Biasing and the Characteristic

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take a battery, a resistor and a p-n junction diode. Connect it one way, and a current of milliamperes flows. Now just flip the battery around. The current falls to microamperes, almost nothing! Same battery, same diode. So why does the diode care which way the battery faces? By the end of this video, you will know why. And in JEE, questions on this come up often. So, let's begin.

0:32Forward bias: p-side to the positive terminal

So, first the diode symbol, which you can see in both circuits. It is an arrow from p, the anode, to n, the cathode. The symbol shows the direction of conventional current in forward bias, just like the orange arrow in the left circuit. Okay? Now remember the junction before we connect any battery. The n-side sits at a higher potential than the p-side, by the barrier potential V nought, the V with a small zero below it. Right? Now the left circuit. The p-side is joined to the positive terminal, the plus sign, and the n-side to the negative terminal, the minus sign. This is called forward bias. So the battery raises the potential of the p-side. Tell me, does the barrier go up, or down? It goes down, yes, exactly. The applied voltage V opposes the barrier. So the barrier falls to V nought minus V, and the depletion region becomes thinner. Clear? With a lower barrier, the majority carriers can cross: holes from the p-side, and electrons from the n-side. They cross by diffusion, and that is the forward current. Got it? But the current becomes large, in milliamperes, the m A on screen, only when V is greater than the knee voltage; that is the greater-than sign. The knee voltage is also called the threshold voltage, or the cut-in voltage. With me?

1:52Reverse bias: p-side to the negative terminal

Now the right circuit. We flip the battery, so the p-side goes to the negative terminal, and the n-side to the positive terminal. This is reverse bias. Okay? Now the battery pulls the p-side down even further. So what happens to the barrier this time? It rises, that is correct. The battery adds to the barrier, so it becomes V nought plus V, and the depletion region becomes wider. Now the majority carriers cannot climb over it, right? So who carries the current now? Only the minority carriers, the electrons on the p-side and the holes on the n-side. The field at the junction sweeps them across, by drift. Got it? This drift current is tiny, in microamperes, the Greek letter mu before the capital A on screen. We call it the reverse saturation current. Now tell me, if I double the reverse voltage, does this current also double? The answer is no, it hardly changes; it is almost independent of the voltage. The number of minority carriers is fixed by thermal generation, and even a small reverse voltage sweeps all of them across. Clear? But keep increasing the reverse voltage. At a large value, V b r, the V with a small b r below it, the current suddenly rises sharply. This is called breakdown. With me?

3:14The I–V characteristic

Now let us put all of this on one graph, current against voltage. In forward bias, the dark blue curve, the current stays small at first, and then rises steeply after the knee. Okay? For silicon, S i on screen, the knee is at about zero point seven volts, and for germanium, G e, at about zero point three volts. The capital V after each number stands for volts. Right? In reverse bias, the orange curve, the current is small and almost flat, until minus V b r, the breakdown voltage. There the curve plunges steeply downward, which means the reverse current grows very large. Now look at the labels on the vertical axis. Are the units the same above and below it? They are not the same, so be careful here. Above the axis the current is in milliamperes, and below it, in microamperes, a thousand times smaller. So never read both branches on one scale; that is a common trap. Got it?

4:12Resistance of a diode

Look at this curve once more. Does a diode obey Ohm's law? No, a diode does not obey Ohm's law. For Ohm's law, the graph must be a straight line through the origin, and this graph clearly is not. So the resistance depends on where you are on the curve. Clear? So we define two resistances at a point. First, the static, or D C, resistance: capital R equals V over I, the voltage over the current at that point. Okay? Second, the dynamic, or A C, resistance, r d, a small r with a small d below it. It equals delta V over delta I, where delta, the Greek capital delta, means a small change. So it is the change in voltage over the change in current. With me? In forward bias, r d is just a few ohms, since a small change in voltage gives a big change in current. In reverse bias, it is many megohms, millions of ohms, because the current hardly changes. Got it?

5:12Example 1: barrier height at 0.5 V forward and 2 V reverse

Now, example one. Here, a junction has V nought equal to zero point seven volts. It is given a forward bias of zero point five volts, and then a reverse bias of two volts. Find the barrier height in each case. Pause the video and try it first. Done? Okay, let's check. Forward bias fights the barrier, so we subtract. Zero point seven minus zero point five gives zero point two volts. Right? Now the reverse bias. Some students write two minus zero point seven, which gives one point three volts. Is that correct? That is the trap, so be careful. In reverse bias, the battery adds to the barrier. Zero point seven plus two gives two point seven volts. And this connects to our opening puzzle. In forward bias the barrier dropped to zero point two volts, and in reverse bias it rose to two point seven volts. That tall barrier is why the current almost stops. Clear?

6:15Example 2: static and dynamic resistance from I-V data

Now example two, a little harder. This time, a diode carries ten milliamperes at zero point seven volts, and twenty milliamperes at zero point eight volts. Find its static resistance at zero point seven volts, and its dynamic resistance in this range. Pause the video and try it first. Done? Okay, let's check. Static resistance is simply V over I at that point. Ten milliamperes is ten times ten to the power minus three amperes, so zero point seven over that gives seventy ohms, written with the Greek capital omega. Okay? Now the dynamic resistance. Many students take the second point, zero point eight volts over twenty milliamperes, which gives forty ohms. Is that the dynamic resistance? That is just another static resistance, so be careful. Dynamic means we use the changes. The change in voltage, delta V, is zero point eight minus zero point seven, which is zero point one volts. The change in current, delta I, is twenty minus ten, which gives ten milliamperes. With me? So r d equals zero point one, over ten times ten to the power minus three, which gives ten ohms. Look, the same diode has seventy ohms static resistance and ten ohms dynamic resistance. Both are correct, because a diode does not follow Ohm's law. Got it?

7:37Quick revision: V0 - V vs V0 + V, R = V/I, rd = dV/dI

So, quick revision. Forward bias, p to positive: the barrier falls to V nought minus V, the depletion region gets thinner, and a large current flows. Reverse bias, p to negative: the barrier rises to V nought plus V, the depletion region gets wider, and only a tiny current flows. And static resistance is V over I, while dynamic resistance is delta V over delta I, okay? Now go and try the practice problems on this page. All the best!