Past JEE Questions: Rectifiers

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle to start. The mains supply in your home runs through fifty complete cycles every second, so its frequency is fifty hertz. Pass it through a full-wave rectifier, and the ripple on the output comes a hundred times every second. So who doubled the frequency, when nobody touched the supply? Today we solve the course's five past JEE questions on rectifiers and filters, and this puzzle is one of them. Questions like these do come up in JEE papers. So, let's begin.

0:34IIT-JEE 1996: whose humps are these?

So, the first question, from IITJEE nineteen ninety six. The output of a full-wave rectifier has four humps in a row, A, B, C and D, in that order. Which of them can come from diode one? On the right is the course's centre-tap rectifier: two diodes, D one and D two, the D with a small one or two below it, and the load, R L, the R with a small L below it. The letters on the far left only mark the transformer coil, not the humps. Okay? Now look at the graphs: the input sine wave on top, against time, t, and the output across R L below it. So tell me, can diode one give two humps in a row? No, it cannot, and here is why. Each diode conducts in alternate half-cycles: D one in one half-cycle, then D two in the next, while D one is reverse biased. Either way, the current in R L flows the same way, along the small orange arrow. Clear? So the humps go D one, D two, D one, D two, just as the output graph labels them. The double arrow below the first hump is T over two, half the input period, the time for one hump. With me? So diode one gives every second hump: humps A and C, or humps B and D, depending on which diode conducts first. Both pairs are possible. Got it?

2:01JEE Main 2021: capacitor or inductor?

Next, a statement question from JEE Main twenty twenty one. Statement one: to get steady d c from a full-wave rectifier, we can connect a capacitor across the output, parallel to the load, R L, the R with a small L below it. Statement two: we can connect an inductor in series with the load. Let us check both, okay? First, the problem. On the right, the dashed curve is the output with no filter: humps falling all the way to zero. That is pulsating d c, a steady part with ripple on it. Right? Now the circuit: after the rectifier box comes the capacitor, C, the two short parallel plates, right across the load, the small rectangle. So, does this capacitor smooth the output? It does smooth it, yes, exactly, if it is large. The capacitor charges up to the peak while the diode conducts, then discharges slowly through the load until the next peak tops it up. That is the orange curve, nearly steady d c, so statement one is true. With me? Now statement two. Many students remember only the capacitor, so they mark this one false. Is it really false? Statement two is also true, and that is the trap. An inductor, L, opposes every change of current. So it holds back the changing ripple, and passes the steady d c. Clear? So both statements are true: the capacitor goes across the load, and the inductor in series with it.

3:37JEE Main 2021: match the four stages

Okay? Now a matching question, also from JEE Main twenty twenty one. Four devices, labelled small a to d, and four jobs, labelled with small Roman numerals, one to four. With me? Here is a neat trick: follow the power from the mains. The course gives the four stages of a d c power supply in order: transformer, rectifier, filter, voltage regulator. Okay? So the a c mains comes in. Which device meets it first, and what does it do? It meets the transformer first, yes, exactly. The transformer steps the a c voltage up or down, so c matches one. Right? Next, the rectifier converts a c voltage into d c voltage, so letter a matches two. Got it? Then the filter removes the ripple, the job our capacitor and inductor just did, so d matches three. Clear? And last, the stabiliser keeps the output voltage constant, even when the input voltage or the load current changes. That is the regulator's job, so b matches four. With me? So the answer is letter a with two, b with four, c with one, and d with three, each dash joining a device to its job. Okay?

4:54Example 1: ripple frequency of full wave rectifier on 50 Hz

Now Example one, our puzzle from the start, from the AIEEE two thousand and five paper. In a full-wave rectifier working from fifty hertz mains, what is the fundamental frequency in the ripple? On screen, capital H and small z stand for hertz. Pause the video and try it first. Done? Okay, let's check. The supply is fifty hertz, so the ripple is fifty hertz too, right? No, no, careful, fifty hertz is the half-wave answer. In a full-wave rectifier, each half-cycle of the input gives one hump, first from D one, then from D two. Okay? So there are two humps in every cycle of the input, and the output repeats after every half period, T over two. Got it? So the ripple frequency is two times fifty, the cross meaning times, which gives one hundred hertz. And that solves our puzzle: nobody touched the supply, the second diode simply fills in the other half of every cycle. Clear?

5:57Example 2: where a bridge rectifier gives its dc output

And now Example two, a little harder, from IITJEE nineteen ninety one. Four diodes are joined in a ring, ABCD, as a full-wave bridge rectifier. The a c input goes across corners B and D, so across which points does the d c output appear? Pause the video and try it first. Done? Okay, let's check. Go round the ring: A, B, C, D, and back to A again. So B and D are not neighbours; they are opposite corners. Clear? Now look at the course's bridge on the right. Its corners have their own letters, so watch the shape, not the letters. The a c source, the circle with a wave, joins the top and bottom corners, marked A and B here, which is one diagonal. The four diodes, D one to D four, make the ring. Okay? Now tell me, where does the load, R L, sit in this bridge? Across the other diagonal, between P and Q, yes, exactly. In each half-cycle two diodes conduct in series, and the current in R L always flows from Q to P, the small orange arrow, so Q is plus and P is minus. With me? Back to our ring, then. The input is across B and D, so the d c output appears across the other diagonal, corners A and C, okay? One diagonal in, the other diagonal out. Remember that picture, and this question takes seconds. Got it?

7:33Quick revision: full wave ripple, filters, bridge

So, quick revision. In a full-wave rectifier the diodes take turns, so the ripple is at twice the mains frequency. To smooth the output, put a capacitor across the load, or an inductor in series with it. And in a bridge, a c goes across one diagonal, and d c comes out from the other. Now go and try the practice problems in this chapter. You can do them, I am sure. All the best!