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

© Electronics ECE 1312 EECE 1312 Classification according to the way materials react to the current when a voltage is applied across them: l Insulators  Materials with very high resistance - current can’t flow  mica, rubber l Conductors  Materials with very low resistance – current can flow easily  copper, aluminum l 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 Materials

© Electronics ECE 1312 EECE 1312 ● 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 Properties

© Electronics ECE 1312 EECE 1312 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 shell 8

© Electronics ECE 1312 EECE 1312 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 Sharing of electrons occurred; and this bond is known as the covalent bond

© Electronics ECE 1312 EECE 1312 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

© Electronics ECE 1312 EECE 1312 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

© Electronics ECE 1312 EECE 1312 Intrinsic 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 n i 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 material E g = the bandgap energy (eV) T = the temperature (Kelvin) remember that K = °C k = Boltzmann’s constant (86 x eV/K)

© Electronics ECE 1312 EECE 1312 Intrinsic Semiconductors Cont. The energy difference between the minimum conduction band energy and the maximum valance band energy is called bandgap energy. Semiconductor Constants MaterialBandgap Energy (eV)B (cm -3 K -3/2 ) Silicon (Si) × Germanium (Ge) × Gallium Arsenide (GaAs) × 10 14

© Electronics ECE 1312 EECE 1312 ● The values of B and E g for several semiconductor materials: EXAMPLE 1 Calculate the intrinsic carrier concentration in silicon at T = 300 K.

© Electronics ECE 1312 EECE 1312 EXAMPLE 2 Find the intrinsic carrier concentration of Gallium Arsenide at T = 300K Answer: 1.8 x 10 6 cm -3 k = Boltzmann’s constant (86 x eV/K)

© Electronics ECE 1312 EECE 1312 EXAMPLE 3 Calculate the intrinsic carrier concentration of Silicon at T = 250K Answer: n i = 1.6 x 10 8 cm -3 k = Boltzmann’s constant (86 x eV/K)

© Electronics ECE 1312 EECE 1312 Extrinsic Semiconductor Since intrinsic concentration, n i 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

© Electronics ECE 1312 EECE 1312 Group V – 5 electrons in the outer shell; Example, Phosphorus, Arsenic The 5 th 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

© Electronics ECE 1312 EECE 1312 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

© Electronics ECE 1312 EECE 1312 The Fermi level shifts due to doping Fermi level

© Electronics ECE 1312 EECE 1312 – 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: n o = the thermal equilibrium concentration of free electrons p o = the thermal equilibrium concentration of holes n i = the intrinsic carrier concentration n o p o = ni 2

© Electronics ECE 1312 EECE 1312 For N-type – electrons are the majority carriers In thermal equilibrium, generally in the n-type semiconductor the donor atom concentration N d is much higher than the intrinsic carrier concentration, called majority charge carrier. And the hole concentration in the n-type semiconductor is called minority charge carrier.

© Electronics ECE 1312 EECE 1312 For P-type – holes are the majority carriers Similarly in thermal equilibrium concentration the majority charge carrier free hole in the p-type semiconductor is And the minority charge carrier electron concentration in the p-type semiconductor is

© Electronics ECE 1312 EECE 1312 Example 1 Calculate the thermal equilibrium electron and hole concentrations. Consider silicon at T = 300 K doped with phosphorous at a concentration of N d = cm -3 and n i = 1.5 x cm -3.

© Electronics ECE 1312 EECE 1312 Example 2 Calculate the majority and minority carrier concentrations in silicon at T = 300K if a)N a = cm -3 b)N d = 5 x cm -3 1.Calculate n i 2.For part (a) – it is p-type 3.For part (b) – it is n-type Answer: a) majority = cm -3 minority = 2.25x 10 3 cm -3 b) majority = 5 x cm -3, minority 4.5 x 10 4 cm -3

© Electronics ECE 1312 EECE 1312 Example 3 A silicon is doped with 5 x arsenic atoms a)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 eV/K) Answer: a)n-type b)n o = 5 x cm -3 and p o = 4.5 x 10 3 cm -3