The Half-Wave Rectifier

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. The diode lets current through one way only, right? Feed it an alternating voltage of peak ten volts, and it passes one half of every cycle. So what is the average output voltage? Half of ten, so five volts? Many students say five, but that is not the answer, and in a few minutes you will see exactly where that five belongs. This circuit, the half-wave rectifier, comes up often in JEE, so let's begin.

0:33One diode, one half of every cycle

So, first, the big idea. Alternating current, written a c on screen, keeps reversing its direction. Converting it into current in one direction only is called rectification, and the little arrow means becomes. The junction diode does it, because it conducts one way only. Okay? Now look at the circuit on the left. The source, the circle with a wavy line, gives V m sine omega t. V m, the V with a small m below it, is the peak voltage, omega, the Greek letter, is the angular frequency, and t is time. Then a single diode, capital D, in series with the load, R L, the R with a small L below it. Right? Take the diode as ideal, a one-way switch. Its arrow points along the top wire, towards the load. So tell me, in which half of each cycle does it conduct? Yes, exactly, it conducts in the positive half. It is forward biased, a closed switch, so the output across R L follows the input. Clear? In the negative half, the diode is reverse biased, an open switch, so the output is zero. Now see the graphs on the right. The blue input swings up and down, but the orange output keeps only the positive humps. With me? The capital T with the double arrow, below the graph, is one period. How many orange pulses do you see in one period? That is correct, just one pulse in each cycle. So the output has the same frequency as the input: fifty hertz in, fifty hertz out. Do not double it, okay?

2:13Average and rms of the output

Alright, now the numbers. Small v is the input voltage at any instant. When the diode conducts, the peak current is capital I with a small m below it, equal to V m over R L, clear? Now, back to our puzzle. The output is on for half of each cycle, so is the average simply V m over two? It is not, and that is the trap, because the average is smaller. Each hump is a rounded curve, not a flat block. Over the full cycle, the average, or d c value, is V d c, with d c written small below the V, equal to V m over pi, the Greek letter pi. And the average current, capital I with d c below it, is the peak current over pi. Right? So for our ten volt peak, V d c is ten over pi, which gives three point one eight volts, not five. Keep that five in mind, it comes back in a moment. This is the r m s value, the root mean square. V r m s, with r m s written small below the V, equals V m over two, and the r m s current is the peak current over two. So a ten volt peak gives five volts r m s: that is where our five was hiding. Never mix up V d c and V r m s, okay? Now the diode itself. In the negative half it is off, so no current flows, and R L has no drop across it. The whole input then sits across the diode, in reverse. The largest such voltage, the peak inverse voltage or PIV, equals V m. Got it? And a real diode needs its knee voltage, V k, the V with a small k below it, before it conducts. For silicon, the S i on screen, it is about zero point seven volts. So the peak output is V m minus V k. Clear?

4:07Efficiency and ripple factor

Now let us ask how good this rectifier is. Efficiency is the d c power in the load, divided by the a c power supplied. Okay? Half of every cycle is thrown away. So tell me, can this efficiency ever reach one hundred percent? It cannot, and here is why. Both powers go into the same load, so the ratio is the peak current over pi, the whole squared, over the peak current over two, the whole squared. That gives four over pi squared, which is forty point six percent at most. Right? With a diode of forward resistance r f, the small r with a small f below it, the efficiency becomes forty point six percent times R L over R L plus r f, a little less. With me? Last, the ripple factor: the square root of, V r m s over V d c, the whole squared, minus one. For the half-wave rectifier, it comes to one point two one. Clear? Since that is above one, the output is a poor d c, far from steady. Got it?

5:13Example 1: 220 V rms supply, find dc output and PIV

Now the first example. Two hundred and twenty volts r m s is applied directly to an ideal half-wave rectifier. Find the d c output voltage and the peak inverse voltage on the diode. Pause the video and try it first. Done? Okay, let's check. Many students write two hundred and twenty over pi, about seventy volts. Is that right? It is not, so be careful here. Two hundred and twenty volts is the r m s value, but our formulas need the peak, V m. Okay? For a sine wave, the peak is the r m s value times root two, the square root sign. So V m is two hundred and twenty times root two, which is three hundred and eleven volts. Right? Then V d c is V m over pi, and three hundred and eleven over pi gives ninety nine volts, not seventy. Clear? And the peak inverse voltage is just V m, three hundred and eleven volts. So the diode must stand three hundred and eleven volts in reverse, much more than the d c output. Got it?

6:23Example 2: Si diode, 1.4 V peak, fraction of cycle on

Now a harder one. An input of peak one point four volts goes to a half-wave rectifier with a silicon diode of knee voltage zero point seven volts. For what fraction of each cycle does the diode conduct? Pause the video and try it first. Done? Okay, let's check. An ideal diode would conduct for half of each cycle. Is it still half here? It is not, and that is the trap. Let theta, the Greek letter, be the phase angle. The diode conducts only while the input exceeds the knee voltage: one point four sine theta greater than zero point seven. So sine theta must be more than one half. Okay? The little circle after each number means degrees. Sine theta equals one half at thirty degrees and at one hundred and fifty degrees, and is bigger in between. So the diode conducts for one hundred and twenty degrees out of three hundred and sixty. Right? So the fraction is one hundred and twenty over three hundred and sixty, which is one-third, not a half. For small inputs, the knee voltage matters. Got it?

7:36Quick revision: Vdc = Vm/pi, Vrms = Vm/2, PIV = Vm, 40.6%

So, quick revision. One diode gives one pulse per cycle, at the input frequency. V d c is V m over pi, V r m s is V m over two, and the peak inverse voltage equals V m. The efficiency is at most forty point six percent, and the ripple factor is one point two one. Now try the practice problems on this page. All the best!