Circuits with Ideal Diodes

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Think of a metro exit gate. It lets you walk out, but it will not let you walk back in, right? An ideal diode treats current the same way, one direction only. Now, there is a diode in a circuit, and no current flows through it at all. So the voltage across it must be zero too, right? Well, it is not that simple. Today we settle that question, and learn a four-step method for any circuit with ideal diodes. Diode circuits come up often in JEE, so, let's begin.

0:34The ideal diode: a one-way switch

So, look at the figure. Each circuit has a battery with its plus and minus terminals, a resistor, the rectangle labelled capital R, and a diode. The diode symbol is an arrow from p, the anode, to n, the cathode, with a bar on the n side. The arrow shows the direction of conventional current in forward bias, okay? And the text below each circuit recalls the last section: the barrier and the depletion width decrease in forward bias, with a diffusion current, and increase in reverse bias, with only a drift current. Now, in the left circuit, the p-side is joined to the plus terminal, and a current, capital I, flows, shown by the orange-red arrow. It is a few milliamperes, which is what the m A in brackets means. So tell me, when exactly do we call a diode forward biased? Yes, exactly, when the p-side is at the higher potential. Notice the word higher, and not positive. That matters in a minute. Right? Now, the course treats the ideal diode as a perfect one-way switch. When it is forward biased, it is a closed switch, with zero resistance and zero voltage drop. V D, the V with a small D written below it, is the voltage across the diode, and here it is zero. Clear? Now look at the right circuit. The p-side is joined to the minus terminal, so this is reverse bias. What does our ideal switch become now? It becomes an open switch, that is correct, with infinite resistance, so no current flows, and capital I equals zero. In the figure, a real diode still lets a tiny current through, in microamperes, the mu A in brackets, where mu is a Greek letter. But the ideal diode takes even that as zero. Got it?

2:25Higher potential, not positive

Now a quick check, taken from the course's own examples. The p-side is at minus five volts, and the n-side is at minus eight volts. Both are negative, okay? So, is this diode forward biased, or reverse biased? Many students say reverse, because of the minus signs, but careful, that is the trap. Minus five is greater than minus eight, and that is what the greater-than sign on screen says. So the p-side is higher, and the diode is forward biased. Only which side is higher matters, never the sign alone. With me?

3:00The method: assume, replace, solve, check

So, how do we solve any circuit with ideal diodes? The course gives a neat four-step method. Step one: assume a state for each diode, forward or reverse. Okay? Step two: replace each forward diode by a plain wire, and each reverse diode by a gap, a break in the circuit. Now there is no diode left at all, only batteries and resistors. Right? Step three: solve the circuit as usual, with Ohm's law, and resistors in series and in parallel. Now tell me, once you have an answer, is the job done? The answer is no, not yet, because step four is the check. Every closed diode must carry its current from p to n, in the direction of its arrow. Clear? And across every open diode, the p-side must be lower than the n-side. If both checks pass, your assumption was right. If a check fails, change the state of that diode and solve again. Got it?

3:58Example 1: diode in series, forward vs reverse bias

Alright, here is Example one. There is a ten volt battery, a one kilo-ohm resistor and an ideal diode in series, just like the circuit in our figure. The k with the Greek capital omega means kilo-ohm, okay? Find the current, and the voltage across the diode, first in forward bias, and then in reverse bias. Pause the video and try it first. Are you done? Okay, let us check. Take forward bias first. By step two, the diode becomes a wire, so V D, the voltage across it, is zero. Right? So only the resistor is left in the loop. The current, capital I, equals ten volts over one kilo-ohm, which gives ten milliamperes, the m A on screen. With me? Now take reverse bias. The diode becomes a gap, so no current flows anywhere in the loop, and capital I equals zero. Okay? And now, here is our puzzle from the start. No current flows through the diode, so the voltage across the diode is zero too, right? That is exactly the trap, so be careful here. The resistor carries no current, so V R, the voltage across the resistor, equals I times R, which is zero. But the battery still gives ten volts, so all ten volts sit across the diode, and V D is ten volts. Clear? This is the course's key rule, and it settles our puzzle. Across an open diode, the whole voltage of that branch appears, even though no current flows. So the diode has zero current, and still ten volts across it. Got it?

5:38Example 2: potential of node N with a diode and 5 V battery

Now a harder one, the kind of problem that comes up in JEE papers. There is a twelve volt battery, joined to a two kilo-ohm resistor, and then to a point called node N, okay? From N, an ideal diode points forward, towards a five volt battery whose other terminal is earthed. And a four kilo-ohm resistor also goes from N to earth. What is the potential of node N? Pause the video and try it first, with our four steps. Alright, for step one, let us assume the diode is open, a gap. Then the two and the four kilo-ohm resistors are in series across twelve volts. So V N, the potential of N, written V with a small N below it, equals twelve times four over six, where the cross on screen means times. That gives eight volts. Okay? Many students stop right here, and write eight volts as the answer. Should we stop here too? No, because we skipped step four, the check. The diode's p-side is at N, at eight volts, and its n-side is on the five volt battery, at five volts. Eight is higher than five, so the p-side is higher, and the diode must conduct. But we assumed it was open, so that assumption fails. Right? So we change our assumption. The diode conducts, and it becomes a wire. Now N is joined straight to the five volt battery, so the diode holds N at five volts. Clear? Now let us check this new state too. Twelve minus five gives seven volts across the two kilo-ohm resistor, and seven over two gives three point five milliamperes into node N, okay? And five over four gives one point two five milliamperes flowing out, through the four kilo-ohm resistor to earth. With me? The rest must go through the diode. Three point five minus one point two five gives two point two five milliamperes, flowing from p to n, just as it should. So the tick mark on screen says this check passes. Okay? So the potential of N is five volts, and not eight. See how the check saved us from the trap? Got it?

7:51Quick revision: wire, gap, and the four-step method

So, quick revision. Forward biased, with the p-side higher, the ideal diode is a wire, with zero voltage drop. Reverse biased, it is a gap, with no current, and the whole branch voltage across it. And the method: assume, replace, solve, then check every diode. Now go and try the practice problems on this page. All the best!