Circuits with Real Diodes

Transcript

0:00Introduction

Hello everyone, let us start with a small puzzle. Take a silicon diode, forward biased, in series with a one hundred ohm resistor, across a cell of zero point five volt. An ideal diode in forward bias is just a closed switch. So the current should be zero point five over one hundred, which gives five milliamperes, right? But build this circuit in the lab, and the current is practically zero. So where did the current go? Let us find out. Diode circuits come up often in JEE, so let's begin.

0:34The constant-drop model

So, here is the characteristic of a silicon diode from an earlier lesson. The current, capital I, goes up in milliamperes, the m A at the top, and the voltage, capital V, goes across in volts. The orange curve on the left is reverse bias: a tiny current in microamperes, written mu A, until breakdown at minus V b r, the breakdown voltage. Now look at the dark blue forward curve on the right. Does a real current flow as soon as V is a little above zero? Hardly any current flows at first. The curve stays almost flat up to about zero point seven volt, the knee, and after the knee the current rises steeply. So here is a better model. Once the diode conducts, the voltage across it stays fixed at V k, a capital V with a small k below it, for knee. Okay? And the values to remember: V k is zero point seven volt for silicon, S i on screen, and zero point three volt for germanium, G e on screen. You can see both in the label under the knee. Got it? Now the important condition. The diode conducts only if the circuit can supply more than V k, even in forward bias. Below that, it is practically off. Right? So, our puzzle is solved. The cell gives only zero point five volt, and the less than sign says that is below zero point seven volt. So the silicon diode never switches on, and the current is practically zero; the wavy equals sign means approximately. Here is a topper's habit: check this condition first, before you use any formula.

2:05Diode in series with a resistor

With me? Now look at the left circuit on screen, forward bias. The triangle with a bar is the diode, with p on the triangle side and n at the bar. It is in series with a resistor, capital R, and a battery with its plus terminal on the p-side. The red arrow is the current, I, in milliamperes. The battery gives a voltage capital V, okay? Where does all of it go? It is shared between the diode and the resistor, yes, exactly. The fixed V k sits across the diode, and the rest, I times R, sits across the resistor. So, V equals V k plus I times R, clear? Now just solve for the current. I equals V minus V k, all over R, the resistance. Take away the knee voltage first, then use Ohm's law on what is left. It is really simple, okay? Some questions also give the forward resistance, a small r with a subscript f. Then the conducting diode acts like a fixed drop V k in series with a small resistor r f, so we add r f to R in the denominator. I equals V minus V k, over R plus r f. With me? And if only r f is given, with no knee voltage, treat the conducting diode as a resistor of value r f.

3:19Reverse bias and the series resistor

Right? Now the right circuit, reverse bias. The battery is turned around, so the p-side goes to the minus terminal. In reverse bias, we take the resistance of the diode as infinite, the infinity sign on screen, like a sideways eight, so the current is zero. Okay? That holds unless the question specifies a reverse current, like the tiny current in microamperes marked in the figure. Then you use that value. Got it? Now a question students often get wrong. In reverse bias, no current flows. So the voltage across the diode must be zero too, right? That is the trap, because the answer is the opposite. With zero current, the resistor drops I times R, which is zero. So the whole battery voltage appears across the diode, and none across the resistor. Clear? One more point. Once the diode conducts, its current rises very steeply, so it could become very large. That is why a series resistor is always needed, to limit the current. Right? Now, the least resistance you can use. If the current must not go above I max, then R min equals V minus V k, over I max. The small words below R and I just mean minimum and maximum. Okay?

4:31Example 1: current with Si and Ge diodes, 1 kilo-ohm, 5 V

Now example one, the easier one. We have a silicon diode of zero point seven volt in series with one kilo-ohm, the k and capital omega on screen, across a five volt battery, forward biased, just like the left circuit in the figure. Find the current, then repeat it for a germanium diode, with zero point three volt. Pause the video and try it first. Have you tried it? Okay, let us check. Five volts is well above zero point seven, so the silicon diode conducts. One kilo-ohm is one thousand ohms, so I equals five minus zero point seven, over one thousand, in amperes. Okay? Five minus zero point seven is four point three. So I is four point three times ten to the power minus three ampere, which is four point three milliamperes. Right? Now germanium. Its knee voltage is only zero point three volt. So will the current be more or less than for silicon? There is more current, yes, exactly, because less voltage is lost across the diode. Five minus zero point three is four point seven, so the current is four point seven milliamperes. Easy, isn't it?

5:42Example 2: Si and Ge diodes in parallel, which one conducts

Now example two, a harder one. Silicon and germanium diodes are connected in parallel, both forward biased. The pair is in series with five kilo-ohms, across twelve volts. Find the voltage across the resistor, and the current in each diode. Pause the video and try it first. Have you got your answer? Okay, let us check. Two diodes side by side, both forward biased. So surely they share the current, maybe half each, right? That is the trap, so be careful here. Both diodes sit across the same two points, so they always have the same voltage. Germanium switches on first, at zero point three volt, and then it holds the pair at zero point three volt. Okay? But silicon needs zero point seven volt to switch on, and it only ever gets zero point three. So silicon stays off, and its current, capital I with a small S i below it, is zero. With me? Now the resistor voltage, capital V with a small R below it. Twelve minus zero point three gives eleven point seven volts. Right? The germanium current is capital I with a small G e below it. Five kilo-ohms is five thousand ohms, so the current is eleven point seven over five thousand, in amperes, and that gives two point three four milliamperes. So germanium takes all the current, simply because its knee voltage is smaller.

7:10Quick revision: knee voltage, rf, reverse bias, Si vs Ge

So, quick revision. Below the knee voltage, no current flows. Above it, I equals V minus V k, over R, with zero point seven volt for silicon and zero point three volt for germanium. With a forward resistance, add r f to R in the denominator. In reverse bias the current is zero, and the whole voltage sits across the diode. And with silicon and germanium in parallel, germanium conducts and silicon stays off. Now go and try the practice problems on this page. All the best!