© Electronics ECE 1312 Chapter 2 Semiconductor Materials and Diodes.

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© Electronics © Electronics ECE 1312 ECE 1312 Chapter 2 Chapter 2 Semiconductor Semiconductor Materials and Materials and Diodes Diodes

Transcript of © Electronics ECE 1312 Chapter 2 Semiconductor Materials and Diodes.

Page 1: © Electronics ECE 1312 Chapter 2 Semiconductor Materials and Diodes.

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Chapter 2Chapter 2Semiconductor Materials Semiconductor Materials

and Diodesand Diodes

Page 2: © Electronics ECE 1312 Chapter 2 Semiconductor Materials and Diodes.

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Classification according to the way materials react to the current when a voltage is applied across them:

Insulators Materials with very high resistance - current can’t flow mica, rubber

Conductors Materials with very low resistance – current can flow easily copper, aluminum

Semiconductors Neither good conductors nor insulators (silicon, germanium) Can be controlled to either insulators by increasing their

resistance or conductors by decreasing their resistance

Classification of MaterialsClassification of Materials

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● An atom is composed of a nucleus, which contains positively charged protons and neutral neutrons, and negatively charged electrons that orbit the nucleus.

● Electrons in the outermost shell are called valence electrons.

Semiconductor Materials and Semiconductor Materials and PropertiesProperties

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A portion of the periodic table in which the more common semiconductors are found

● Elemental Semiconductors Silicon (Si) and germanium (Ge) are in group IV. Hence, they have 4 electrons in their outer shells

Do you still remember?A stable atoms need ? electrons at its outermost shell8

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• Si have 4 electrons in their outer shells

• needs another 4 to become stable

• So, when there are 4 other Si nearby = 4 electrons:

● Atoms come into close proximity to each other and so the valence electrons interact to form a crystal.

Si SiSi

Si

Si

Sharing of electrons occurred; and this bond is known as the covalent bond

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BANDGAP ENERGY, Eg

• Now, in order to break the covalent bond, a valence electron must gain enough energy to become free electrons.

• The minimum energy required is known as the bandgap energy, Eg

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ILLUSTRATION WHEN A VALENCE ELECTRON IS FREE

1. Becomes free electron

2. Becomes empty

3. Electron moves to fill space

4. Becomes empty

5. Electron moves to fill space

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Intrinsic SemiconductorIntrinsic Semiconductor

● Intrinsic Semiconductor A single-crystal semiconductor material with no other types of

atoms within the crystal.

The densities of electrons and holes are equal.

The notation ni is used as intrinsic carrier concentration for the concentration of the free electrons as well as that of the hole:

B = a coefficient related to the specific semiconductor materialEg = the bandgap energy (eV)T = the temperature (Kelvin) remember that K = °C + 273.15 k = Boltzmann’s constant (86 x 10-6 eV/K)

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● The values of B and Eg for several semiconductor materials:

Example: Calculate the intrinsic carrierconcentration in silicon at T = 300 K.

Intrinsic SemiconductorIntrinsic Semiconductor

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• Example 2Find the intrinsic carrier concentration of Gallium Arsenide at temperature = 300K

Answer: 1.8 x 106 cm-3

k = Boltzmann’s constant (86 x 10-6 eV/K)

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Example 3

Answer: 1.4 eV

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Extrinsic SemiconductorExtrinsic Semiconductor

• Since intrinsic concentration, ni is very small, so, very small current is possible

• So, to increase the number of carriers, impurities are added to the Silicon/Germanium.

• The impurities will be from Group V and Group III

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• Group V – 5 electrons in the outer shell; Example, Phosphorus, Arsenic

• The 5th electron are loosely bound to the Phosphorus atom• Hence, even at room temperature, the electron has enough

energy to break away and becomes free electron.• Atoms from Group V are known as donor impurity (because

it donates electrons)

Group V + Si = n-type semiconductor

Extra electron

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• Group III – 3 electrons in the outer shell; Example, Boron• The valence electron from outer shells are attracted to fill

the holes added by the insertion of Boron• Hence, we have movement of holes• Atoms from Group III are known as acceptor impurity

(because it accept electrons)

Group III + Si = p-type semiconductor

Extra hole

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– The materials containing impurity atoms are called extrinsic semiconductors, or doped semiconductors.

– Effects of doping process• controls the concentrations of free electrons and holes• determines the conductivity and currents in the materials.

– The relation between the electron and hole concentrations in thermal equilibrium:

no = the thermal equilibrium concentration of free electrons

po = the thermal equilibrium concentration of holes ni = the intrinsic carrier concentration

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For N-type – electrons are the For N-type – electrons are the majority carriersmajority carriers

At room temperature (T = 300 K), each donor atom donates a free electron to the semiconductor.

• If the donor concentration Nd is much larger than the intrinsic concentration, approximately:

• Then, the hole concentration:

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For P-type – holes are the For P-type – holes are the majority carriersmajority carriers

Similarly, at room temperature, each acceptor atom accepts a valence electron, creating a hole.

• If the acceptor concentration Na is much larger than the intrinsic concentration, approximately:

• Then, the electron concentration:

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Example 1Calculate the thermal equilibrium electron and hole concentrations.

Consider silicon at T = 300 K doped with phosphorous at a concentration of Nd = 1016 cm-3 and ni = 1.5 x 1010 cm-3.

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Example 2Calculate the majority and minority carrier concentrations in silicon at T = 300K if

a) Na = 1017cm-3

b) Nd = 5 x 1015cm-3

1. Calculate ni

2. For part (a) – it is p-type

3. For part (b) – it is n-type

Answer: a) majority = 1017cm-3 minority 2.25x 103 cm-3

b) Majority 5 x 1015cm-3, minority 4.5 x 104 cm-3

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• EXAMPLE 1Calculate the intrinsic carrier concentration of Silicon at T = 250K

• EXAMPLE 2A silicon is doped with 5 x 1016

arsenic atomsa) Is the material n-type or p-type?b) Calculate the electrons and holes

concentration of the doped silicon at T=300K

k = Boltzmann’s constant (86 x 10-6 eV/K)

Answer: ni = 1.6 x 108 cm-3

Answer: a)n-typeb)no = 5 x 1016 cm-3 and po = 4.5 x 103 cm-3