IIT-JEE Questions: Junction Diode

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle from an old IITJEE paper. Take two identical p-n junction diodes, same everything, and connect them in series with a battery. Surely each diode takes half the voltage, right? Well, that is not always true. Connect them one way, and one diode takes nearly all the voltage! Today we solve the course's seven past JEE questions on the junction diode, and this puzzle is the last one. Questions on biasing come up often in JEE, so these are worth doing carefully. Let's begin.

0:37IIT-JEE 1998: a junction on its own

So, the first question, from IITJEE nineteen ninety eight. Here, the p-n junction diode is not connected to any circuit. Which statement is correct? Option A says the potential is the same everywhere, and option B says the p-side is higher. Options C and D say there is a field at the junction, from n to p, or from p to n. On the right is the course's picture: holes, the small open circles, on the p-side, and electrons, the black dots, on the n-side. Now, holes are positive, and they sit on the p-side. So the p-side must be at the higher potential, right? No, no, careful, that is exactly the trap. Holes near the junction diffused across and recombined, leaving uncovered acceptor ions, the circled minus signs, on the p-side part of the depletion region. In the same way, electrons left uncovered donor ions, the circled plus signs, on the n-side part. So the n-side part is positive. Okay? Now tell me, which way does the electric field at the junction point? It points from positive to negative, so from the n-side to the p-side, yes, exactly! That is the orange arrow, marked capital E for the field. So option C is correct, and option D is wrong. Got it? And look at the graph below the junction. The potential rises from the p-side to the n-side, by the barrier V nought, the V with a small zero below it. So the potential is not the same everywhere, and the p-side is lower, not higher. Options A and B are both wrong. With me?

2:09IIT-JEE 1988, 1990, 1997: biasing

Now we connect a battery, as in the figure on the right. The diode symbol is an arrow from p to n, and R is a resistor. IITJEE nineteen eighty eight asked where the p end goes in forward bias. Look at the left circuit: it goes to the positive terminal, the plus sign. Okay? The same question also asks this: in which direction does the current flow through the diode? From the p-side to the n-side, yes, exactly. That is the orange arrow, capital I, in milliamperes, the m A on screen. And the diode symbol points the same way, from p to n. Right? Next, IITJEE nineteen ninety. Which biasing offers a high resistance to the flow of current? Reverse bias, that is correct. The barrier rises, and only a tiny current flows, in microamperes, the mu A in the right circuit, where mu is a Greek letter. The second part asks about the connection: in reverse bias, the p-side goes to the negative terminal, the minus sign. Got it? Now IITJEE nineteen ninety seven, for a silicon junction. Which process mainly moves the charge carriers, diffusion or drift? In forward bias, and in reverse bias? In forward bias it is diffusion: the barrier is lower, so the majority carriers diffuse across. In reverse bias it is drift: the field at the junction sweeps the few minority carriers across. With me?

3:39IIT-JEE 1997: two diodes with resistance

Now a numerical one, from IITJEE nineteen ninety seven. Each diode has a forward resistance of fifty ohms, the capital Greek omega, and an infinite backward resistance. One diode, forward biased, is in series with one hundred and fifty ohms; the other, reverse biased, sits in a parallel branch. All this is in series with one hundred ohms, across a six volt battery, V for volts. So first, which of the two parallel branches carries current? Only the forward branch, yes, exactly. The reverse biased diode has infinite resistance, so its branch carries no current. It simply drops out of the circuit. Okay? So the total resistance is one hundred and fifty plus one hundred, which makes two hundred and fifty ohms, right? Oh no, careful, the forward diode is not a plain wire here. Its forward resistance of fifty ohms is part of the circuit, so we treat the diode as a fifty ohm resistor. Got it? So the current, capital I, is six over the total resistance, which is fifty plus one hundred and fifty plus one hundred. That is six over three hundred. Clear? And six over three hundred gives zero point zero two ampere, where capital A stands for ampere. That is the current through the one hundred ohm resistor. With me?

4:59Example 1: net flow of holes from n to p, which bias

Time for Example one, from IITJEE nineteen ninety three. In which bias of a p-n junction is the net flow of holes from the n-region to the p-region? Pause the video and try it first. Done? Okay, let's check. Many students say forward bias, because that is when a big current flows, right? Wait, careful, that is the trap here. In forward bias, the holes are majority carriers on the p-side, and they diffuse from p to n. That is the opposite direction. Okay? Now, the n-region does have a few holes, but there they are minority carriers. With me? In reverse bias, the field at the junction sweeps these minority holes across, by drift, from n to p. So the answer is reverse bias. Clear?

5:49Example 2: two identical junctions in series, equal drops

And now, our puzzle from the start, from IITJEE nineteen eighty nine. Two identical p-n junctions are connected in series with a battery, in three ways. In the first, one is forward and one is reverse biased. In the second, both are forward biased. In the third, both are reverse biased. In which arrangements are the potential drops across the two junctions equal? Pause the video and try it first. Done? Okay, let's check. The junctions are identical, so the drops are equal in all three arrangements, right? No, no, careful, identical junctions are not enough. Start with what is surely true: in series, the same current, capital I, flows through both junctions. Okay? The drop across each junction is that current times its resistance, where the cross sign means times. So equal drops need equal resistances. Right? In the second and third arrangements, both junctions are in the same state, both forward, or both reverse. Identical junctions in the same state have equal resistances, so the drops are equal. Got it? Now the first arrangement, with one forward and one reverse. Which junction takes more of the voltage? The reverse biased one, yes, exactly. The forward biased junction has a low resistance, and the reverse biased junction has a very high resistance. Clear? So the reverse biased junction takes nearly all the battery voltage, and the drops are far from equal. That is our puzzle, solved. So the answer is the second and the third arrangements. See how one idea, the same current in series, settled the whole question?

7:32Quick revision: unbiased junction, forward and reverse bias

So, quick revision. In an unbiased junction, the n-side is higher by the barrier V nought, and the field points from n to p. In forward bias, the p-side goes to the positive terminal, majority carriers diffuse, and the current flows from p to n. In reverse bias, the p-side goes to the negative terminal, minority carriers drift, and the resistance is very high. Now go and try the practice problems on this page. You can do them, I am sure. All the best!