SEMICONDUCTOR MATERIALS

Material Conductor Semi-conductor Semi-conductor Insulator

UNITS

Example Copper Germanium Silicon Mica  ( 10 0.5

-6 500 10 10 m) Resistivity,  is given by:  = ( RA)/L =  m 2 / m =  m Conductivity, G is given by: G = 1/  =  -1 m -1 = S (Siemens) 1

Germanium and Silicon

Have the same crystal structure as diamond.

Both can be made to purity levels of 1 in 10 billion (1 in 10 10 ) Can significantly change properties by "doping". Just 1 in 10 6 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 5

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. 8

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 E g 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 11

n type semiconductor

Antimony impurity in n-type material 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.

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Energy

n type semiconductor

E g = 0.05 eV (Si) 0.01 eV (Ge) Donor energy level E g as before Valence Band Valence Band Effect of donor impurities on the energy band structure A discrete energy level (the

donor level

appears in the forbidden gap). Since the E g 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 14

p type semiconductor

Energy Acceptor energy level Conduction Band E g as before E g = 0.05 eV (Si) 0.01 eV (Ge) Valence Band Effect of acceptor impurities on the energy band structure A discrete energy level (the

acceptor level

appears in the forbidden gap). Since the E g for this level is small electrons can easily be

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

Donor atoms (Phosphorus) P P _ _ + _ _ P + _ Majority carriers Acceptor atoms (Boron) + B _ + _ + B + B Minority carriers n type semiconductor p type semiconductor 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 (<10 -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

I majority = 0 I S + P + N + depletion zone

+

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 20 leakage current I S .

Forward Bias

+ P + I majority I S ] I D = I majorit y - I S N depletion zone I D

+ -

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 21 electrons crossing the barrier.

Diode Current vs Diode Voltage

The current I D through a diode is calculated using: I D = where: I S = reverse saturation current K  1) k = 11,600/  for Ge and  with  low current levels,  = 1 = 2 for Si at = 1 for both at high current levels T K = T C + 273 o 22

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, V Z .

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 200 o C, Ge diodes up to 100 o 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 I S will double for every 10 o C increase in temperature.

Threshold voltage levels decrease with increasing temperature i.e forward characteristics become more ideal.

Effect of temperature on Si diode characteristics 25

Temperature Effects

Ge has much higher increase in I S with temperature increase. Thus Ge is not good at higher temperatures.

Effect of temperature on Si diode characteristics 26

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 = V D + V R = V D + I D R or, solving for I D, gives the load-line equation in terms of current:  E - V D R 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 V D = 0 then I D = E/R and if I D = 0 then V D = 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: I D = K  1) the equation for the device and, E = V D + I D R 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 V D = 0 then I D = E/R = 10V/1k  = 10 mA and, if I D = 0 then V D = E = 10V 34

Example 1

The coordinates of the Q point are shown on the graph. For greater accuracy need a larger scale plot.

To obtain V R can use: V R = I R = (9.25 mA)(1 k  ) = 9.25 V or, V R = E – V D = 10 – 0.78 V = 9.22 V Difference in results is due to the accuracy in reading the graph. 35

Example 2

Repeat analysis for

Example 1

using

R = 2 k

 If V D = 0 then I D = E/R = (10 V)/(2 k  ) = 5 mA and, if I D = 0 then V D = E = 10V. Construct a new load line as shown in next slide: 36

Example 2

Intersection of load line with curve gives new Q point.

Then, V R = I R = (4.6 mA)(2 k  ) = 9.2 V or, V R = E – V D = 10 – 0.7 V = 9.3 V Again, difference in results is due to the accuracy in reading the graph. 37

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 >> V T .

Throughout the course we will almost always use the practical model.

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