Set 1

Transcript

0:00Introduction

Hello everyone! One quick puzzle first. Germanium doped with indium becomes p-type, and holes, which behave like positive charges, are the majority carriers. So the crystal carries a net positive charge, right? Hold that thought; problem two of today's set answers it. This is mixed set one on electronic devices, in JEE Main style. So, let's pick up the tools, and spot the traps.

0:29Tool 1: light and the band gap

Tool number one is light and the band gap, E g, the capital E with a small g below. Light can make an electron hole pair only if the photon energy E is greater than or equal to E g; that is the sign on screen. Okay? And here is the quick link: lambda, the Greek letter for wavelength, in nanometres, equals twelve hundred and forty over E in electron volts, the e V on screen. Turned round, the energy is twelve hundred and forty over lambda. Right? Now problem one. Light of wavelength seven hundred and seventy five nanometres falls on gallium arsenide, whose band gap is one point four three electron volts. Can it create electron hole pairs? Twelve hundred and forty over seven hundred and seventy five gives one point six electron volts. That is more than the band gap of one point four three, so yes, it can. Clear? The same tool works the other way for an LED, problem twelve. Its photon energy, h nu, is about the band gap; h is Planck's constant, nu, the Greek letter like a v, is the frequency, and the wavy equals sign means approximately. For yellow light, twelve hundred and forty over five hundred and ninety gives about two point one electron volts. Got it?

1:45Tool 2: doping keeps the crystal neutral

Tool number two is doping; look at the figure. In n-type, on the left, a pentavalent dopant gives a donor level, E D, just below the conduction band, and many electrons. In p-type, on the right, a trivalent dopant gives an acceptor level, E A, just above the valence band, and many holes. Indium is in group thirteen, so it is trivalent, and germanium doped with indium is p-type. Okay? So holes are the majority carriers. And each indium atom that has accepted an electron becomes a fixed negative ion. Right? Now, back to our opening question. So many positive holes in there. Is the crystal positively charged? It is not charged at all, and this is the classic trap. Each hole made by an acceptor is balanced by its fixed negative ion, and each thermally made hole by its free electron, so the crystal as a whole stays neutral. So the answer to problem two is p-type, holes, and negative fixed ions. Clear? Now the most useful line: n e times n h equals n i squared. Here n e and n h, the n with a small e or h below, are the numbers of electrons and holes per unit volume, and n i is the intrinsic carrier concentration. At a given temperature it holds for doped and pure crystals alike, so raising one kind of carrier lowers the other. With me? Try problem three. Here n i is ten to the power sixteen per cubic metre, and doping raises n e to four times ten to the power twenty. What is n h now? So, n h is ten to the power sixteen, squared, over four times ten to the power twenty. That gives two point five times ten to the power eleven per cubic metre, the m to the power minus three on screen. Got it?

3:37Tool 3: the diode in a circuit

Tool number three is the diode in a circuit. The figure is the characteristic of a silicon diode: capital I, the current, against capital V, the voltage. An ideal diode is a perfect one-way switch. Forward biased, it is a closed switch with no voltage drop; reverse biased, it is an open switch with no current. Okay? But a real diode conducts only after its knee voltage, V k, the V with a small k below, at zero point seven in the figure. In series with a resistance R across a battery V, and taking r f, the diode's own forward resistance, the small r with an f below, as zero, capital I equals V minus V k, over R, okay? V k is zero point seven volts for silicon, S i, and zero point three for germanium, G e. Right? Now problem eight. Two ideal diodes are in series, pointing the same way, with one kilo-ohm, across a five volt battery that forward biases them. What is the current, and what is it if one diode is turned round? Both diodes are closed switches, so the whole five volts falls across the resistor; the Greek capital omega is the ohm, so k omega is a kilo-ohm. So the current is five milliamps, the m A on screen. Clear? Now turn one diode round. Many students halve the current here, but it becomes zero, because that diode is now reverse biased, an open switch, and it breaks the whole series path. With me?

5:07Example 1: silicon diode current, 10 V and 470 ohm

Now, example one, problem seven. Here, a silicon diode is forward biased, in series with four hundred and seventy ohms, across a ten volt battery. Find the current. Pause the video and try it first. Done? Okay, let's check. Ten volts over four hundred and seventy ohms gives about twenty one point three milliamps, right? That answer forgets the diode, so be careful here. Ten volts is far more than the knee voltage of silicon, zero point seven volts, so the diode conducts, with zero point seven volts across it. Clear? So capital I equals V minus V k, over R, which is ten minus zero point seven, over four hundred and seventy. That is nine point three over four hundred and seventy amperes. Right? Nine point three over four hundred and seventy gives about zero point zero one nine eight amperes, which is nineteen point eight milliamps. Very good if you got it!

6:05Example 2: Zener current, 15 V input, 1 kilo-ohm load

Now a harder one, example two, problem eleven. On the right of the figure, the Zener diode, the triangle with a bent bar, is reverse biased, in parallel with the load R L, and R S is the series resistor; the small letters below are labels. V in is fifteen volts, R S two hundred and fifty ohms, V Z, the Zener voltage, ten volts, and R L one kilo-ohm. Find the Zener current. Pause the video and try it first. Done? Okay, let's check. Before any formula, we must test one thing. Is the Zener diode actually in breakdown? Remove the Zener and find the voltage across the load: fifteen times one thousand, over one thousand two hundred and fifty, gives twelve volts. Twelve volts is more than ten, so the diode is in breakdown, and the output is held at ten volts. Okay? Now capital I, the current through R S, is V in minus V Z, over R S, okay? Fifteen minus ten, over two hundred and fifty, gives twenty milliamps. Right? So the Zener diode carries all twenty milliamps, right? No, the load takes its share first, and that is the trap. The load current, I L, is V Z over R L: ten over one thousand, which is ten milliamps. Clear? At the top junction the current splits, so the Zener current, I Z, is capital I minus I L: twenty minus ten, equals ten milliamps. Excellent work if you got that!

7:37Quick revision: band gap, ne nh = ni squared, diode tests

So, quick revision. Light makes pairs only if twelve hundred and forty over lambda, in nanometres, is at least the band gap in electron volts. Doping obeys n e times n h equals n i squared, and the crystal stays neutral. For every diode, test its state first; then capital I is V minus V k, over R, and the Zener current is capital I minus the load current. Now try all fifteen problems on this page yourself, before you check the answers. All the best!