Diode

DEE3253/6533 – ANALOGUE ELECTRONICS FUNDAMENTAL

Diodes

Subtopics

1.0 Semiconductor diodes (3 hours)1.1 Introduction to semiconductors materials1.2 Introduction to diode

1.3 Introduction to Zener diode and LED 2.0 Diode applications ( 6 hours)

2.1 Load line analysis and diode approximation 2.2 Series-Parallel Configuration 2.3 Half-wave and Full-wave rectification 2.4 Clippers and Clampers

2.5 Zener diode application

Early Diodes

Thermionic diodes are thermionic valve devices (also known as vacuum tubes)

Electrodes surrounded by a vacuum within a glass envelope, similar in appearance to incandescent light bulbs.

Semiconductor Diodes

Most modern diodes are based on semiconductor p-n junctions In a p-n diode, conventional current can flow from the p-type

side (the anode) to the n-type side (the cathode), but cannot flow in the opposite direction.

Diode symbol

Semiconductor Materials

Semiconductor means the elements having a conductivity between a conductor and an insulator

Commonly used:– Germanium (Ge)– Silicon (Si)– Gallium Arsenide (GaAs)

Single-crystal

Compound

Atomic Structure

Every atom is composed of 3 basic particles: electron, proton & neutron

The outermost orbit are called valence electrons Semiconductor have only 3 – 5 valence electrons

Carbon Silicon Germanium

Atomic Structure

Covalent bonding is the bonding of atoms, strengthened by the sharing of electrons

Ex: covalent bonding of silicon atoms

Energy Levels

There are specific energy levels associated with each orbiting electron

To become a conductor, electron from valence band must absorb energy to across the energy gap and into the conduction band

The energy levels are different for every element

Extrinsic Materials

n-type materials– Extra electrons in the

covalent bond gets from n-type materials such as antimony, arsenic and phosphorus

Extrinsic Materials

p-type materials– Extra holes in the

covalent bond gets from p-type materials such as boron, gallium and indium

Semiconductor Diodes

By simply joining the n-type and p-type material together, a semiconductor diode was born

depletionregion

Semiconductor Diodes

Connect the diode to a source:

The electrons in n-type material will attracts to +ve terminal of the source

The holes in p-type material will attracts to –ve terminal of the source

So, the depletion region becomes bigger and electrons cannot pass, hence no current flows

This is called REVERSE BIAS

Semiconductor Diodes

Connect the diode to a source:

The electrons in n-type material will attracts to +ve terminal of the source and jump into p-type material

The holes in p-type material will attracts to –ve terminal of the source and jump into n-type material

So, the depletion region becomes lesser/gone and electrons can pass, hence current will flows

This is called FORWARD BIAS

Semiconductor Diodes

To conclude, diode can be represented as a switch (but not ideal)

For a forward-bias, 0.7V (knee voltage, VD) have to be applied for the diode (for silicon)

For germanium, VD = 0.2 ~ 0.3V For gallium arsenide, VD = 1.2 ~ 1.4V

Semiconductor Diodes

Comparison of Ge, Si and GaAs diodes

Zener region

Zener Diode

A special type of diode that is supposed to be reversed biased Zener diode works in zener region where the diode start to

breakdown at breakdown avalanche voltage (VZ), and the current is avalanche current (IZ)

It limits a voltage to a certain point to pass through the zener diode

LED (Light-Emitting Diode)

In a forward-biased p-n junction, recombination of the holes and electrons requires energy possessed by the unbound free electrons

In Si and Ge, most of the energy is dissipated in the form of heat and photons

But in other material such as GaAs, the energy generate light but it is invisible for the eye to see (infrared)

Other materials that emit light during forward-bias operation

Color Construction Forward Voltage

AmberBlueGreenOrangeRed WhiteYellow

AlInGaPGaNGaP

GaAsPGaAsP

GaNAlInGaP

2.15.02.22.01.84.12.1

LED (Light-Emitting Diode)

How an LED works

Diode Approximation

Diode equivalent circuits:– Ideal Equivalent Circuit– Simplified Equivalent Circuit– Piecewise-Linear Equivalent Circuit

Purpose: to represent diode

Piecewise-Linear Equivalent Circuit

Approximation representation of the actual diode Diode have VD and rav for the slope region

Simplified Equivalent Circuit

Assume straight vertical line of ID at VD

No rav

Ideal Equivalent Circuit

Diode as an ideal switch No VD or rav

Load-Line Analysis

A simple analysis which used the diode characteristic to obtain the Q-point (operation point)

A series diode circuit and characteristic:

Load-Line Analysis

For For Connect a line between E / R and E The overlap of the lines becomes the Q-point of the

diode and IDQ and VDQ will be obtained

RDRD IVVVE

REIRIRIEV DDDD ,0 ,0

EVVRVEI DDDD ,)0( ,0

Load-Line Analysis

Problem 2.1a

Determine ID, VD & VR

The circuit:

Problem 2.1a

The diode characteristic:

Problem 2.1a

Solution:– The circuit representation:– For

– For

RIVVVE DDRD

mA 24.2433.080 ,0

kREI

RIEV

D

DD

V 8)0( ,0

EVRVEI

D

DD

Problem 2.1a

The load-line analysis becomes:

RE

E

Q-point

DQI

DQV

Problem 2.1a

From the analysis:– VDQ = VD ≈ 0.9 V– IDQ = ID ≈ 21.5 mA– For VR,

V 095.7)33.0)(5.21(

kmRIV DR

Problem 2.1b

Re-do Problem 2.1a using approximate (simplified) model for diode and compare the result

The diode characteristic becomes:

Problem 2.1b

The solution:

Problem 2.1b

VD is always 0.7 V, so VD = VDQ = 0.7 V From the graph, IDQ = ID ≈ 22 mA So, we get VR = (22m)(0.33k) = 7.26 V ≈ 7.3 V Using Kirchoff’s voltage law, E = VD + VR

The answer are the same

V 3.77.08

R

R

VV

Problem 2.1c

Re-do Problem 2.1a using ideal model for diode and compare the result

The diode characteristic becomes:

Problem 2.1c

The solution

Problem 2.1c

VD is always 0 V, so VD = VDQ = 0 V It acts like an ideal switch From the graph, IDQ = ID ≈ 24.24 mA So, we get VR = (24.24m)(0.33k) = 7.9992 V ≈ 8 V Using Kirchoff’s voltage law, E = VR

The answer are the same

V 8 RV

Series-Parallel Configuration

Diode can be applied to any circuits Usually diode is represented as an

approximated (simplified) model diode To keep the calculation simple, just use the

Kirchoff’s voltage & current law Hint: it is easier to use nodal analysis

technique for circuit representation Important: strong knowledge in CIRCUIT

THEORY!!!!!

Problem 2.5a

Find I The circuit:

Problem 2.5a

For Si, VD = 0.7 V Notice that the diode is in reverse-bias

configuration So, no current will flow, I = 0 A

Problem 2.5b

Find I The circuit:

Problem 2.5b

Solution:– Using nodal analysis, node V is equal to the

voltage supplied, so V = 20

V

Problem 2.5b

Using the simple Ohm’s law:

A 965.020

7.020

RVI

Problem 2.5c

Find I The circuit:

Problem 2.5c

Solution:– One of the diode is in reverse-bias resulting in

open circuit for that part

Problem 2.5c

So, by using the simple Ohm’s law:

A 11010

RVI

Problem 2.7a

Find Vo

The circuit:

Problem 2.7a

Both the diode are in forward-bias, so both are short-circuited For Si, VD = 0.7 V For Ge, VD = 0.3 V The circuit becomes:

0.7 V 0.3 V19 V

SIMPLIFIED

Problem 2.7a

Solution:– Using nodal analysis, voltage at Vo:

V 5.922

19

0

00

VkV

kV

Problem 2.7b

Find Vo

The circuit:

Problem 2.7b

The circuit becomes:

0.7 V

Problem 2.7b

Solution:– The nodal analysis of node Vo:

V 77.4

)2(2.17.010

0

00

Vk

VkV

Problem 2.11a

Find Vo & I: The circuit:

Problem 2.11a

For Si, VD = 0.7 V For Ge, VD = 0.3 V Because of this, current

will flow in the Ge’s diode route

Naturally, current will select the easiest/fastest route

So, the circuit becomes:

0.3 V

Problem 2.11a

Solution:– Using Kirchoff’s voltage law:

– For I, by using basic Ohm’s law:

V 7.93.0100 V

mA 7.91

7.90 kR

VI

Problem 2.11b

Find Vo & I: The circuit:

Problem 2.11b

For the same type of diode, the circuit will becomes:

Because there is no resistor exist in the parallel route of the diode, current will flow in only one of the diode’s route

0.7 V0.7 V

0.7 V

Problem 2.11b

Solution:– Using Kirchoff’s voltage law:

– For I, by using Ohm’s law:

V 6.147.07.0160 V

mA 553.07.4

126.14

kI

Problem 2.13

Find Vo & ID

The circuit:

Problem 2.13

The circuit becomes:

0.7 V

0.7 V

9.3 V

SIMPLIFIED 9.3 V

SIMPLIFIED

Problem 2.13

The solution:– The nodal analysis for node Vo:

– For ID:

V 2.621

3.9

0

00

VkV

kV

mA 55.12

2.63.9

kID

Rectification

Rectify means improvement, cure healing (pembaikan, penambahbaikan)

For a sinusoidal waveform or any supply that has a variation of input value, diode can be used for rectification

Rectification are used to modified the input value to become only the signal that we want

Half-Wave Rectification

For a full cycle of a sinusoidal or continuous waveform, only half of the waveform is taken to be rectified

Half-Wave Rectification

For the period 0 T/2, the sinusoidal input will give a forward bias supply to the circuit

The diode will “on” and current will pass through Assume that the diode is ideal

Half-Wave Rectification

For the period T/2 T, the sinusoidal input will give a reverse bias supply to the circuit

The diode will “off” and no current can pass through Assume that the diode is ideal

Half-Wave Rectification

For a continuous periodic waveform, the rectified waveform will become:

Where as:

mdc VV 318.0

Problem 2.25

Sketch Vo and determine Vdc:

Problem 2.25

Solution:– To obtain Vm from Vrms:

– The output Vo will be:

– Vdc will be:

V 56.155)110(2

2

rmsm VV

V 47.49)56.155(318.0318.0

mdc VV

Problem 2.26

Sketch Vo

Problem 2.26

Solution:– For the positive input supply:

– The circuit becomes:For Vi < 0.7V: For Vi ≥ 0.7V:

0.7 V

Problem 2.26

– The output for the positive input supply becomes:

– For the negative input supply:

Problem 2.26

– The circuit becomes:

– For maximum Vo: - The output becomes:

+

-

VVkV

kV

o 091.9101

10 00

Problem 2.26

– Combine both the output becomes:

PIV or PRV

Peak Inverse Voltage (PIV) or Peak Reverse Voltage (PRV)

It is a rating to make sure for the reverse-bias operation, the diode didn’t enter the Zener region

PIV is set according to the circuit and the input voltage

mV rating PIV for half-wave rectifier

Full-Wave Rectification

The whole cycle of input signal is used and rectified

Two commonly types of full-wave rectifier:– Bridge Network– Center-Tapped (CT) Transformer

The dc level from a sinusoidal input can be improved 100%. So the Vdc becomes:

mdc VV 636.0

Full-Wave Rectifier: Bridge Network

The most commonly bridge network configuration are build with 4 diodes

Full-Wave Rectifier: Bridge Network

For the positive input supply, the current will take the route as shown below, and the output will becomes:

Full-Wave Rectifier: Bridge Network

For the negative input supply, the current will take the route as shown below, and the output will becomes:

Full-Wave Rectifier: Bridge Network

Combine both of the output becomes:

Full-Wave Rectifier: Bridge Network

Due to the maximum voltage from the input supply is Vm, to keep the diode away from the Zener region, the PIV rating is:

mV rating PIV for full-wave rectifier: bridge network

Full-Wave Rectifier: Center-Tapped (CT) Transformer

It is constructed with 2 diodes and a center-tapped transformer The transformer ratio is 1:2

Full-Wave Rectifier: Center-Tapped (CT) Transformer

For the positive input supply:

Full-Wave Rectifier: Center-Tapped (CT) Transformer

For the negative input supply:

Problem 2.28

The circuit:

1 kΩ120 Vrms

All diodes are silicon

Problem 2.28a

Determine DC voltage for output Solution:

– Vm:– Vo:– So, Vdc:

V 71.16912022 rmsm VVV 31.1687.07.071.1690 V

)31.168(636.0.6360 0 VVdc

Problem 2.28b

Determine the required PIV rating for each diode from problem 2.28

Solution:– PIV: V 01.1697.031.168(load) PIV Dm VV

Problem 2.28c

Find the maximum current through each diode

Solution:– ID(max): mA 31.168

131.168(max)0

(max) kR

VI

LD

Problem 2.28d

What is the required power for each diode? Solution:

mW 82.1787.031.168(max) mVIP DD

Problem 2.31

Sketch Vo and determine Vdc

The input and circuit:

Problem 2.31

Solution:– For the positive input supply:– Simplified the circuit:

vi

+

-

vovi

+

-

vo

SIMPLIFIED

1.1 kΩ

Problem 2.31

– Vo(peak):

– The same for negative input supply– So the output will becomes:

V 67.561.12.2

170

0

00

Vk

VkV

56.67 V

Clippers

Configuration that employ diodes to “clip” away a portion of an input signal without distorting the remaining part of the applied waveform

Mainly, there are two types of configuration– Series– Parallel

Clippers

Example of series configuration and the output waveform:

Clippers

Example of parallel configuration and the output waveform:

Clippers

Notice something? Is the configuration similar to something?

Half-wave rectifieris a part ofCLIPPERS

configuration

Example 2.18

Sketch vo

For positive input cycle:

The output will be the sum of vi and +5V

Example 2.18

The output waveform will become:

50 ivv

Example 2.18

For negative input cycle:– For vi ≤ 5: - For vi ≥ 5:

+

-

ivv 50

+

-

00 v

Example 2.18

The output waveform will becomes:

Example 2.18

By comparing the input with the whole output:

Example 2.20

Sketch vo

Example 2.20

For positive input cycle:– For vi ≤ 4 - For vi ≥ 4

ivv 0

+

-

40 v

Example 2.20

The output waveform will become:

Example 2.20

The output for negative input cycle will always +4V due to the external supply of 4V series with the diode

The diode will always be in the “on” mode– The circuit becomes: - The output waveform:

40 v

Example 2.20

By comparing the input with the whole output:

Clampers

Construct of a diode, a resistor and a capacitor

It will shift the waveform to a different level without changing the appearance of the original input signal

The capacitor and resistor ( ) must be large to ensure it doesn’t discharge during the interval that the diode is non-conducting

RC

Clampers

The circuit:

Clampers

Steps for clampers analysis:1. Start the analysis with the condition where the

diode is in forward bias2. The capacitor will charge up instantaneously

during the interval of +ve or –ve input supply where the diode is in forward-bias condition

3. The capacitor will discharge during the next interval of +ve or –ve input supply where the diode is in reverse-bias condition

4. Check that the total swing of the output is the same with the input

Example 2.22

Sketch vo

Example 2.22

Just to check whether the capacitor is appropriate for clamper’s configuration:

For the input given:

For every half interval (+ve or –ve input cycle):

This shows that the capacitor is capable of charging and discharging according to the clamper’s configuration requirement

ms 10)1.0)(100( kRC

ms 11000

1f1T

ms 5.02ms 1

2T

2T

Example 2.22

To start the analysis with the diode in forward-bias mode, the negative input cycle has to be inserted first into the circuit

– The circuit: - The output waveform:

The capacitor will charge up to 25V

Example 2.22

For the next half input cycle that is the +ve cycle:– The circuit: - The output waveform:

The capacitor will discharge thevoltage of 25V

Example 2.22

The whole output waveform will become:

Checking the total swing of the output must match the input:

The total swing of the output is the same with the input that is 30 V

Zener Diodes

The application of Zener diodes have been explained in Subtopic 1.3

The analysis of Zener diodes can be divided into 3 categories:– Fixed Vi and RL

– Fixed Vi, variable RL

– Variable Vi, fixed RL

To make the analysis simple, the analysis will be explain directly from the examples

Example 2.26a (Fixed Vi and RL)

Determine VL, VR and IZ

Example 2.26a (Fixed Vi and RL)

To check whether VZ is in the “on” or “off” mode, the value of VL must be determine first

To do that, take out the Zener diode from the diode The circuit become:

Example 2.26a (Fixed Vi and RL)

By doing a nodal analysis for the node VL

As we can see, the value of VL is smaller than VZ, so the Zener diode is in the “off” mode

Which will result in:

And:

V 73.82.11

16

L

LL

Vk

VkV

A 0ZI

V 27.773.816

R

LRi

VVVV

Example 2.26b (Fixed Vi and RL)

Repeat Example 2.26a with RL = 3kΩ

3 kΩ

Example 2.26b (Fixed Vi and RL)

The same analysis is repeated from Example 2.26a where the Zener diode is taken out to examine the value of VL

The circuit becomes:

3 kΩ

Example 2.26b (Fixed Vi and RL)

By doing a nodal analysis for the node VL

As we can see, the value of VL is larger than VZ, so the Zener diode is in the “on” mode

When the Zener diode is in the “on” mode, it will maintain the voltage of 10V. Because of that VL becomes:

And VR becomes:

V 1231

16

L

LL

VkV

kV

V 01 ZL VV

V 61016 RV

Example 2.26b (Fixed Vi and RL)

Using current divider theory:

mA 67.2310

16

kk

RV

RV

III

L

LR

LiZ

Example 2.27 (Fixed Vi, Variable RL)

Determine the range of RL and IL that will result in VL being maintained at 10 V

Example 2.27 (Fixed Vi, Variable RL)

To maintain VL at 10 V, the Zener diode must be in the “on” mode

For IZM = 32 mA, the current at load:

The load would be:

mA 8

321

1050

mk

III ZMRL

k 25.1810mI

VRL

LL

Example 2.27 (Fixed Vi, Variable RL)

For IZ(min), the Zener diode are assume “off” but the voltage VZ are maintained at 10 V

The load current would be:

The load would be:

mA 401

1050

k

II RL

2504010mI

VRL

LL

Example 2.27 (Fixed Vi, Variable RL)

Retrieve back all the IL and RL value:

250mA 40k 25.1mA 8

LL

LL

RIRI

IL (min)

IL (max) RL (min)

RL (max)

Example 2.28 (Variable Vi, Fixed RL)

Determine the range of Vi that will maintain the Zener diode in the “on” mode

Example 2.28 (Variable Vi, Fixed RL)

To maintain Zener diode in “on” mode, VZ must equal to VL:

Taking the maximum current of the Zener diode, input current becomes:

The input voltage will become:

V 20 LZ VV

mA 67.762.1

2060

km

III LZMR

V 87.36)220)(67.76(20

i

Ri

VmRIV

Example 2.28 (Variable Vi, Fixed RL)

For IZ(min), the Zener diode are assume “off” but the voltage VZ are maintained at 20 V

Using nodal analysis at node VL:

Retrieve back all the value of Vi:

V 67.232.1

20220

20

i

i

Vk

V

V 67.23V 87.36 ii VV

Vi (max) Vi (min)

Voltage multiplier circuits use a combination of diodes and capacitors to step up the output voltage of rectifier circuits.

• Voltage Doubler• Voltage Tripler• Voltage Quadrupler

Voltage Multiplier Circuits

Voltage Doubler

This half-wave voltage doubler’s output can be calculated as

Vout = VC2 = 2Vm

Vm = peak secondary voltage of the transformer.

Operation of a Voltage Doubler Circuit

The 1st capacitor charges up to Vm during the positive half of the cycle,then the 2nd capacitor charges up to Vm in the same polarity as the 1st capacitor,finally the output is the sum of the voltages across both capacitors:

Vout = 2Vm

Voltage Tripler and Quadrupler Circuits

By adding more diode-capacitor networks the voltage can be increased.

Practical Applications of Diode Circuits

Rectifier CircuitsConversions of AC to DC for DC operated circuitsBattery Charging Circuits

Simple Diode CircuitsProtective Circuits against OvercurrentPolarity ReversalCurrents caused by an inductive kick in a relay circuit

Zener CircuitsOvervoltage ProtectionSetting Reference Voltages