PN Junction Diode - tiiciiitm.com

16-03-2015

1

PN Junction

Diode

Table of Contents

What are diodes made out of?____________________slide 3

N-type material_________________________________slide 4

P-type material_________________________________slide 5

The pn junction_________________________________slides 6-7

The biased pn junction___________________________slides 8-9

Properties of diodes_____________________________slides 10-11

Diode Circuit Models ____________________________slides 12-16

The Q Point____________________________________slides 17-18

Dynamic Resistance_____________________________slides 19-20

Types of diodes and their uses ___________________ slides 21-24

Sources_______________________________________slide 25

What Are Diodes Made Out Of?

• Silicon (Si) and Germanium (Ge) are the two most

common single elements that are used to make Diodes.

A compound that is commonly used is Gallium

Arsenide (GaAs), especially in the case of LEDs

because of it’s large bandgap.

• Silicon and Germanium are both group 4 elements,

meaning they have 4 valence electrons. Their

structure allows them to grow in a shape called the

diamond lattice.

• Gallium is a group 3 element while Arsenide is a group

5 element. When put together as a compound, GaAs

creates a zincblend lattice structure.

• In both the diamond lattice and zincblend lattice, each

atom shares its valence electrons with its four closest

neighbors. This sharing of electrons is what ultimately

allows diodes to be build. When dopants from groups

3 or 5 (in most cases) are added to Si, Ge or GaAs it

changes the properties of the material so we are able

to make the P- and N-type materials that become the

diode.

Si

+4

Si

+4

Si

+4

Si

+4

Si

+4

Si

+4

Si

+4

Si

+4

Si

+4

The diagram above shows the

2D structure of the Si crystal.

The light green lines

represent the electronic

bonds made when the valence

electrons are shared. Each Si

atom shares one electron with

each of its four closest

neighbors so that its valence

band will have a full 8

electrons.

16-03-2015

2

N-Type Material

N-Type Material: When extra valence electrons are introduced

into a material such as silicon an n-type

material is produced. The extra valence

electrons are introduced by putting

impurities or dopants into the silicon. The

dopants used to create an n-type material

are Group V elements. The most commonly

used dopants from Group V are arsenic,

antimony and phosphorus.

The 2D diagram to the left shows the extra

electron that will be present when a Group V

dopant is introduced to a material such as

silicon. This extra electron is very mobile.

+4 +4

+5

+4

+4 +4 +4

+4 +4

P-Type Material

P-Type Material: P-type material is produced when the dopant

that is introduced is from Group III. Group

III elements have only 3 valence electrons

and therefore there is an electron missing.

This creates a hole (h+), or a positive charge

that can move around in the material.

Commonly used Group III dopants are

aluminum, boron, and gallium.

The 2D diagram to the left shows the hole

that will be present when a Group III dopant

is introduced to a material such as silicon.

This hole is quite mobile in the same way the

extra electron is mobile in a n-type material.

+4 +4

+3

+4

+4 +4 +4

+4 +4

The PN Junction Steady State1

P n

- - - - - -

- - - - - -

- - - - - -

- - - - - -

- - - - - -

+ + + + + +

+ + + + + +

+ + + + + +

+ + + + + +

+ + + + + +

Na Nd

Metallurgical

Junction

Space Charge Region ionized

acceptors ionized

donors

E-Field

+ + _

_

h+ drift h+ diffusion e- diffusion e- drift = =

16-03-2015

3

The PN Junction Steady State

P n

- - - - -

- - - - -

- - - - -

- - - - -

+ + + + +

+ + + + +

+ + + + +

+ + + + +

Na Nd

Metallurgical

Junction

Space Charge Region ionized

acceptors ionized

donors

E-Field

+ + _

_

h+ drift h+ diffusion e- diffusion e- drift = = = =

When no external source

is connected to the pn

junction, diffusion and

drift balance each other

out for both the holes

and electrons

Space Charge Region: Also called the depletion region. This region includes

the net positively and negatively charged regions. The space charge region

does not have any free carriers. The width of the space charge region is

denoted by W in pn junction formula’s.

Metallurgical Junction: The interface where the p- and n-type materials meet.

Na & Nd: Represent the amount of negative and positive doping in number of

carriers per centimeter cubed. Usually in the range of 1015 to 1020.

The Biased PN Junction

P n

+ _

Applied

Electric Field

Metal

Contact

“Ohmic

Contact”

(Rs~0)

+ _

Vapplied

I

The pn junction is considered biased when an external voltage is applied.

There are two types of biasing: Forward bias and Reverse bias.

These are described on then next slide.

The Biased PN Junction

Forward Bias: In forward bias the depletion region shrinks slightly in width.

With this shrinking the energy required for charge carriers to cross the

depletion region decreases exponentially.

Therefore, as the applied voltage increases, current starts to flow

across the junction.

The barrier potential of the diode is the voltage at which appreciable

current starts to flow through the diode.

The barrier potential varies for different materials.

Reverse Bias: Under reverse bias the depletion region widens.

This causes the electric field produced by the ions to cancel out the

applied reverse bias voltage.

A small leakage current, Is (saturation current) flows under reverse bias

conditions.

This saturation current is made up of electron-hole pairs being

produced in the depletion region.

Saturation current is sometimes referred to as scale current because of

it’s relationship to junction temperature.

Vapplied > 0

Vapplied < 0

16-03-2015

4

Properties of Diodes Figure – The Diode Transconductance Curve

• VD = Bias Voltage

• ID = Current through

Diode. ID is Negative

for Reverse Bias and

Positive for Forward

Bias

• IS = Saturation

Current

• VBR = Breakdown

Voltage

• V = Barrier Potential

Voltage

VD

ID (mA)

(nA)

VBR

~V

IS

Properties of Diodes The Shockley Equation

• The transconductance curve on the previous slide is characterized by

the following equation:

ID = IS(eVD/VT – 1)

• As described in the last slide, ID is the current through the diode, IS is

the saturation current and VD is the applied biasing voltage.

• VT is the thermal equivalent voltage and is approximately 26 mV at room

temperature. The equation to find VT at various temperatures is:

VT = kT q

k = 1.38 x 10-23 J/K T = temperature in Kelvin q = 1.6 x 10-19 C

• is the emission coefficient for the diode. It is determined by the way

the diode is constructed. It somewhat varies with diode current.

• For a silicon diode is around 2 for low currents and goes down to

about 1 at higher currents

Diode Circuit Models

The Ideal Diode

Model

The diode is designed to allow current to flow in only one

direction.

The perfect diode would be a perfect conductor in one

direction (forward bias) and a perfect insulator in the other

direction (reverse bias).

In many situations, using the ideal diode approximation is

acceptable. Example: Assume the diode in the circuit below is ideal. Determine the

value of ID if a) VA = 5 volts (forward bias) and b) VA = -5 volts (reverse

bias)

+

_ VA

ID

RS = 50 a) With VA > 0 the diode is in forward bias

and is acting like a perfect conductor so:

ID = VA/RS = 5 V / 50 = 100 mA

b) With VA < 0 the diode is in reverse bias

and is acting like a perfect insulator,

therefore no current can flow and ID = 0.

16-03-2015

5


The Ideal Diode with

Barrier Potential This model is more accurate than the simple ideal

diode model because it includes the approximate

barrier potential voltage.

Remember the barrier potential voltage is the

voltage at which appreciable current starts to

flow. Example: To be more accurate than just using the ideal diode model

include the barrier potential. Assume V = 0.3 volts (typical for a

germanium diode) Determine the value of ID if VA = 5 volts (forward bias).

+

_ VA

ID

RS = 50 With VA > 0 the diode is in forward bias

and is acting like a perfect conductor

so write a KVL equation to find ID:

0 = VA – IDRS - V

ID = VA - V = 4.7 V = 94 mA

RS 50

V

+

V

+


The Ideal Diode

with Barrier

Potential and

Linear Forward

Resistance

This model is the most accurate of the three.

It includes a linear forward resistance that is calculated from the

slope of the linear portion of the transconductance curve.

However, this is usually not necessary since the RF (forward

resistance) value is pretty constant.

For low-power germanium and silicon diodes the RF value is

usually in the 2 to 5 ohms range, while higher power diodes have

a RF value closer to 1 ohm.

Linear Portion of

transconductance

curve

VD

ID

Δ VD

Δ ID

RF = ΔVD

Δ ID

+ V

RF


The Ideal Diode

with Barrier

Potential and

Linear Forward

Resistance

Example: Assume the diode is a low-power diode

with a forward resistance value of 5 ohms. The

barrier potential voltage is still: V = 0.3 volts (typical

for a germanium diode) Determine the value of ID if

VA = 5 volts.

+

_ VA

ID

RS = 50

V

+

RF

Once again, write a KVL equation

for the circuit:

0 = VA – IDRS - V - IDRF

ID = VA - V = 5 – 0.3 = 85.5 mA

RS + RF 50 + 5

16-03-2015

6


Values of ID for the Three Different Diode Circuit Models

Ideal Diode

Model

Ideal Diode

Model with

Barrier

Potential

Voltage

Ideal Diode

Model with

Barrier

Potential and

Linear Forward

Resistance

ID 100 mA 94 mA 85.5 mA

These are the values found in the examples on previous

slides where the applied voltage was 5 volts, the barrier

potential was 0.3 volts and the linear forward resistance

value was assumed to be 5 ohms.

The Q Point

The operating point or Q point of the diode is the quiescent or no-signal condition.

The Q point is obtained graphically and is really only needed when the applied

voltage is very close to the diode’s barrier potential voltage.

The example below that is continued on the next slide, shows how the Q point is

determined using the transconductance curve and the load line.

+

_ VA

= 6V

ID

RS = 1000

V

+

First the load line is found by substituting in different

values of V into the equation for ID using the ideal

diode with barrier potential model for the diode. With

RS at 1000 ohms the value of RF wouldn’t have much

impact on the results.

ID = VA – V

RS

Using V values of 0 volts and 1.4 volts we obtain ID

values of 6 mA and 4.6 mA respectively.

Next we will draw the line connecting these two points

on the graph with the transconductance curve. This

line is the load line.

The Q Point

ID (mA)

VD (Volts)

2

4

6

8

10

12

0.2 0.4 0.6 0.8 1.0 1.2 1.4

The

transconductance

curve below is for a

Silicon diode. The

Q point in this

example is located

at 0.7 V and 5.3 mA.

4.6

0.7

5.3

Q Point: The intersection of the

load line and the

transconductance curve.

16-03-2015

7

Dynamic Resistance

The dynamic resistance of the diode is mathematically determined as the

inverse of the slope of the transconductance curve. Therefore, the

equation for dynamic resistance is:

rF = VT

ID

The dynamic resistance is used in determining the voltage drop across

the diode in the situation where a voltage source is supplying a

sinusoidal signal with a dc offset (when a waveform has unequal amounts of

signal in the positive and negative domains).

The ac component of the diode voltage is found using the following

equation:

vF = vac rF

rF + RS The voltage drop through the diode is a combination of the ac and dc

components and is equal to:

VD = V + vF

Dynamic Resistance

Example: Use the same circuit used for the Q point example but change

the voltage source so it is an ac source with a dc offset. The source

voltage is now, vin = 6 + sin(wt) Volts. It is a silicon diode so the barrier

potential voltage is still 0.7 volts.

+

vin

ID

RS = 1000

V

+

The DC component of the circuit is the

same as the previous example and

therefore ID = 6V – 0.7 V = 5.2 mA

1000

rF = VT = 1 * 26 mV = 4.9

ID 5.3 mA

= 1 is a good approximation if the dc

current is greater than 1 mA as it is in this

example.

vF = vac rF = sin(wt) V 4.9 = 4.88 sin(wt) mV

rF + RS 4.9 + 1000

Therefore, VD = 700 + 4.9 sin (wt) mV (the voltage drop across the

diode)

Types of Diodes and Their Uses

PN Junction

Diodes:

Are used to allow current to flow in one direction

while blocking current flow in the opposite

direction. The pn junction diode is the typical diode

that has been used in the previous circuits.

A K

Schematic Symbol for a PN

Junction Diode

P n

Representative Structure for

a PN Junction Diode

Zener Diodes: Are specifically designed to operate under reverse

breakdown conditions. These diodes have a very

accurate and specific reverse breakdown voltage.

A K

Schematic Symbol for a

Zener Diode

16-03-2015

8


Schottky

Diodes:

These diodes are designed to have a very fast

switching time which makes them a great diode for

digital circuit applications. They are very common

in computers because of their ability to be switched

on and off so quickly. A K


Schottky Diode

Shockley

Diodes:

The Shockley diode is a four-layer diode while other

diodes are normally made with only two layers.

These types of diodes are generally used to control

the average power delivered to a load.

A K


four-layer Shockley Diode


Light-Emitting

Diodes:

Light-emitting diodes are designed with a very large

bandgap so movement of carriers across their

depletion region emits photons of light energy.

Lower bandgap LEDs (Light-Emitting Diodes) emit

infrared radiation, while LEDs with higher bandgap

energy emit visible light. Many stop lights are now

starting to use LEDs because they are extremely

bright and last longer than regular bulbs for a

relatively low cost.

A K


Light-Emitting Diode

The arrows in the LED

representation indicate

emitted light.


Photodiodes: While LEDs emit light, Photodiodes are sensitive to

received light. They are constructed so their pn

junction can be exposed to the outside through a

clear window or lens.

In Photoconductive mode the saturation current

increases in proportion to the intensity of the

received light. This type of diode is used in CD

players.

In Photovoltaic mode, when the pn junction is

exposed to a certain wavelength of light, the diode

generates voltage and can be used as an energy

source. This type of diode is used in the

production of solar power.

A K

A K

Schematic Symbols for

Photodiodes

16-03-2015

9

Diodes Circuits

Key Words:

Diode Limiter

multi diode Circuits

Rectifier Circuits

+ +

-

vi vo vo

-

+

R D

t

t

Von

vi

vo

When vi > Von , D on vo vi；

vi < Von， D off vo = 0。

multiple diodes

Circuits

R

+5V

V2

V1 Vo

D1

D2

V1(V)

V2(V)

Vo(V)

Logic

output 0

0

0.7

0

5

0

0.7

0

0

5

0.7

0

5

5

5

1

16-03-2015

10

Rectifier Circuits

One of the most important applications of diodes is in the

design of rectifier circuits. Used to convert an AC signal into

a DC voltage used by most electronics.

Rectifier Circuits

Simple Half-Wave Rectifier

What would the

waveform

look like if not an ideal

diode?

Rectifier Circuits

Bridge Rectifier

Looks like a Wheatstone bridge. Does not require a center

tapped transformer.

Requires 2 additional diodes and voltage drop is double.

16-03-2015

11

Diode Circuits:

Applications

Applications – Rectifier Circuits

Half-Wave Rectifier Circuits

Applications – Rectifier Circuits

Battery-Charging Circuit

16-03-2015

12

Half-Wave Rectifier with Smoothing Capacitor

Large Capacitance

i=dq/dt or Q = IL T

Q = Vr C

then

C ~ (ILT) / Vr

Half-Wave Rectifier with Smoothing Capacitor

Large Capacitance

Forward bias

charge cycle

Reverse bias

discharge cycle

Start

Vr Peak-to-peak riple voltage

i=dq/dt or Q = IL T

Q = Vr C

then

C ~ (ILT) / Vr

typically :VL ~V m- (Vr /2)

Full-Wave rectifier Circuits

The sources are out of phase

16-03-2015

13

Wave Shaping Circuits Clipper Circuits

Batteries replaced by Zener diodes

+ 600

mV

I flow

below 600 mV

I flow

Above 600 mV

- 600

mV

Half-Wave Limiter Circuits

Current flows thru the resistor until +600 mV is reached, then

flows thru the Diode.

The plateau is representative of the voltage drop of the diode

while it is conducting.

Voltage

divider

Linear Small Signal Equivalent Circuits (1)

When considering electronic circuits in which dc supply voltages are used to bias a nonlinear devices at their operating points and a small ac signal is injected into the circuit to find circuit response:

Split the analysis of the circuit into two parts:

(a)analyze the dc circuit to find the operating point

(b)consider the small ac signal

16-03-2015

14


Since virtually any nonlinear ch-tic is approximately linear (straight) if we consider a sufficiently small

segment

THEN

We can find a linear small-signal equivalent circuit for the nonlinear device to use in the ac analysis

The small signal diode circuit can be substituted by a single equivalent resistor.


dc supply voltage results in operation at Q

An ac signal is injected into the circuit and

swings the instantaneous point of operation

slightly above and below the Q point

For small changes

D

QD

DD v

dv

dii

iD –the small change in diode current from the Q-point

vD –the small change in diode voltage from the Q-point

(diD/dvD) – the slope of the diode ch-tic evaluated at the point Q


dc supply voltage results in operation at Q

An ac signal is injected into the circuit and

swings the instantaneous point of operation

slightly above and below the Q point

For small changes

D

QD

DD v

dv

dii

iD –the small change in diode current from the Q-point

vD –the small change in diode voltage from the Q-point

(diD/dvD) – the slope of the diode ch-tic evaluated at the point Q

1

QD

DD

dv

dir

Dynamic resistance of the diode

D

D

Dr

vi

16-03-2015

15

From small signal diode analysis

q

kTV

nV

vIi

T

T

d

sD

1exp Differentiating

the Shockley eq.

T

D

T

S

D

D

nV

v

nVI

dv

diexp

1

… and following the math on p.452 we can write that dynamic resistance of the diode is

DQ

T

DI

nVr


T

DQ

sDQnV

vII exp~

where

Example - Voltage-Controlled Attenuator

DC control signal

C1, C2 – small or large ?

C in dc circuit – open circuit

C in ac circuit –short circuit

Find the operating point and perform the small signal analysis to obtain the small signal voltage gain

CjZC

1


DC control signal

Dc circuit for Q point (IDQ, VDQ)

DQ

T

DI

nVr Compute at the Q

point (IDQ, VDQ)

16-03-2015

16


The dc voltage source is equivalent to a short circuit for ac signals.

Voltage gain RR

R

v

vA

p

p

in

v

0

Sources

Dailey, Denton. Electronic Devices and Circuits, Discrete and Integrated. Prentice Hall, New

Jersey: 2001. (pp 2-37, 752-753)

2 Figure 1.10. The diode transconductance curve, pg. 7

Figure 1.15. Determination of the average forward resistance of a diode, pg 11

3 Example from pages 13-14

Liou, J.J. and Yuan, J.S. Semiconductor Device Physics and Simulation. Plenum Press,

New York: 1998.

Neamen, Donald. Semiconductor Physics & Devices. Basic Principles. McGraw-Hill,

Boston: 1997. (pp 1-15, 211-234)

1 Figure 6.2. The space charge region, the electric field, and the forces acting on

the charged carriers, pg 213.

Documents

PN Junction Diode - tiiciiitm.com