SEMICONDUCTOR MATERIALS

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SEMICONDUCTOR MATERIALS
Material Example r (? m) Conductor Copper 10-
6 Semi-conductor Germanium 0.5 Semi-conductor Sil
icon 500 Insulator Mica 1010
UNITS Resistivity, ? is given by r (RA)/L
? m2 / m ? m Conductivity, G is given by
G 1/r ?-1m-1 S (Siemens)
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Germanium and Silicon
Have the same crystal structure as diamond.
Both can be made to purity levels of 1 in 10
billion (1 in 1010) Can significantly change
properties by "doping". Just 1 in 106 impurity
atoms can substantially increase the conductivity.
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Atomic Structure
The atom is composed of three basic particles
electrons, protons, and neutrons The neutrons
and protons form the nucleus. The electrons
revolve around the nucleus in a fixed orbit.
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Atomic Structure
Germanium has 32 orbiting electrons
Silicon has 14 orbiting electrons
(b)
Bohr models of (a) Germanium and (b) Silicon The
potential (ionization potential) required to
remove any of these four valence electrons is
lower than that required for any other electron
in the structure.
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Covalent Bonding
The four valence electrons are bonded to four
adjoining atoms. This bonding of atoms by sharing
of electrons is called covalent bonding.
Covalent bonding of the silicon atom
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Intrinsic semi-conductors
Silicon and germanium when carefully refined to
reduce impurities to a very low level, are called
intrinsic semiconductors. The conductivity for
both pure materials is quite low. An increase in
temperature causes a substantial increase in the
number of free electrons, thus increasing
conductivity. These materials thus have a
negative temperature coefficient of resistance,
(i.e. resistance decreases with temperature).
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Energy Levels
The more distant the electron from the nucleus
the higher the energy state. Any electron that
has left its parent atom to become a "free"
electron has a higher energy state that any
electron in the atomic structure.
Energy levels for isolated atoms
Between the discrete energy levels are gaps in
which no electrons in the isolated atomic
structure can appear.
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Energy Bands
As the atoms of a material are brought closer
together to form the crystal lattice structure
the discrete levels of each atom will merge into
bands.
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Energy Bands
There is a forbidden region between the valence
band and ionization level (conduction
band). Since, W Q V 1 eV (1.6 x 10-19 C)
(1 V) 1.6 x 10-19 J Since Eg is less for Ge
than Si a larger number of valence electrons for
Ge will have sufficient energy to cross the
forbidden gap and become free. Thus Ge will have
a greater conductivity than Si.
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Extrinsic materials
A semiconductor that has been subjected to
"doping" is called an extrinsic material.
Doping can be as low as 1 part in 10 million.
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Extrinsic materials
There are two types of extrinsic material n
type impurity atoms have 5 valence
electrons e.g. antimony, arsenic, and
phosphorus p type impurity atoms have 3
valence electrons e.g. boron, gallium, and
indium
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n type semiconductor
There is an additional free electron for each
impurity atom that is unassociated with any
particular covalent bond. This is a "free"
electron. The structure is electrically neutral
since each Sb atom has an additional proton to
balance the extra electron.
Antimony impurity in n-type material
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n type semiconductor
Effect of donor impurities on the energy band
structure A discrete energy level (the donor
level appears in the forbidden gap). Since the Eg
for this level is small electrons can easily be
excited to the conduction band.
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p type semiconductor
There are now an insufficient number of electrons
to complete the covalent bonds of the new
lattice. The resulting vacancy is called a hole
and is represented by a small circle or positive
sign since the vacancy can accept a free
electron. Again the structure is electrically
neutral.
Boron impurity in p type semiconductor
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p type semiconductor
Effect of acceptor impurities on the energy band
structure
A discrete energy level (the acceptor level
appears in the forbidden gap). Since the Eg for
this level is small electrons can easily be
excited from the valence band, leaving holes in
the valence band.
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Majority and minority carriers
Intrinsic state Free electrons in Ge or Si are
due only to those few electrons in the valence
band that have acquired enough energy to break
the covalent bond or to the few impurities that
could not be removed. Vacancies left behind in
the valence band will be approximately same as
number of free electrons.
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Majority and minority carriers
Extrinsic state For n type number of holes about
same as intrinsic but far more free electrons.
For p type number of holes far outweighs number
of electrons. In n type material the electron is
the majority carrier and the hole the minority
carrier. In p type material the electron is the
minority carrier and the hole the majority carrier
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Semiconductor diode
If the n type and p type are joined together the
electrons and holes in the region of the junction
combine forming a depletion layer.
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Semiconductor diode
This very thin layer (lt10-3 mm) blocks drifting
of electrons and holes. It acts as an
insulator. In order to pass through the deletion
zone, electrons need energy. Thus, the zone acts
as a voltage barrier. This is the junction
voltage which is 0.7 V for Si and 0.3 V for Ge.
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Reverse Bias
Electrons and holes are attracted away from the
junction so it widens. Some minority carriers
will cross the junction and will get a very small
reverse saturation or leakage current IS.
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Forward Bias
If the battery voltage is greater than the
junction voltage majority carriers will cross the
junction and the depletion regions narrows
considerably. As applied bias increases
depletion region gets smaller and get a flood of
electrons resulting in an exponential increase in
the number of electrons crossing the barrier.
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Diode Current vs Diode Voltage
The current ID through a diode is calculated
using where IS reverse saturation
current k 11,600/? with ? 1 for Ge and ? 2
for Si at low current levels, ? 1 for both at
high current levels TK TC 273o
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Zener Region
As the reverse bias voltage across the diode
increases reach a point where valence electrons
absorb sufficient energy to reach ionization and
get an avalanche breakdown. The voltage where
this occurs is called the Zener Voltage, VZ.
The maximum reverse bias potential that can be
applied before entering the Zener region is
called the peak inverse voltage (PIV) or peak
reverse voltage (PRV).
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Silicon vs Germanium
PIV ratings for Si diodes are about 1000 V and
about 400 V for Ge diodes. Si diodes can be used
up to 200o C, Ge diodes up to 100o C. However,
Si has a junction voltage of 0.7 V, and Ge has a
junction voltage of 0.3 V.
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Temperature Effects
The reverse saturation current IS will double for
every 10oC increase in temperature. Threshold
voltage levels decrease with increasing
temperature i.e forward characteristics become
more ideal.
Effect of temperature on Si diode
characteristics
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Temperature Effects
Ge has much higher increase in IS with
temperature increase. Thus Ge is not good at
higher temperatures.
Effect of temperature on Si diode
characteristics
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Diode Resistance
As operating point of diode changes from one
region to another resistance changes
dramatically. (Very high in reverse bias, low in
forward bias). Also, due to non-linear nature of
curve resistance changes.
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Diode Resistance
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Diode Equivalent Circuits
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Load Line Analysis
Diode and resistor are in a series circuit with
DC power supply E. Using the Kirchoff voltage
law E VD VR VD IDR or, solving
for ID, gives the load-line equation in terms of
current
The diode characteristic curve is shown for
forward bias at the left.
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Load Line Analysis
As the load-line equation is linear we need only
two points If VD 0 then ID E/R and if ID
0 then VD E this gives two points on the load
line, and is shown in the next slide.
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Load Line Analysis
The point of intersection between the network
line and the device curve is the operating point
Q for the device and the circuit. The Q point is
the quiescent point
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Load Line Analysis
Alternatively, can find Q mathematically by
solving two equations
the equation for the
device and, E VD IDR the load
line equation This mathematics involves
non-linear methods which is complex and time
consuming. Load-line analysis is not only much
simpler but provides a "pictorial" solution for
finding the Q point values. We will use this
method again for transistors later on in the
course.
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Example 1
To construct the load line need two points on
axes. If VD 0 then ID E/R 10V/1k? 10
mA and, if ID 0 then VD E 10V
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Example 1
The coordinates of the Q point are shown on the
graph. For greater accuracy need a larger scale
plot.
To obtain VR can use VR I R (9.25 mA)(1 k?)
9.25 V or, VR E VD 10 0.78 V
9.22 V Difference in results is due to the
accuracy in reading the graph.
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Example 2
Repeat analysis for Example 1 using R 2 k? If
VD 0 then ID E/R (10 V)/(2 k?) 5 mA and,
if ID 0 then VD E 10V. Construct a new
load line as shown in next slide
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Example 2
Intersection of load line with curve gives new Q
point.
Then, VR I R (4.6 mA)(2 k?) 9.2 V or,
VR E VD 10 0.7 V 9.3 V Again,
difference in results is due to the accuracy in
reading the graph.
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Example 3
Repeat Example 1 using the practical diode model.
Q point values obtained are almost the same.
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Example 4
Repeat Example 2 using the practical diode model.
Q point values obtained are almost the same.
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Example 5
Repeat Example 1 using the ideal diode model.
Q point values obtained are not as good.
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Summary
Practical model gives quite accurate results and
is quite simple. Although even simpler, the
ideal model is only good for when E gtgt
VT. Throughout the course we will almost always
use the practical model.