Earlier JEE Main Questions: Special Diodes

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. One hundred and twenty volts is fed through a five kilo-ohm resistor to a fifty volt Zener diode, with a ten kilo-ohm load across it. The resistor carries fourteen milliamperes, so the Zener carries all fourteen, right? Hold on to your answer; it is a JEE Main question from twenty nineteen. Today we solve earlier JEE Main questions on special diodes. So, let's begin.

0:31Idea 1: the photodiode works in reverse bias

So, idea number one, the photodiode. In the left graph, capital I is the current and capital V is the voltage. The photodiode is used in reverse bias, the left half of this graph, okay? The right graph is the solar cell, with V o c, its open-circuit voltage, and I s c, its short-circuit current; we come back to it soon. If the photon energy, h nu, is greater than E g, the band gap, light makes electron hole pairs at the junction. Here h is Planck's constant, nu, the Greek letter, is the frequency, and E g is the E with a small g below it. The junction field separates the pairs, giving the photocurrent. The labels I one, I two and I three are three light intensities, with I three the brightest, as the greater-than signs show. And the sign below the graph means proportional to: the photocurrent is proportional to the intensity. Right? Now a JEE Main question from twenty twenty. We increase the biasing voltage on a photodiode. How does the size of the photocurrent change? Start where a curve crosses the voltage axis, and move left, past zero and into reverse bias. The size of the current grows at first, and then the line goes flat. So the photocurrent increases at first, and then saturates. With me? It saturates because the bias only sweeps across the carriers that the light makes. Soon all of them are collected, and then more voltage cannot give more current; only more light can. Clear?

2:07Idea 2: name the device from its graph

Now idea number two, from JEE Main twenty sixteen. Four devices are given by their I V characteristics, current against voltage, and we name each. In part a, current flows in the forward direction only, rising after a knee, so it is a simple diode. Okay? Part b is the same, but with a sharp breakdown at a definite reverse voltage. That is the Zener diode, built to work at breakdown. Right? Now part c is a curve in the fourth quadrant, where the voltage is positive but the current is reversed. Which device gives a curve like that? Yes, exactly, it is the solar cell. It needs no external bias, and it supplies current to the load instead of drawing it. Got it? And part d is a straight line through the origin, whose slope depends on the light. That is a resistor whose value the light sets, a light-dependent resistance. So the order is simple diode, Zener diode, solar cell, light-dependent resistance. With me?

3:11Idea 3: LEDs of different colours

Now idea number three, from JEE Main twenty thirteen. The I V curves of red, yellow, green and blue L E Ds are drawn on the same axes. An LED, a light-emitting diode, works in forward bias, so all four curves lie in the first quadrant. Okay? Like any diode, each curve rises steeply after a threshold voltage, its knee. Right? Now the real question. Which colour needs the highest threshold voltage, red or blue? It is blue that needs the most. The threshold rises in the order red, yellow, green, blue, as the less-than signs show. Clear? The reason is that the photon energy, h nu, is about equal to the band gap, E g; the wavy equals sign means about. Blue photons carry the most energy, so a blue LED needs the largest band gap, and so the highest threshold. Got it?

4:06Idea 4: the Zener regulator, test first

Now idea number four, the Zener diode. On the left of the figure is its characteristic, with the breakdown at minus V Z, where V Z is the Zener voltage; on the right is the Zener regulator. The small letters below V, R and I are labels: in for input, S for series, L for load, Z for Zener. In breakdown, the series current, capital I, is V in minus V Z, over R S, okay? The load current, I L, is V Z over R L, and the Zener current, I Z, is capital I minus I L, clear? But does the Zener diode always break down, whatever the resistances? It does not, so always test first. Remove the Zener and find the load voltage, V in times R L, over R S plus R L, okay? If it is less than V Z, the arrow tells us the diode is off, and I Z is zero. Right? Here is a JEE Main twenty nineteen question on exactly this. Twelve volts, through five hundred ohms, feeds a ten volt Zener; the Greek capital omega is the ohm. Clear? Across the diode are two resistors of one thousand five hundred ohms in parallel, which make seven hundred and fifty ohms. Now remove the Zener. What voltage does the load get? Twelve times seven hundred and fifty, over one thousand two hundred and fifty, gives seven point two volts. That is less than ten, so the diode never breaks down, and the Zener current is zero. With me?

5:40Example 1: band gap of a photodiode detecting 400 nm

Now Example one, from JEE Main twenty twenty. Here, a semiconductor photodiode can detect a photon with a maximum wavelength of four hundred nanometres, the n m on screen. Find its band-gap energy. Pause the video and try it first. Done? Okay, let's check. Each photon makes a pair only if its energy, h c over lambda, is at least the band gap; lambda, the Greek letter, is the wavelength, and c is the speed of light. So only lambda less than or equal to h c over E g is detected, and E g equals h c over lambda max, the longest wavelength. Right? There is a shortcut worth knowing by heart. What number turns a wavelength in nanometres into an energy in electron volts? Yes, it is one thousand two hundred and forty: the band gap in electron volts is about that number over the wavelength in nanometres. Okay? So E g is one thousand two hundred and forty over four hundred, which gives three point one electron volts, the e V on screen. Got it?

6:46Example 2: Zener current, 120 V supply, 10 kohm load

And now Example two, from JEE Main twenty nineteen, our opening puzzle. One hundred and twenty volts feeds, through five kilo-ohms, a Zener diode of breakdown voltage fifty volts, with a ten kilo-ohm load in parallel; the small k with the Greek capital omega means kilo-ohm. Find the current through the Zener diode. Pause the video and try it first. Done? Okay, let's check, and first, the test. With the Zener removed, the load gets one hundred and twenty times ten, over five plus ten, which gives eighty volts. That is more than fifty, as the greater-than sign says, so the diode is in breakdown, okay? So the series current is one hundred and twenty minus fifty, which is seventy volts, over five thousand ohms. That is seventy over five thousand amperes, which gives fourteen milliamperes, the m A on screen. Right? Now, does the Zener diode carry all fourteen milliamperes? No, no, wait, the load takes its share first. The load has fifty volts across it, so I L is fifty over ten thousand ohms, which gives five milliamperes. Clear? So the Zener current is fourteen minus five, which gives nine milliamperes. So the answer is nine, not fourteen, since the series current splits between the diode and the load. Got it?

8:09Quick revision: photodiode, solar cell, LEDs, Zener test

So, quick revision. The photodiode works in reverse bias, its photocurrent rises and then saturates, and E g in electron volts is about one thousand two hundred and forty over lambda in nanometres. The solar cell works in the fourth quadrant, and the LED threshold rises from red to blue. For the Zener, test first; then I Z equals capital I minus I L, okay? Now try the other past questions on this page yourself. All the best!