Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Your transformer gives six volts r m s, and you pass it through a single ideal diode to get direct current. How much direct voltage comes out? Six volts, right? Hold on to that guess; our second example settles it. This is part A of our electronic devices problem bank, at JEE Main level. Three quick tools first, then two examples. So, let's begin.

0:30Tool 1: doping and the carrier product

So, tool number one. Here, silicon, with five times ten to the power twenty eight atoms per cubic metre, is doped with one phosphorus atom per ten to the power seven silicon atoms. Find n e and n h, the numbers of electrons and of holes per unit volume, the n with a small e or h below. With me? Phosphorus has five valence electrons, so it is pentavalent. It donates its fifth electron, so it is a donor, and the silicon becomes n-type, with electrons as the majority carriers. Okay? Each phosphorus atom gives one free electron, so n e is about N D, the capital N with a small D below, the donor concentration; the wavy equals sign means approximately. Five times ten to the power twenty eight, over ten to the power seven, gives five times ten to the power twenty one per cubic metre, written m to the power minus three. Right? Now the holes. In pure silicon, n e equals n h equals n i, the intrinsic carrier concentration, which is one point five times ten to the power sixteen per cubic metre. So after doping, are there still that many holes? No, there are far fewer, and that is the trap. At equilibrium, n e times n h equals n i squared, for pure and doped silicon alike. Here n i is for S i, silicon, at three hundred kelvin, the capital K on screen. So raising the electrons lowers the holes. Clear? Now, n i squared is two point two five times ten to the power thirty two. Divide by five times ten to the power twenty one, and n h comes out as four point five times ten to the power ten per cubic metre. That is tiny compared with the electrons.

2:26Tool 2: the Zener regulator

Got it? Next, tool number two. Here, an eight volt Zener diode is connected to a twelve volt supply through one hundred ohm, the Greek capital omega, with no load. Find the Zener current and the power dissipated in the diode. Okay? Remember, a Zener regulator works in reverse bias, in the breakdown region. There, the voltage across the diode stays at V Z, the V with a small Z below, the Zener voltage, so eight volts, where capital V means volt. With no load, the twelve volt supply is more than eight volts, so the diode really is in breakdown. Right? The rest of the supply drops across R S, the R with a small S below, the series resistor. So the current I equals V in, the supply voltage, minus V Z, over R S, as on screen. Twelve minus eight, over one hundred, gives zero point zero four ampere, the capital A on screen. With me? With no load, all of this current flows through the Zener, so I Z, the Zener current, is forty milliamperes, the m A on screen. Clear? Now the power. Many students multiply the twelve supply volts by zero point zero four ampere. Is that the power in the diode? It is not, because only eight volts sit across the diode; the other four volts heat the resistor. So P Z, the power in the Zener, equals V Z times I Z, eight times zero point zero four, which gives zero point three two watt, the W on screen. Got it?

4:02Tool 3: NAND and NOR

Now, tool number three, from logic gates. For which input combinations do a two-input NAND gate and a two-input NOR gate give the same output? The inputs are A and B, and Y is the output. NAND is AND followed by NOT, so Y equals A dot B with a bar on top, and the bar means NOT. NOR is OR followed by NOT, so Y equals A plus B with a bar on top. Okay? There are four input pairs, written in brackets: zero zero, zero one, one zero and one one. Let us write both outputs for each pair. Right? NAND gives zero only when both inputs are one. So for the four pairs, NAND gives one, one, one and zero. Clear? Now it is your turn to try. NOR gives one only when both inputs are zero. So what are its four outputs? Yes, exactly, NOR gives one, zero, zero and zero. Now compare the two rows, place by place. With me? They match in the first place, where both give one, and in the last place, where both give zero. So the answer is the input pairs zero zero, and one one. Got it?

5:19Example 1: silicon diode with 40 ohm on a 1.5 V cell

Alright, example one, an easy one. Here, a silicon diode is forward biased, in series with forty ohm, across a one point five volt cell. Find the current. Pause the video and try it first. Done? Okay, let's check. Many students write one point five over forty, which gives thirty seven point five milliamperes. Is that right? It is not, and that is the trap. Remember, a conducting silicon diode keeps a fixed voltage across it, V k, the V with a small k below, which is zero point seven volt. Our cell gives one point five volts, more than that, so the diode does conduct. Okay? So only the rest of the voltage drives the resistor. I equals V minus V k, over R, with V the cell voltage and R the resistance, which is one point five minus zero point seven, over forty, so zero point eight over forty. Right? That gives zero point zero two ampere, which is twenty milliamperes, our final answer.

6:26Example 2: half-wave rectifier dc output from 6 V rms

Clear? Now our puzzle from the start, a slightly harder one. Here, a half-wave rectifier with an ideal diode is fed from a six volt r m s secondary, the output coil of a transformer. Find the d c, or direct current, output. Pause the video and try it first. Done? Okay, let's check. In the figure, the source with the wavy sign gives V m sine omega t, where V m is the peak and omega, the Greek letter, is the angular frequency. D is the diode, and R L, the box, is the load. On the right, the input is a full sine wave against time t, but the output across R L keeps only the positive halves, one pulse in each period, the capital T on screen. With me? Now tell me, is the peak, V m, equal to six volts? No, it is not, and that is the trap. The question says r m s, root mean square, and for a sine wave the peak is root two times the r m s value. So V m is six root two, about eight point four nine volts. Right? Only the positive halves get through, as the figure shows. So for a half-wave rectifier with an ideal diode, the d c output, V d c, the V with a small d c below, is V m over pi. Okay? So V d c is eight point four nine over pi, which gives about two point seven volts. So our puzzle's answer is only about two point seven volts, not six! Got it?

8:05Quick revision: carriers, diode drops, gates, rectifier

So, quick revision. First, at equilibrium, n e times n h equals n i squared, for pure and doped material alike. Second, a conducting silicon diode drops zero point seven volt, and the power in a Zener in breakdown is V Z times I Z, okay? Third, write the truth tables for NAND and NOR, and for a half-wave rectifier, V d c is V m over pi. You solved JEE Main level questions today, well done. Now try the practice problems on this page. All the best!