The Zener Diode as a Voltage Regulator

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. You have a twelve volt supply and a ten volt Zener diode. The Zener holds its voltage fixed, so the output must be ten volts, right? Many students say ten. But by the end of this video, you will see a circuit where the output is only nine volts, and the Zener does nothing at all. The Zener regulator comes up often in JEE, so let's begin.

0:28Zener diode: built for breakdown

So, what is a Zener diode? It is a p-n junction, heavily doped, and made to work in the reverse breakdown region. In the circuit on the right of the figure, see its symbol: a triangle with a bent bar, like a little letter Z, okay? Heavy doping makes the depletion layer very thin, less than ten to the power minus six metre, and the small m is for metre. Right? Now put just a few volts of reverse bias across such a thin layer. The field becomes about five times ten to the power six volts per metre, or more. The wavy equals sign means about, and V m to the power minus one is read as volts per metre. That is strong enough to pull electrons out of their bonds. Clear? Now the graph on the left of the figure. Capital I is the current, and capital V is the voltage. In reverse bias, the orange curve, almost no current flows until minus V Z, the V with a small capital Z below it, the Zener voltage. There, the reverse current rises sharply, and the curve plunges straight down. Okay? Look at that steep orange line. The current changes a lot along it, so what stays fixed? Yes, the voltage, which stays almost constant at V Z however much the current changes. The label on the graph says just that: V stays at V Z while capital I changes. This one property is why we use a Zener diode. Got it? And in forward bias, the dark blue curve marked forward, a Zener diode behaves just like an ordinary diode. With me?

2:06The Zener regulator circuit

Now the circuit on the right of the figure. The supply is V in, with the small word in below the V, the unregulated input. The Zener is in reverse bias, its bar facing the positive side, and in parallel with the load, R L, the R with a small capital L below it. And R S, with a small capital S, is the series resistor that drops the extra voltage. Okay? Now, here is a question. The load is in parallel with the Zener. So while the diode is in breakdown, what is the voltage across the load? Yes, exactly, it is just V Z, the same as across the diode. So capital I, the current through R S, shown by the orange arrow at the top, is V in minus V Z, over R S, okay? And the load current, I L, is V Z over R L, right? At the top black dot, the current splits in two. So the Zener current, I Z, is capital I minus I L, okay? And P Z, the power in the diode, is V Z times I Z, clear? Now a common trap, okay? Does the diode always break down, whatever the resistances? It does not, so always test first. Remove the Zener, and find the voltage across the load: V in times R L, over R S plus R L, right? If this is below V Z, the diode never reaches breakdown. It is off, I Z is zero, and the output is simply this voltage. Got it?

3:37Why the output stays at VZ

So why is this a regulator? Say the input V in rises by a few volts. The output is fixed at V Z, so where do the extra volts go? Yes, exactly, they appear across R S, the series resistor. So capital I and the Zener current both rise, while the output stays put. Okay? Now keep the input steady, and let the load take more current. Capital I is still V in minus V Z, over R S, so it does not change. So the Zener current, I Z, falls by the same amount. With me? Either way, the output stays at V Z, nice and steady. That is voltage regulation. Clear? But the diode has a power rating, P max, with max written small below the letter. So the largest safe Zener current, I Z max, is P max over V Z, okay? And when is the Zener current largest? With no load and the highest input, because then the whole current goes through the diode. So the least series resistance that protects it is V in max minus V Z, over I Z max. Right?

4:44Example 1: find I, IL, IZ and Zener power for a 20 V input

Now the first example. V in is twenty volts, R S is one kilo-ohm, V Z is ten volts, and R L is two kilo-ohms. The small k with the Greek capital omega means kilo-ohm. Find the currents I, I L and I Z, and the power in the Zener diode. Pause the video and try it first. Done? Okay, let's check, and first, the test. Remove the Zener, and the load gets twenty times two over one plus two, which gives about thirteen point three volts. The greater-than sign says that is more than ten volts, so the diode is in breakdown. Okay? Now the current through the series resistor. Twenty minus ten is ten volts, over one thousand ohms, which gives ten milliamperes, the m A on screen. Right? The load current is ten volts over two thousand ohms, which gives five milliamperes. So the Zener current is ten minus five, which also gives five milliamperes. Clear? Now the power in the diode. Do we multiply the Zener current by V Z, or by V in? We use V Z, the voltage across the diode. Ten volts times five milliamperes gives fifty milliwatts, the m W on screen. Using twenty volts would give a hundred milliwatts, which is wrong. Got it?

6:07Example 2: 12 V supply and 10 V Zener, but output is 9 V

Now a harder one, and it is our puzzle from the start. Twelve volts is connected through R S of five hundred ohms to a Zener diode with V Z of ten volts, in parallel with R L of one thousand five hundred ohms. The Greek capital omega alone means ohm. Find the output voltage and the Zener current. Pause the video and try it first. Done? Okay, let's check. It is a ten volt Zener, so the output is ten volts, right? Let us use the formulas blindly. Capital I would be twelve minus ten, over five hundred, which gives four milliamperes. And I L would be ten over one thousand five hundred, about six point seven milliamperes. Okay? Then I Z, which is I minus I L, comes out negative; the less-than sign says it is below zero. That is impossible, so the guess of ten volts is wrong. Right? So we do the test first, as we learnt. Remove the Zener, and the load gets twelve times one thousand five hundred over two thousand, which gives nine volts. And nine volts is less than V Z, ten volts. Clear? So the diode never breaks down. The output is nine volts, and the Zener current is zero. Puzzle solved: with these resistors, the Zener does nothing at all. Got it?

7:31Quick revision: regulator currents and the test-first rule

So, quick revision. The Zener diode is heavily doped and works in reverse breakdown, where its voltage stays at V Z, okay? In the regulator, the current I is V in minus V Z, over R S, the load current is V Z over R L, and I Z is the difference. And always test first, because without breakdown, the Zener current is zero. Now try the practice problems on this page. All the best!