Filters

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Take a full-wave rectifier on the fifty hertz mains. The current through the load flows in one direction only, so it is dc, right? But put its output on a screen, and it falls all the way to zero, one hundred times every second. You would not want to run a radio, or an amplifier, on that; you would hear a loud hum, right? Today we see how one capacitor turns those humps into nearly steady dc, and how to find the ripple that is left. Questions on ripple come up often in JEE, so let's begin.

0:37The capacitor filter: hold the voltage up

So, look at the graph on the right. The dashed curve is the output without a capacitor: a row of humps, each rising to a peak and falling back to zero. So the output is pulsating: a steady part, with a ripple riding on top. Okay? Now, a filter is a circuit that removes that ripple. So let me ask you this. Between two peaks, the rectifier output drops towards zero. What could hold the output voltage up during that gap? Something that stores charge, yes, exactly, a capacitor! Look at the circuit on the left. The capacitor, capital C, the two parallel lines, sits right across the load resistance, R L, the R with a small capital L below it. So the capacitor is in parallel with the load. Right? Now follow the solid curve. While the diode conducts, the capacitor charges right up to the peak voltage, V m, the V with a small m below it. Then the input falls, and the capacitor feeds the load itself, discharging slowly through R L, until the next peak tops it up again. That is why the label says nearly steady dc. With me? Now tell me, for the smoothest output, should this capacitor be large or small? It should be large, of course, and here is why. R L times C is the time constant of the discharge, and it must be much greater than T, the time period; the double greater-than sign means much greater than. Then the capacitor loses very little charge before the next peak, and the output stays close to V m, the peak. Clear?

2:13How big is the ripple?

Alright, now the most useful part: how big is the ripple? Between two peaks, the capacitor alone supplies the load current, capital I, for some time, small t. Okay? So how long is that gap between the peaks? Think about a half-wave rectifier first, and then a full-wave one. The half-wave output has one pulse per cycle, so the gap is capital T, a whole time period. The full-wave output has twice the input frequency, so the peaks come twice as often, and t is T over two. Right? Now, the charge lost is the current times the time, I times t. And charge divided by capacitance gives the voltage. So the ripple, peak to peak, is delta V equals I times t, over C, okay? The capital delta, the little triangle, means the change in V. Now put t equal to T, and remember that T is one over f, where f is the frequency of the mains. So for half-wave, delta V equals I, over f times C, got it? For full-wave, the time is half as long, so delta V equals I, over two times f times C, which is half the ripple for the same current and capacitor. Forgetting this two is a common slip, so first ask which rectifier it is. And to make the ripple smaller, use a bigger capacitor, or draw less current; this formula tells you both. Clear?

3:41Inductor filter and the complete power supply

Now, the capacitor is not the only filter. We can also put an inductor, capital L, in series with the load. Okay? Why would that help? Think about what an inductor does whenever the current through it tries to change. It opposes every change of current, yes! The steady dc does not change, so it passes through easily. But the ripple keeps changing, so the inductor blocks it. Right? Finally, the full picture. Here is a complete power supply that works from the mains, in order, following the arrows: the transformer, then the rectifier, then the filter, and then the voltage regulator. This order is worth remembering. With me?

4:25Example 1: full-wave ripple, 100 mA and 1000 microfarad

Now let's do Example one. Here, a full-wave rectifier on fifty hertz mains supplies one hundred milliamperes, the m A on screen, with a one thousand microfarad filter capacitor. The Greek letter mu with the F is the microfarad. Estimate the ripple. Pause the video and try it first. Done? Okay, let's check. It is full-wave, so we use delta V equals I, over two times f times C, okay? Now, the capacitor is one thousand microfarads. Can we just put one thousand straight in for C? No, no, careful, with the current in amperes, that would give a ripple of a millionth of a volt, far too good to be true! Change to S I units first. One hundred milliamperes is zero point one ampere, and one thousand microfarads is ten to the power minus three farad. Right? Now, two times fifty is one hundred, and one hundred times ten to the power minus three is zero point one. So delta V is zero point one over zero point one, which is one volt. And if you forgot the two, you would get two volts, which is the answer for a half-wave rectifier. Clear?

5:40Example 2: capacitor for 1% ripple with a 600 ohm load

And now Example two, a harder one. Here, a capacitor filter charges to twelve volts at each peak of a fifty hertz full-wave rectifier, and feeds a six hundred ohm load; the capital omega is the ohm. What capacitance keeps the fall in voltage between peaks within one percent? Pause the video and try it first. Done? Okay, let's check. One percent of what? Of the twelve volts, so surely we need the twelve volts in the working, right? Let us see. The load current is about V over R L; the wavy equals sign means approximately. The fall is I times T over two, divided by C, from our ripple formula. Divide by V, and V cancels, so delta V over V is about T over two, over R L times C, okay? So the twelve volts drops out completely, and only the fraction matters. With me? Now for the time between peaks. One over fifty is zero point zero two second, and since it is full-wave, the gap is half of that, zero point zero one second. Using the full period T here is the usual slip, and it doubles the answer. Okay? One percent is zero point zero one. So zero point zero one, over six hundred times C, must be less than or equal to zero point zero one; that is the less-than sign with a line under it. Right? C is in the denominator, so when we solve for it, the sign turns around. So C is greater than or equal to zero point zero one, over zero point zero one times six hundred, which is one over six hundred farad. Clear? One over six hundred is about one point six seven times ten to the power minus three farad. That is about one thousand seven hundred microfarads. Got it?

7:38Quick revision: ripple formulas, filters, power supply

So, quick revision. Capacitor filter: a large capacitor across the load, and the output stays near V m if R L times C is much greater than the time period. Ripple: delta V equals I, over f times C, for half-wave, and I, over two times f times C, for full-wave. An inductor in series passes the dc and blocks the ripple. And a power supply goes transformer, rectifier, filter, voltage regulator. Now go and try the practice problems on this page. All the best!