The Zener Diode and its Breakdown Voltage

Transcript

0:00Introduction

Hello everyone, here is a small puzzle before we start. Suppose a diode is joined in reverse, through a resistor, to a nine volt supply. In reverse, a diode should block the current, right? But this one carries a current, and the voltmeter across it reads five point one volts. Turn the supply up, and that reading hardly moves. So where does the rest of the voltage go, and why does the diode not burn out? That diode is a Zener diode, and today we study its breakdown voltage. It is part of the experimental skills in JEE, and questions on it do come up. So, let's begin.

0:38A diode built to work in reverse

First, look at the circuit on the right of the figure. The symbol in the middle, a triangle with a bar whose ends are bent, is the Zener diode. It is heavily doped, and we operate it in reverse bias. Okay? Now the graph on the left, with capital V across and capital I up; the orange curve is the reverse side. Up to the breakdown voltage, V Z, with a small capital Z below, the reverse current is negligible. The graph marks this Zener voltage as minus V Z, on the reverse side. So what happens to the current when we reach V Z? If you said it shoots up, that is correct: at V Z the current rises sharply, while the voltage across the diode stays almost constant. That is the breakdown region written on the graph. Clear? And in forward bias, the dark blue curve marked forward, the Zener diode behaves just like an ordinary diode. Got it?

1:35Reading V_Z in the experiment

In the experiment, we take the reverse readings and plot the reverse characteristic. Then V Z is read from it as the voltage at which the curve turns almost parallel to the current axis. With me? Now let me ask you this. Why does a Zener diode break down at such a low voltage, and at such a sharply defined one? The answer is the heavy doping, because heavy doping makes the depletion layer very thin, okay? So even a small reverse voltage produces an electric field strong enough to pull electrons out of the covalent bonds. That is why the breakdown comes at a low and sharply defined voltage. Right? In breakdown we also use the dynamic resistance, capital delta V over capital delta I; the little triangle is capital delta, a small change. The voltage hardly changes while the current changes a lot, so this resistance is only a few ohms. Clear? Now a practical one, for a six volt Zener. Should the voltmeter be zero to one volt, or zero to ten volts? And should you use a microammeter or a milliammeter? Very good if you said a zero to ten volt voltmeter and a milliammeter. The breakdown current is of the order of milliamperes, the m A on screen. Got it?

2:51The series resistance R_S keeps it safe

Now back to the circuit on the right. On its left is the supply, V in, the input voltage. At the top is a resistor, R S, with a small capital S below, and the orange arrow, capital I, is the current. On the far right, R L is a load for the voltage regulator, with its own current, I L, and the formulas under the circuit include it. We do not need the load today. Now, in breakdown the current rises sharply, so what stops it from growing without limit? Yes, exactly, the series resistance, R S, limits the current. The diode holds V Z, so the rest of the supply, V in minus V Z, is across the series resistance. So capital I equals V in minus V Z, over R S, okay? Every Zener also has a power rating, P max, with max written small below. The largest safe Zener current is P max over V Z, and the current must stay below it. Right? An ordinary diode may be damaged by breakdown. So why is a Zener diode not damaged, even though it works in breakdown all the time? The breakdown of a Zener is reversible, as long as the power, V Z times I Z, stays within the rating; I Z, with the small capital Z below, is the Zener current. And the series resistance keeps the current low enough. So that is why the diode in our puzzle does not burn out. With me?

4:21Example 1: 5.1 V Zener, 200 ohm, current at 9 V and 4 V

Now the examples, the easier one first. Here, a five point one volt Zener has a series resistance of two hundred ohms; the Greek capital omega is the ohm. Find the current through it when the supply is nine volts. Then the supply is set to four volts, so find the current, and the voltage across the Zener. Pause the video and try it first. Done? Okay, let's check. At nine volts the diode is in breakdown, so it holds five point one volts. The rest, nine minus five point one, which is three point nine volts, is across the resistor. Three point nine over two hundred gives nineteen point five milliamperes. So that is where the rest of the voltage in our puzzle went. Okay? Now the supply is set to four volts. So we use the same formula, four minus five point one, over two hundred, right? But that gives a negative current. Does that make sense? No, it does not, so be careful here. Four volts is less than V Z, as the less-than sign shows, so the diode is below breakdown, and practically no current flows. Clear? And with no current, there is no drop across the series resistance. So the whole four volts appear across the Zener. Got it? Now plot the voltage across the Zener against the supply voltage. It is a straight line of slope one up to V Z, and then a horizontal line at V Z, the Zener voltage. That horizontal part is exactly why, in our puzzle, the reading hardly moved. Right?

6:01Example 2: 0.5 W Zener, max current and least resistance

Now the harder one. Here, a five point one volt, zero point five watt Zener is studied with a supply that can go up to twelve volts; capital W is the watt. Find the largest safe current, and the least series resistance. Pause the video and try it first. Done? Okay, let's check. The largest safe current, I max, is P max over V Z, the power rating over the Zener voltage. Zero point five over five point one gives about zero point zero nine eight ampere, which is ninety eight milliamperes. Okay? Now the series resistance. The supply can be anywhere up to twelve volts. So which value of the supply should we use to find the least safe resistance? Yes, exactly, the largest one, twelve volts, because that is when the current is largest. Clear? So R S equals twelve minus five point one, over zero point zero nine eight. That is six point nine over zero point zero nine eight, which gives about seventy ohms; the wavy equals sign means approximately. With me? It is the least resistance because any smaller R S lets the current go above ninety eight milliamperes. So the series resistance must be at least about seventy ohms. If you got both answers on your own, very well done!

7:32Quick revision: V Z, series resistance, safe current

So, quick revision. First, the Zener is heavily doped and works in reverse bias: at V Z the current rises sharply, while the voltage stays almost constant. Second, V Z is read where the reverse curve turns almost parallel to the current axis. Third, above V Z the current is V in minus V Z, over R S, and it must stay below P max over V Z, okay? Now go and try the practice problems on this page. All the best!