AIEEE and Earlier JEE Main Questions: Junction Diode

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Take a germanium diode and a silicon diode, side by side in parallel, both forward biased. The pair is in series with a resistor, across a twelve volt battery. Now we turn the germanium diode around, so it is reverse biased. Surely a reversed diode blocks the current, so the voltage across the resistor falls to zero, right? Hold on to that answer; it is a JEE Main question from twenty nineteen, and we solve it today. In this video we go through A I triple E and earlier JEE Main questions on the junction diode. So, let's begin.

0:41Idea 1: forward bias means the p-side is higher

So, idea number one: which way is forward bias? Look at the left circuit. The triangle with a bar is the diode, with p on the triangle side and n at the bar. In forward bias, the p-side goes to the plus terminal of the battery, so the p-side is at the higher potential. Okay? In the right circuit the battery is turned around. The p-side goes to the minus terminal, so it is at the lower potential, and that is reverse bias. Right? Now a question from the A I triple E two thousand four paper. When a junction diode is forward biased, what happens to the depletion region, and to the barrier height? Let us see. The applied voltage, capital V, opposes the barrier potential, V nought, the V with a small zero below it. So the barrier falls to V nought minus V, and the depletion region becomes thinner. With me? So the answer is that both are reduced. And in reverse bias it is the other way round: the barrier rises to V nought plus V, and the depletion region gets wider. Clear?

1:46Idea 2: a real diode keeps its own drop

Now idea number two, the real diode. On this characteristic, the current, capital I, stays tiny until the knee, and then rises steeply. So once a real diode conducts, it keeps a fixed voltage across it, V k, the V with a small k below it, for knee. It is zero point seven volt for silicon, S i, and zero point three volt for germanium, G e. The orange part on the left is reverse bias, where only a tiny current flows, in microamperes, the mu A, until breakdown at minus V b r. We will not need that part today. Okay? So, with a battery of voltage V and a resistor R in series, take away the drop first, and use Ohm's law on the rest. I equals V minus V k, all over R, okay? We always put a resistor in series, to limit the current. If the current must not go above I max, then R min, the least resistance, equals V minus V k, over I max. The small words below R and I just mean minimum and maximum. Got it? Now a JEE Main twenty twenty question. Here, a diode has a forward drop of zero point five volt, and its safe current is ten milliamperes, the m A on screen. It is used with a battery of e m f one point five volts. Find the least resistance to put in series with it. One point five minus zero point five is one volt, and ten milliamperes is zero point zero one ampere. One over zero point zero one gives one hundred ohms; the capital omega is the ohm. And if you got one hundred and fifty ohms, you forgot to take away the diode's drop. With me?

3:23Idea 3: an ideal diode is a one-way switch

Now idea number three. An ideal diode is a perfect one-way switch. Forward biased, it is a closed switch, a plain wire with no drop. Reverse biased, it is an open switch, a gap. Okay? Now the A I triple E two thousand six paper. Here, a twelve volt battery is connected through a four ohm resistor to two parallel branches. One has an ideal diode, D one, the D with a small one below it, with three ohms. The other has D two, connected the opposite way, with two ohms. D one is reverse biased. Find the current in the circuit. D one is reverse biased, so it is a gap, and its branch carries no current at all. And D two points the other way, so it is forward biased, a plain wire. Right? So only four ohms and two ohms are in the circuit, in series. Twelve over six gives two amperes, the capital A on screen. Adding all three resistors is the usual mistake here. Clear?

4:27Example 1: ammeter reading with a silicon diode, 3 V

Now, Example one, from JEE Main twenty eighteen. Here, a silicon diode is connected in the forward direction, in series with a two hundred ohm resistor and an ammeter, across a three volt battery. It is just like the left circuit in the figure, with an ammeter added. Find the ammeter reading. Pause the video and try it first. Done? Okay, let's check. The quick answer is three over two hundred, which gives fifteen milliamperes, right? No, no, wait, that treats the diode as a plain wire. It is a silicon diode, so it keeps zero point seven volt across it. Three volts is greater than zero point seven, as the sign between them says, so it does conduct. Okay? So the current is three minus zero point seven, over two hundred, which is two point three over two hundred amperes. Right? Two point three over two hundred gives zero point zero one one five ampere, which is eleven point five milliamperes. So the ammeter reads eleven point five milliamperes, not fifteen. Got it?

5:34Example 2: Ge and Si diodes in parallel, change in output

And now Example two, from JEE Main twenty nineteen, and yes, it is our opening puzzle. Germanium and silicon diodes start conducting at zero point three volt and zero point seven volt. They are in parallel, both forward biased, in series with a five kilo-ohm resistor across twelve volts. The output, V nought, is taken across the resistor; here V nought is just the name of the output, not the barrier. By how much does V nought change if the germanium diode is reversed? Pause the video and try it first. Done? Okay, let's check. First, before anything is reversed. Both diodes are forward biased, so which one conducts? Germanium does, because it switches on first, at zero point three volt. Both diodes sit across the same two points, so the pair is held at zero point three volt, and silicon, which needs zero point seven, stays off. With me? The rest of the twelve volts sits across the resistor. Twelve minus zero point three gives eleven point seven volts. Okay? Now turn the germanium diode around, just as in our puzzle. Does the output fall to zero? That is the trap, because the reversed germanium diode is only a gap, while silicon is still forward biased, and now it conducts at zero point seven volt. Right? So now the resistor gets twelve minus zero point seven, which gives eleven point three volts. Clear? So the change is eleven point seven minus eleven point three, which is zero point four volt. Only zero point four volt, and no fall to zero at all. Reversing one diode just hands the job to the other one, and notice that we never even needed the five kilo-ohms. Got it?

7:19Quick revision: bias, knee voltage, ideal and real diodes

So, let us do a quick revision. Forward bias means the p-side is at the higher potential; the barrier falls to V nought minus V, and the depletion region gets thinner. For a real diode, take away the drop first: I equals V minus V k, over R, with zero point seven volt for silicon and zero point three volt for germanium. An ideal diode is a wire or a gap, and with real diodes in parallel, the one with the lower knee voltage conducts first. Now go through the other past questions on this page, and try each one yourself first. All the best!