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# Monthly Archives: April 2016

## GATE Previous Questions on Zener Diode

GATE-2013

1)In the circuit shown below, the knee current of ideal zener diode is 10 mA. To maintain 5 volts across the load RL, the minimum value of RL in Ω’s and the minimum power rating of the zener diode in mW respectively are. Given $V_z=5V$

1. 125 and 125
2. 125 and 250
3. 250 and 125
4. 250 and 250

GATE 2004

2) In the voltage regulator shown, the load current can vary from 100 mA to 500 mA. Assuming that the zener diode is ideal (i.e. the zener knee current is negligibly small and zener resistance is zero in the breakdown region), the value of R is

GATE 1994

3) A Zener diode works on the principle of

1. Tunneling of charge carriers across the junction
2. Thermionic emission
3. Diffusion of charge carriers across the junction
4. Hopping of charge carriers across the junction

GATE 2006

4) In the circuit shown below , assume that Zener diode is ideal with a breakdown voltage $V_z$=6v. The waveform across load R looks like

## Difference between Zener breakdown and Avalanche breakdown

In reverse biased  P-N junction diode carries only a small current called reverse saturation current. However , if the reverse voltage across the device increases, eventually after certain voltage limit, breakdown occurs and a large current is observed.

This breakdown phenomenon can be due to two effects

1. Zener effect
2. Avalanche effect

Based on the effects , the breakdown  can be named as Zener breakdown or Avalanche breakdown.

Zener breakdown (Between 3-8 v)

The depletion region in a P-N junction is devoid of any carriers. However, a high electric field in this region will be sufficient  enough to detach remaining  electrons  from the covalent bond. Once freed, the electrons are accelerated by  the field and swept to n side of the junction. This effect occurs at strength of about 10^6 V/cm.

In order to create such high field, a narrow depletion region is required which translates to high doping levels on both sides of the junction. This type of breakdown occurs at a reverse voltages in the order of 3-8 V.

Avalanche breakdown

Junctions with moderate/low doping levels  won’t exhibit Zener breakdown. But as the reverse bias voltage across such devices increases, avalanche effect takes place. Each carrier entering the depletion region experiences a very high electric field and hence a large acceleration as denoted by the equation below

$a$=$\frac{Eq}{m}$

These accelerated electrons will have enough force to break the other electrons from their covalent bonds. This phenomenon can lead to avalanche i.e each electron freed by the impact may itself speed up in the electric field and can detach few more electrons from their respective covalent bonds. Now these 2 electrons may collide further atoms and displace few more carriers as depicted in the picture below.

Temperature effects on these breakdowns.

1. As the temperature increases, the breakdown voltage decreases in case of Zener diodes( Zener effect shows negative temperature coefficient). For example, if the Zener effect was visible at V=4v, then this would be visible at 3V if the temperature is increased.
2. As the temperature increases, the breakdown voltage increases in case of Avalanche diodes.(Avalanche effecct shows positive temperature coefficient).

## GATE : Multiple choice questions for Diodes

1)Tunnel diode and Photo diode are operated in _____ and ____ bias  respectively

a. Reverse and Forward

b. Forward and Reverse

c. Reverse and Reverse

d. Forward and Forward

2)  Avalanche breakdown in a diode occurs when

a. Potential barrier is reduced to zero

b. Forward current exceeds certain value

c. Reverse bias exceeds a certain value

d. None of these

3) In a Zener diode

a. Only the P-region is heavily doped

b. Only the N-region is heavily doped

c. Both P and N region are heavily doped

d. Both P and N region are lightly doped

4)Which of these diodes is also called hot carrier diode.

a. PIN diode

b. LED

c. Photo diode

d. Schottkey diode

5) What is the voltage across the load resistor R2 in the below diagram

a) 6V            b) 5V        c) 0v         d) 4v

## GATE : Different types of Diodes questions

There will be many “Match the following” questions related to diodes in GATE and IES examinations every year. So, it is very important to get these marks.

In this post we will see different types of diodes and their applications and a small quiz at the end.

Tunnel Diode :

Tunnel Diode is a high doped p-n junction diode

Applications

• In high speed switches, switching speed is in the order of nano or pico seconds due to a phenomenon called “Tunneling”
• Microwave oscillator
• Amplifier
• High Speed/Low power

Detailed analysis can be found in the video lecture.

PIN Diode :

PIntrinsic-N diode is a diode with a wide, lightly doped near intrinsic semiconductor region between P and N type semiconductor region. P and N type are heavily doped because they are used as Ohmic contacts. Mainly used in Radio frequency applications.

• Used as a Variable attenuator/resistor in RF and microwave frequency range
• High speed switches
• Used in RF Modulator circuit and in phase shifter circuit.
• It is a current controlled device.

Photo Diode

Under reverse biased conditions, the current flowing through a P-N junction diode is very small and it is called reverse saturation current. It is in the order of microamperes and it is due to flow of minority carriers. If light is focused on the junction, the light photons generate electron-hole pairs on both sides of the junction. The minority charge carriers will increase and so does the saturation current. Increase in the current is directly proportional to intensity of light. This phenomenon is used in photo diode.

• Dark current in photo diode refers to reverse saturation current without any “LIGHT”.
• Photo diodes find application in high speed counting and solar energy systems

LIGHT EMITTING DIODES (LED)

By far, the most popular diode in the world. Every one who has studied electronics has used LED’s in one of their digital electronics project for sure.

Working : In a forward biased P-N junction diode, recombination of electrons and holes takes place at the junction, after recombination, energy is released in the form of heat.

In case of Gallium Arsenide the energy of electron is converted into light energy in the form of photons. This process of light emission in p-n junction is called electroluminescence. This is the principle behind light emitting diodes.

A LED is a P-N junction diode, which when sufficiently forward biased causes electrons to recombine with holes which then releases  energy in the form of photons . The color of the light depends on the band gap of the semiconductor.

Depending on the energy, different colors are emitted.

$\Delta E$=hf= $\frac {hc}{\lambda}$

For example

A LED made up of gallium phospide doped with nitrogen emits green light and A LED made up of gallium arsenide phospide produces red light.

IES Question :

Q) Match List -1 (Diode) with List-2 (common application)

List -1                                                                                         List-2

A. Tunnel Diode                                                  1. Reading of film sound track,

B. PIN Diode                                                       2. High frequency oscillator circuits.

C. Zener Diode                                                    3. Very high frequency switching  circuits

D. Photo Diode                                                   4. Reference voltage

## Derivation of Characteristic Impedance of Transmission line

We talked in great detail about Transmission lines in the introductory lecture. Now we will see how to derive characteristic impedance from transmission line equations.

$V$=$V^+ e^{-\gamma x}$$V^- e^{+\gamma x}$

$I$ =$I^+ e^{-\gamma x}$$I^- e^{+\gamma x}$

As discussed in the video, if lumped  circuit analysis has to be valid at all frequencies, the length of the sub section must tend to Zero.(Because even if the sub section is very small, after certain frequency, the effect of transit time will be significant)… (It is discussed in great detail in the video lecture)

$\lim_{\Delta x \to 0} \frac{\Delta V}{\Delta x} =\frac{dV}{dx}$ = $-(R+j\omega L) I$

$\lim_{\Delta x \to 0} \frac{\Delta I}{\Delta x}=\frac{dI}{dx}$ = $-(G+j \omega C) V$

Substituting the value of V in the above equation, we get.

$\frac {d}{dx}$ {$V^+ e^{-\gamma x}$$V^- e^{+\gamma x}$} =$-(R+j\omega L)$ { $I^+ e^{-\gamma x} + I^- e^{+\gamma x}$}

==> $- \gamma V^+ e^{-\gamma x} + \gamma V^- e^{+\gamma x}$ =$-(R+j\omega L)$ { $I^+ e^{-\gamma x} + I^- e^{+\gamma x}$}

Co-efficient of $e^{-\gamma x}$    :  – $\gamma V^+$$-(R+j\omega L) I^+$

Co-efficient of $e^{\gamma x}$    :  $\gamma V^-$$-(R+j\omega L) I^-$

Since $\gamma = \sqrt{ (R+j\omega L) (G+ j\omega C) }$

$\frac{V^+}{I^+}$ = $\frac{R+j\omega L}{\gamma}$ = $\sqrt{\frac{R+j\omega L}{G+ j\omega C}}$

$\frac{V^-}{I^-}$ = $- \frac{R+j\omega L}{\gamma}$$\sqrt{\frac{R+j\omega L}{G+ j\omega C}}$

It can be seen that, the above equations has the dimensions of Impendance(Ratio of Voltage to Current) and is a function of primary constants of the line and operating frequency. It is therefore called the “Characteristic Impendance” of the transmission line , often denoted by $Z_o$.

$Z_o$$\sqrt{\frac{R+j\omega L}{G+ j\omega C}}$

The formula of $\gamma$ and $Z_o$ of a transmission line are very important. Many GATE memory based questions will be based on this formula. So remember  the equation and solve few questions related to the characteristic impedance of transmission line.