down at all. Remove the Zener, and R S and R L share the fifteen volts, with two hundred and fifty plus one thousand, that is one thousand two hundred and fifty ohms, in total. So the load gets fifteen times one thousand over one thousand two hundred and fifty, which gives twelve volts; the cross means times. That is more than ten, so the diode is in breakdown, and the output is ten volts. Okay? So the current through R S is fifteen minus ten, over two hundred and fifty, which gives zero point zero two amperes, the capital A, or twenty milliamperes, the m A on screen. Does all of it go through the Zener? No, because the load takes its share. The load current, I L, is ten volts over one thousand ohms, which gives zero point zero one amperes, or ten milliamperes. Right? So the Zener current, I Z, is twenty minus ten, which gives ten milliamperes. In the figure, that is the split at the top dot. Clear?

7:50Quick revision: band gap, doping, rectifiers, Zener test

So, quick revision. Light makes pairs only if h nu is at least E g, and n e times n h equals n i squared, even after doping. For diodes, check conduction first, and a full-wave output has frequency two f, twice the input. For a Zener, test for breakdown first, then I Z is capital I minus I L, okay? Now open Set one, give it twenty five minutes, then check the solutions. All the best!