FAD1022 L34-L38 — Semiconductors & Op-Amps

Comprehensive coverage of semiconductor physics, diode operation, transistor biasing, and operational amplifier configurations.

Lecture Files

Lecture 34 — Semiconductor Theory and Materials For Diode
Lecture 35 — Transistor (Fixed & Emitter Stabilized Bias)
Lecture 36 — Transistor (Voltage Divider Bias)
Lecture 37 — Op-Amp (Inverting)
Lecture 38 — Op-Amp (Non-inverting)

Lecture 34 — Semiconductor Theory and Materials For Diode

Semiconductor Materials

Semiconductors are a special class of elements with conductivity between that of a good conductor and an insulator. Their conductivity can be controlled in ways impossible for conductors.

Energy Bands

Conduction band: energy band consisting of free electrons. These free electrons originate from the valence band when energized/excited (receive energy). They are highly mobile and responsible for electrical conductivity.
Insulator: large band gap — electrons cannot easily jump to conduction band.
Semiconductor: moderate band gap — some electrons can be excited to conduction band.
Conductor: overlapping or very small gap — free electrons readily available.

Doping the Intrinsic Semiconductor

Adding impurity atoms to a semiconductor material to improve conductivity, producing extrinsic semiconductors.

N-Type

Add element with five valence electrons into Si crystallite structure
Examples: antimony (Sb), arsenic (As), phosphorus (P)
Electrons are majority charge carriers; holes are minority carriers

P-Type

Add element with three valence electrons into Si crystallite structure
Examples: boron (B), gallium (Ga), indium (In)
Holes are majority charge carriers; electrons are minority carriers

Both holes and electrons drive current.

Charge Carriers: Holes and Electrons

A hole is the "opposite" of an electron. Unlike an electron which has negative charge, holes have positive charge equal in magnitude but opposite in polarity. Holes are not physical particles; they are the absence of an electron in an atom. Holes move from positive to negative, whereas electrons move from negative to positive.

Diode

A solid-state device created by joining p-type and n-type material.

P-N Junction Formation

At the instant the two materials are joined:

Electrons from the n-region diffuse across the junction and combine with holes
Combination results in a lack of carriers near the junction — the Depletion Region
Filling a hole makes a negative ion and leaves behind a positive ion on the n-side
A space charge builds up, creating a depletion region which inhibits further electron transfer unless helped by applying a forward bias

Biasing

Three possibilities:

Bias Condition	Connection	Diode State
No bias	—	—
Reverse bias ($V_D < 0$)	p-region to negative terminal, n-region to positive terminal	OFF
Forward bias ($V_D > 0$)	p-region to positive terminal, n-region to negative terminal	ON

Knee Voltage (Cut-in Voltage)

Minimum voltage at which the forward-biased diode starts conducting current.

Semiconductor	Knee Voltage (V)
Ge	0.3
Si	0.7
GaAs	1.5

Circuit must be supplied with knee voltage (or more) for current to conduct. Also referred to as offset, threshold, or firing potential ($V_D$).

Temperature Effects on Knee Curve

When temperature increases:

Thermal energy of electrons and holes within the silicon crystal increases
Easier for charge carriers to overcome the potential barrier at the p-n junction
The knee of the curve shifts to the left — diode turns on at a lower forward voltage

Diode Circuit Configurations

DC Series Configuration

Forward bias (ON state):

Replace Si diode with 0.7 V source (knee voltage)
Apply Kirchhoff's voltage rule: $E - V_R - V_D = 0$

Reverse bias (OFF state):

Replace diode with open circuit
$I_D = I_R = 0$ A
$V_R = I_R R = 0$
$V_D = E - V_R = E$

Parallel Configuration

Voltage for parallel connection is always the same. When diode is in forward bias, it acts as a voltage limiter — potential difference across parallel connection is limited to knee voltage (0.7 V for Si).

Multiple Diodes in Parallel

When two diodes of different materials are in parallel (e.g., Si and Ge):

The diode with lower knee voltage turns ON first and maintains its voltage
The other diode never reaches its required knee voltage and remains OFF
Example: Ge (0.3 V) turns ON; Si (0.7 V) remains OFF

Diode with AC Inputs

Half-Wave Rectifier

Process of removing one-half the input signal to establish a DC level.

Positive cycle: current flows clockwise; $V_m - V_D - V_O = 0$ → $V_O = V_m - V_D$ Negative cycle: no current flows; $V_O = 0$

Average DC value: $V_{DC} = 0.318(V_m - V_D)$

Clippers

Clippers have the ability to "clip" off a portion of the input signal.

Output taken across diode
Can limit positive or negative portions depending on configuration

AC Equivalent Visualization

During positive cycle: visualize as +ve DC power supply with current clockwise During negative cycle: visualize as +ve DC power supply with current clockwise (but opposite polarity)

Capacitor in Semiconductor Circuits

By introducing capacitor at certain position in the circuit, we can block DC current from flowing through one of the resistors. When used for this purpose, it is known as a blocking capacitor.

Topics Not Covered in L34

Full-wave rectification
Load line (operation line)
Clampen
Filter
Zener diode
Avalanche breakdown

Lecture 35 — Transistor (Fixed & Emitter Stabilized Bias)

Introduction to Bipolar Junction Transistor (BJT)

BJT = Bipolar Junction Transistor, a three-terminal semiconductor device (emitter, base, collector).
NPN: Current flows from collector terminal to emitter terminal. Base is P-type sandwiched between N-type emitter and collector.
PNP: Current flows from emitter terminal to collector terminal. Base is N-type sandwiched between P-type emitter and collector.
Base acts as a gate controlling the flow of current.
This lecture focuses ONLY on NPN transistors in the common-emitter configuration.

Transistor as Amplifier

A transistor amplifies the amplitude of the input signal.
The transistor does not change the frequency of the signal.
The desired output preserves the same ripple waveform as the input, only scaled in amplitude.
Two criteria must be fulfilled for the transistor to conduct:
1. Small current must flow into the Base.
2. Voltage between Base and Emitter, $V_{BE}$, must be more than 0.7 V (for silicon-based transistors).

Why Biasing?

Relying on the input signal alone to provide base current and voltage is not reliable for amplifier applications; the current may fluctuate.
For use as a switch, direct input control may be acceptable.
For use as an amplifier, if the signal is too small the transistor may fail to turn "ON" properly.
Biasing introduces fixed DC current and voltage to the base, establishing a stable operating point so the transistor remains in the active region regardless of how small the AC input signal is.

Fundamental Current and Gain Relationships

KCL (Kirchhoff's Current Law): $$I_E = I_B + I_C$$

Current Gain (Beta, β): $$\beta = \frac{I_C}{I_B} \quad \Rightarrow \quad I_C = \beta I_B$$

Beta is the factor by which current is amplified.
For a device with $\beta = 100$, the collector current is 100× the base current.
In transistor specification sheets, beta is referred to as $h_{FE}$ (for AC) or $h_{fe}$ (for DC).
Beta is temperature-dependent, which causes Q-point instability.

Alpha (α): $$\alpha = \frac{I_C}{I_E}$$

Ratio of collector current to emitter current.

Emitter Current in terms of Base Current: $$I_E = I_B(\beta + 1)$$

Transistor Operating Regions

Region	Emitter Junction	Collector Junction	Behavior	Key Formula
Active (Linear)	Forward biased	Reverse biased	Amplifier	$I_C = \beta I_B$
Saturation	Forward biased	Forward biased	Closed switch	$I_{C(sat)}$ (max current)
Cutoff	Open / Reverse	Open / Reverse	Open switch	$I_C = 0$, $I_B = 0$

Active region: The transistor acts best as an amplifier. $I_C$ is controlled by $I_B$.
Saturation region: Collector and emitter behave as if shorted. $I_C$ reaches maximum and is no longer controlled by $I_B$.
Cutoff region: Collector and base are effectively open. All currents ($I_C$, $I_B$, $I_E$) are approximately zero.

1. Fixed-Bias Circuit

Circuit: Base resistor $R_B$ connects $V_{CC}$ to the base. Collector resistor $R_C$ connects $V_{CC}$ to the collector. Emitter is grounded. AC input couples through $C_1$ to the base; AC output couples through $C_2$ from the collector.

Base-Emitter Loop Equation: $$V_{CC} - I_B R_B - V_{BE} = 0$$ $$I_B = \frac{V_{CC} - V_{BE}}{R_B}$$

Collector-Emitter Loop Equation: $$V_{CC} - I_C R_C - V_{CE} = 0$$ $$V_{CE} = V_{CC} - I_C R_C$$

Saturation Current: $$I_{C(sat)} = \frac{V_{CC}}{R_C}$$

Q-Point Instability: Since $I_C = \beta I_B$ and $I_B$ is fixed by $R_B$, the collector current is directly proportional to $\beta$. If $\beta$ changes (e.g., due to temperature), $I_C$ changes by the same proportion, causing the Q-point to shift dramatically.

Lecture Example: $V_{CC} = 12,V$, $R_B = 240,k\Omega$, $R_C = 2.2,k\Omega$, $\beta = 50$

$I_B = \frac{12 - 0.7}{240,k\Omega} = 47.08,\mu A$
$I_C = (50)(47.08,\mu A) = 2.35,mA$
$V_{CE} = 12 - (2.35,mA)(2.2,k\Omega) = 6.83,V$
$I_{C(sat)} = \frac{12}{2.2,k\Omega} = 5.45,mA$
If $\beta$ changes to 100: $I_C = 4.708,mA$ (nearly 100% increase)

2. Emitter-Stabilized Bias

Circuit: Adds an emitter resistor $R_E$ between the emitter and ground. This provides improved stability over fixed-bias because $R_E$ introduces negative feedback against $\beta$ variations.

Base-Emitter Loop Equation: $$V_{CC} - I_B R_B - V_{BE} - I_E R_E = 0$$ Substituting $I_E = I_B(\beta + 1)$: $$I_B = \frac{V_{CC} - V_{BE}}{R_B + (\beta + 1)R_E}$$

Collector-Emitter Loop Equation: $$V_{CC} - I_C R_C - V_{CE} - I_E R_E = 0$$ Using the approximation $I_E \approx I_C$ (valid when $\beta \gg 1$): $$V_{CE} = V_{CC} - I_C(R_C + R_E)$$

Saturation Current: $$I_{C(sat)} = \frac{V_{CC}}{R_C + R_E}$$

Other Related Equations:

$V_E = I_E R_E$
$V_C = V_{CE} + V_E = V_{CC} - I_C R_C$
$V_B = V_{BE} + V_E = V_{CC} - I_B R_B$

Stability Mechanism: The term $(\beta + 1)R_E$ appears in the denominator of the $I_B$ equation. As $\beta$ increases, the denominator increases, causing $I_B$ to decrease. This partially compensates for the $\beta$ increase, keeping $I_C$ more stable than in fixed-bias.

A useful design approximation for strong stability is when $(\beta + 1)R_E \gg R_B$ (commonly $(\beta + 1)R_E \geq 10 R_B$). Under this condition: $$I_C \approx \frac{V_{CC} - V_{BE}}{R_E}$$ making $I_C$ largely independent of $\beta$.

Lecture Example 1: $V_{CC} = 20,V$, $R_B = 430,k\Omega$, $R_C = 2,k\Omega$, $R_E = 1,k\Omega$, $\beta = 50$

$I_B = \frac{20 - 0.7}{430,k\Omega + (51)(1,k\Omega)} = \frac{19.3}{481,k\Omega} = 40.1,\mu A$
$I_C = (50)(40.1,\mu A) = 2.01,mA$
$V_{CE} = 20 - (2.01,mA)(2,k\Omega + 1,k\Omega) = 13.97,V$

Lecture Example 2 (same circuit, $\beta = 100$):

$I_B = \frac{19.3}{430,k\Omega + (101)(1,k\Omega)} = \frac{19.3}{531,k\Omega} = 36.3,\mu A$
$I_C = (100)(36.3,\mu A) = 3.63,mA$
$V_{CE} = 20 - (3.63,mA)(3,k\Omega) = 9.11,V$

Fixed-Bias vs Emitter-Stabilized Comparison

Configuration	$\beta = 50$ ($I_C$)	$\beta = 100$ ($I_C$)	Change
Emitter-Stabilized ($R_E = 1,k\Omega$)	2.01 mA	3.63 mA	+80.59%
Fixed-Bias (no $R_E$)	2.24 mA	4.49 mA	+100%

Conclusion: Adding $R_E$ reduces the sensitivity of $I_C$ to $\beta$ variations, providing better Q-point stability. $I_C$ for fixed-bias changes 100% when $\beta$ doubles, whereas emitter-stabilized shows only ~80.6% change for the same $\beta$ swing.

Lecture 36 — Transistor (Voltage Divider Bias)

Circuit Overview

The voltage divider bias configuration adds a resistive voltage divider ($R_{B1}$, $R_{B2}$) to the base network to provide a stable base voltage $V_B$ and base-emitter voltage $V_{BE}$. This gives more control and thermal stability than fixed-bias or emitter-stabilized bias.

Approximate vs Exact Analysis

There are two ways to analyze the voltage divider bias loop:

Exact analysis — uses Thevenin equivalent ($V_{TH}$, $R_{TH}$)
Approximate analysis — simpler, but only valid under a specific condition

This lecture focuses only on approximate analysis.

Approximation Condition

For the approximate method to be valid with high accuracy:

$$\beta R_E \geq 10 R_{B2}$$

Equivalently, this can be written as $R_{TH} \leq 0.1\beta R_E$ (where $R_{TH} = R_{B1} \parallel R_{B2}$).

Key stability insight: In the approximate analysis, the formula to determine $I_B$ does not depend on $\beta$. This makes the voltage divider bias far more stable against transistor parameter variations than fixed-bias or emitter-stabilized bias.

Approximate Analysis Procedure

Step	Formula	Description
1	$V_B = \frac{R_{B2} V_{CC}}{R_{B1} + R_{B2}}$	Base voltage from divider rule
2	$V_E = V_B - V_{BE}$	Emitter voltage
3	$I_E = \frac{V_E}{R_E}$	Emitter current
4	$I_C \cong I_E$	Collector current (approx.)
5	$V_{CE} = V_{CC} - I_C(R_C + R_E)$	Collector-emitter voltage

Worked Example

Given: $V_{CC} = 22,V$, $R_{B1} = 39,k\Omega$, $R_{B2} = 3.9,k\Omega$, $R_C = 10,k\Omega$, $R_E = 1.5,k\Omega$, $\beta = 140$

Step 1 — Check condition: $$\beta R_E = (140)(1.5,k\Omega) = 210,k\Omega$$ $$10 R_{B2} = 10(3.9,k\Omega) = 39,k\Omega$$ $$210,k\Omega \geq 39,k\Omega \quad \text{(satisfied)}$$

Step 2 — Calculate $V_B$: $$V_B = \frac{(3.9,k\Omega)(22,V)}{39,k\Omega + 3.9,k\Omega} = 2,V$$

Step 3 — Calculate $V_E$: $$V_E = V_B - V_{BE} = 2,V - 0.7,V = 1.3,V$$

Step 4 — Calculate $I_C$: $$I_C \cong I_E = \frac{V_E}{R_E} = \frac{1.3,V}{1.5,k\Omega} = 0.867,mA$$

Step 5 — Calculate $V_{CE}$: $$V_{CE} = V_{CC} - I_C(R_C + R_E) = 22,V - (0.867,mA)(11.5,k\Omega) = 12.03,V$$

Output: $I_C = 0.867,mA$, $V_{CE} = 12.03,V$

Bias Type Comparison

Quantity	Fixed Bias	Emitter-Stabilized	Voltage Divider (Approx.)
$I_B$	$\frac{V_{CC} - V_{BE}}{R_B}$	$\frac{V_{CC} - V_{BE}}{R_B + (\beta+1)R_E}$	Depends on divider (not directly calculated)
$I_C$	$\beta I_B$	$\beta I_B$	$\cong I_E = \frac{V_B - V_{BE}}{R_E}$
$V_{CE}$	$V_{CC} - I_C R_C$	$V_{CC} - I_C(R_C + R_E)$	$V_{CC} - I_C(R_C + R_E)$

Lecture 37 — Op-Amp (Inverting)

Introduction to Op-Amps

An op-amp is a high-gain voltage amplifier integrated into a single IC, composed of many transistors, diodes, and resistors.
It amplifies weak electric signals (both DC and AC).
The name "operational amplifier" reflects its ability to perform mathematical operations: summation, subtraction, integration, differentiation, and division.

Symbol & Power Supply

Inverting Input (-): Output is 180° out of phase (inverted) with respect to the input.
Non-Inverting Input (+): Output is in phase (0° phase shift) with the input.
Vcc: Positive power supply terminal.
Vee: Negative power supply terminal.
Simplified diagrams often omit power supply pins, but they are always present.

Two Golden Rules

No current flows into the input pins — the op-amp has extremely high input impedance (ideally infinite). In practice, a negligible amount of current may flow.
The output voltage adjusts to bring the input pins to the same voltage — the voltage difference between $V_{in(+)}$ and $V_{in(-)}$ becomes zero. This rule applies only when there is feedback.

Feedback

Feedback is the process of taking a portion of the output signal and feeding it back to the input.
Negative Feedback: Output is fed back to the inverting (-) input. This is used in linear amplifier configurations.
Positive Feedback: Output is fed back to the non-inverting (+) input.

Inverting Amplifier

The input signal is applied to the inverting input through an input resistor $R_1$, while the non-inverting input is grounded.
Virtual Ground: Because the non-inverting input is at 0 V and Rule 2 forces both inputs to the same potential, the inverting input node is also at 0 V — this point is called virtual ground.
Since no current enters the op-amp input (Rule 1), the current through $R_1$ equals the current through the feedback resistor $R_f$: $$I = \frac{V_{in}}{R_1} = -\frac{V_{out}}{R_f}$$
Rearranging gives the inverting amplifier output: $$V_{out} = -\frac{R_f}{R_1} V_{in}$$

Characteristics

Amplifies the input signal.
Inverts the output with respect to the input (applies to both AC and DC signals).

Worked Example

Given an inverting amplifier with $R_1 = 100 \text{ k}\Omega$ and $R_f = 500 \text{ k}\Omega$, find the input voltage required to produce $V_{out} = -10 \text{ V}$.

$$V_{out} = -\frac{R_f}{R_1} V_{in}$$

$$-10 = -\frac{500}{100} V_{in}$$

$$V_{in} = 2 \text{ V}$$

Lecture 38 — Op-Amp (Non-inverting)

Lecturer: Zainal Abidin (ZAA)

Revision: Voltage Division in Series Circuits

The lecture begins by revisiting series resistor circuits. For resistors in series with a voltage source, the voltage across each resistor is proportional to its resistance. This concept is essential for analyzing the non-inverting amplifier's feedback network.

Non-Inverting Amplifier Configuration

When the input signal is connected to the non-inverting (+) input pin of the op-amp, the circuit is called a non-inverting amplifier. The output is amplified and remains in phase with the input (not inverted).

Circuit operation:

The input (e.g., a battery, sensor, or microphone) is applied to the non-inverting (+) pin.
Negative feedback is provided from the output to the inverting (-) pin through a feedback resistor $R_f$.
A resistor $R_1$ connects the inverting (-) pin to ground.

Analysis using ideal op-amp rules:

Virtual Short (Rule 2): The voltage at the inverting (-) pin equals the voltage at the non-inverting (+) pin. If $V_{in}$ is applied to (+), then the voltage at (-) is also $V_{in}$.
No Input Current: No current flows into the op-amp inputs. Therefore, the current flowing through $R_1$ to ground must be the same current flowing through $R_f$ from the output.

The voltage across $R_1$ is $V_{in}$ (since the inverting pin is at $V_{in}$ and the other end is grounded). The current through $R_1$ is: $$I = \frac{V_{in}}{R_1}$$

Since the same current flows through $R_f$, the voltage across $R_f$ is: $$V_{R_f} = I \cdot R_f = V_{in} \cdot \frac{R_f}{R_1}$$

The output voltage is the sum of the voltage at the inverting pin and the voltage across $R_f$: $$V_{out} = V_{in} + V_{R_f} = V_{in} + V_{in}\frac{R_f}{R_1} = V_{in}\left(1 + \frac{R_f}{R_1}\right) = \left(\frac{R_1 + R_f}{R_1}\right)V_{in}$$

Characteristics

Amplifies the input signal (DC or AC).
Will NOT invert the output signal — the output remains in phase with the input.

Example

Determine the output voltage of a non-inverting amplifier for $V_1 = 2\text{ V}$, $R_1 = 100\text{ kΩ}$, and $R_f = 500\text{ kΩ}$:

$$V_{out} = \left(\frac{100\text{ kΩ} + 500\text{ kΩ}}{100\text{ kΩ}}\right)(2\text{ V}) = 6 \times 2\text{ V} = \mathbf{+12\text{ V}}$$

The positive result confirms the output is not inverted.

Comparison: Inverting vs Non-Inverting Amplifier

Using the same resistor values ($R_1 = 100\text{ kΩ}$, $R_f = 500\text{ kΩ}$) and input ($V_1 = 2\text{ V}$):

Configuration	Gain Formula	Output
Inverting	$V_{out} = -\frac{R_f}{R_1}V_{in}$	$-10\text{ V}$ (inverted)
Non-inverting	$V_{out} = \left(\frac{R_1 + R_f}{R_1}\right)V_{in}$	$+12\text{ V}$ (non-inverted)

Both configurations amplify the signal, but the inverting amplifier produces a negative (inverted) output while the non-inverting amplifier produces a positive (non-inverted) output.

Key Concepts

Semiconductors & Diodes — band theory, doping, p-n junction
Transistors & Biasing — BJT operation, biasing circuits
Operational Amplifiers — ideal op-amp characteristics
Semiconductor Theory — energy bands, intrinsic vs extrinsic
Doping — n-type and p-type semiconductors
P-N Junction — depletion region, forward and reverse bias
Diodes — I-V characteristics, applications
Bipolar Junction Transistors (BJT) — NPN and PNP structures
Biasing Circuits — fixed bias, emitter-stabilized, voltage divider
DC Analysis — Q-point stabilization
Operational Amplifiers — differential inputs, high gain
Inverting Amplifier — virtual ground, gain calculation
Non-inverting Amplifier — gain formula, buffer applications

Diagrams

BJT Operating Regions

graph TB
    A[("BJT Operating Regions")] --> B["Active Region"]
    A --> C["Saturation Region"]
    A --> D["Cutoff Region"]
    
    B --> B1["Emitter FB, Collector RB"]
    B --> B2["Amplifier: Ic = beta Ib"]
    C --> C1["Both Junctions FB"]
    C --> C2["Closed Switch: Ic = Ic(sat)"]
    D --> D1["Both Junctions RB"]
    D --> D2["Open Switch: Ic = 0"]
    
    B -.->|"Ib too large"| C
    C -.->|"Reduce Ib"| B
    B -.->|"Ib = 0"| D
    D -.->|"Vbe > 0.7V"| B
    
    style A fill:#e7f5ff,stroke:#1971c2
    style B fill:#d3f9d8,stroke:#2f9e44
    style C fill:#ffe3e3,stroke:#c92a2a
    style D fill:#f8f9fa,stroke:#868e96

Inverting vs Non-Inverting Op-Amp

graph TB
    subgraph inverting["Inverting Amplifier"]
        I1["Input to (-) Pin"] --> I2["Virtual Ground at 0V"]
        I2 --> I3["Vout = -(Rf/R1) Vin"]
    end
    subgraph noninverting["Non-Inverting Amplifier"]
        N1["Input to (+) Pin"] --> N2["Virtual Short Follows Vin"]
        N2 --> N3["Vout = (1 + Rf/R1) Vin"]
    end
    
    inverting ~~~ noninverting

Summary

This module bridges physics and electronics, covering semiconductor device operation. Students learn the physics of p-n junctions, analyze transistor biasing circuits for stable operation, and understand op-amp configurations. Practical circuit analysis skills are developed for both DC biasing and AC signal amplification.

Lecturer

Zainal Abidin (ZAA) — PASUM Physics Lecturer

FAD1022 - Basic Physics II — main course page
AC Circuits — circuit analysis foundation
Electrostatics — electric field concepts in depletion regions