Set 3

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take a bridge rectifier made of silicon diodes, with a peak input of twelve volts. The peak output is also twelve volts, right? Work it out, and you get only ten point six volts. So where did one point four volts go? Hold on to your answer. This is Set three of the mixed problems on electronic devices, and problems like these come up often in JEE Main. Let's begin.

0:31Tool 1: a conducting diode keeps its drop

Let us start with tool one. The figure shows the current against the voltage for a silicon diode: in forward bias, almost no current flows until the knee, and then it rises steeply. On the left, the orange part, in reverse bias only a tiny current in micro-amperes flows, until breakdown. In our simple model, a conducting diode keeps a fixed voltage V k, the V with a small k below, the knee voltage. That is zero point seven volt for silicon, the S i, and zero point three volt for germanium, the G e, with capital V meaning volt. Okay? With a resistor R in series across a battery of voltage V, only the rest drives the current. So I equals V minus V k, all over R, as on screen. And the diode conducts only if V is more than V k, the greater-than sign. With me? Now a problem from this set. Here, a silicon diode and a germanium diode are in series, both forward biased, with one kilo-ohm, the k and the Greek capital omega, across six volts. Six volts over one thousand ohms gives six milliamperes, right? Be careful, because each diode keeps its own drop. Six, minus zero point seven, minus zero point three, leaves five volts for the resistor, and six volts is well above the total drop of one volt, so both diodes do conduct. Five over one thousand is zero point zero zero five ampere, the capital A, which is five milliamperes, the m A on screen. Clear?

2:04Tool 2: the bridge rectifier

Now tool two, the bridge rectifier. It has four diodes, D one to D four, the D with a small number below. The a c supply, the circle with a wavy line, is across the diagonal A B, and the load R L, the R with a small L below, sits across the other diagonal, between P and Q, in the middle. So when A is positive, how many diodes does the current pass through? It passes through two diodes, exactly. When A is positive, D two and D three conduct, and when B is positive, D four and D one conduct. Either way, two diodes conduct in series, and the current in R L always flows from Q to P, the orange arrow, so Q is the plus end and P is the minus end. Okay? Now back to our puzzle. Each conducting silicon diode keeps zero point seven volt, so two of them take two times zero point seven, which is one point four volts. Twelve minus one point four gives a peak output of ten point six volts. That is where the one point four volts went. Got it? Now an assertion and a reason from this set. The assertion says the output of a full-wave rectifier is easier to filter than that of a half-wave rectifier. Okay? The reason says its ripple has twice the frequency, since both halves of every cycle send current through the load, and the ripple is smaller compared with the d c. So, are both true, and does the reason explain the assertion? Both are true, and the reason explains it. With a capacitor filter, the capacitor feeds the load for only half a period of the a c between peaks, not a full period, so less ripple is left. Clear?

3:46Tool 3: a truth table from an expression

Now tool three, logic gates. The figure shows the five basic gates with their symbols, Boolean expressions and truth tables, with A and B the inputs, and Y the output. The plus sign means OR, the dot means AND, and the bar on top means NOT, and so does the small circle at the output of the NAND, NOR and NOT symbols. With me? Now a problem from this set. Here, a circuit gives Y equals A B, plus A bar B bar, and letters side by side also mean AND. Find Y for all four input combinations, and name the function. Okay? So, let me ask you this. When the inputs A and B are both zero, what is the output Y? The output is one. The term A B is zero, but A bar and B bar are both one, so the second term is one dot one, which is one. Zero plus one gives one. Right? The rest work the same way. On screen, zero one means A is zero and B is one. When the two inputs differ, both terms are zero, so Y is zero. And when both inputs are one, the term A B is one, so Y is one. Got it? So the outputs are one, zero, zero and one. The output is one when the inputs are the same, and that is the exclusive NOR gate. Clear?

5:11Example 1: series resistor for a 3 V blue LED on 12 V

Now, Example one, the easier one. Here it is: a blue LED drops three volts at fifteen milliamperes. Find the series resistance needed on a twelve volt supply. Pause the video and try it first. Done? Okay, let's check. Remember tool one, because an LED is also a diode in forward bias. So, of the twelve volts, how much is left for the resistor? The resistor gets only what the LED does not keep. It keeps V F, the V with a small capital F below, its forward voltage, which is three volts here. So R equals V minus V F, over the current I, as on screen. With me? Twelve minus three leaves nine volts. And fifteen milliamperes is fifteen times ten to the power minus three ampere, which is zero point zero one five ampere. Okay? Nine over zero point zero one five gives six hundred ohms. So always take away the LED voltage first. Got it?

6:15Example 2: peak and rms current of a full-wave rectifier

Now Example two, a little harder. Here it is: an ideal full-wave rectifier delivers one ampere of d c load current. Find the peak and r m s values of the load current. Pause the video and try it first. Done? Okay, let's check. Here, I sub m, the I with a small m below, is the peak current, and I d c, with a small d c below, is the d c current. So I sub m equals pi, the Greek letter, times I d c, and with the wavy equals sign meaning about, that is about three point one four amperes, right? Oh no, careful, because that is the half-wave formula. In a full-wave rectifier, both halves of each cycle send current through the load, so I d c equals two I sub m, over pi. And I r m s, with a small r m s below, equals I sub m over root two, the square root sign on screen. Okay? So I sub m equals pi over two, times I d c. Pi over two, times one ampere, gives about one point five seven amperes. That is the peak. With me? And I r m s is one point five seven over root two, which gives about one point one one amperes. Both values are more than the d c current of one ampere. Clear?

7:33Quick revision: diode drops, bridge, rms current, XNOR

So, quick revision. Each conducting diode keeps its drop, zero point seven volt for silicon and zero point three volt for germanium, and an LED needs a series resistor of V minus V F, over I, as on screen. In a bridge, two diodes conduct in each half-cycle, and the output frequency doubles. For an ideal full-wave rectifier, I d c is two I sub m over pi, and I r m s is I sub m over root two. And the exclusive NOR gives one when the inputs are the same. Now try the rest of this set on this page, and all the best!