Semiconductor Basics - Ccsu

woundcallousSemiconductor

Nov 1, 2013 (4 years and 12 days ago)

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Semiconductor Basics

Chapter 1

Atomic Structure


Elements are made of atoms


110 Elements; each has an atomic
structure


Today,
quarks

and leptons, and their
antiparticles, are candidates for being
the
fundamental

building blocks from
which all else is made!


Bohr Model


Atoms have planetary structure


Atoms are made of
nucleus

(Protons
(+) & Neutrons) and
electrons

(
-
)





110 th element is called Darmstadtium (Ds)

Atomic Structure


Atoms go around the nucleolus in their
orbits



discrete distances


Each orbit has some energy level


The closer the orbit to the nucleus the less
energy it has


Group of orbits called
shell


Electrons on the same shell have similar energy
level


Valence shell

is the outmost shell


Valence shell has
valence electrons

ready to
be freed


Number of electrons (Ne) on each shell (n)




First shell has 2 electrons


Second shell has 8 electrons (not shown here)



Ne = 2n
2

Valence Shell


Atoms are made of
valence
shell

and
core


Core includes nucleus and
other inner shells


For a Carbon atom the atomic
number is 6


Core charge = 6 P + 2 e =
(+6)+(
-
2)=(+4)


Remember the first shell has 2
electrons

Elements


Basic categories


Conductors


Examples: Copper, silver


One valence electron , the
e

can easily be freed


Insulators


Valence electrons are tightly
bounded to the atom


Semiconductors


Silicon, germanium (single
element)


Gallium arsenide, indium
phosphide (compounds)


They can act as conductors
or

insulators


Conduction band

is
where the electron
leaves the valence shell
and becomes free

Valence band

is where
the outmost shell is

Always free
electrons

Free
electrons

Semiconductors


Remember the further away from the
nucleus the less energy is required
to free the electrons


Germanium is less stable


Less energy is required to make the
electron to jump to the conduction
band




When atoms combine to form a
solid
,
they arrange themselves in a
symmetrical patterns


Semiconductor atoms (silicon) form
crystals


Intrinsic

crystals have no impurities



Conduction Electrons and Holes


Electrons exist only within
prescribed
energy bands



These bands are separated by
energy gaps



When an electron jumps to the
conduction band it causes a
hole


When electron falls back to its initial
valence
recombination

occurs


Consequently there are two
different types of currents


Hole current (electrons are the
minority carriers)


Electron current (holes are the
minority carriers)

Remember: We are interested in electrical current!

Doping


By adding impurities to the intrinsic
semiconductor we can change the
conductivity of the material


this is called
doping


N
-
type doping


P
-
type doping


N
-
type:
pentavalent
(atom with 5 valence
electrons) impurity atoms are added


[Sb(Antimony) + Si]


Negative charges (electrons) are generated




N
-
type has lots of free electrons


P
-
type: trivalent (atom with 3 valence
electrons) impurity atoms are added


[B(Boron) + Si]


Positive charges (holes) are generated


P
-
type has lots of holes




Diodes


N region has lots of free electrons


P region has lots of holes


At
equilibrium
: total number positive and negative
charges is the same (@ room temp)


At the pn junction the electrons and holes with
different charges form an
electric field


In order to move electrons through the electric field
(generate current) we need some force (voltage)


This potential difference is called
barrier voltage


When enough voltage is applied such that electrons
are moved then we are
biasing

the diode


Two layers of positive and negative charges for
depletion region



the region near the pn
-
junction is
depleted of charge carriers)


Biasing Types of a Diode


Forward bias


Bias voltage V
Bias

> barrier voltage
V
Bar


Reduction in
+

and


ions


smaller
depletion region



V
Bar
Depends on material, doping,
temp, etc. (e.g., for silicon it is 0.7 V)


Reverse bias


Essentially a condition that prevents
electrons to pass through the diode


Very small reverse break down current


Larger

depletion region is generated


Cathode

n region

Anode

p region

Connected to the
negative
side of
the battery

Connected to the
positive
side of
the battery

A

K

Biasing Types of a Diode (Forward)

Cathode

n region

Anode

p region

A

K

Moving

electrons

Small dynamic resistance

VBias

n

p

Conventional
Current Flow

Conventional
Current Flow

I (Forward)

Very Small

Moving

Electrons:

Reverse Current)

Biasing Types of a Diode (Reverse)

Cathode

n region

Anode

p region

A

K

Large resistance

VBias

n

p

Conventional
Current Flow

Holes are left
behind; large
depletion region

Instant pull of
electrons

I
-
V Characteristic of a Diode


Forward bias: current passes through


The
knee

is where VBias=VBar


At point B VBias < VBar


Very little current


Note that at the knee the current increases rapidly but V(forward)
stays almost the same



Reveres bias: No current passes through


When VBias < VBar


Very little current (mu or nano Amp)


At the
knee
, the reverse current increases rapidly but the reverse
voltage remains almost the same


Large reverse current can result in overheating and possibly
damaging the diode (V=50V or higher typically)



Overheating results from high
-
speed electrons in the p
-
region knocking out electrons of atoms in n
-
region from their
orbit to the conduction band


Hence, we use limiting resistors




Electrons moving

from n to p region

Modeling a Diode (Forward Biasing)

Use r’d

(internal resistance)

-

Not linear!

Complete Modeling of a Diode

Note that IF is the
actual

direction of electron current

Forward bias: V
Bias

= V
F

+ I
F
(R
LIMIT
+r
d
); r
d

is typically given, V
F

typically is 0.7 V

Reverse bias: V
Bias

= V
R

+ I
R

* R
LIMIT
; I
R

is typically given


V
R

V
F

Showing the Actual electron direction

Example

Find the current through the diode and the
voltage across the resistor.

Assume r
d

= 10 ohm


Biasing?

Forward bias: V
Bias

= V
F

+ I
F
(R
LIMIT
+r
d
)

10 = 0.7 + IF(RLIMIT+10)


I
F
=9.21 mA

V
F
=0.7+I
F
*r
d
= 792 mV

V
RLIMIT

= I
F

* R
LIMIT

= 9.21V

V
F

Forward biased

Example

Find the current through the diode and the
voltage across the resistor.

Assume I
R

= I uA


Note:
Reverse biased

V
R

Reverse bias: V
Bias

= V
R

+ I
R

* R
LIMIT

V
RLIMIT

= I
R
*V
RLI MIT

= 1mA

V
R
=V
BIAS
-
V
RLIMIT
=4.999 V

Forward Bias

Calculate the voltage across the resistor.

Reverse Bias

Calculate the voltage across the resistor.

Do this example on your own:

Make sure you can calculate and

find all currents
-

Hint: find Vn, first

Vn

Vn

i
1

i
2

i
3

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