Operational Amplifier Inverting Amplifier Op Amp Configurations Summing Amplifier Non - Inverting Amplifier If you don’t remember the formulas, remember these two characteristics and perform nodal analysis vo = Aod ( v2 – v1) Operational Amplifier Open loop mode v1 = v2 to create Aod =  Two main characteristics CHAPTER 8 No current going into the op-amp

Chapter 2 Semiconductor Materials and Diodes

Classification of Materials 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

Semiconductor Materials and Properties 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.

Elemental Semiconductors 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

Si have 4 electrons in their outer shells needs another 4 to become stable So, when there are 4 other Si nearby = 4 electrons: Si Sharing of electrons occurred; and this bond is known as the covalent bond Si Si Si Si Atoms come into close proximity to each other and so the valence electrons interact to form a crystal.

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

ILLUSTRATION WHEN A VALENCE ELECTRON IS FREE 1. Becomes free electron 3. Electron moves to fill space 5. Electron moves to fill space 2. Becomes empty 4. Becomes empty

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 material Eg = the bandgap energy (eV) T = the temperature (Kelvin) remember that K = °C + 273.15 k = Boltzmann’s constant (86 x 10-6 eV/K)

Intrinsic Semiconductors Cont. The energy difference between the minimum conduction band energy and the maximum valance band energy is called bandgap energy. Semiconductor Constants Material Bandgap Energy (eV) B (cm-3 K-3/2) Silicon (Si) 1.1 5.23 × 1015 Germanium (Ge) 0.66 1.66 × 1015 Gallium Arsenide (GaAs) 1.4 2.10 × 1014  

The values of B and Eg for several semiconductor materials: EXAMPLE 1 Calculate the intrinsic carrier concentration in silicon at T = 300 K.

EXAMPLE 2 Find the intrinsic carrier concentration of Gallium Arsenide at T = 300K k = Boltzmann’s constant (86 x 10-6 eV/K) Answer: 1.8 x 106 cm-3

Calculate the intrinsic carrier concentration of Silicon at T = 250K k = Boltzmann’s constant (86 x 10-6 eV/K) EXAMPLE 3 Calculate the intrinsic carrier concentration of Silicon at T = 250K Answer: ni = 1.6 x 108 cm-3

Extrinsic 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

Group V + Si = n-type semiconductor Extra electron 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

Group III + Si = p-type semiconductor Extra hole 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

The Fermi level shifts due to doping

Effects of doping process 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 po = ni2 no = the thermal equilibrium concentration of free electrons po = the thermal equilibrium concentration of holes ni = the intrinsic carrier concentration

For N-type – electrons are the majority carriers In thermal equilibrium, generally in the n-type semiconductor the donor atom concentration Nd 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.

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

Example 1 Calculate 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.

Example 2 Calculate the majority and minority carrier concentrations in silicon at T = 300K if Na = 1017cm-3 Nd = 5 x 1015cm-3 Calculate ni For part (a) – it is p-type 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

A silicon is doped with 5 x 1016 arsenic atoms k = Boltzmann’s constant (86 x 10-6 eV/K) Example 3 A silicon is doped with 5 x 1016 arsenic atoms Is the material n-type or p-type? Calculate the electrons and holes concentration of the doped silicon at T=300K Answer: n-type no = 5 x 1016 cm-3 and po = 4.5 x 103 cm-3