Set 2

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. You have a Zener diode rated six volts and zero point three watts, and a nine volt supply. Connect it straight across the supply, with nothing to limit the current, and it can be damaged. So what one small part keeps it safe, and how big must it be? Hold on to your answer. This is Set two of the mixed problems on electronic devices, and problems like these come up often in JEE Main. Let's begin.

0:31Tool 1: carriers in a doped crystal

So, tool one. The figure shows n-type silicon on the left, full of electrons, and p-type on the right, full of holes. On screen, n e times n h equals n i squared: n e and n h, the n with a small e or h below, are the numbers of free electrons and holes per unit volume, and n i is the intrinsic carrier concentration. At a given temperature, it holds for pure and doped crystals alike. Okay? In p-type, holes are the majority, so n h is about capital N A, the capital N with a small capital A below, the number of acceptor atoms per unit volume, and it is much bigger than n e. The wavy sign means approximately, and the double greater-than sign means much bigger than. With me? Now a statement from this set. Each hole acts as a positive charge. So a p-type crystal, full of holes, carries a net positive charge, right? Be careful here, because that statement is false. Each hole is balanced by a fixed acceptor ion, which is negative, so the crystal as a whole stays neutral. Clear? Now a problem from this set. Silicon, the S i on screen, has n i equal to one point five times ten to the power sixteen per cubic metre. It is doped with two times ten to the power twenty one boron atoms per cubic metre, the m to the power minus three on screen, the unit of both numbers. Boron is trivalent, so the crystal is p-type with capital N A equal to that number. So what is n e now? So n e is n i squared over n h, with n h equal to capital N A, okay? One point five squared is two point two five, so that is two point two five times ten to the power thirty two, over two times ten to the power twenty one. With me? That gives about one point one times ten to the power eleven electrons per cubic metre. Doping raised the holes and pushed the electrons far down. Got it?

2:34Tool 2: the real diode in a circuit

Now tool two, the real diode. In the figure, in forward bias, almost no current flows until the knee, and then it rises steeply. The knee voltage, V k, the V with a small k below, is zero point seven volt for silicon and zero point three volt for germanium, the G e, with capital V meaning volt. Okay? So the diode keeps its V k, and only the rest drives the current: I equals V minus V k, over R plus r f, with R the resistor in series and r f, the small r with a small f below, the forward resistance. Current flows only if V is more than V k, the greater-than sign. Got it? Now a problem from this set. The knee voltage is zero point six volt and the forward resistance ten ohms, the Greek capital omega meaning ohm. The diode is forward biased, in series with ninety ohms, across a one point six volt cell. Ninety plus ten is one hundred ohms, so the current is one point six over one hundred, which gives sixteen milliamperes, right? Be careful, because the diode keeps its zero point six volt. Only one point six minus zero point six, which is one volt, drives the current through ninety plus ten, which is one hundred ohms. That gives zero point zero one ampere, the capital A, which is ten milliamperes, the m A on screen.

3:54Tool 3: the Zener diode as a regulator

Clear? Now tool three, the Zener diode. In the circuit on the right of the figure, see its symbol: a triangle with a bent bar. To hold the output voltage steady, should it be in forward bias or in reverse bias? Yes, it must be in reverse bias, in the breakdown region. On the graph on the left, along the steep part of the orange curve, the current changes a lot, but the size of the voltage stays at V Z, the V with a small capital Z below, the Zener voltage. Okay? In the circuit, the Zener is in parallel with the load, R L, the R with a small capital L below, and the series resistor, R S, the R with a small capital S below, drops the excess voltage. That resistor is the small part from our opening puzzle, right? The current I in R S is V in, the input voltage, minus V Z, over R S, as on screen. The load takes I L, equal to V Z over R L, and the Zener takes the rest, so I Z equals I minus I L, okay? These hold only while the Zener is in breakdown. Got it?

5:00Example 1: longest wavelength for a 0.67 eV band gap

Now, Example one, the easier one. Here it is: a semiconductor has a band gap of zero point six seven electron volt, the e V on screen. Find the longest wavelength it can absorb. Pause the video and try it first. Done? Okay, let's check. To be absorbed, a photon must lift an electron across the band gap. So as the wavelength gets longer, does the photon carry more energy or less? It carries less, exactly, because the photon energy is h c over lambda, with h Planck's constant, c the speed of light, and lambda, the Greek letter, the wavelength. It must be greater than or equal to E g, the E with a small g below, so lambda is less than or equal to h c over E g. With me? And here is the shortcut: the wavelength in nanometres, the n m, equals one thousand two hundred and forty, over the energy in electron volts. Right? So the longest wavelength, lambda max, the lambda with a small max below, is one thousand two hundred and forty over zero point six seven. That gives about one thousand eight hundred and fifty nanometres. Clear?

6:10Example 2: least series resistor for a 6 V, 0.3 W Zener

Now Example two, our opening puzzle. Here it is: a Zener diode is rated six volts and zero point three watts, the capital W meaning watt. Find the least series resistance that protects it on a nine volt supply. Pause the video and try it first. Done? Okay, let's check. First, the safe current. Power is V Z times I Z, so the largest safe current, I Z max, is P max, the power rating, over V Z, the Zener voltage. Zero point three over six gives zero point zero five ampere, or fifty milliamperes. With me? With no load, all of the current I through R S flows through the Zener, so that is the worst case. Okay? Now the resistor. Nine volts over zero point zero five ampere gives one hundred and eighty ohms, right? Be careful, because the Zener holds its six volts. With V in equal to nine, only nine minus six, which is three volts, sits across the resistor. So R S is three over zero point zero five, which gives sixty ohms. Got it? Any smaller resistor lets more than fifty milliamperes through, the greater-than sign on screen, and that is not safe. So the answer is sixty ohms, and that solves our opening puzzle. Clear?

7:29Quick revision: mass action law, real diode, Zener, 1240/Eg

So, quick revision. n e times n h equals n i squared, and a doped crystal stays neutral. For a real diode, I equals V minus V k, over R plus r f, only if V is more than V k. The Zener works in reverse bias, its safe current is P max over V Z, and the least safe R S, with no load, is V in minus V Z, over that current. And the longest wavelength, in nanometres, is one thousand two hundred and forty over E g in electron volts. Now try the rest of this set on this page. All the best!