Diodes with Varying Inputs

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take a sine wave that swings up to plus twelve volts and down to minus twelve volts. Now you are given just a resistor, a diode and a five volt battery. Can you build a circuit whose output never goes above plus five volts, but still goes all the way down to minus twelve? Hold on to that question. Today the input keeps changing, so the diode keeps switching on and off, and that is the whole secret. Questions like this do appear in JEE, so let's begin.

0:35Diode in series: one polarity passes

So, first the simplest circuit. The diode, capital D, is in series with the load, R L, the R with a small L below it. Take it as an ideal diode, a perfect one-way switch, facing so that a positive input forward biases it. Now feed it a sine wave. Which half of the wave reaches the load? The positive half, yes, exactly. When the input is positive, the diode is forward biased, so it is on, and an ideal diode is just a wire. So the load gets the input. Right? In the negative half, the diode is reverse biased, so it is off, a gap. No current flows, and the load gets zero. With me? So one diode in series with the load passes only one polarity of the input. Keep this picture in mind, okay?

1:23Diode across the output: clipping

Now let us move the diode. Put a resistor, capital R, in series with the input, and connect the diode across the output, between the two terminals where we read the output voltage. When this diode conducts, it acts like a wire joining those two terminals. So what is the output voltage then? It is held near zero, that is correct. The conducting diode shorts the output, so the input voltage now drops across the resistor instead. Right? And when the diode is off, it is a gap. No current flows through the resistor, so there is no drop across it, and the output equals the input. Got it? So whenever the diode conducts, that part of the wave is cut away, flat at zero. We say that part of the wave is clipped. Nice and simple, okay?

2:11A battery moves the clipping level

Now one small change. Put a battery, V B, a capital V with a small B below it, in series with the diode, facing so that it opposes conduction. So now the diode has to beat the battery first. Tell me, when can it switch on? Yes, exactly, it switches on only when the output tries to rise above V B, the battery's voltage. Below that, the battery keeps the diode off. Right? So with an ideal diode, the wave is now clipped at V B, instead of at zero. The battery has moved the clipping level. Clear? But real diodes are not ideal. If the diode is made of silicon, is the clipping level still exactly V B? Not exactly, so be careful here. For a real diode, add the diode drop, which is about zero point seven volts for silicon, the S i on screen. So the level becomes V B plus about zero point seven volts. Got it?

3:09Any input: find the switching instants

Now here is one method that works for any input, a sine wave, a square wave or a triangle. First, mark the instants at which the diode switches on or off. Okay? When it is on, treat it as a wire, or as a fixed drop for a real diode. When it is off, it is a gap. So one changing input becomes a few simple steady circuits. That is the whole trick. Right?

3:33Capacitor with a diode: clamping

One last idea, and now we bring in a capacitor. Together with a diode, a capacitor can hold the peak of the input. Clear? Or, it can shift the whole waveform up or down. This is called clamping. With me? These two words sound alike, so students often mix them up. So tell me, clipping or clamping, which one changes the shape of the wave? Clipping changes the shape, yes, exactly. Clipping cuts part of the wave away, while clamping keeps the whole wave and only shifts it up or down. Got it?

4:09Example 1: 12 V sine wave clipped at 5 V by a diode

Now, example one. Here, a sine wave of peak twelve volts is applied through a resistor to an ideal diode connected across the output, with its p-side towards the resistor. And a five volt battery is in series with the diode, opposing its conduction. Between what limits does the output vary? Look, this is exactly our opening puzzle. Pause the video and try it first. Done? Okay, let's check. The p-side faces the resistor, so the diode wants to conduct when the output is positive, but the five volt battery pushes back. So the diode conducts only while the input exceeds five volts. Now, without the battery, this circuit cut the positive half flat at zero. So is the top cut at zero again? Not at zero this time, and that is the trap. Until the input passes five volts, the diode is off, and the output rises with the input. Above five volts, the diode conducts, and the battery holds the output flat at plus five volts. Right? For the rest of the cycle, the input is below five volts, so the diode is off. No current flows through the resistor, so the output equals the input, all the way down to minus twelve volts. With me? So the output varies from minus twelve volts to plus five volts. And that solves our opening puzzle. The battery puts a ceiling at five volts, and everything below it passes untouched. Clear?

5:40Example 2: triangle wave, peak load voltage, conduction time

Now example two, a little harder. This time, a triangular wave of peak six volts, symmetrical about zero, is applied to an ideal diode in series with a load, and with a two volt battery that opposes conduction. Find the peak load voltage, and the fraction of each cycle for which the diode conducts. Pause the video and try it first. Done? Okay, let's check. First, use our method and find when the diode switches on. The battery pushes back with two volts, so the diode is on only while v, the input voltage, is greater than two volts, as the greater-than sign shows. Okay? At the very top, the input is six volts. The ideal diode takes nothing, and the battery takes two volts. So the peak load voltage is six minus two, which gives four volts. Right? For what fraction of each cycle does the diode conduct? Here is a hint: a triangle rises and falls in straight lines. Look at the positive half first. The input climbs in a straight line from zero to six volts, and comes back down the same way. Because the lines are straight, the time is in proportion to the voltage covered. So the input stays above two volts for six minus two, over six, which equals two-thirds of the positive half. With me? So the diode conducts for two-thirds of the cycle, right? That is the trap, so be careful. Two-thirds is the fraction of the positive half only, and in the negative half the diode never conducts. The positive half is one half of the cycle, so two-thirds times one half gives one-third. Got it?

7:22Quick revision: series diode, clipping and clamping

So, quick revision. One diode in series with the load passes only one polarity, and for any input, first find the instants when it switches. Put the diode across the output, and it clips the wave, and a battery V B in series with it moves the clipping level. And a capacitor with a diode gives clamping, which shifts the whole wave up or down. Now go and try the practice problems on this page. All the best!