P-n diodes

7/30/2019 P-n diodes

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Semiconductor p-n junction diodes

p

n


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p-type material

Semiconductor material

doped with acceptors.

Material has high hole

concentration

Concentration of free

electrons in p-type material

is very low.

n-type material

Semiconductor material

doped with donors.

Material has high

concentration of free

electrons.

Concentration of holes in

n-type material is very

low.

p-n junction formation


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p-type material

Contains

NEGATIVELY

charged acceptors

(immovable) and

POSITIVELY chargedholes (free).

Total charge = 0

n-type material

Contains

POSITIVELY charged

donors (immovable)

and NEGATIVELY

charged free electrons.

Total charge = 0


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p-type material

Contains

NEGATIVELY

charged acceptors

(immovable) and

POSITIVELY chargedholes (free).

Total charge = 0

n-type material

Contains

POSITIVELY charged

donors (immovable)

and NEGATIVELY

charged free electrons.

Total charge = 0

What happens if n- and p-type materials are in close contact?



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p- n junction formation


Being free particles, electrons start diffusing from n-type material into p-material

Being free particles, holes, too, start diffusing from p-type material into n-material

Have they been NEUTRAL particles, eventually all the free electrons

and holes had uniformly distributed over the entire compound crystal.

However, every electrons transfers a negative charge (-q) onto the p-

side and also leaves an uncompensated (+q) charge of the donor on the

n-side.

Every hole creates one positive charge (q) on the n-side and (-q) on thep-side


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p- n junction formation


Electrons and holes remain staying close to the p-n junction because

negative and positive charges attract each other.

Negative charge stops electrons from further diffusion

Positive charge stops holes from further diffusion

The diffusion forms a dipole charge layer at the p-n junction interface.

There is a built-in VOLTAGE at the p-n junction interface that prevents

penetration of electrons into the p-side and holes into the n-side.

p-type n-type


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p- n junction current voltage characteristics

What happens when the voltage is applied to a p-n junction?

The polarity shown, attracts holes to the left and electrons to the right.

According to the current continuity law, the current can only flow if all

the charged particles move forming a closed loop

However, there are very few holes in n-type material and there are

very few electrons in the p-type material.There are very few carriers available to support the current through the

junction plane

For the voltage polarity shown, the current is nearly zero

p-type n-type


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p- n junction current voltage characteristics

What happens if voltage of opposite polarity is applied to a p-n junction?

The polarity shown, attracts electrons to the left and holes to the right.

There are plenty of electrons in the n-type material and plenty of holes in

the p-type material.

There are a lot of carriers available to cross the junction.

When the voltage applied is lower than the built-in voltage,

the current is still nearly zero

p-type n-type

When the voltage exceeds the built-in voltage, the current can flow through

the p-n junction


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Diode current voltage (I-V) characteristics

= 1

kTqVII S exp

p n

Semiconductor diode consists of a p-n junction with two

contacts attached to the p- and n- sides

IS is usually a very small current, IS

10-17

10-13

AWhen the voltage V is negative (reverse polarity) the exponential term -1;

The diode current is IS ( very small).

0V

When the voltage V is positive (forward polarity) the exponential term

increases rapidly with V and the current is high.

k Boltzmann constantT junction temperature (K)

Note that at room temperature,

(kT/q) 0.026 V


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The I-V characteristic of the diode

IS

= 1

kT

qVII S exp

exp 1

0.026

S

VI I

=

At room temperature, a simplified form of the I-V characteristic can be used:

where V is in [Volts] and

the term kT/q was substituted with 0.026V


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The experimental I-V characteristic of a Si diode


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p- n diode circuit notation

p

n

1p

kT

qV

IS

1p

kT

qV

IS

When plus is applied to the p-side,

the current is high. This voltage

polarity is called FORWARD.

When plus is applied to the n-side,

the current is nearly zero. This voltage

polarity is called REVERSE.


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p- n diode applications:

current rectifiers

1p

kT

qV

IS

1p

kT

qV

IS

+-

Time

Voltage

+

-

Time

Current


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Why do we need rectifiers?

Hydroelectric Power Stations produce AC voltage

A flux of water

rotates the coils of

the AC generators

and produces AC

voltage


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Many appliances need DC voltage sources


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Diode Rectifiers


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Light emitters

P-n junction can emit the

light when forward biased

p-type n-type

+

-

+

-

Electrons drift into p-material and find plenty of holes there. They

RECOMBINE by filling up the empty positions.

Holes drift into n-material and find plenty of electrons there. They also

RECOMBINE by filling up the empty positions.

The energy released in the process of annihilation produces

PHOTONS the particles of light


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Photon Energy Light wavelength Light Color

Visible light - that which is detectable by the human eye - consists

of wavelengths ranging from approximately

0.780 micrometer down to 0.390 micrometer.

The photon energy - wavelength relation:

][

1.24][

meVEPH

=


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Semiconductor Bandgap Light Color


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Lighting the Future by Solid-State Lighting


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+

-


Photodetectors

P-n junction can detect light

when reverse biased

p-type n-type

When the light illuminates the p-n junction, the photons energy RELEASES free

electrons and holes.

They are referred to as PHOTO-ELECTRONS and PHOTO-HOLES

The applied voltage separates the photo-carriers attracting electrons toward

plus and holes toward minus

As long as the light is ON, there is a current flowing through the p-n junction


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Photodetector applications- Optical

communications


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Photodetector applications- Security systems


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Solar Cells

Solar revolution" is the idea that one day we will all

use free electricity from the sun.

On a bright, sunny day, the sun shines approximately

1,000 watts of energy per square meter of the planet's surface


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Solar cells (contined)

1967 - Soyuz 1 is the first manned spacecraft to be powered by solar

cells

The spacecraft crashed during its return to Earth. One of

the reasons: the left solar panel deployment failure


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Electrical circuit with p-n diodes

Typical LED bias circuit

(forward bias applied)

E

RB

0 0.2 0.4 0.6 0.80

0.005

0.01

0.015

0.02

0.025

0.03

0.035

VD

I

P-n diode I-V characteristic

The p-n diode I-V does not follow the Ohms law. The KVL:

E = VR + VD = I R + VD

1DSqV

I IkT

exp

=

VD

VR

1DSqV

I IkT

exp

=


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Solving a circuit with a p-n diode

method 1: varying VD

E = VR + VD = I R + VD

E

RB

VD

VR

1DSqV

I IkT

exp

=

Assuming we know the diode voltage VD,

we can calculate the diode current.

Knowing the current we can find the VR.

Then we can find VR +VD

The sum must be equal to E.

If it is not, then the assumed VD was wrong.

We need to change the VD and try again.


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method 1: varying VDMATLAB code and results

% Circuit with p-n Diode

% Method 1: varying VD

clear all

close all

%Device and circuit parameters

E=10;

RB=400;

Is=2e-12;

kTq=0.026;

%Solution

Vmax=0.75;

VD=0:0.001:Vmax;

I=Is*(exp(VD/kTq)-1);

VR=I*RB;

Vtot=VD+VR;

dE=abs(E-Vtot);

[Eop,m]=min(dE);

%Results

VDop=VD(m)

Iop=I(m)

VRop=I(m)*RB

%Validation

Eop=VRop+VDop

Results:

VDop = 0.6030

Iop = 0.0236

VRop = 9.4489

Eop = 10.0519


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method 2: Load line

E

RB

VD

VR

1DD SqV

I IkT

exp

=

Considering VD as a variable

we can calculate the diode current ID.

Assuming we know VD, we can find the

resistor current IR using KVL.

R DR

B B

V E VI

R R

= =

In the load line approach two currents: ID and

IR are plotted as functions of VD

In the series circuit, IR

= ID

The intercept of the two plots provides the

operating current.


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method 2: Load line

MATLAB code

% Circuit with p-n Diode

% Method 2: Load Line

clear all

close all

%Device and circuit parameters

E=2;

RB=20;

Is=2e-12;

kTq=0.026;

Vmax=E;

VD=0:0.001:Vmax;

ID=Is*(exp(VD/kTq)-1);

% Load-line

IR=(E-VD)/RB;

dI=abs(IR-ID);

[dImin,m]=min(dI);

IDop=ID(m)

IRop=IR(m)

VDop=VD(m)

plot(VD,ID,VD,IR,VDop,IDop,'ro')

xlabel('Voltage, V')

ylabel('Current, A')

axis([0 E 0 0.25])

grid on


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method 2: Load line

MATLAB results

0 0.5 1 1.5 20

0.05

0.1

0.15

0.2

0.25

Voltage, V

Current

,A

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P-n diodes