Chapter Notes

Introduction

Electronic circuits are built using devices that can control the flow of electrons. For a long time, these devices were vacuum tubes (also called valves), which were bulky, consumed a lot of power, operated at high voltages (~100 V), and were not very reliable.

The era of modern electronics began with the development of solid-state semiconductor devices. Unlike vacuum tubes where electrons flow through a vacuum, in semiconductor devices, the flow of charge carriers happens entirely within the solid material.

Key Differences: Vacuum Tubes vs. Semiconductor Devices

Simple factors like light, heat, or a small applied voltage can change the number of mobile charges in a semiconductor, allowing us to control the current. This led to the creation of devices like diodes and transistors, which are the foundation of all modern electronics, from smartphones to computers.

Classification of Metals, Conductors and Semiconductors

Solids can be classified based on their ability to conduct electricity (conductivity) or their internal energy structure (energy bands).

On the basis of conductivity

Materials are categorized based on their electrical resistivity ($\rho$) or its inverse, conductivity ($\sigma = 1/\rho$).

• Metals: Have very low resistivity and high conductivity.
  • $\rho \sim 10^{-2} - 10^{-8} \, \Omega \text{m}$
  • $\sigma \sim 10^{2} - 10^{8} \, \text{S m}^{-1}$
• Insulators: Have very high resistivity and low conductivity.
  • $\rho \sim 10^{11} - 10^{19} \, \Omega \text{m}$
  • $\sigma \sim 10^{-11} - 10^{-19} \, \text{S m}^{-1}$
• Semiconductors: Have resistivity and conductivity values between those of metals and insulators.
  • $\rho \sim 10^{-5} - 10^{6} \, \Omega \text{m}$
  • $\sigma \sim 10^{5} - 10^{-6} \, \text{S m}^{-1}$

Semiconductors can be further classified:

• Elemental semiconductors: Made of a single element, like Silicon (Si) and Germanium (Ge).
• Compound semiconductors: Made of two or more elements, such as Gallium Arsenide (GaAs) or Cadmium Sulphide (CdS).

On the basis of energy bands

In an isolated atom, electrons occupy specific energy levels. However, when atoms come together to form a solid, these discrete energy levels merge into continuous ranges of energy called energy bands.

• Valence Band: The energy band containing the valence electrons (the outermost electrons of an atom). At absolute zero temperature, this band is completely filled with electrons.
• Conduction Band: The energy band just above the valence band. Electrons in this band are free to move and conduct electricity. It is normally empty.
• Energy Band Gap ($E_g$): The energy difference between the top of the valence band ($E_V$) and the bottom of the conduction band ($E_C$). The size of this gap determines whether a material is a conductor, insulator, or semiconductor.

Case I: Metals (Conductors) In metals, the valence band and conduction band overlap, or the conduction band is only partially filled. There is no energy gap ($E_g \approx 0$). This allows electrons to move freely into the conduction band, making metals excellent conductors of electricity.

Case II: Insulators Insulators have a very large energy gap ($E_g > 3 \text{ eV}$). It takes a huge amount of energy to move an electron from the valence band to the conduction band. Under normal conditions, the conduction band is empty, so no current can flow.

Case III: Semiconductors Semiconductors have a small but finite energy gap ($E_g < 3 \text{ eV}$). At absolute zero (0 K), they behave like insulators. However, at room temperature, thermal energy is sufficient to excite some electrons from the valence band across the gap into the conduction band. This allows a small amount of current to flow.

Intrinsic Semiconductor

A semiconductor in its purest form is called an intrinsic semiconductor. Silicon (Si) and Germanium (Ge) are the most common examples.

• Structure: Both Si and Ge have four valence electrons. In a crystal, each atom forms four covalent bonds with its neighbors by sharing electrons. At low temperatures, all electrons are locked in these bonds.
• Electron-Hole Pair Generation: As temperature increases, some electrons gain enough thermal energy to break free from their covalent bonds.
  • A freed electron becomes a mobile negative charge carrier in the conduction band.
  • The vacancy left behind in the covalent bond is called a hole. A hole acts as a mobile positive charge carrier.
• Charge Carriers: In an intrinsic semiconductor, free electrons and holes are always created in pairs. Therefore, the number of free electrons ($n_e$) is equal to the number of holes ($n_h$). This concentration is called the intrinsic carrier concentration ($n_i$).
  • $n_e = n_h = n_i$
• Current Conduction: When a voltage is applied, both electrons and holes move, contributing to the total current.
  • Free electrons move towards the positive terminal.
  • Holes move towards the negative terminal (which is equivalent to bonded electrons hopping into the vacancies).
  • The total current ($I$) is the sum of the electron current ($I_e$) and the hole current ($I_h$): $I = I_e + I_h$
• Recombination: A free electron can also fall back into a hole, eliminating both charge carriers. At equilibrium, the rate of electron-hole generation equals the rate of recombination.

At absolute zero temperature ($T = 0 \text{ K}$), no electrons have enough energy to break their bonds. The valence band is full, and the conduction band is empty. The material behaves like a perfect insulator. At temperatures above absolute zero ($T > 0 \text{ K}$), some electron-hole pairs are generated, allowing for some conductivity.

Solution

Carbon (C), Silicon (Si), and Germanium (Ge) are all in Group 14 of the periodic table and have four valence electrons. However, these electrons are in different orbits (2nd for C, 3rd for Si, 4th for Ge). The energy required to remove an electron from its atom (the ionization energy or energy gap $E_g$) depends on how tightly it is bound.

• For Carbon (diamond), the electrons are held very tightly, resulting in a large energy gap ($E_g \approx 5.4 \text{ eV}$).
• For Silicon, the gap is smaller ($E_g \approx 1.1 \text{ eV}$).
• For Germanium, the gap is the smallest ($E_g \approx 0.7 \text{ eV}$).

At room temperature, there is not enough thermal energy to create a significant number of free electrons in Carbon, so it acts as an insulator. In Si and Ge, the smaller energy gaps allow thermal energy to create enough electron-hole pairs for them to act as semiconductors.

Extrinsic Semiconductor

The conductivity of an intrinsic semiconductor is very low and often not useful for practical devices. To improve conductivity, a process called doping is used.

Doping is the deliberate addition of a small, controlled amount of a specific impurity to a pure semiconductor. The impurity atoms are called dopants, and the resulting material is an extrinsic semiconductor.

The dopant atoms must be of a similar size to the semiconductor atoms (e.g., Si or Ge) so they can fit into the crystal lattice without distorting it. Dopants are typically chosen from groups 13 or 15 of the periodic table.

n-type semiconductor

An n-type semiconductor is created by doping a pure semiconductor like Si or Ge with a pentavalent impurity (an element with 5 valence electrons), such as Arsenic (As), Phosphorus (P), or Antimony (Sb).

• Mechanism: When a pentavalent atom replaces a Si atom in the crystal lattice, four of its valence electrons form covalent bonds with the neighboring Si atoms. The fifth electron is very weakly bound to its parent atom.
• Donors: It takes very little energy (~0.01 eV for Ge, ~0.05 eV for Si) to free this fifth electron, allowing it to move into the conduction band. Because the pentavalent impurity donates a free electron, it is called a donor impurity.
• Charge Carriers: In an n-type semiconductor, the vast majority of charge carriers are electrons from the donor atoms. Thermally generated holes are also present but in much smaller numbers.
  • Majority carriers: Electrons ($n_e \gg n_h$)
  • Minority carriers: Holes
• Energy Band Diagram: The donor atoms create a new energy level, the donor energy level ($E_D$), which lies just below the conduction band. Electrons can easily be excited from $E_D$ to the conduction band.

p-type semiconductor

A p-type semiconductor is created by doping Si or Ge with a trivalent impurity (an element with 3 valence electrons), such as Boron (B), Aluminum (Al), or Indium (In).

• Mechanism: When a trivalent atom replaces a Si atom, it can only form three covalent bonds. The fourth bond has a vacancy for an electron, which is a hole.
• Acceptors: An electron from a neighboring Si atom can easily jump into this hole with very little energy, creating a new hole in its original position. This makes the hole available for conduction. Because the trivalent impurity accepts an electron to complete its bonds, it is called an acceptor impurity.
• Charge Carriers: In a p-type semiconductor, the majority of charge carriers are holes created by the acceptor atoms.
  • Majority carriers: Holes ($n_h \gg n_e$)
  • Minority carriers: Electrons
• Energy Band Diagram: The acceptor atoms create a new energy level, the acceptor energy level ($E_A$), which lies just above the valence band. Electrons from the valence band can easily be excited to this level, leaving behind holes in the valence band.

In any semiconductor at thermal equilibrium, the product of electron and hole concentrations is constant and equal to the square of the intrinsic carrier concentration: $n_e n_h = n_i^2$

Given

• Number of Si atoms = $5 \times 10^{28} \text{ m}^{-3}$
• Doping concentration = 1 ppm (part per million) = $10^{-6}$
• Intrinsic carrier concentration, $n_i = 1.5 \times 10^{16} \text{ m}^{-3}$
• The dopant is pentavalent Arsenic (As), so it's an n-type semiconductor.

To Find

• Number of electrons, $n_e$
• Number of holes, $n_h$

Formula

$N_D = (\text{doping concentration}) \times (\text{number of Si atoms})$ $n_e \approx N_D$ $n_e n_h = n_i^2$

Solution

First, calculate the number of donor atoms ($N_D$). Since each pentavalent As atom donates one electron, this is approximately the number of free electrons.

$N_D = (10^{-6}) \times (5 \times 10^{28} \text{ m}^{-3}) = 5 \times 10^{22} \text{ m}^{-3}$

The number of electrons from doping ($5 \times 10^{22} \text{ m}^{-3}$) is much larger than the number of thermally generated electrons ($n_i \approx 10^{16} \text{ m}^{-3}$). Therefore, we can assume the total electron concentration is approximately equal to the donor concentration.

$n_e \approx N_D = 5 \times 10^{22} \text{ m}^{-3}$

Now, use the mass action law to find the number of holes.

$n_h = \frac{n_i^2}{n_e} = \frac{(1.5 \times 10^{16})^2}{5 \times 10^{22}}$ $n_h = \frac{2.25 \times 10^{32}}{5 \times 10^{22}} = 4.5 \times 10^{9} \text{ m}^{-3}$

Final Answer The number of electrons is $n_e \approx 5 \times 10^{22} \text{ m}^{-3}$ and the number of holes is $n_h = 4.5 \times 10^{9} \text{ m}^{-3}$. As expected for an n-type semiconductor, $n_e \gg n_h$.

p-n Junction

A p-n junction is formed when a p-type semiconductor and an n-type semiconductor are brought into contact, typically by doping one side of a single semiconductor crystal. This junction is the fundamental building block for most semiconductor devices, including diodes and transistors.

p-n junction formation

Two key processes occur immediately after the junction is formed:

1. Diffusion: Due to the difference in concentration, holes from the p-side (where they are abundant) diffuse across the junction to the n-side. Similarly, electrons from the n-side (where they are abundant) diffuse to the p-side. This movement of charge carriers constitutes a diffusion current.

2. Drift:

   • When an electron diffuses from the n-side to the p-side, it leaves behind a positively charged, immobile donor ion on the n-side.
   • When a hole diffuses from the p-side to the n-side, it leaves behind a negatively charged, immobile acceptor ion on the p-side.
   • This creates a thin layer on both sides of the junction that is depleted of free charge carriers. This region is called the depletion region or space-charge region.
   • The layer of positive ions on the n-side and negative ions on the p-side creates an electric field directed from the n-side to the p-side.
   • This electric field causes minority charge carriers (electrons on the p-side, holes on the n-side) that wander near the junction to be swept across. This movement constitutes a drift current, which flows in the opposite direction to the diffusion current.

Initially, the diffusion current is large and the drift current is small. As more charges diffuse, the depletion region widens, the electric field gets stronger, and the drift current increases. Equilibrium is reached when the drift current becomes equal and opposite to the diffusion current, resulting in zero net current across the junction.

The electric field in the depletion region creates a potential difference across the junction, known as the barrier potential ($V_0$). This potential barrier opposes the further diffusion of majority carriers.

Semiconductor Diode

A semiconductor diode is a p-n junction with metallic contacts at its ends, allowing an external voltage to be applied. It is a two-terminal device. The symbol for a diode has an arrow pointing from the p-side to the n-side, indicating the direction of conventional current flow when the diode is "on".

The behavior of a diode changes drastically depending on how an external voltage (bias) is applied.

p-n junction diode under forward bias

A diode is forward biased when the p-side is connected to the positive terminal of a battery and the n-side is connected to the negative terminal.

• Effect on Barrier Potential: The applied voltage ($V$) opposes the built-in barrier potential ($V_0$). This reduces the effective barrier height to ($V_0 - V$) and narrows the depletion region.
• Current Flow: With a lower barrier, majority carriers (holes from the p-side and electrons from the n-side) can easily diffuse across the junction. This process is called minority carrier injection.
• Result: A significant current flows through the diode. The magnitude of this forward current is typically in milliamperes (mA) and increases exponentially as the applied voltage increases.

p-n junction diode under reverse bias

A diode is reverse biased when the p-side is connected to the negative terminal of a battery and the n-side is connected to the positive terminal.

• Effect on Barrier Potential: The applied voltage ($V$) adds to the built-in barrier potential ($V_0$). This increases the effective barrier height to ($V_0 + V$) and widens the depletion region.
• Current Flow: The high potential barrier prevents the flow of majority carriers across the junction. However, a very small current, called the reverse saturation current, flows due to the drift of minority carriers across the junction.
• Result: The current is extremely small, typically in microamperes ($\mu$A), and remains almost constant regardless of the applied reverse voltage.

If the reverse voltage is increased to a critical value called the breakdown voltage ($V_{br}$), the reverse current increases sharply. Operating a general-purpose diode in this region can permanently destroy it due to overheating.

V-I Characteristics of a p-n Junction Diode

The graph of current (I) versus voltage (V) for a diode is its V-I characteristic.

• Forward Bias Region: The current is almost zero until the applied voltage crosses a certain value called the threshold voltage or cut-in voltage. Beyond this point, the current increases rapidly (exponentially).
  • For Germanium diodes, threshold voltage $\approx 0.2 \text{ V}$.
  • For Silicon diodes, threshold voltage $\approx 0.7 \text{ V}$.
• Reverse Bias Region: A very small, nearly constant reverse saturation current flows.

This characteristic shows that a diode essentially allows current to flow in only one direction (when forward biased), acting like a one-way valve for electricity.

The dynamic resistance of a diode is the ratio of a small change in voltage to the corresponding small change in current. $r_d = \frac{\Delta V}{\Delta I}$

Given

The V-I characteristic curve from Figure 14.17.

To Find

(a) Diode resistance when $I_D = 15 \text{ mA}$ (forward bias). (b) Diode resistance when $V_D = -10 \text{ V}$ (reverse bias).

Formula

Resistance can be calculated as a static value ($R = V/I$) or a dynamic value ($r = \Delta V / \Delta I$). Here we will calculate dynamic resistance for forward bias and static resistance for reverse bias.

Solution

(a) Forward Bias Resistance at $I_D = 15 \text{ mA}$ We calculate the dynamic resistance ($r_{fb}$) in the region around $15 \text{ mA}$. From the graph:

• At $I = 10 \text{ mA}$, the voltage is $V = 0.7 \text{ V}$.
• At $I = 20 \text{ mA}$, the voltage is $V = 0.8 \text{ V}$.

The change in current is $\Delta I = 20 \text{ mA} - 10 \text{ mA} = 10 \text{ mA} = 10 \times 10^{-3} \text{ A}$. The change in voltage is $\Delta V = 0.8 \text{ V} - 0.7 \text{ V} = 0.1 \text{ V}$.

$r_{fb} = \frac{\Delta V}{\Delta I} = \frac{0.1 \text{ V}}{10 \times 10^{-3} \text{ A}} = 10 \, \Omega$

Answer for part (a) = $10 \, \Omega$

(b) Reverse Bias Resistance at $V_D = -10 \text{ V}$ From the graph, at a reverse voltage of $V = -10 \text{ V}$, the reverse current is $I = -1 \, \mu\text{A} = -1 \times 10^{-6} \text{ A}$. We calculate the static resistance ($r_{rb}$) here.

$r_{rb} = \frac{V}{I} = \frac{10 \text{ V}}{1 \times 10^{-6} \text{ A}} = 1.0 \times 10^7 \, \Omega$

Answer for part (b) = $1.0 \times 10^7 \, \Omega$

Application of Junction Diode as a Rectifier

A rectifier is a circuit that converts alternating current (AC) into direct current (DC). This is one of the most common applications of a p-n junction diode, which allows current to flow in only one direction.

Half-wave Rectifier

A half-wave rectifier converts only one half of the AC input cycle into DC.

• Circuit: It consists of a transformer, a single diode, and a load resistor ($R_L$).
• Operation:
  1. During the positive half-cycle of the AC input, the diode is forward biased and conducts current. A voltage appears across the load resistor.
  2. During the negative half-cycle of the AC input, the diode is reverse biased and does not conduct. No current flows, and the output voltage is zero.
• Output: The output is a pulsating DC voltage that is present for only half of the input cycle.

Full-wave Rectifier

A full-wave rectifier converts both halves of the AC input cycle into a pulsating DC output, making it more efficient than a half-wave rectifier. A common design uses a centre-tap transformer and two diodes.

• Circuit: The p-sides of two diodes ($D_1$ and $D_2$) are connected to the ends of the transformer's secondary winding. The n-sides are connected together, and the output is taken between this common point and the center tap of the transformer.
• Operation:
  1. During the positive half-cycle, point A is positive, so diode $D_1$ is forward biased and conducts. Current flows through the load resistor $R_L$. Diode $D_2$ is reverse biased and off.
  2. During the negative half-cycle, point B is positive, so diode $D_2$ is forward biased and conducts. Current flows through the load resistor $R_L$ in the same direction as before. Diode $D_1$ is reverse biased and off.
• Output: The output is a continuous series of positive pulses, utilizing both halves of the AC input wave.

Filter Circuits

The output of a rectifier is a pulsating DC, not a steady DC like that from a battery. To smooth out these pulses, a filter circuit is used.

The simplest filter is a large capacitor connected in parallel with the load resistor ($R_L$).

• How it works:
  1. When the rectified voltage is rising, the capacitor charges up to the peak voltage.
  2. When the rectified voltage starts to fall, the capacitor begins to discharge slowly through the load resistor.
  3. Before the capacitor can discharge significantly, the next voltage pulse arrives and recharges it to the peak value.
• Result: The capacitor filter smooths out the pulsating output, producing a much steadier DC voltage with a small ripple. The larger the capacitor, the smoother the output DC will be.