Full-Wave Rectifiers

Transcript

0:00Introduction

Hello everyone! Remember the half-wave rectifier? It simply throws away one half of every cycle of the mains. Today we rescue that half, and make both halves push current the same way through the load. Now a small puzzle: we feed this full-wave rectifier from fifty hertz mains. What is the frequency of the ripple in its output? Fifty hertz, surely, right? Hold on to your answer; we will check it soon. Rectifiers come up often in JEE, so let's begin.

0:32Centre-tap: two diodes take turns

So, here is the centre-tap full-wave rectifier. On the left is the secondary coil of a transformer, with ends A and B, and a centre tap in the middle, marked C on screen. Diodes D one and D two, the D with a small one or two below it, join the ends to the load, R L, the R with a small capital L below it. With me? Now, in the first half-cycle, end A is positive with respect to the centre tap. Tell me, which diode conducts, and which one stays off? Yes, exactly, D one conducts, because it is forward biased, while D two is reverse biased and stays off. In the next half-cycle it flips: end B is positive, so D two conducts and D one is off. Okay? And here is the beautiful part. In both halves, the current goes through R L in the same direction, shown by the red arrow. Now look at the graphs on the right, both plotted against time, t. The upper one is the input, and the lower one is the output across R L, the load. In the output, every half-cycle gives a hump above the axis, one from D one, the next from D two, as the labels show. Clear? Now back to our puzzle. The input is fifty hertz, the H z on screen, meaning cycles per second. So the output is also fifty hertz, isn't it? That is the trap, so let us count the humps. Below the output graph, the arrow marks T over two, where T is the time period of the input. So a hump comes every half period, two humps in every input cycle. The output frequency is twice the input: two times fifty gives one hundred hertz, the ripple frequency. Got it?

2:21The averages double (ideal diodes)

Now the numbers, taking the diodes as ideal, okay? V m, the V with a small m below it, is the peak voltage applied to the load in each half-cycle. For half-wave, the d c or average output was V m over pi, with pi the Greek letter, on the badge. Now there is a hump in every half-cycle. So what is the average now? Yes, exactly, it doubles. V d c, the V with small d c below it, equals two V m over pi. The same goes for the current I d c, which equals two times I sub m, over pi, where I sub m, the capital I with a small m, is the peak current. Right? Next, the r m s values, the small r m s meaning root mean square. V r m s equals V m over root two, with the root sign over the two, and the current I r m s equals I sub m over root two. These match the plain a c input, because flipping a hump does not change its square. With me? Now the efficiency, the d c power in the load divided by the a c power supplied. For full-wave it is eight over pi squared, which is about eighty one point one percent at most. Most books round this to eighty one point two percent, so you will often see that number too. Okay? And the ripple factor, which tells us how much ripple is left, is zero point four eight, against one point two one for half-wave. So the full-wave output is much smoother. Clear?

3:54The bridge: four diodes, no centre tap

Now the second design, the bridge rectifier. Four diodes, D one to D four, sit in a diamond, with no centre tap. The a c supply, the circle with a wave inside, goes across one diagonal, between A and B, and the load R L across the other, between P and Q, just as the note on the right says, okay? Now follow the arrows of the diodes. When terminal A is positive, the current goes from A through D two to Q, through R L from Q to P, and back through D three to terminal B, okay? So D two and D three conduct, as the first note on the right says. With me? In the other half, B is positive, so D four and D one conduct. Still the current goes through the load from Q to P, shown by the red arrow, with the plus sign at Q and the minus sign at P, as shown. That is what the note on the right says, the current in R L is always from Q to P, in both halves. Got it? So let me ask you this: in each half-cycle, how many diodes does the current pass through on its way round? Yes, exactly, two diodes, in series with the load. So with real diodes, the load loses two diode drops, not one. Remember that, okay?

5:13Peak inverse voltage: the classic trap

Now a point that comes up often in exams, the peak inverse voltage, or PIV for short. It is the largest reverse voltage across a diode while it is off, so the diode must be able to stand it. Clear? In the centre-tap circuit, V m is the peak of only half the secondary, from one end to the centre tap. So what is the PIV of each diode here? Just V m, like the half-wave rectifier, or more? The answer is two V m, because the diode that is off faces the whole secondary, both halves together. Many students write just V m here, so do not fall for it. Okay? In the bridge there is no centre tap, so V m is the peak of the whole secondary, and the PIV of each diode is just V m. With me?

6:02Example 1: Vdc of a full wave rectifier with peak 10 V

Now example one, an easy warm-up. The peak voltage across the load of an ideal full-wave rectifier is ten volts. Find the d c output voltage. Pause the video and try it first. Done? Okay, let's check. V d c equals two V m over pi, and V m is ten volts, so two times ten, over pi, that is twenty over pi. Okay? And twenty over pi gives six point three seven volts, the answer. Right? If you got three point one eight volts, you used ten over pi, the half-wave formula. In full-wave both halves count, so the average is double. Clear?

6:43Example 2: 12-0-12 V centre tap, Vdc and PIV of each diode

Now a harder one, with a trap inside. Here, a centre-tap rectifier uses a secondary of twelve, zero, twelve volts r m s. Find the d c output, and the PIV of each diode, taking the diodes as ideal. Pause the video and try it first. Done? Okay, let's check. Twelve, zero, twelve means twelve volts r m s from each end to the centre tap, and twenty four volts across the whole coil. So which peak do we use, twelve root two, or twenty four root two? We use twelve root two, because in a centre-tap circuit, V m is the peak of only half the secondary. Peak equals r m s times root two, so V m is twelve root two, about seventeen volts. Okay? Now V d c equals two V m over pi. Two times seventeen, over pi, gives ten point eight volts. Right? Then the PIV is two V m, so two times seventeen, that is thirty four volts in reverse on each diode. If you wrote seventeen volts, you used the bridge rule by mistake. Got it?

7:56Quick revision: frequency doubles, Vdc, PIV 2Vm vs Vm

So, quick revision. Both halves are used, so the output frequency is twice the input. V d c is two V m over pi, the efficiency is about eighty one percent at most, and the ripple factor is zero point four eight. And the PIV is two V m for centre-tap, but V m for the bridge. Now go and try the practice problems on this page. All the best!