Introduction to Electronics

Definition of Electronics

Electronics is the branch of science that deals with the study of the flow and control of
electrons (electricity) and their behavior in vacuums, gases, and semiconductors.

History of Electronics

● 1895 - H.A. Lorentz postulated the existence of electrons.

● 1897 - J.J. Thomson proved the existence of electrons.
● 1904 - Fleming invented the vacuum diode (valve).
● 1906 - De Forest introduced the Triode, an improved vacuum tube.
● 1947 - Shockley invented the junction transistor (BJT).
● 1959 - Kilby announced the concept of Integrated Circuits (ICs).
● 1971 - Intel developed the first 4-bit microprocessor.
● 1980s-2000s - Evolution of Very Large Scale Integration (VLSI) and Ultra Large
Scale Integration (ULSI) in microprocessors.

Difference Between Electrical and Electronics

Category Electrical Engineering Electronics Engineering

Conductors Metals (Copper, Aluminum) Non-metals (Silicon, Germanium)

Voltage & High Voltage (kV), High Low Voltage (mV, V), Low Amps
Current Amps (mA)

Frequency Low (50Hz/60Hz) High (kHz, MHz, GHz)

Basic Electrical Units & Definitions

1. Passive Components

● Operate without external power.

● Examples: Resistors, Capacitors, Inductors, Diodes.

2. Active Components

● Require external power to operate.

● Examples: Transistors (BJT, MOSFET), ICs, LEDs.

3. Direct Current (DC)

● Current flows in one direction only.

● Conventional current flow: Positive to Negative.

4. Alternating Current (AC)

● Current flows in both directions cyclically.

● Frequency (Hz) determines the number of cycles per second.
5. Voltage (V)

● Unit: Volts (V), Symbol: V or U

6. Current (I)

● Unit: Amperes (A), Symbol: I

7. Resistance (R)

● Unit: Ohms (Ω), Symbol: R

8. Capacitance (C)

● Unit: Farads (F), Symbol: C

9. Inductance (L)

● Unit: Henrys (H), Symbol: L

10. Impedance (Z)

● Unit: Ohms (Ω), Symbol: Z

11. Decibels (dB)

● Unit: Bel (B), Decibel (dB)

Electronics Components

1. Resistors

● Function: Limit current flow.

● Color-coded for value identification.

2. Capacitors

● Function: Store and release charge.

● Types: Electrolytic, Ceramic, Tantalum.

3. Inductors

● Function: Store energy in a magnetic field.

● Used in: Filters, Transformers, Chokes.

4. Diodes

● Function: Allow current to flow in one direction only.

● Types:
○ PN Junction Diode - Basic rectifier diode.
○ Zener Diode - Provides voltage regulation.
○ Light Emitting Diode (LED) - Emits light when current flows.
5. Transistors

● Function: Amplification & Switching.

● Types:
○ Bipolar Junction Transistor (BJT)
○ Field Effect Transistor (FET)

Formulas and Example Problems

1. Ohm’s Law Application

Formula:

V=I*R

Problem: Find the current flowing through a 100Ω resistor connected to a 10V source.

Solution:

I = V / R = 10V / 100Ω = 0.1A (100mA)

2. Capacitance Charge Calculation

Formula:

Q=C*V

Problem: A 10µF capacitor is charged to 5V. Find the stored charge.

Solution:

Q = 10 * 10^(-6) F * 5V = 50 * 10^(-6) C = 50 µC

3. Inductive Voltage Calculation

Formula:

V = L * (dI/dt)

Problem: An inductor of 2H experiences a change in current of 3A/s. Find the induced voltage.

Solution:

V = 2H * 3A/s = 6V

4. Impedance Calculation

Formula:

Z = R + j(XL - XC)

where:
XL = 2 * π * f * L (Inductive Reactance)
XC = 1 / (2 * π * f * C) (Capacitive Reactance)
f = Frequency (Hz)

5. Decibel Calculation

Formula:

For Power:

dB = 10 * log10(P2 / P1)

For Voltage or Current:

dB = 20 * log10(V2 / V1)

ASIC ELECTRONICS REVIEWER - SEMICONDUCTORS

Introduction to Semiconductors

Definition of a Semiconductor

A semiconductor is a material with electrical conductivity between that of a conductor (e.g.,

copper) and an insulator (e.g., glass). Its resistivity decreases with increasing temperature,
unlike metals. The conductivity of semiconductors can be altered through a process called
doping, where impurities are introduced into the crystal structure.

Atomic Structure of Silicon and Germanium

● Silicon (Si) and Germanium (Ge) are the most common semiconductor materials.
● Both have four valence electrons and form covalent bonds.
● Silicon has 14 protons, and Germanium has 32 protons.
● Silicon is more stable at high temperatures, making it the preferred material for
semiconductor applications.
● Germanium has some applications in solar cells and detectors due to recent
advancements in porous germanium.

Covalent Bonding in Semiconductors

● Covalent bonds occur when atoms share electrons.

● In a pure silicon crystal, each atom shares four valence electrons with neighboring
atoms, forming a stable structure.
● A crystal lattice structure is formed, and a pure silicon crystal without impurities is
called an intrinsic semiconductor.

Conduction in Semiconductors

Energy Bands in Semiconductors

 ● Energy Gap (Bandgap): The difference in energy between the valence band and the
conduction band.
● Intrinsic silicon has enough thermal energy at room temperature for some electrons to
jump from the valence band to the conduction band, creating free electrons and
holes.
● Electron-Hole Pair: When an electron gains energy and moves to the conduction band,
it leaves behind a hole in the valence band.
● Recombination: Occurs when a free electron loses energy and falls back into a hole.

Types of Current in Semiconductors

1. Electron Current: Free electrons move through the conduction band toward a positive
voltage.
2. Hole Current: Electrons in the valence band jump between adjacent atoms, creating the
illusion that holes move.

Comparison of Semiconductors, Conductors, and Insulators

Material Type Energy Gap Free Conductivity

Electrons

Conductor (e.g., copper) No gap (overlapping bands) Many Very high

Semiconductor (e.g., Small gap (~1.1 eV for Si) Moderate Moderate

silicon)

Insulator (e.g., glass) Large gap (>2 eV) Very few Very low

Doping: N-Type and P-Type Semiconductors

What is Doping?

Doping is the process of adding impurities to intrinsic silicon to increase conductivity. The two
types of doped semiconductors are N-type and P-type.

N-Type Semiconductor (Negative Charge Carriers)

● Formed by adding pentavalent atoms (5 valence electrons) such as Phosphorus (P),

Arsenic (As), or Antimony (Sb).
● The extra valence electron becomes a free electron in the conduction band.
● Majority carriers: Free electrons.
● Minority carriers: Holes.

P-Type Semiconductor (Positive Charge Carriers)

● Formed by adding trivalent atoms (3 valence electrons) such as Boron (B), Aluminum
(Al), or Gallium (Ga).
● These elements create holes in the lattice where electrons are missing.
● Majority carriers: Holes.
● Minority carriers: Electrons.
Comparison Between N-Type and P-Type Semiconductors
Property N-Type Semiconductor P-Type Semiconductor

Doping Element Pentavalent (P, As, Sb) Trivalent (B, Al, Ga)

Majority Carrier Electrons Holes

Minority Carrier Holes Electrons

Charge Type Negative Positive

Conductivity Higher due to free electrons Higher due to holes

Formulas and Example Problems

1. Conductivity and Resistivity Relationship

Formula:

σ=1/ρ

where:

● σ = Conductivity (Siemens per meter, S/m)

● ρ = Resistivity (Ohm-meter, Ω·m)

Example: If the resistivity of silicon is 2.3 × 10⁻³ Ω·m, find its conductivity.

σ = 1 / (2.3 × 10⁻³)
σ = 434.78 S/m

2. Current Density Formula

Formula:

J=σ*E

where:

● J = Current Density (A/m²)

● σ = Conductivity (S/m)
● E = Electric Field (V/m)

Example: If the conductivity of a semiconductor is 500 S/m and the applied electric field is 200
V/m, find the current density.

J = 500 * 200
J = 100,000 A/m²
3. Intrinsic Carrier Concentration

Formula:

n_i = A * e^(-Eg / (2kT))

where:

● n_i = Intrinsic carrier concentration

● A = Constant
● Eg = Energy gap (eV)
● k = Boltzmann’s constant (8.617 × 10⁻⁵ eV/K)
● T = Temperature (K)

4. Mobility of Charge Carriers

Formula:

μ = v_d / E

where:

● μ = Mobility (cm²/V·s)
● v_d = Drift velocity (m/s)
● E = Electric field (V/m)

Example: If the drift velocity of electrons is 0.02 m/s under an electric field of 500 V/m, find
the mobility.

μ = 0.02 / 500
μ = 4 × 10⁻⁵ cm²/V·s

Introduction to Diodes

Definition of a Diode

A diode is a semiconductor device that allows current to flow in one direction only while
restricting current in the opposite direction. Diodes are commonly known as rectifiers because
they convert alternating current (AC) into direct current (DC).

Properties of a Diode

● Composed of P-type and N-type semiconductor materials.

● Functions based on the PN junction.
● Rated by voltage and current capacity.
● Two main operational states: Forward Bias and Reverse Bias.

PN Junction and Depletion Region

Formation of the PN Junction

A PN junction is created when P-type and N-type semiconductors are joined together,
forming a junction where charge carriers interact.

● N-region: Contains many free electrons.

● P-region: Contains many holes.
● When the junction is formed, electrons from the N-region diffuse into the P-region
and recombine with holes, creating a region depleted of charge carriers known as the
depletion region.

Depletion Region and Barrier Potential

● The depletion region contains immobile ions, which create an electric field.
● This forms a barrier potential that prevents further carrier movement across the
junction.
● Barrier potential values:
○ Silicon: ~0.7V
○ Germanium: ~0.3V

Biasing of a Diode

1. Forward Bias (Conducting State)

● Definition: A condition that allows current to flow through the diode.

● Connection:
○ Negative terminal of the battery → N-region
○ Positive terminal of the battery → P-region
● Effects:
○ Reduces the depletion region.
○ Overcomes the barrier potential.
○ Allows electrons to flow from N to P, recombining with holes.
● Voltage requirement:
○ Silicon diode: Minimum 0.7V required to conduct.
○ Germanium diode: Minimum 0.3V required to conduct.

2. Reverse Bias (Non-Conducting State)

● Definition: A condition that prevents current flow through the diode.

● Connection:
○ Positive terminal of the battery → N-region
○ Negative terminal of the battery → P-region
● Effects:
○ Increases the depletion region.
○ Prevents charge carriers from moving across the junction.
○ Only a small leakage current (due to minority carriers) flows.

3. Reverse Breakdown (Avalanche Effect)

● If reverse voltage is increased beyond a certain limit, the diode enters reverse
breakdown.
● Process:
○ Minority carriers accelerate and knock valence electrons into the conduction
band.
○ A chain reaction occurs, creating a large avalanche of free electrons.
 ● Most diodes are damaged in this condition, except Zener diodes, which are
designed to operate in reverse breakdown for voltage regulation.

Formulas and Example Problems

1. Diode Current in Forward Bias

Formula:

I = I_s * (e^(V / ηVT) - 1)

where:

● I = Diode current (A)

● I_s = Reverse saturation current (A)
● V = Applied voltage (V)
● η = Ideality factor (typically 1-2)
● VT = Thermal voltage (~25mV at room temperature)

Example: If I_s = 1 × 10⁻¹² A, V = 0.7V, and η = 1, find the diode current.

I = (1 × 10⁻¹²) * (e^(0.7 / 0.025) - 1)

I ≈ 10 mA

2. Reverse Breakdown Voltage

Formula:

V_Z = V_B

where:

● V_Z = Zener voltage (V)

● V_B = Breakdown voltage (V)

Example: A Zener diode is rated at 5.1V. What is its breakdown voltage?

V_Z = 5.1V

3. Power Dissipation in a Diode

Formula:

P=V*I

where:

● P = Power (W)
● V = Voltage drop across the diode (V)
● I = Current through the diode (A)

Example: A diode conducts 0.02A at 0.7V. Find the power dissipation.

P = 0.7V * 0.02A
P = 0.014W or 14mW

Diode Characteristics and Behavior

V-I Characteristic Curve of a Diode

● The V-I characteristic curve represents the relationship between voltage (V) and
current (I) in a diode.
● Forward Bias Region:
○ Below barrier potential: Very little current flows.
○ At barrier potential: Current starts to increase rapidly.
○ Beyond barrier potential: Current increases significantly and must be limited
by a series resistor.
○ Barrier potential values:
■ Silicon diode: ~0.7V
■ Germanium diode: ~0.3V
● Reverse Bias Region:
○ Very little reverse current flows until the breakdown voltage (V_BR) is
reached.
○ After breakdown, a large reverse current flows, which can damage the diode
unless it is a Zener diode.

Diode Symbol and Terminal Identifications

● Standard diode schematic symbol:

○ Anode (A): Connected to P-region.
○ Cathode (K): Connected to N-region.
○ Arrowhead in the symbol indicates conventional current direction
(A → K).
● Biasing conditions:
○ Anode positive w.r.t cathode → Forward-biased (conducts current).
○ Anode negative w.r.t cathode → Reverse-biased (no conduction).

Diode Approximations (Models)

1. Ideal Diode Model

● Forward Bias: Acts as a closed switch (ON).

● Reverse Bias: Acts as an open switch (OFF).
● Simplifications:
○ No voltage drop in forward bias.
○ No reverse current in reverse bias.
○ Used for basic troubleshooting and simplified circuit analysis.

2. Practical Diode Model

● Forward Bias:
○ Acts as a closed switch in series with a small “battery” representing the barrier
potential (0.7V for Si, 0.3V for Ge).
○ Requires at least the barrier potential voltage to conduct.
 ● Reverse Bias:
○ Acts as an open switch (similar to the ideal model).

3. Complete Diode Model

● Forward Bias:
○ Includes barrier potential and low forward resistance (dynamic resistance,
r_f).
● Reverse Bias:
○ Includes high internal resistance (reverse resistance, r_r).
● More accurate but used only in detailed circuit analysis.

Formulas and Example Problems

1. Diode Current in Forward Bias

Formula:

I = I_s * (e^(V / ηVT) - 1)

where:

● I = Diode current (A)

● I_s = Reverse saturation current (A)
● V = Applied voltage (V)
● η = Ideality factor (typically 1-2)
● VT = Thermal voltage (~25mV at room temperature)

Example: If I_s = 1 × 10⁻¹² A, V = 0.7V, and η = 1, find the diode current.

I = (1 × 10⁻¹²) * (e^(0.7 / 0.025) - 1)

I ≈ 10 mA

2. Reverse Breakdown Voltage

Formula:

V_BR = V_Z

where:

● V_BR = Breakdown voltage (V)

● V_Z = Zener voltage (V) for Zener diodes

Example: A Zener diode is rated at 5.1V. What is its breakdown voltage?

V_BR = 5.1V

3. Power Dissipation in a Diode

Formula:

P=V*I

where:

● P = Power (W)
● V = Voltage drop across the diode (V)
● I = Current through the diode (A)

Example: A diode conducts 0.02A at 0.7V. Find the power dissipation.

P = 0.7V * 0.02A
P = 0.014W or 14mW

Diode Rectifiers

1. Half-Wave Rectifier

● A half-wave rectifier allows only one half of an AC cycle to pass through.

● Operation:
○ During the positive half-cycle, the diode is forward-biased, allowing current to
flow through the load resistor (R_L).
○ During the negative half-cycle, the diode is reverse-biased, blocking current.
○ The result is a pulsating DC output with only positive half-cycles.
● Peak Inverse Voltage (PIV):
○ The maximum reverse voltage the diode must withstand.
○ PIV = V_p(input) (peak input voltage).
● Average Output Voltage:

Vavg = Vp(out) / π

2. Full-Wave Rectifier

● A full-wave rectifier converts both halves of the AC input into DC.

● Advantages:
○ Provides smoother DC output compared to a half-wave rectifier.
○ Higher average output voltage.
● Average Output Voltage:

V_avg = 2 * V_p(out) / π

Types of Full-Wave Rectifiers

(a) Center-Tapped Full-Wave Rectifier

● Uses a center-tapped transformer with two diodes.

● Operation:
○ During the positive half-cycle, one diode conducts, and the other is reverse-
biased.
○ During the negative half-cycle, the roles reverse.
 ○ The output remains unidirectional.
● Peak Inverse Voltage (PIV):

PIV = 2 * Vp(out)

● Output Voltage:

Vp(out) = Vp(secondary) / 2

(b) Bridge Full-Wave Rectifier

● Uses four diodes instead of a center-tapped transformer.

● Operation:
○ During both halves of the AC cycle, two diodes conduct, allowing current flow
through the load resistor.
○ Provides a continuous pulsating DC output.
● Peak Inverse Voltage (PIV):

PIV = Vp(out)

● Output Voltage:

Vp(out) = Vp(secondary)

Transformer Turns Ratio and Output Voltage

● The turns ratio (n) of a transformer affects the secondary voltage.

● Formula:

Vsec = n * Vprimary

● For a full-wave rectifier:

Vp(out) = Vp(sec) / 2

Example Problems

1. Half-Wave Rectifier Output Calculation

Problem: If a half-wave rectifier has a peak input voltage of 20V, find its average output
voltage.

Solution:

V_avg = V_p(out) / π
V_avg = 20V / 3.14
V_avg ≈ 6.37V

2. Full-Wave Rectifier Output Calculation

Problem: If a full-wave rectifier has a peak input voltage of 40V, find its average output voltage.
Solution:

V_avg = (2 * V_p(out)) / π
V_avg = (2 * 40V) / 3.14
V_avg ≈ 25.48V

3. Bridge Rectifier PIV Calculation

Problem: A bridge rectifier has a peak secondary voltage of 50V. Find the required PIV rating
for each diode.

Solution:

PIV = V_p(out)
PIV = 50V

PART 5: POWER SUPPLIES

1. Basic DC Power Supply

● Converts 120V, 60Hz AC to a constant DC voltage.

● Components:
○ Transformer: Steps up or steps down voltage.
○ Rectifier: Converts AC to pulsating DC.
○ Filter (Capacitor): Smooths fluctuations in rectified voltage.
○ Regulator: Maintains a steady DC output.

2. Capacitor-Input Filter

● Charges quickly during the positive cycle.

● Discharges slowly through the load resistor when input voltage decreases.
● Reduces ripple voltage, the fluctuation in output DC voltage.
● Ripple Factor Formula:

r = V_r / V_DC

where:

● V_r = Ripple voltage

● V_DC = Average DC output voltage

3. IC Voltage Regulators

● Integrated Circuit (IC) voltage regulators stabilize voltage.

● Examples:
○ 78XX series (Positive Regulators): e.g., 7812 = +12V output.
○ 79XX series (Negative Regulators): e.g., 7912 = -12V output.
○ LM317 (Adjustable Regulator): Output from 1.2V to 37V.
● Line Regulation Formula:
% Line Regulation = (ΔV_out / ΔV_in) * 100

● Load Regulation Formula:

% Load Regulation = [(V_NL - V_FL) / V_FL] * 100

where:

● V_NL = No-load voltage

● V_FL = Full-load voltage

PART 6: SPECIAL-PURPOSE DIODES

1. Zener Diodes

● Used for voltage regulation.

● Operates in reverse breakdown without damage.
● Types of Breakdown:
○ Zener Breakdown (<5V): High doping, narrow depletion layer.
○ Avalanche Breakdown (>5V): Electron collisions create more free carriers.
● Zener Voltage Formula:

Vin = V_Z + I_Z * R_s

where:

● V_in = Input voltage

● V_Z = Zener voltage
● I_Z = Zener current
● R_s = Series resistor

2. Varactor Diodes (Voltage-Controlled Capacitors)

● Acts as a variable capacitor.

● Capacitance decreases as reverse voltage increases.
● Used in:
○ Tuned circuits (e.g., radio frequency tuning).
○ Frequency modulators.

3. Light-Emitting Diodes (LEDs)

● Emits light when forward-biased.

● Materials & Colors:
○ GaAs (Gallium Arsenide): Infrared.
○ GaAsP (Gallium Arsenide Phosphide): Red, orange.
○ GaN (Gallium Nitride): Blue, white.
● LED Current Formula:

I = (V_s - V_LED) / R

where:
 ● V_s = Supply voltage
● V_LED = LED voltage
● R = Series resistor

4. Photodiodes

● Operates in reverse bias.

● Generates current when exposed to light.
● Used in light sensors, counters, and optical communication.

PART 7: TROUBLESHOOTING TECHNIQUES

1. APM Approach (Analysis, Planning, Measurement)

● Analysis: Identify symptoms and eliminate obvious causes.

● Planning: Choose a troubleshooting method:
○ Input-to-output tracing: Start at input and move forward.
○ Output-to-input tracing: Start at output and move backward.
○ Half-splitting method: Start at the middle and move towards the fault.
● Measurement: Use DMM, oscilloscope to check voltages.

2. Common Faults in Power Supplies

● Open Diode (Half-Wave Rectifier): No output voltage.

● Open Diode (Full-Wave Rectifier): Increased ripple voltage.
● Shorted Diode (Bridge Rectifier): Potential fuse blowout.
● Faulty Filter Capacitor:
○ Open: Full-wave rectified voltage instead of DC.
○ Shorted: Zero output, potential diode failure.
○ Leaky: Increased ripple voltage.
● Transformer Issues:
○ Open winding: No output voltage.
○ Shorted winding: Changes in output voltage.

Example Problems

1. Zener Diode Voltage Regulation

Problem: A 10V Zener diode regulates voltage with a 1kΩ series resistor and 20mA current.
Find the input voltage.

V_in = V_Z + I_Z * R_s

V_in = 10V + (20mA * 1kΩ)
V_in = 10V + 20V
V_in = 30V

2. LED Resistor Calculation

Problem: Find the resistor needed for a 5V supply and a 2V LED with 20mA current.
R = (V_s - V_LED) / I
R = (5V - 2V) / 20mA
R = 3V / 0.02A
R = 150Ω

3. Load Regulation Calculation

Problem: A regulator outputs 24.8V at no load and 23.9V at full load. Find load regulation.

% Load Regulation = [(V_NL - V_FL) / V_FL] * 100

% Load Regulation = [(24.8V - 23.9V) / 23.9V] * 100
% Load Regulation ≈ 3.77%