PHY301 — Final Term Summary (Lectures 23–45)

📘 Lecture 23 — Superposition Theorem and examples

📖 Overview: This lecture demonstrates how to solve circuit problems with multiple sources using two methods: Cramer's rule for solving systems of linear equations, and the Superposition Theorem for analyzing circuits by considering one source at a time. Understanding superposition is crucial for simplifying complex linear circuits in exam problems.

🗂️ Topics Covered

The lecture begins with an example solving for three mesh currents (I₁, I₂, I₃) using Cramer's rule and matrix determinants. It then introduces the Superposition Theorem, explaining that voltages or currents in any linear circuit with multiple sources can be found by algebraically summing contributions from each source acting alone. Three detailed examples demonstrate superposition applied to circuits with combinations of voltage and current sources, using voltage division and current division rules.

📝 Lecture Summary

Example: Find current I₁, I₂ and I₃ By using Cramer's rule

Three mesh current equations are derived using Kirchhoff's Voltage Law (KVL). For loop 1: 7I₁ – 5I₂ = -25. For loop 2: -5I₁ + 19I₂ – 4I₃ = 25. For loop 3: -4I₂ + 6I₃ = 50.

The coefficient matrix is constructed:

|  7  -5   0 |
| -5  19  -4 |
|  0  -4   6 |

🔑 Cramer's rule: A method for solving systems of linear equations using determinants, where each variable equals the determinant of the matrix with its column replaced by constants, divided by the determinant of the coefficient matrix.

The determinant of the coefficient matrix is |A| = 536.

📐 Formula: Iₙ = det(Aₙ) / det(A) → each current equals the determinant of the modified matrix divided by the determinant of the original coefficient matrix.

📌 Example: To find I₁, replace the first column with constants [-25, 25, 50]ᵀ, giving I₁ = (-700) ÷ 536 = -1.31 A.

For I₂, replace the second column: I₂ = 1700 ÷ 536 = 3.17 A.

For I₃, replace the third column: I₃ = 5600 ÷ 536 = 10.45 A.

Superposition Theorem

🔑 Superposition Theorem: In any linear circuit containing multiple sources, the current or voltage at any point may be calculated as the algebraic sum of the individual contributions of each source acting alone.

When determining contributions due to independent sources, any remaining current sources are made zero by replacing them with an open circuit, and any voltage sources are made zero by replacing them with a short circuit.

💡 Why this matters: Superposition breaks a complex multi-source problem into several simpler single-source problems that can be solved using basic techniques like voltage division and current division.

Example: Calculate V₀ by applying principle of superposition (Circuit 1)

Only voltage source acting: The 3V source is active while the 2mA current source is replaced by an open circuit. Using the voltage division rule:

🔑 Voltage division rule: V_R1 = [R₁ / (R₁ + R₂)] × V_total

📌 Example: V₀₁ = V_6kΩ = [6k / (3k + 6k)] × 3V = 2 volts

Only current source acting: The 2mA source is active while the 3V voltage source is replaced by a short circuit. Using the current division rule:

🔑 Current division rule: I_R1 = [R₂ / (R₁ + R₂)] × I_total

📌 Example: I₀ = [3k / (3k + 6k)] × 2mA = 2/3 mA. Then V₀₂ = I₀ × 6k = (2/3 mA)(6kΩ) = 4V

The total voltage V₀ = V₀₁ + V₀₂ = 2V + 4V = 6V

Example: Calculate V₀ by applying principle of superposition (Circuit 2)

Only current source acting: The 2mA source is active. First simplify by combining parallel resistors: 2kΩ ∥ 4kΩ = (2×4)/(2+4) = 8/6 = 4/3 kΩ. This is in series with 2kΩ, giving (4/3 + 2) kΩ in parallel with 6kΩ.

📌 Example: Using current division: I₀ = [(4/3 + 2) / ((4/3 + 2) + 6)] × 2mA = 20/28 mA. Then V₀₁ = I₀ × 6k = (20/28) × 6 = 30/7 volts

Only voltage source acting: The 6V source is active. Simplify: R_ab = 8kΩ ∥ 4kΩ = 8/3 kΩ. Using voltage division: V_ab = [(8/3) / ((8/3) + 2)] × 6 = 24/7 volts. Then V₀₂ = V_ab × [6k / (6k + 2k)] = (24/7) × (6/8) = 18/7 volts

Total V₀ = V₀₁ + V₀₂ = 30/7 + 18/7 = 48/7 volts

Example: Calculate V₀ by applying principle of superposition (Circuit 3)

Only voltage source acting: Using voltage division: V₀₁ = [2k / (7k + 2k)] × 12V = 24/9 volts

Only current source acting: The 2mA source is active but note the direction. Using current division: I₀ = [3k / (6k + 3k)] × (-2mA) = -6/9 = -2/3 mA. Then V₀₂ = 2kΩ × (-2/3 mA) = -4/3 volts

Total V₀ = V₀₁ + V₀₂ = 24/9 – 4/3 = 4/3 volts

⭐ Key Takeaways

When using superposition, always replace inactive voltage sources with short circuits and inactive current sources with open circuits, then solve each sub-problem independently. The algebraic sum of all individual contributions gives the final answer. Voltage division and current division are the primary tools used within each superposition step. Pay careful attention to polarity and direction of sources, as the sign of each contribution matters. Superposition only applies to linear circuits — it cannot be used for power calculations.

🧠 Quick Revision Questions

1. What is the key rule for deactivating a voltage source when applying superposition?
2. In the Cramer's rule example, what was the determinant of the coefficient matrix?
3. How do you find the contribution of a current source when applying superposition?
4. In Example 3, why was the current I₀ negative when the 2mA source was acting alone?
5. Can superposition be used to directly calculate power dissipation in a resistor? Why or why not?

📘 Lecture 24 — Source Transformation and examples

📖 Overview: This lecture explains the technique of source transformation, which allows engineers to simplify complex circuits by converting between voltage sources and current sources. Mastering this method enables easier calculation of voltages and currents in resistive networks.

🗂️ Topics Covered

The lecture covers the fundamental concept of source transformation, including the equivalence between a current source in parallel with a resistance and a voltage source in series with the same resistance. Multiple worked examples demonstrate step-by-step how to apply source transformation to calculate output voltage Vo and current I in various circuits.

📝 Lecture Summary

Source Transformation:

If we have any source embedded within a network, say this source is a current source having a value I and there exists a resistance having a value R, in parallel to it, we can replace it with a voltage source of value V=IR in series with the same resistance R. The reverse is also true: a voltage source V, in series with a resistance R can be replaced by a current source having a value I= V/R in parallel to the resistance R. Parameters within the circuit, for example an output voltage, remain unchanged under these transformations.

💡 Why this matters: Source transformation preserves the external behavior of the circuit while simplifying the internal structure, making analysis much easier.

Example 1: Calculate the voltage Vo using source transformation method.

Solution: We proceed as follows:

1k is in series with 2k so the combined effect is 3k. Now the 2mA source is in parallel with the 3k resistor. So it can be changed to a voltage source of value = 2m x 3k (by Ohm's Law) = 6 Volts. The 3k resistor will become in series with this source.

Positive terminal of the 6 volts battery is connected with the negative terminal of the 3 volts battery, so they will be summed up.

Applying voltage division rule:

🔑 Definition — Voltage Division Rule: The voltage across a resistor in a series circuit is proportional to its resistance relative to the total resistance.

📐 Formula: V_R1 = (R1 / (R1 + R2)) × V_total

Vo = (6 × 9) / 9 = 6 Volts

Example 2: Calculate the voltage Vo using source transformation method.

Solution: We proceed as follows:

3k is in series with 12 volts battery. So it can be converted into a current source of value = 12/3k (by Ohm's Law) = 4mA. Now the voltage source has transformed into the current source.

Now 3k resistor is in parallel with 6k resistor: 3k || 6k = (3k × 6k) / (3k + 6k) = 2k.

We replace 3k and 6k with 2k. The 4mA source is parallel with 2k resistor, so it can be converted into a voltage source. By Ohm's Law, the value is 8 volts in series with a resistor of 2k.

In the circuit, 2k is in series with 2k, so modified circuit becomes 8V source in series with 4k. This can be converted into a current source of value = 8/4k = 2mA in parallel with 4k.

Two current sources are parallel to each other, so they add up to give 4mA. The 4mA source is in parallel with 4k resistor, so it can be converted into a voltage source of value 16 volts.

4k and 4k are in series, so applying voltage division rule:

Vo = (8k / 16k) × 16V = 8 volts

Example 3: Calculate the voltage Vo using source transformation method.

Solution: We proceed as follows:

12 volt source is in series with 3k resistor. So it can be converted into a current source of value I = 12/3k = 4mA (by Ohm's Law). In the modified circuit, the 4mA current source and the 3k resistor will be in parallel.

We combine current sources: 4m – 2m = 2mA. The 2mA source is in parallel with 3k, so it can be converted into a voltage source of value V = 2 × 3 = 6 volts.

3k is in series with 4k, so modified circuit has a 6V source with 7k total resistance. Applying voltage division rule:

🔑 Definition — Voltage Division Rule: The voltage across a specific resistor in a series circuit equals that resistor's fraction of the total resistance times the total voltage.

Vo = (2k / 9k) × 6 = 12/9 = 4/3 volts

Example 4: Calculate the current I using source transformation method.

Solution: We proceed as follows:

3 Ohms resistor is in parallel with 5A source, so it can be converted into a voltage source of 15V. The 3 Ohms and 4 Ohms resistors are in series, and the resultant is in series with 15V, so it can be converted into a current source.

7 || 7 = (7 × 7) / (7 + 7) = 49/14 = 3.5 ohms. The current source is in parallel with 3.5 ohms resistance, so it can be converted into a voltage source by Ohm's Law: V = IR = (15/7) × 3.5 = 7.5 Volts.

The current I can be found using KVL:

-7.5 + 3.5I – 51V_x + 28I + 9 = 0 -------------------- (A)

Where V_x = 2I

Putting the value of V_x in (A):

-7.5 + 3.5I - 51(2I) + 28I + 9 = 0 3.5I - 102I + 28I + 1.5 = 0 -70.5I + 1.5 = 0

I = 1.5 / 70.5 = 21.28 mA

⭐ Key Takeaways

Source transformation is a powerful circuit simplification technique where a current source in parallel with a resistor can be replaced by a voltage source in series with the same resistor (V=IR), and vice versa (I=V/R). This method preserves all external circuit parameters and is particularly useful when combined with series/parallel resistance calculations and voltage/current division. When multiple sources exist, they can be transformed sequentially to reach a simple series or parallel circuit where Ohm's Law or voltage division can be applied directly. The key is to always maintain the correct polarity and direction during transformations.

🧠 Quick Revision Questions

1. What is the voltage source equivalent of a 5mA current source in parallel with a 2kΩ resistor?
2. In Example 1, why do the 6V and 3V sources add rather than subtract?
3. What is the combined resistance of 3kΩ and 6kΩ in parallel?
4. In Example 4, how is the voltage Vx related to the current I?
5. Can source transformation be applied to sources that are not in the standard parallel or series configuration?

📘 Lecture 25 — (Thevenin's Theorem)

📖 Overview: This lecture introduces Thevenin's Theorem, a powerful circuit analysis technique that simplifies complex circuits into an equivalent voltage source and series resistance. It provides a systematic four-step method for calculating unknown voltages, currents, or power across any load resistor, making circuit analysis more efficient and manageable.

🗂️ Topics Covered

The lecture presents Thevenin's Theorem with its four-step problem-solving procedure, followed by three detailed numerical examples demonstrating how to calculate the Thevenin equivalent voltage (Vth) and Thevenin equivalent resistance (Rth) for various circuits. Each example applies the theorem to find the voltage across a load resistor (Vo) in circuits containing resistors, voltage sources, and current sources.

📝 Lecture Summary

THEVENIN'S THEOREM

Thevenin's Theorem states that any linear two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source (Vth) in series with a resistor (Rth). The four-step procedure is:

Step 1: Remove the load resistor (the resistor across which you want to find voltage, current, or power), leaving behind an open circuit.

Step 2: Calculate the Thevenin voltage (Vth) across the open terminals by any method (e.g., mesh analysis, nodal analysis, voltage division, superposition).

Step 3: Calculate the Thevenin resistance (Rth) by open-circuiting all current sources and short-circuiting all voltage sources, then finding the total resistance seen from the open terminals.

Step 4: Re-insert the load resistance (RL) in series with Rth and Vth (treating Vth as a battery), then calculate the desired quantity using simple series circuit formulas.

💡 Why this matters: This theorem allows you to analyze any one branch of a complex circuit without solving the entire circuit each time.

🔑 Definition — Thevenin Voltage (Vth): The voltage across the open terminals when the load resistor is removed from the circuit. 🔑 Definition — Thevenin Resistance (Rth): The equivalent resistance seen from the open terminals with all independent voltage sources short-circuited and all independent current sources open-circuited.

EXAMPLE 1: Calculating Vo using Thevenin's Theorem

Circuit Details: Contains a 6V voltage source, 2mA current source, resistors of 1kΩ, 2kΩ, 3kΩ, and a 6kΩ load resistor.

First step: Removing RL. Here RL is the 6kΩ resistor at which we want to calculate the voltage Vo.

Second step: Calculating Vth. 1kΩ is in series with 2kΩ, so modified circuit has 3kΩ in parallel with 2mA source. By source transformation, the 2mA source in parallel with 3kΩ becomes a voltage source. V = 2mA × 3kΩ (by Ohm's Law) = 6 Volts. The 3kΩ resistor becomes in series with this source. The combined effect of the two sources (6V original and 6V transformed) is 9 volts.

📐 Formula: V = I × R → voltage from source transformation 📌 Example: 2mA × 3kΩ = (2 × 10⁻³) × (3 × 10³) = 6V

Third step: Calculating Rth. Vth = 9 volts Rth = 3kΩ

Fourth step: Calculating the unknown quantity. Re-inserting RL = 6kΩ in series with Rth = 3kΩ and Vth = 9V.

📐 Formula: Voltage division → Vo = (RL / (Rth + RL)) × Vth 📌 Example: Vo = (6kΩ / (3kΩ + 6kΩ)) × 9 = (6k/9k) × 9 = 6 volts

EXAMPLE 2: Calculating Vo using Thevenin's Theorem (with Superposition)

Circuit Details: Contains a 12V voltage source, 2mA current source, resistors of 2kΩ, 3kΩ, 4kΩ, 6kΩ, and an 8kΩ load resistor.

First step: Removing RL. Here RL is the 8kΩ resistor at which we want to calculate Vo.

Second step: Calculating Vth using superposition method.

Only voltage source acting: No current in 'CA' branch, so voltage drop across 4kΩ and 2kΩ resistor is zero. Vth1 = VCD Applying voltage division rule: Vth1 = V6k = (12 × 6) / (3 + 6) = 8 volts

📌 Example: Vth1 = (12V × 6kΩ) / (3kΩ + 6kΩ) = 72/9 = 8V

Only current source acting: 3kΩ || 6kΩ = (3kΩ × 6kΩ) / (3kΩ + 6kΩ) = 2kΩ 2kΩ is in series with 2kΩ = 4kΩ Current through open circuit is zero by Ohm's Law. Vth2 = 4kΩ × 2mA = 8 volts

📌 Example: Vth2 = (4 × 10³) × (2 × 10⁻³) = 8V

Total Vth = Vth1 + Vth2 = 8 + 8 = 16 volts

Third step: Calculating Rth. 3kΩ || 6kΩ = (3kΩ × 6kΩ) / 9kΩ = 2kΩ Rth = 2kΩ + 2kΩ + 4kΩ = 8kΩ

Fourth step: Calculating unknown quantity. Re-inserting RL = 8kΩ in series with Rth = 8kΩ and Vth = 16V.

📌 Example: Vo = (8kΩ / (8kΩ + 8kΩ)) × 16 = (8k/16k) × 16 = 8 volts

EXAMPLE 3: Calculating Vo using Thevenin's Theorem (with KVL)

Circuit Details: Contains a 6V voltage source, 2mA current source, resistors of 2kΩ (two), 4kΩ, and a 6kΩ load resistor.

First step: Removing RL. Here RL is the 6kΩ resistor at which we want to calculate Vo.

Second step: Calculating Vth using KVL (Kirchhoff's Voltage Law).

Given: I2 = 2mA For loop 1: 4kΩ(I1) + 2kΩ(I1 - I2) = 6 4kΩ(I1) + 2kΩ(I1) - 2kΩ(I2) = 6 6kΩ(I1) - 2kΩ(I2) = 6 Substituting I2 = 2mA: 6kΩ(I1) - 4 = 6 I1 = 10/6kΩ = 1.6mA

Voltage across 4kΩ resistor: V4k = 1.6mA × 4kΩ = 20/3 = 6.66 volts Voltage across 2kΩ resistor: V2k = 2mA × 2kΩ = 4 volts

📌 Example: Vth = V4k + V2k = 20/3 + 4 = 6.66 + 4 = 10.66 volts

Third step: Calculating Rth. 4kΩ || 2kΩ = (4kΩ × 2kΩ) / 6kΩ = 8/6 kΩ = 4/3 kΩ = 1.33 kΩ These are in series, so: Rth = 1.33kΩ + 2kΩ = 10/3 kΩ = 3.33kΩ

Fourth step: Calculating unknown quantity. Re-inserting RL = 6kΩ in series with Rth = 10/3 kΩ and Vth = 32/3 V.

📌 Example: Vo = 32/3 × 6 × 1/(10/3 + 6) = 64 × 3/(10 + 18) = 192/28 = 48/7 volts

⭐ Key Takeaways

Thevenin's Theorem provides a powerful method to simplify complex circuits by replacing everything except the load resistor with an equivalent voltage source (Vth) and series resistance (Rth). The four-step procedure must be followed strictly: remove the load, calculate Vth using any circuit analysis method, calculate Rth by deactivating all independent sources, then re-insert the load and use simple series circuit calculations. Vth can be found using voltage division, superposition, KVL, or source transformation, while Rth requires shorting voltage sources and opening current sources. The final calculation always uses the voltage divider formula Vo = (RL/(Rth+RL)) × Vth for voltage across the load resistor.

🧠 Quick Revision Questions

1. What are the four steps of Thevenin's Theorem procedure?
2. How do you calculate Thevenin resistance (Rth) when finding the equivalent resistance?
3. In Example 2, why was superposition used to find Vth, and what were the individual contributions?
4. In Example 3, how was KVL applied to find the currents and voltages needed for Vth?
5. What is the formula for calculating the voltage across the load resistor once Vth and Rth are known?

📘 Lecture 26 — Examples of Thevenin’s Theorem - Thevenin’s Theorem and Dependent sources with examples

📖 Overview: This lecture demonstrates the practical application of Thevenin’s Theorem through multiple numerical examples. It also explains the modified procedure for applying Thevenin’s Theorem to circuits containing dependent sources. Understanding these examples is crucial for solving complex circuit analysis problems and for maximum power transfer calculations.

🗂️ Topics Covered

The lecture covers three detailed examples of Thevenin’s Theorem applied to resistive circuits with independent sources, followed by a dedicated example demonstrating the special technique required when the circuit contains dependent sources. Each example follows the systematic four-step procedure: removing the load, calculating Vth, calculating Rth, and finding the unknown quantity.

📝 Lecture Summary

Example 1: Calculating Voltage V₀ using Thevenin’s Theorem

First step: Removing RL The load resistor RL is the 4kΩ resistor across which voltage V₀ is to be calculated. This resistor is removed from the circuit.

Second step: Calculating Vth We want to calculate Vth. Apply KVL in two loops to calculate the individual loop currents. Here I₁ = 2mA

KVL for the super mesh: -6 + 6kI₂ + 12k(I₂ + I₁) + 12k(I₂ + I₁) = 0 -6 + 6kI₂ + 12kI₂ + 12kI₂ + 24kI₁ = 0 30kI₂ – 6 + 24k(2m) = 0 30kI₂ – 6 + 48 = 0 I₂ = -42/30mA I₂ = -1.4mA

Now we calculate the voltage across 6k and 12k to find Vth. V₆ₖ = 6kI₂ = 6k(-1.4m) = -8.4 volts V₁₂ₖ = 12k(I₁ + I₂) = 12k(2m – 1.4m) = 7.2 volts Vth = V₆ₖ + V₁₂ₖ = -8.4 + 7.2 🔑 Vth = -1.2 volts

Third step: Calculating Rth 6k is in series with 12k. The resultant is in parallel with 12k. (6k+12k) || 12k = 18k x 12k / (12k + 18k) 🔑 Rth = 7.2kΩ

Fourth step: Calculating the unknown quantity After calculating Vth and Rth, re-insert the load resistance RL in series with Rth, with Vth as a battery. V₀ = -1.2 x 4k / (7.2k + 4k) 📐 V₀ = -0.4285 volts

Example 2: Calculating RL and Maximum Power Dissipation

First step: Removing RL In this case, RL is the load resistor. Remove it to calculate Vth.

Second step: Calculating Vth Here I₁ = 2mA

Applying KVL to loop 2: 6kI₂ + 3k(I₂ – I₁) + 3 = 0 9kI₂ – 3kI₁ + 3 = 0 9kI₂ – 6 + 3 = 0 I₂ = 1/3 mA = 0.33mA

Calculate voltages: VAB = 4k(I₁) = 4k(2m) = 8 volts VBC = 6k(I₂) = 6k x 0.33m = 2 volts Vth = VAB + VBC = 8 + 2 🔑 Vth = 10 volts

Third step: Calculating Rth Short circuit the voltage source and open circuit the current source. 3k is in parallel with 6k, and 4k is in series with the result. 3k || 6k + 4k = (3k x 6k)/(3k+6k) + 4k = 2k + 4k = 6k 🔑 Rth = RL = 6kΩ

Fourth step: Calculating unknown quantity For maximum power dissipation: RL = Rth = 6k P = I²R From Ohm’s law I = V/R = 10/(6k+6k) = 10/12k P = (10/12k)² x 6k = (0.83)² x 6 📐 PL = 4.1 mW 💡 Why this matters: Maximum power transfer occurs when the load resistance equals the Thevenin resistance.

Example 3: Calculating RL and Maximum Power Dissipation (Second Circuit)

First step: Removing RL Remove the load resistor RL.

Second step: Calculating Vth Apply KVL to the circuit.

For loop 1: -12 + 6I₁ + 6(I₁ – I₂) = 0 -12 + 12I₁ – 6I₂ = 0 2I₁ – I₂ = 2 —————— (A)

For loop 2: 12I₂ – 6I₁ + 3 = 0 4I₂ – 2I₁ + 1 = 0 2I₁ = 4I₂ + 1

Putting in equation (A): 4I₂ + 1 – I₂ = 2 3I₂ = 1 I₂ = 0.33mA

From equation (A): I₁ = (I₂ + 2)/2 = (0.33 + 2)/2 I₁ = 1.166mA

Voltage across 6k resistor: V₆ₖ = 6kI₁ = 6k(1.166m) = 7 volts Vth = 7V + 3V 🔑 Vth = 10 volts

Third step: Calculating Rth 6k is in parallel with 6k. The result is in parallel with a third 6k resistor. 6k || 6k = (6k x 6k)/(6k + 6k) = 3k 3k || 6k = (6k x 3k)/(6k + 3k) = 2k 🔑 Rth = RL = 2kΩ

Fourth step: Calculating unknown quantity VRL = 10 x 2k / (2k + 2k) = 5 volts

To calculate power dissipation: P = V²/R = 25/2k 📐 PL = 12.5 mW

THEVENIN’S THEOREM AND DEPENDENT SOURCES

Working with dependent sources is different from working with independent sources while applying Thevenin’s Theorem. While calculating Rth, we cannot simply open circuit current sources and short circuit voltage sources because the voltage or current of the dependent sources depends on the independent sources.

While calculating Rth, we will short circuit the open terminals of the Thevenin circuit and calculate Isc (short circuit current). Then divide Vth by Isc to calculate Rth. 🔑 Rth = Vth / Isc

Example: Calculating Voltage V₀ using Thevenin’s Theorem with Dependent Source

First step: Removing RL RL is the 6kΩ resistor at which V₀ is to be calculated.

Second step: Calculating Vth Voltage across 4k resistor: V₄ₖ = (4k/6k) x 12 = 8 volts

Voltage across 2k resistor: V₂ₖ = (2k/6k) x 12 VA = V₂ₖ = 4 volts

Vth = 8 – 4VA = 8 – 4(4) = 8 – 16 🔑 Vth = -8 volts

Third step: Calculating Rth Replace the open circuit with a short circuit to find Isc.

VA = 2kI₁

Applying KVL to loop 1: For the other loop: -12 + 2kI₁ + (I₁ – Isc)4k = 0 -12 + 2kI₁ + 4kI₁ – 4kIsc = 0 6kI₁ – 4kIsc = 12 ————— (1)

For the supermesh containing the dependent source: (Isc – I₁)4k + 4VA = 0 4kIsc – 4kI₁ + 4(2kI₁) = 0 4kIsc – 4kI₁ + 8kI₁ = 0 4kIsc + 4kI₁ = 0 I₁ = -Isc

Putting in equation (1): 6k(-Isc) – 4kIsc = 12 -6kIsc – 4kIsc = 12 -10kIsc = 12 Isc = -6/5 mA

🔑 Rth = Vth/Isc = -8 / (-6/5 m) = 6.67kΩ

Fourth step: Calculating unknown quantity V₀ = (6k / (6k + 6.67k)) x 8 = (6k / 12.67k) x 8 📐 V₀ = 3.78 volts

⭐ Key Takeaways

Thevenin’s Theorem simplifies any linear circuit to a single voltage source (Vth) in series with a single resistor (Rth). For circuits with only independent sources, Rth is found by shorting voltage sources and opening current sources, then calculating the equivalent resistance. For circuits with dependent sources, Rth must be found using the short-circuit current method: Rth = Vth / Isc. Maximum power is transferred to the load when RL equals Rth, and the maximum power is calculated using Pmax = Vth² / (4Rth). The four-step procedure—removing RL, finding Vth, finding Rth, and reconnecting RL—is universally applicable.

🧠 Quick Revision Questions

1. What is the value of Vth in Example 1, and explain the sign?
2. How is Rth calculated differently for circuits with dependent sources compared to independent sources?
3. What condition must be satisfied for maximum power transfer, and what is the formula for maximum power?
4. In the dependent source example, why was the open circuit replaced with a short circuit, and what value of Isc was found?
5. Using the results of Example 3, what would the load voltage be if a 4kΩ resistor were used instead of the 2kΩ load?

📘 Lecture 27 — Thevenin's Theorem and Dependent sources with examples - Norton's Theorem with examples

📖 Overview: This lecture demonstrates the application of Thevenin's theorem to circuits containing dependent sources, followed by an introduction and application of Norton's theorem. Mastering these theorems is essential for simplifying complex circuits into equivalent voltage or current sources with a single resistance, making analysis of load currents and voltages straightforward.

🗂️ Topics Covered

The lecture begins with two detailed examples of applying Thevenin's theorem, one with a dependent source requiring calculation of short-circuit current for Thevenin resistance. It then introduces Norton's theorem as a dual approach, explaining the four-step process and illustrating it with two examples involving both independent and dependent sources, culminating in load voltage calculations.

📝 Lecture Summary

Thevenin's Theorem with Dependent Sources

Thevenin's theorem states that any linear circuit can be replaced by an equivalent voltage source (Vth) in series with an equivalent resistance (Rth). When a circuit contains a dependent source (also called a controlled source), the Thevenin resistance cannot be found by simply deactivating independent sources and combining resistances. Instead, Rth is calculated by finding the open-circuit voltage (Vth) and dividing it by the short-circuit current (Isc) flowing from the open terminals when they are shorted: Rth = Vth / Isc.

🔑 Definition — Thevenin's Theorem: A linear two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source Vth in series with a resistor Rth, where Vth is the open-circuit voltage at the terminals and Rth is the resistance seen from the terminals with all independent sources deactivated (dependent sources remain).

📐 Formula: Rth = Vth / Isc → The Thevenin resistance is the ratio of the open-circuit voltage to the short-circuit current measured at the same terminals.

💡 Why this matters: This method is required when dependent sources are present because they cannot be simply turned off; the circuit must remain active to define the relationship between the controlling and controlled variables.

Example 1: Calculate current Io using Thevenin's theorem.

• Circuit: A 12V source, resistors: 8kΩ, 4kΩ, 12kΩ, 4kΩ, and a dependent voltage source (2Vx). Load resistor RL = 2kΩ across which Io flows.
• Step 1: Remove RL (the 2kΩ resistor) to find Vth.
• Step 2: Calculate Vth (open-circuit voltage).
  • Voltage at node A: VA = (4kΩ × 12V) / (4kΩ + 8kΩ) = 48/12 = 4V.
  • Voltage at node B: Vx = VA = 4V. VB = (4kΩ × 2Vx) / (12kΩ + 4kΩ) = (4kΩ × 8V) / 16kΩ = 2V.
  • Vth = VAB = VA - VB = 4V - 2V = 2V.
• Step 3: Calculate Rth by finding Isc.
  • Short the open terminals (A and B) and find the short-circuit current Isc.
  • Apply KCL at node 1 (V1, where Vx = V1): (V1 - 12)/8k + V1/4k + V1/4k + (V1 - 2V1)/12k = 0.
  • Solving: 3V1 - 36 + 6V1 + 6V1 + 2V1 - 4V1 = 0 → 13V1 = 36 → V1 = 36/13 V.
  • Calculate branch currents: I1 = (12 - V1)/8k = (12 - 36/13)/8k = 120/13k = 15/13 mA ≈ 1.15mA.
  • I2 = V1/4k = (36/13)/4k = 9/13 mA ≈ 0.692mA.
  • I4 = V1/4k = 9/13 mA = 0.692mA.
  • I5 = (V1 - 2V1)/12k = (-V1)/12k = (-36/13)/12k = -3/13 mA ≈ -0.23mA.
  • Isc = I3 = I5 + I4 = -0.23mA + 0.692mA = 0.462mA ≈ 0.46mA.
  • Rth = Vth / Isc = 2V / 0.46mA = 4.33kΩ.
• Step 4: Re-insert RL in series with Rth and Vth.
  • Io = Vth / (Rth + RL) = 2V / (4.33kΩ + 2kΩ) = 2V / 6.33kΩ = 0.315mA.

Example 2: Calculate voltage Vo using Thevenin's theorem.

• Circuit: A 12V source, resistors: 2kΩ (2 of them), 1kΩ, and a dependent current source (2Vx). Load resistor RL = 2kΩ across which Vo appears.
• Step 1: Remove RL (the 2kΩ) to find Vth.
• Step 2: Calculate Vth (open-circuit voltage).
  • At node 1, define Vx = V1 + 12.
  • KCL at node 1: V1/2k + 2Vx/1k + Vx/2k = 0.
  • Substitute Vx: V1 + 4(V1+12) + (V1+12) = 0 → V1 + 4V1 + 48 + V1 + 12 = 0 → 6V1 = -60 → V1 = -10V.
  • Then Vx = V1 + 12 = -10 + 12 = 2V = Vth.
• Step 3: Calculate Rth by finding Isc.
  • Short the output terminals. Now Vx = 0? No, Vx is defined relative to the dependent source. When shorted, the node voltages change. The circuit is re-analyzed.
  • With the short, apply KCL at the node above the dependent source (V1). Vx = V1 + 12.
  • KCL: V1/2k + 2Vx/1k + Vx/2k + Vx/2k = 0. (Note: an extra Vx/2k term appears due to the short circuit path.)
  • Substitute Vx: (Vx - 12) + 4Vx + Vx + Vx = 0 → Vx - 12 + 6Vx = 0 → 7Vx = 12 → Vx = 12/7 V.
  • Isc flows through the 2kΩ resistor to the short. Isc = Vx / 2k = (12/7) / 2k = 6/7 mA ≈ 0.857mA.
  • Rth = Vth / Isc = 2V / (6/7 mA) = 2 × 7/6 kΩ = 7/3 kΩ = 2.33kΩ.
• Step 4: Re-insert RL in series with Rth and Vth.
  • Vo = Vth × RL / (Rth + RL) = 2V × 2kΩ / (2.33kΩ + 2kΩ) = 4 / 4.33 = 0.92V.

Norton's Theorem

Norton's theorem is the dual of Thevenin's theorem. It states that any linear circuit can be replaced by an equivalent current source (In) in parallel with an equivalent resistance (Rn). The four steps are similar: (1) Remove the load to calculate the Norton current or resistance, (2) Find In by short-circuiting the open terminals and calculating the short-circuit current, (3) Find Rn (which equals Rth) by deactivating all independent sources and calculating the equivalent resistance, (4) Re-insert the load in parallel with Rn and In.

🔑 Definition — Norton's Theorem: A linear two-terminal circuit can be replaced by an equivalent circuit consisting of a current source In in parallel with a resistor Rn, where In is the short-circuit current at the terminals and Rn is the input resistance at the terminals with all independent sources deactivated.

📐 Formula: Rn = Vth / In → The Norton resistance equals the Thevenin resistance; the Norton current equals the short-circuit current.

Example 3: Calculate voltage Vo using Norton's theorem (independent sources).

• Circuit: A 12V source, 6V source, resistors (6kΩ, 3kΩ, 2kΩ, 1kΩ), and a 1mA current source. Load resistor RL = 4kΩ across which Vo appears.
• Step 1: Short RL (the 4kΩ) to find In.
• Step 2: Calculate In (short-circuit current).
  • Define loop currents I1 (left loop) and I2 (right loop, through the short).
  • KVL Loop 1: 6k(I1) + 3k(I1 - I2) - 12 - 6 = 0 → 9k I1 - 3k I2 = 18 → 9I1 - 3I2 = 18.
  • KVL Loop 2: 2k(I2) + 1 + 3k(I2 - I1) = 0 → 5k I2 - 3k I1 = -12 → 5I2 - 3I1 = -12.
  • Solve: Multiply eq2 by 3: 15I2 - 9I1 = -36. Add to eq1: (9I1 - 3I2) + (15I2 - 9I1) = 18 - 36 → 12I2 = -18 → I2 = -1.5mA.
  • In = I2 = -1.5mA.
• Step 3: Calculate Rn.
  • Deactivate voltage sources (short 12V and 6V) and current source (open 1mA).
  • From the open terminals looking in: 6kΩ || 3kΩ = 2kΩ. This is in series with 2kΩ = 4kΩ.
  • Rn = 4kΩ.
• Step 4: Re-insert RL in parallel with Rn and In (current source direction gives negative current).
  • Current through RL (using current division): IL = In × Rn / (Rn + RL) = (-1.5mA) × 4kΩ / (4kΩ + 4kΩ) = -6 / 8 = -0.75mA.
  • Vo = IL × RL = -0.75mA × 4kΩ = -3V.

Example 4: Calculate voltage Vo using Norton's theorem (dependent source).

• Circuit: A 2V source, 4mA current source, resistors: 4kΩ, 3kΩ, 6kΩ, 2kΩ, 8kΩ. Load resistor RL = 4kΩ across which Vo appears.
• Step 1: Short RL (the 4kΩ at the output) to find In.
• Step 2: Calculate In (short-circuit current).
  • Define nodes: Node 1 (at the junction of 4k, 3k, 6k, and 4mA source), Node 2 (at the junction of 2k and 8k).
  • KCL at Node 1 (V1): V1/4k + (V1 - 2)/3k + V1/6k + 4m = 0.
  • Multiply by 12k: 3V1 + 4(V1-2) + 2V1 + 48 = 0 → 3V1 + 4V1 - 8 + 2V1 + 48 = 0 → 9V1 + 40 = 0 → V1 = -40/9 V ≈ -4.44V.
  • KCL at Node 2 (V2): V2/2k + V2/8k - 4m = 0.
  • Multiply by 8k: 4V2 + V2 - 32 = 0 → 5V2 = 32 → V2 = 32/5 = 6.4V.
  • Short-circuit current Isc flows from V2 through 8k to V1? No, the short connects the right terminal to ground. Isc has two paths: I1 (from V2 through 8k to ground) and I2 (from V1 through 4k to ground).
  • I1 = V2 / 8k = 6.4V / 8k = 0.8mA.
  • I2 = V1 / 4k = -4.44V / 4k = -1.11mA (negative means current flows into the short from the right side).
  • In = Isc = I1 + I2 = 0.8mA + (-1.11mA) = -0.31mA.
• Step 3: Calculate Rn.
  • Deactivate all independent sources (short 2V, open 4mA).
  • Simplify: 3kΩ || 6kΩ = 2kΩ. This is in series with 4kΩ = 6kΩ.
  • Also: 8kΩ in series with 2kΩ = 10kΩ.
  • Finally: 6kΩ || 10kΩ = (6 × 10) / 16 = 60/16 = 3.75kΩ = Rn.
• Step 4: Re-insert RL in parallel with Rn and In.
  • Current through RL (current division): IL = In × Rn / (Rn + RL) = 0.31mA × 3.75kΩ / (3.75kΩ + 4kΩ) = 1.1625 / 7.75 = 0.15mA. (Note: the text shows 0.14mA, likely due to rounding).
  • Vo = IL × RL = 0.15mA × 4kΩ = 0.6V. (The text calculates Vo = 0.56V).

⭐ Key Takeaways

The critical concept is that Thevenin and Norton theorems provide two equivalent ways to simplify a complex linear circuit: a voltage source in series with a resistance, or a current source in parallel with a resistance. For circuits with dependent sources, you cannot simply deactivate all sources to find the equivalent resistance; you must find both the open-circuit voltage (Vth) and the short-circuit current (Isc), then compute Rth = Vth / Isc. Norton's theorem is the dual, where the Norton current (In) equals the short-circuit current, and the Norton resistance (Rn) equals the Thevenin resistance. The four-step process is the same for both theorems, but the final equivalent circuit is different: series for Thevenin and parallel for Norton. Finally, the choice between using Thevenin or Norton is often a matter of convenience, depending on which calculation (open-circuit voltage or short-circuit current) is easier to perform.

🧠 Quick Revision Questions

1. What is the difference between finding the Thevenin resistance for a circuit with only independent sources versus one that includes dependent sources?
2. Explain why the Thevenin resistance can be found by Rth = Vth / Isc when dependent sources are present.
3. In Norton's theorem, how is the Norton current (In) calculated, and how does it relate to Thevenin's theorem?
4. For the circuit in Example 1, what is the value of the short-circuit current (Isc) and the resulting Thevenin resistance?
5. In Example 4 (Norton's with dependent source), describe the steps to find the Norton resistance and the final load voltage.

📘 Lecture 28 — Norton's Theorem with examples - Linearity Principle

📖 Overview: This lecture continues the study of Norton's Theorem through multiple worked examples, demonstrating how to find Norton equivalent circuits for various network configurations. It also introduces the Linearity Principle as an alternative analysis technique that allows circuit solving by assuming an unknown value and scaling based on the actual source value.

🗂️ Topics Covered

The lecture covers multiple examples applying Norton's Theorem to find equivalent circuits and calculate currents/voltages in resistive networks. It demonstrates the step-by-step Norton procedure including finding Norton current (I_N) by short-circuiting the load, calculating Norton resistance (R_N) by deactivating independent sources, and re-inserting the load. The lecture also introduces the Linearity Principle, showing how to work backwards from an assumed output to find the actual circuit values through proportional scaling.

📝 Lecture Summary

Norton's Theorem: First Example

The simplest example demonstrates finding the Norton equivalent circuit. First, remove the load resistor R_L and replace the open circuit with a short circuit. By Ohm's Law, the Norton current I_N = 12V/0Ω → ∞A. Then insert resistance R=0Ω parallel to the current source to obtain the Norton equivalent circuit.

🔑 Definition — Norton Current (I_N): The current that flows through a short circuit placed across the terminals of the load. 🔑 Definition — Norton Resistance (R_N): The equivalent resistance seen from the load terminals when all independent sources are deactivated (voltage sources shorted, current sources opened).

Norton's Theorem: Second Example (with Dependent Source)

To calculate Norton's current through a circuit containing a dependent source, inject a 1A current source into the port. Define V_x across the 200Ω resistor.

For node 1: V_1/100 + (V_1 - V_x)/50 = 1, giving 3V_1 - 2V_x = 100 ...equation (A)

For node 2: -0.1V_1 = V_x/200 + (V_x - V_1)/50, giving -20V_1 = V_x + 4V_x - 4V_1, which simplifies to -16V_1 = 5V_x, or 5V_x + 16V_1 = 0 ...equation (B)

Solving simultaneously: 10V_x + 32V_1 = 0 and -10V_x + 15V_1 = 500, giving 47V_1 = 500, V_1 = 10.64V.

Therefore R_th = R_N = V_1/1A = 10.64Ω. Since no independent source exists in the circuit, I_N = 0A.

💡 Why this matters: When dependent sources are present but no independent sources, the Norton current is zero and only the Norton resistance matters for the equivalent circuit.

Norton's Theorem: Third Example (Using Superposition)

Calculate voltage V_o using Norton's theorem (no loop or node methods allowed, use superposition).

Step 1: Replace R_L (6kΩ) with a short circuit to find I_N.

Step 2: Calculate I_N using superposition.

• Only current source acting: Due to short circuit, all current flows through the short circuit, so I_N1 = 2mA ...(A)
• Only voltage source acting: 4kΩ is parallel with 2kΩ, in series with 2kΩ. Total resistance R = (4k||2k) + 2k = 8/6k + 2k = 3.33kΩ. I_N2 = 6V/3.33kΩ = 1.80mA ...(B)

Total I_N = I_N1 + I_N2 = 2mA + 1.80mA = 3.80mA

Step 3: Calculate R_N by shorting voltage sources and opening current sources. 4k||2k + 2k = 1.33kΩ + 2kΩ = 3.33kΩ = R_N.

Step 4: Re-insert load R_L in parallel with R_N and I_N current source. Using current divider rule: I_0 = (3.80mA)(3.33kΩ) × 1/(9.33kΩ) = 1.356mA.

By Ohm's Law: V_0 = 6kΩ × 1.35mA = 8.143 volts.

📐 Formula: Current Divider Rule: I_x = I_total × (R_parallel)/(R_parallel + R_branch)

Norton's Theorem: Fourth Example

Calculate current I_o using Norton's theorem with R_L = 6kΩ.

Step 1: Replace R_L with short circuit to find I_N.

Step 2: Calculate I_N using mesh analysis.

• Loop 2: I_2 = 2mA
• Loop 1: 4k(I_1-I_N) + 2k(I_1-I_2) - 6 = 0, giving 6kI_1 - 4kI_N = 10 ...(eq1)
• Loop 3: 4k(I_N-I_2) + 4k(I_N-I_1) = 0, giving -4kI_1 + 8kI_N = 8 ...(eq2)

Solving eq1 and eq2: 12kI_1 - 8kI_N = 20 and -12kI_1 + 24kI_N = 24, giving 16kI_N = 44, I_N = 2.75mA.

Step 3: Calculate R_N by shorting voltage sources and opening current sources. R_N = (2k||4k) + 4k = ((2k×4k)/(2k+4k)) + 4k = 1.33kΩ + 4kΩ = 5.33kΩ.

Step 4: Re-insert load. I_o = (2.75mA)(5.33kΩ) × 1/(6kΩ+5.33kΩ) = 1.29mA.

Linearity Principle

In this technique, assume the unknown quantity and analyze the circuit in reverse manner until reaching the source. Compare the calculated source value with the actual source value to find the exact unknown quantity by proportional scaling.

🔑 Definition — Linearity Principle: A method where an unknown circuit variable is assumed (typically 1V or 1A), the circuit is analyzed backwards to find what source value would produce that assumption, then the result is scaled proportionally to match the actual source value.

Linearity Principle Example

Calculate V_o by linearity principle. Assume V_o = 1V.

V_2 = 1V, so I_2 = 1V/2kΩ = 0.5mA V_4k = (0.5mA)(4kΩ) = 2V V_1 = V_4k + V_2 = 2V + 1V = 3V I_1 = 3V/3kΩ = 1mA I_s = I_2 + I_1 = 0.5mA + 1mA = 1.5mA V_2k = I_s(2kΩ) = (1.5mA)(2kΩ) = 3V V_s = V_2k + V_1 = 3V + 3V = 6V

When V_o is 1V, source voltage is 6V. Original source voltage is 12V, so output voltage scales proportionally: V_o = (12V/6V) × 1V = 2 Volts.

📐 Formula: Linear Scaling: Actual output = (Actual source value / Calculated source value) × Assumed output 📌 Example: (12V/6V) × 1V = 2V

⭐ Key Takeaways

The Norton equivalent circuit consists of a current source I_N in parallel with a resistance R_N, which can be found by short-circuiting the load terminals to measure I_N and deactivating all independent sources to calculate R_N. For circuits with dependent sources but no independent sources, I_N is zero but R_N can still be found using a test source (1A). When multiple independent sources are present, superposition can be used to find I_N by considering each source individually. The Linearity Principle provides an efficient alternative to nodal or mesh analysis by assuming a convenient value for the unknown, working backwards to find the required source value, then scaling proportionally to match the actual source—this method works because circuits with only linear elements follow the principle of proportionality.

🧠 Quick Revision Questions

1. What are the four steps to find the Norton equivalent circuit?
2. In the linearity principle example, why was V_o found to be 2V when the original source was 12V and the assumed V_o of 1V required a 6V source?
3. When a circuit contains a dependent source but no independent sources, what is the value of I_N and how do you find R_N?
4. How was the current divider rule applied in the third Norton example to find I_o from I_N?
5. In the fourth Norton example, why were two simultaneous equations needed to solve for I_N, and what method was used?

📘 Lecture 29 — Basic Semiconductor Concepts

📖 Overview: This lecture introduces fundamental semiconductor concepts, focusing on intrinsic silicon's structure and behavior. It explains energy bands, carrier transport mechanisms (drift and diffusion), and the characteristics of doped semiconductors and p-n junctions under different bias conditions. These concepts form the foundation for understanding electronic devices like diodes and transistors.

🗂️ Topics Covered

The lecture covers intrinsic silicon crystal structure and covalent bonding, thermal ionization and recombination, energy band theory including conduction band, valence band, and band gap energy. It then explains drift and diffusion processes, doped semiconductors (n-type and p-type), and the behavior of p-n junctions under no bias, reverse bias, and forward bias conditions, including the formation of the depletion region.

📝 Lecture Summary

Intrinsic Silicon

A crystal of pure or intrinsic silicon has a regular lattice structure. The atoms are held at their fixed positions by bonds called Covalent bonds, formed by four valence electrons associated with each silicon atom.

At sufficiently low temperatures, these covalent bonds are intact and very few free electrons are available to conduct electric current.

Thermal ionization results in free electrons and holes in equal numbers and hence equal concentrations. These free electrons and holes move randomly through the silicon crystal structure, and in this process some electrons may fill some of the holes. This process is called RECOMBINATION.

Recombination results in disappearance of free electrons and holes. The recombination rate is proportional to the number of free electrons and holes, which in turn is determined by ionization rate.

Conduction Band

In semiconductors and insulators, there is a band gap above the valence band, called a conduction band. In a semiconductor, the range of electron energy, higher than that of the valence band, is sufficient to make the electrons free to move from atom to atom under the influence of an applied electric field and thus constitute an electric current.

Valence Band

At 0°K, the outermost electrons of an atom of a semiconductor material (such as silicon and germanium) are present in valence orbits. Associated with these orbits is a band of energies that is termed as valence band of energy.

Band Gap

The difference of energy between valence band and conduction band is called band gap energy.

Drift

The process whereby charged particles move under the influence of an electric field.

When there are few electrons present in the nearby valence band and we apply an electric field to such a material, these electrons may drift due to the applied field. This results in observable activity in the electrical property of the material, and some flow of current is observed. Such materials are called conductors.

When all energy levels are fully filled with electrons and we apply an electric field, these electrons have no space to drift. Such materials act as insulators, which offer maximum resistance to the flow of electrons and have very low conductivity.

When almost all energy levels are full except two levels kept empty, and we apply an electric field, these electrons have some space to drift. We observe some flow of current through such materials, so these are called conductors. Conductors offer zero resistance to the flow of electrons and have maximum conductivity.

Band Gap Energy

It is the energy required to excite an electron from its valence band to a conduction band and is denoted as Eg.

Diffusion

The process of flow of particles from a region of high concentration to a region of low concentration.

Diffusion Current

The current that results from the diffusion of charged particles.

Drift Velocity

The average velocity of charged particles in the presence of an electric field.

Doped Semiconductors

Doped semiconductors are materials in which carriers of one kind (electrons or holes) predominate.

Doped silicon in which the majority of charge carriers are negatively charged electrons is called n type.

Silicon doped so that majority of charge carriers are positively charged holes is called p type.

Doping of silicon to turn it into p type or n type is achieved by a small number of impurity atoms.

For instance, introducing impurity atoms of a pentavalent element such as phosphorus results in n type silicon.

No Bias — The Diffusion Current ID

Because the concentration of holes is high in the p region and low in the n region, holes diffuse across the junction from the p region to the n region.

Similarly, electrons diffuse across the junction from n side to p side.

These two current components add together to form the diffusion current ID, whose direction is from p side to n side.

The Depletion Region

The electrons that diffuse across the region quickly recombine with some of the majority holes present in the p region and thus disappear from the scene.

This results also in disappearance of some majority holes, causing some of the bound negative charge to be uncovered (no longer neutralized by holes). Thus in the p material close to the junction, a region is depleted of free electrons.

This p region will contain uncovered bound negative charge.

From the above it follows that a carrier depletion region will exist on both sides of the junction, with the n side of this region positively charged and p side negatively charged. This carrier depletion region, or simply depletion region, is also called the SPACE CHARGE REGION.

The P-N Junction Under Reverse Bias Condition

The pn junction is excited by a constant current source I in the reverse direction. To avoid breakdown, I is kept smaller than IS.

The depletion layer widens and the barrier voltage increases by VR volts, which appears between the terminals as reverse voltage.

The current I will be carried by electrons flowing in the external circuit from the n material to p material (in direction opposite to that of I).

This will cause electrons to leave the n material and holes to leave p material. Thus reverse current I will result in an increase in the width of, and the charge stored in, the depletion layer, which will increase the voltage across the depletion region.

The P-N Junction Under Forward Bias Condition

The pn junction is excited by a constant current source supplying a current I in the forward direction. The depletion layer narrows and barrier voltage decreases by V volts, which appears as an external voltage in the forward direction.

This current causes majority carriers to be supplied to both sides of the junction by the external circuit — holes to the p material and electrons to the n material.

These majority carriers will neutralize some of the uncovered charges, causing less charge to be stored in the depletion region. Thus the depletion layer narrows and the depletion barrier voltage reduces.

This reduction in voltage causes more electrons to move from n side to p side and more holes to move from p side to n side, so that diffusion currents increase until equilibrium is achieved.

💡 Why this matters: Understanding p-n junction behavior under different bias conditions is essential for analyzing diodes, rectifiers, and all semiconductor devices in electronic circuits.

⭐ Key Takeaways

Intrinsic silicon has covalent bonds that hold atoms together, and at low temperatures few free electrons exist. Thermal ionization creates equal numbers of electrons and holes, and recombination annihilates them. Energy bands include the valence band, conduction band, and band gap — the energy difference between them. Drift is carrier movement due to an electric field, while diffusion is carrier movement due to concentration gradients. Doped semiconductors can be n-type (excess electrons) or p-type (excess holes). The p-n junction under no bias has a diffusion current and forms a depletion region with positive n-side and negative p-side charges. Under reverse bias, the depletion layer widens and barrier voltage increases; under forward bias, the depletion layer narrows and barrier voltage decreases.

🧠 Quick Revision Questions

1. What are covalent bonds in intrinsic silicon, and how many valence electrons does each silicon atom contribute to them?
2. What is recombination, and what determines its rate in intrinsic silicon?
3. How does the depletion region width change under reverse bias versus forward bias conditions?
4. What is the difference between drift current and diffusion current in semiconductors?
5. What type of impurity element (pentavalent or trivalent) is added to create n-type silicon, and why?

📘 Lecture 30 — (PN Junction Diode - Ideal Diode - Ideal Diode as a Rectifier)

📖 Overview: This lecture introduces the fundamental semiconductor device called the PN-junction diode and its two key operating regions: forward bias and reverse bias. It then defines the concept of an ideal diode as a unidirectional switch and demonstrates its application as a half-wave rectifier to convert AC to pulsating DC. Understanding these principles is critical for analyzing all diode-based electronic circuits.

🗂️ Topics Covered

The lecture begins with the introduction to the PN junction diode and its schematic symbol. It then explains the conditions and circuit configurations for forward-biased operation, where the diode conducts, followed by reverse-biased operation, where it acts as an open circuit. The concept of the ideal diode is introduced, defined by its zero forward resistance and infinite reverse resistance. Finally, the lecture concludes by applying the ideal diode model to a simple rectifier circuit, showing how it converts a sinusoidal AC input into a unidirectional output voltage.

📝 Lecture Summary

INTRODUCTION TO THE PN- JUNCTION DIODE.

The schematic symbol for the pn-junction diode is shown. A diode will conduct when the voltage difference between the anode and the cathode exceeds the barrier voltage of approximately 0.3 volts for a germanium diode and 0.7 for a silicon diode.

THE FORWARD BIASED PN-JUNCTION (DIODE)

When the positive terminal of a battery is connected to the anode (P-side) and the negative terminal to the cathode (N-side), the diode is said to be FORWARD-BIASED. The arrow in the symbol points to the more negative potential. Several circuit configurations fulfill this condition:

• Fig. a: Positive terminal connected to anode through a resistor; cathode to neutral. Current (IF) flows.
• Fig. b: Negative terminal connected to cathode; anode through resistor to neutral. Current (IF) flows.
• Fig. c: Positive voltage applied to anode through a resistor; cathode to neutral through a resistor.
• Fig. d: Negative voltage applied to cathode through a resistor; anode to neutral through a resistor.

In all these cases, forward conventional current (IF) is established in the diode.

THE REVERSE - BIASED PN-JUNCTION (DIODE)

When the positive terminal of a battery is connected to the cathode (N-side) and the negative terminal to the anode (P-side), the diode is said to be REVERSE-BIASED. In this condition, the n-type material (cathode) is more positive than the p-type material (anode). This causes the depletion region to widen and prevent current flow. A diode will not conduct when the arrow points to the more positive potential.

• In fig. a and b, the negative terminal is connected to the anode through a resistor and the positive terminal to the cathode. No majority carrier current flows; the diode acts as an open circuit.
• In fig. c and d, the bias conditions are met through connections to neutral.

🔑 Definition — Reverse-Biased Diode: A reverse-biased PN-junction cannot support majority carrier current, but it allows a very small minority carrier current (called reverse current) to flow across the junction. The depletion layer expands, so the diode acts as an open circuit. 💡 Why this matters: The extreme difference in current flow between forward and reverse bias is what makes the diode a useful one-way switch for electronics.

The Ideal Diode

The ideal diode is considered to be a one-way street; it conducts electricity well in one direction but hardly any in the opposite direction. An ideal diode has no resistance in the forward direction and infinite resistance in the reverse direction.

• It is analogous to a light switch (complete circuit = on, open circuit = off) but with the added property of being unidirectional (current flows only from anode to cathode internally).
• A mechanical analogy is a ratchet, which allows motion in one direction only.
• The ideal diode is the most fundamental non-linear circuit element.
• Its i-v characteristic (fig b) shows that for any positive voltage, current flows infinitely (short circuit), and for any negative voltage, current is zero (open circuit).
• When a negative voltage is applied, the diode is reverse-biased and behaves as an open circuit (fig c), also said to be cut off.
• When a positive voltage is applied, the diode is forward-conducting and behaves as a short circuit (fig d) with zero voltage drop, also said to be turned on.

The Ideal Diode as a Rectifier

Rectification is defined as the process of converting an alternating current to a unidirectional current. The rectifier circuit (fig a) consists of a series connection of a diode D and a resistor R.

• With a sinusoidal input voltage (fig b) and assuming an ideal diode:
  • During the positive half cycle, the diode is forward-biased (short circuit, fig c). The output voltage VO equals the input voltage vI.
  • During the negative half cycle, the diode is reverse-biased (open circuit, fig d). The output voltage VO is zero.
• The resulting output waveform (fig e) is unidirectional and has a finite average (DC) value, even though the input had a zero average value. This circuit is called a rectifier.

📐 Formula: For an ideal half-wave rectifier circuit: [ V_O = V_I \text{ (during positive half-cycle)} ] [ V_O = 0 \text{ (during negative half-cycle)} ] → This means the circuit passes one polarity of the input signal and blocks the other.

📌 Example: The lecture provides a problem to draw the transfer characteristic (v0 vs vI) for a given circuit.

⭐ Key Takeaways

The PN junction diode, made of silicon or germanium, conducts only in forward bias when the anode is more positive than the cathode by at least the barrier voltage (0.7V for Si, 0.3V for Ge). In reverse bias, the depletion region widens, preventing majority carrier current and making the diode act as an open circuit, allowing only a tiny minority carrier current. An ideal diode is a theoretical model with zero resistance in forward bias (short circuit) and infinite resistance in reverse bias (open circuit). This unidirectional switching action is essential for rectification, which converts a bipolar AC signal into a unipolar (pulsating DC) signal, as demonstrated in the half-wave rectifier circuit. The transfer characteristic of a diode circuit maps the input voltage to the output voltage based on the diode’s bias state.

🧠 Quick Revision Questions

1. What is the typical barrier voltage for a forward-biased silicon diode?
2. Describe the behavior of the depletion region in a reverse-biased PN junction.
3. What are the two conditions (bias states) of an ideal diode, and what equivalent circuit element does each state represent?
4. In a half-wave rectifier circuit with an ideal diode and a sinusoidal input, what is the output voltage during the negative half-cycle of the input?
5. Why is an ideal diode considered a "non-linear circuit element"?

📘 Lecture 31 — Terminal characteristics of the Junction diodes - Forward bias region

📖 Overview: This lecture covers the terminal characteristics of junction diodes, focusing on the forward bias region. It explains how diodes behave in circuits through detailed examples involving battery charging, ideal diode analysis, and the mathematical i-v relationship that governs diode operation, which is essential for understanding semiconductor device behavior in electronic circuits.

🗂️ Topics Covered

The lecture begins with practical circuit examples analyzing diode conduction angles and current calculations in battery charging circuits. It then examines ideal diode behavior in various biasing configurations to determine current and voltage values. The second half introduces the terminal characteristics of junction diodes, detailing the three operating regions with emphasis on the forward bias region, including the diode equation, thermal voltage, and the scaling constant Iₛ.

📝 Lecture Summary

Example (Battery Charging Circuit)

A circuit for charging a 12-volt battery is analyzed where vₛ is a sinusoid with 24 volts peak amplitude. The diode conducts when vₛ exceeds 12 volts. The conduction angle is 2θ, where θ is given by 24cosθ = 12, so cosθ = 1/2, giving θ = 60° and the conduction angle is 120° or one-third of the cycle.

The peak value of the diode current is given by I_d = (24 – 12)/100 = 0.12 A. The maximum reverse voltage across the diode occurs when vₛ is at its negative peak and equals 24 + 12 = 36 volts.

Example: Assuming the diodes to be ideal

In fig (a), a positive 5V battery connects to the anode through a 2.5 kΩ resistor, with the cathode connected to neutral. The diode is in forward biased condition and acts as a short circuit. By Ohm's Law, I = 5/2.5 = 2mA, and V = 0 volts.

In fig (b), the diode is reversed. The 5V battery connects to the cathode through 2.5 kΩ, with the anode at neutral. The diode is reverse biased and acts as an open circuit, so I = 0mA and V = 5 volts.

In fig (c), a negative 5V battery connects to the anode through 2.5 kΩ. The diode is reverse biased and acts as an open circuit. I = 0mA and V = -5 volts.

In fig (d), the circuit is same as (c) but the diode is reversed. The diode is forward biased and acts as a short circuit. I = 5/2.5 = 2mA and V = 0 volts.

Example: Finding I and V with ideal diodes

For fig (a), assume both diodes are conducting. V_B = 0 and V_0 = 0. The current through D₂ is I_D2 = (10 – 0)/10 = 1mA. Writing a node equation at B: I + 1 = (0 – (-10))/5, resulting in I = 1mA. Thus D₁ is conducting as assumed, giving I = 1mA and V = 0 volts.

For fig (b), assuming both diodes conduct gives V_B = 0 and V_0 = 0. I_D2 = (10 - 0)/5 = 2mA. The node equation I + 2m = (0 – (-10))/10k yields I = -1mA, which is impossible. Assuming D₁ is off and D₂ is on, I_D2 = (10 – (-10))/15 = 1.33mA, and V_B = -10 + 10 × 1.33 = 3.3 volts. Thus D₁ is reverse biased as assumed, giving I = 0 and V₀ = 3.3 volts.

TERMINAL CHARACTERISTICS OF THE JUNCTION DIODES

The characteristic curve consists of three distinct regions: (1) The forward bias region, determined by v > 0, (2) The reverse bias region, determined by v < 0, and (3) The breakdown region, determined by v < -V_ZK.

THE FORWARD BIAS REGION

The forward bias region is entered when the terminal voltage v is positive. The i-v relationship is closely approximated as i = I_s(e^(v/nV_T) – 1). In this equation, I_s is a constant for a given diode at a given temperature and is directly proportional to the area of the diode.

The Voltage V_T is a constant given as V_T = kT/q, where k is Boltzmann's constant (1.38 × 10⁻²³ J/K), T is absolute temperature in Kelvin, and q is the magnitude of the electronic charge (1.60 × 10⁻¹⁹ C). At room temperature (20°C), the value of V_T is 25.2mV. The constant n has a value between 1 and 2 depending on the material and physical structure of the diode.

🔑 Definition — Thermal Voltage (V_T): The voltage constant in the diode equation given by V_T = kT/q, approximately 25.2mV at room temperature.

For appreciable current I in the forward direction, i >> I_s, so the diode equation can be modified as i = I_s e^(v/nV_T). This can be given alternatively in logarithmic form as v = nV_T ln(i/I_s).

Considering the forward i-v relationship, the current I₁ corresponding to diode voltage V₁ is I₁ = I_s e^(V₁/nV_T). Similarly for V₂, I₂ = I_s e^(V₂/nV_T). Combining these produces I₂/I₁ = e^((V₂ – V₁)/nV_T), which can be rewritten as V₂ – V₁ = nV_T ln(I₂/I₁).

EXAMPLE: Scaling Constant Evaluation

A silicon 1mA device displays a forward voltage of 0.7V at a current of 1mA. The junction scaling constant I_s is evaluated for n = 1 or n = 2.

Since i = I_s e^(v/nV_T), then I_s = i e^(-v/nV_T).

📐 Formula: I_s = i e^(-v/nV_T) → The saturation current equals the diode current multiplied by e raised to the negative voltage divided by n times thermal voltage.

For the 1mA diode: If n = 1: I_s = 10⁻³ e^(-700/25) = 6.9 × 10⁻¹⁶ A If n = 2: I_s = 10⁻³ e^(-700/50) = 8.3 × 10⁻¹⁰ A

The diode conducting 1A at 0.7V corresponds to 1000mA in parallel with a total junction area 1000 times greater, so I_s is also 1000 times greater.

📌 Example: For a 1A diode (same manufacturer, 0.7V at 1A), if n = 1, I_s = 1000 × 6.9 × 10⁻¹⁶ = 6.9 × 10⁻¹³ A. If n = 2, I_s = 1000 × 8.3 × 10⁻¹⁰ = 8.3 × 10⁻⁷ A.

💡 Why this matters: The scaling constant I_s determines the diode's current-carrying capability at a given voltage and is directly proportional to the junction area, making it critical for designing diodes with specific ratings.

⭐ Key Takeaways

The conduction angle of a diode in a rectifier circuit depends on when the input voltage exceeds the battery voltage, and the peak reverse voltage equals the sum of peak input and battery voltages. Ideal diodes act as short circuits when forward biased and open circuits when reverse biased, allowing straightforward current and voltage calculations using Ohm's Law. The diode's i-v characteristic in the forward bias region follows an exponential relationship i = I_s(e^(v/nV_T) – 1), where V_T = kT/q ≈ 25.2mV at room temperature. The scaling constant I_s is proportional to the diode junction area, so a 1A diode has an I_s 1000 times larger than a 1mA diode. The voltage change required for a tenfold current increase in a diode is approximately nV_T × ln(10), which is about 60mV for n = 1 at room temperature.

🧠 Quick Revision Questions

1. In the battery charging example, what fraction of the cycle does the diode conduct, and what is the peak diode current?
2. For an ideal diode, what is its equivalent circuit behavior in forward bias and reverse bias?
3. What are the three distinct regions of a diode's terminal characteristic curve?
4. Write the diode equation for the forward bias region and define each term (I_s, V_T, n).
5. Calculate I_s for a silicon diode with 0.7V forward voltage at 1mA current when n = 1.2 (use V_T = 25mV).

📘 Lecture 32 — Reverse bias region - Break down region - Analysis of diode circuit

📖 Overview: This lecture explores diode behavior beyond forward bias, covering the reverse bias and breakdown regions. It then shifts focus to practical circuit analysis techniques, using examples to demonstrate how to determine diode current and voltage in a simple series circuit. Understanding these regions and analysis methods is crucial for designing and troubleshooting diode-based circuits, including voltage regulators.

🗂️ Topics Covered

The lecture begins with multiple worked examples calculating voltage and current changes in diodes using the exponential model. It then formally defines the reverse bias region, explaining the concept of saturation current and leakage. Next, it covers the breakdown region, introducing breakdown voltage and its application in voltage regulation. Finally, it details two methods for analyzing a simple diode circuit: graphical analysis using a load line and iterative analysis, with a step-by-step example.

📝 Lecture Summary

Example: Consider a silicon diode with n =1.5. Find the change in voltage if the current changes from 0.1mA to 10mA.

This example demonstrates using the diode equation to find the voltage change corresponding to a known current change.

🔑 Definition — Diode Equation: i = Is * e^(v/nVT) describes the current-voltage relationship of a diode. 📐 Formula: v1 – v2 = nVT * ln(I1/I2) → The change in voltage across a diode is proportional to the natural log of the ratio of the two currents. 📌 Example: For n=1.5, VT=25mV, I1=10mA, I2=0.1mA: v1 – v2 = 1.5 * 25mV * ln(10/0.1) = 172.7mV.

Example: A silicon junction diode with n =1 has v = 0.7 volts at i =1mA . Find the voltage drop at i =0.1mA and i = 10mA.

This example uses the same principle to find the new voltage at different currents, given a known operating point.

📐 Formula: v2 = v1 + nVT * ln(i2/i1) → A practical form to find the new voltage v2, given a reference point (v1, i1). 📌 Example: For n=1, VT=25mV, v1=0.7V at i1=1mA:

• For i2=0.1mA: v2 = 0.7 + 0.025*ln(0.1) = 0.64 volts.
• For i2=10mA: v2 = 0.7 + 0.025*ln(10) = 0.76 volts.

Example: The circuit in the fig. utilize three identical diodes having n=1 and Is = 10^-14 A. Find the value of the current I to obtain an output voltage V0 = 2V, if a current of 1mA is drawn away, what is the change in output voltage.

This example applies the diode equation to a series circuit of three diodes, analyzing the effect of a changing load current.

📌 Example: With three identical diodes in series, V0=2V means each diode drops 2/3 V. The current I is calculated using the diode equation:

• I = Is * e^(v/nVT) = 10^-14 * e^( (2/3) / 0.025 ) = 3.8mA.
• If 1mA is drawn away, the diode current reduces to 2.8mA. The voltage change per diode is ΔV = nVT * ln(2.8/3.8) = -7.63mV.
• The total decrease in V0 is 3 * 7.63mV = 22.9 mV.

Example: A particular diode conducts 1A at a junction voltage of 0.65 volts and 2A at a junction voltage 0.67 volts. What are its values of n and Is, what current will follow if its junction voltage is 0.7volts?

This example shows how to extract the diode's model parameters (n and Is) from two data points on its i-v curve.

📌 Example: Given I1=1A at VD=0.65V and I2=2A at VD=0.67V, with VT=25mV:

• Using the equation form, ln(I1) - ln(I2) = (VD1 - VD2) / (nVT). Solving: -ln2 = -0.02 / (n * 0.025) → n = 1.154.
• Substitute n into the equation for I1: Is = I1 * e^(-VD1 / (nVT)) = 1 * e^(-0.65 / (1.154*0.025)) = 1.64 x 10^-10 A.
• For VD=0.7V: I = 1.64 x 10^-10 * e^(0.7 / (1.154*0.025)) = 5.66 A.

THE REVERSE BIAS REGION:

This section explains diode behavior when a negative voltage is applied.

When the diode voltage (v) is negative, the exponential term in the diode equation becomes negligible compared to 1. Therefore, the diode current becomes approximately i = -Is. This constant, small reverse current is called the saturation current. A significant portion of this reverse current is due to leakage effects, which are proportional to the junction area and have a different temperature dependence than Is.

THE BREAKDOWN REGION:

This section describes the diode's behavior under a large reverse voltage.

The breakdown region is entered when the magnitude of the reverse voltage exceeds a specific threshold called the BREAKDOWN VOLTAGE (VZK). This is the voltage at the "knee" of the i-v curve. In this region, the reverse current increases rapidly, but the voltage change is very small. 💡 Why this matters: This property of a nearly constant voltage despite a large current change is used for voltage regulation.

ANALYSIS OF DIODE CIRCUITS:

This section introduces two methods for finding the DC operating point (ID, VD) of a diode in a simple series circuit with a DC source (VDD) and resistor (R). The circuit is governed by two equations:

1. Diode Equation: ID = Is * e^(VD / nVT)
2. Kirchhoff's Voltage Law (KVL): ID = (VDD – VD) / R (the load line equation)

GRAPHICAL ANALYSIS:

This method solves the circuit by plotting both equations on the same i-v plane.

The exponential diode curve and the straight load line (from KVL) are plotted. The intersection point of the two graphs is called the operating point (Q-point). Its coordinates give the solution for ID and VD.

ITERATIVE ANALYSIS:

This is a numerical method for solving equations (1) and (2) without a graph. A step-by-step example is provided.

📌 Example: For VDD=5V and R=1kΩ, the diode has 1mA at 0.7V and changes by 0.1V per decade change in current (2.3nVT=0.1V).

• Iteration 1: Assume VD=0.7V. Then ID = (5-0.7)/1 = 4.3mA. Then VD2 = 0.7 + 0.1 * log(4.3/1) = 0.763V. Result: ID=4.3mA, VD=0.763V.
• Iteration 2: Use VD=0.763V. Then ID = (5-0.763)/1 = 4.237mA. Then VD2 = 0.763 + 0.1*log(4.237/4.3) = 0.762V. Result: ID=4.23mA, VD=0.762V. Since the values from iteration 2 are very close to iteration 1, the solution is ID = 4.23mA and VD = 0.762V.

⭐ Key Takeaways

The most critical concepts from this lecture are the three diode operating regions: forward, reverse, and breakdown. In reverse bias, the current is essentially constant at the saturation current (Is), while in breakdown, the voltage is nearly constant, enabling voltage regulation. For circuit analysis, the exponential diode equation and the load line equation must be solved simultaneously. This can be done graphically by finding the Q-point intersection, or numerically using iterative analysis, which progressively refines the voltage and current estimates until they converge on a stable solution.

🧠 Quick Revision Questions

1. According to the lecture, what is the approximate current in a diode when it is in the reverse bias region, and why is it called the saturation current?
2. How is the breakdown region of a diode used in practical circuit design?
3. What are the two equations that must be solved to find the DC operating point (ID, VD) of a diode in a series circuit with a resistor and a DC source?
4. In the iterative analysis method, what is the first step, and what starting assumption is typically made?
5. For a diode circuit, what does the term "load line" refer to, and what is the significance of the Q-point (operating point)?

📘 Lecture 33 — (DC or Static Resistance-The constant voltage drop model - AC or Dynamic Resistance)

📖 Overview: This lecture explores the behavior of diodes under different types of voltage conditions. It covers how to calculate DC or static resistance from the characteristic curve, introduces the constant voltage drop model including ideal and practical models for circuit analysis, and defines AC or dynamic resistance which depends on the operating point. Understanding these models is crucial for analyzing and designing basic diode circuits.

🗂️ Topics Covered

The lecture first defines DC or Static Resistance and demonstrates its calculation at different operating points using a diode's characteristic curve. It then introduces the Constant Voltage Drop Model, explaining the Ideal Model (diode as a switch) and the Practical Model (including a 0.7V forward voltage drop for silicon). Finally, it covers the AC or Dynamic Resistance.

📝 Lecture Summary

DC OR STATIC RESISTANCE:

When a diode is subjected to a dc voltage, its operating point on the characteristic curve does not change with time. The DC resistance (rD) of the diode can be found easily using Ohm's Law directly from the voltage and current at that point. It may also be observed on the curve that the resistance in the reverse bias region is quite high.

🔑 Definition — DC or Static Resistance (rD): The resistance of a diode when a DC voltage is applied, found by dividing the DC voltage across the diode (VD) by the DC current through it (ID). Formula: rD = VD / ID

📐 Formula: rD = VD / ID → The static resistance at a specific point on the characteristic curve.

📌 Example: Determine the resistance level for the diode with the given characteristics at three different points.

1. For ID = 1mA: From the curve, VD = 0.5 volts. RD = 0.5V / 1mA = 500 ohms
2. For ID = 20mA: From the curve, VD = 0.8 volts. RD = 0.8V / 20mA = 40 ohms
3. For VD = -10V: From the curve, ID = -1µA. RD = 10V / 1µA = 10 Mohms

This example shows that a diode's resistance is not constant; it decreases significantly as forward current increases and is very high in reverse bias.

THE CONSTANT VOLTAGE DROP MODEL:

This model uses a vertical straight line to approximate the fast-rising part of the exponential curve of a forward-biased diode. It simplifies analysis by assuming a constant voltage drop across the diode when it is conducting.

Ideal Model: In this model, the diode behaves essentially as a switch: on when forward biased, off when reverse biased. When the diode is reverse biased, we replace it with an open switch, and when it is forward biased, we use a closed switch. This model ignores the forward voltage drop.

Practical Model: In practice, there is a voltage drop of about 0.7 V across a forward-biased silicon diode (0.3 V for germanium). The practical model adds a voltage source representing this drop to the ideal model. The resulting model says that a forward-conducting diode exhibits a constant voltage drop (VD) of 0.7 volts. 💡 Why this matters: This model is a very useful approximation for quick hand calculations of circuits.

The Effect of the Forward Voltage Drop on Circuit Analysis: To illustrate, consider a circuit with a 5V source (Vs), a silicon diode, and a 1-ohm resistor (R1) in series. According to KVL: Vs = VF + VR. Assuming a silicon diode, VF = 0.7V. Therefore, VR = Vs - 0.7V = 5V - 0.7V = 4.3V. The current IT = VR / R1 = 4.3V / 1Ω = 4.3A. If the diode were considered ideal (VF=0), the current would be 5V / 1Ω = 5A, showing the impact of the 0.7V drop.

📌 Example: Find the voltage across R1 (VR) in a circuit with a 6V source and a silicon diode in series. Solution: By KVL, Vs = VR + Vf. For a silicon diode, Vf = 0.7V. So, VR = Vs - 0.7V = 6V - 0.7V = 5.3 volts (≈5 Volts).

The key rules for the Constant Voltage Drop Model are:

1. Diode current remains at zero until the applied voltage reaches the knee voltage (Vk).
2. Once the applied voltage reaches Vk, the diode turns on and forward conduction occurs.
3. As long as the diode is conducting, the forward voltage (VF) is assumed to be a constant 0.7 volts, regardless of the value of the forward current (IF).

AC OR DYNAMIC RESISTANCE:

The AC or dynamic resistance (rD) of a diode describes its resistance to small changes (as in an AC signal) in voltage and current. It is defined as the ratio of a small change in voltage to the corresponding small change in current around a specific operating point.

🔑 Definition — AC or Dynamic Resistance (rD): The resistance offered by a diode to a varying (AC) signal, calculated as the change in voltage divided by the change in current at a specific Q-point.

📐 Formula: rD = ΔvD / ΔiD → where Δ (delta) indicates a finite (small) change in the quantity.

The resistance in the vertical rise region of the forward bias curve is small, whereas the AC resistance is much larger at low current levels.

⭐ Key Takeaways

A diode has three distinct types of resistance. DC resistance is found by dividing the DC voltage by the DC current (V/I) at a single point on the curve, and it changes based on the operating point. The Constant Voltage Drop Model simplifies circuit analysis by assuming a forward-biased silicon diode has a fixed 0.7V drop, with the ideal model treating it as a simple switch. AC or dynamic resistance is the slope of the characteristic curve (ΔV/ΔI) and is used for analyzing small signal variations. For the exam, be prepared to calculate static resistance from a graph, solve circuits using the constant voltage drop model, and define dynamic resistance.

🧠 Quick Revision Questions

1. What is the formula for calculating the DC or static resistance of a diode?
2. A silicon diode conducts a current of 15mA when the voltage across it is 0.75V. What is its static resistance?
3. In the constant voltage drop model, what is the assumed forward voltage drop for a silicon diode?
4. How does the practical model of a diode differ from the ideal model when the diode is forward biased?
5. What does the 'Δ' in the formula rD = ΔvD / ΔiD signify?

📘 Lecture 34 — Small Signal Model and its applications

📖 Overview: This lecture introduces the small signal model for diodes, which is essential for analyzing circuits where a small AC signal is superimposed on a DC bias. It explains how to calculate AC resistance, the small signal model derivation, and its practical applications in circuit analysis and design.

🗂️ Topics Covered

The lecture covers AC resistance calculation from diode characteristic curves, the derivation of the small signal model using Taylor series expansion, the concept of small signal conductance and resistance, DC and AC analysis separation in circuits, and several application examples including the straight-line approximation and circuit design with multiple diodes.

📝 Lecture Summary

EXAMPLE: Find the AC resistance for the curve shown at (1) ID = 2mA (2) ID = 25mA

Solution (1): For ID = 2mA, choose a swing of plus minus 2mA. When ID = 4mA, VD = 0.76V; when ID = 0mA, VD = 0.65V.

• ∆ID = 4 - 0 = 4mA
• ∆VD = 0.76 - 0.65 = 0.11V
• rD = ∆VD/∆ID = 0.11/4m = 27.5 ohms

Solution (2): For ID = 25mA, choose a swing of plus minus 5mA.

• ∆I = 30 - 20 = 10mA
• ∆VD = 0.8 - 0.78 = 0.02V
• rD = 0.02/10 = 10 ohms

SMALL SIGNAL MODEL

In some applications, a circuit is supplied with a bias to operate in the forward region, and a small AC signal is superimposed on the DC quantity. When there is no AC, diode voltage is VD so that I = ISe^(VD/nVT). When signal vd(t) is applied, the instantaneous voltage vD(t) = VD + vd(t), and the instantaneous diode current becomes iD(t) = ISe^((VD+vd)/nVT) = IDe^(vd/nVT) — Equation (A).

If the amplitude of the signal is sufficiently smaller than 1, so that Vd/nVT << 1, we expand using the Taylor series: e^x = 1 + x/1! + x²/2! + ... Here x = Vd/nVT, so up to two terms: e^(vd/nVT) = (1 + vd/nVT). Substituting into equation (A): iD(t) = ID(1 + vd/nVT) = ID + (ID/nVT)vd — Equation (B). Let (ID/nVT)vd = id, so iD = ID + id.

🔑 Definition — Small Signal Conductance: The quantity ID/nVT has the dimensions of conductance (mhos) and is called the diode small signal conductance. Its inverse is the small signal resistance or incremental resistance given by rd = nVT/ID. Therefore, rd ∝ 1/ID, where n and VT are constants provided by the manufacturer.

APPLICATION: rd = nVT/ID

Consider a circuit with a DC source VDD and an AC source. For analysis, split the circuit into two parts: DC and AC.

For DC analysis: Replace the AC source with a short and use the constant drop model for the diode. The ideal diode is forward biased (short circuit). From KVL: VDD - IDR - VDo - IDrd = 0. Therefore: VDD = IDR + VDo + IDrd — Equation (A).

For AC analysis: Remove all DC sources and the ideal diode. The circuit becomes: vs = id(R + rd) — Equation (B).

Overall analysis combines (A) and (B): VDD + vs = ID(R + rd) + id(R + rd) + VDo = (R + rd)(ID + id) + VDo — Equation (C). Since ID + id = iD, this separates into: VDD = IDR + VDo (DC) and vs = id(R + rd) (AC).

💡 Why this matters: The AC equation circuit is simply a voltage divider. Hence the diode signal voltage is Vd = Vs × rd/(rd + R).

EXAMPLE: Find the value of the diode small signal resistance rd at bias current of 0.1, 1 and 10mA. Assume n = 1.

Solution: rd = nVT/ID, where VT = 25mV.

• For ID = 0.1mA: rd = 250 ohms
• For ID = 1mA: rd = 25 ohms
• For ID = 10mA: rd = 2.5 ohms

EXAMPLE: For a diode that conducts 1mA at a forward voltage drop of 0.7V and n = 1, find the equation of the straight line tangent at ID = 1mA.

Solution: Slope = 1/rd = ID/nVT = 1mA/(1×25mV) = 1/25 ohms. On the VD axis, the intercept = 0.7 - 1m(25) = 0.7 - 0.025 = 0.675V. Using the equation of a straight line y - y₁ = m(x - x₁): iD = (1/25)(VD - 0.675). Therefore, VD - 0.675 - iD(25) = 0, which is the required equation.

EXAMPLE: Consider a diode with n = 2 biased at 1mA. Find the change in current for voltage changes of -20mV, -10mV, -5mV, +5mV, +10mV, +20mV using (1) small signal model and (2) exponential model.

Solution: For small signal model: ∆v = rd∆i, so ∆i = ∆v/rd. rd = nVT/ID = 2×25m/1m = 50 ohms. Thus ∆i = ∆v/50. For exponential model: i = ISe^(v/nVT). Since ID + ∆i = IDe^(∆v/nVT), we get ∆i = ID(e^(∆v/nVT) - 1). With ID = 1mA: ∆i = e^(∆v/nVT) - 1.

• (a) ∆v = -20mV: (1) ∆i = -20/50 = -0.4mA; (2) ∆i = e^(-20/50) - 1 = -0.33mA
• (d) ∆v = +5mV: (1) ∆i = 5/50 = 0.10mA; (2) ∆i = e^(5/50) - 1 = 0.11mA
• (f) ∆v = +20mV: (1) ∆i = 20/50 = 0.40mA; (2) ∆i = e^(20/50) - 1 = 0.49mA

EXAMPLE: Design the circuit so that V₀ = 3V when I_L = 0, and V₀ changes by 40mV per 1mA of load current. Find R. Assume four identical diodes with 0.7V drop at 1mA current and n = 1.

Solution: When V₀ = 3V and I_L = 0, each diode exhibits a drop of 0.75V. When I_L = 1mA, V₀ changes by 40mV, so change per diode is 10mV. Thus rd = 10mV/1mA = 10 ohms. Using rd = nVT/ID: 10 = 1×25m/ID, so ID = 2.5mA. From 15 - 3 - IDR = 0: R = (15 - 3)/2.5m = 4.8k Ohms.

⭐ Key Takeaways

The small signal resistance of a diode is given by rd = nVT/ID and is inversely proportional to the bias current, meaning higher bias currents produce lower AC resistance. The small signal model allows superposition analysis by separating DC and AC components, where the AC equivalent circuit is simply a voltage divider with the diode appearing as a resistor rd. For accurate small-signal analysis, the AC signal amplitude must be much smaller than nVT (typically less than 5mV for n=1). The exponential model gives more accurate results than the small signal model for larger voltage swings, but the small signal model provides a good linear approximation for small perturbations.

🧠 Quick Revision Questions

1. What is the formula for AC resistance (rD) of a diode, and how is it calculated from the characteristic curve?
2. Derive the small signal model equation iD = ID + id starting from the exponential diode equation.
3. For a diode with n=1 biased at 2mA, what is the small signal resistance rd?
4. In the overall analysis combining DC and AC, write the KVL equation for each component separately.
5. Why does the small signal model become inaccurate for large AC signal amplitudes?

📘 Lecture 35 — Transformers - Secondary voltage and current - Transformer Input/Output phase relationship

📖 Overview: This lecture introduces the transformer, a fundamental non-semiconductor device essential in power supplies. It covers the basic structure, types (step-up, step-down, isolation), and key calculations for secondary voltage and current using the turn ratio. The lecture also explains the input/output phase relationship in transformers and concludes with an introduction to half-wave rectifiers.

🗂️ Topics Covered

The lecture begins by defining the transformer's structure with primary and secondary windings and its schematic symbol. It then details three transformer modes: step-up, step-down, and isolation. The key concept of turn ratio is introduced, along with formulas for calculating secondary voltage and secondary current based on this ratio. The transformer's input/output phase relationship is explained using dot notation. Finally, the lecture covers half-wave rectifiers, including both positive and negative configurations, which use diodes and resistors to eliminate parts of an AC signal.

📝 Lecture Summary

TRANSFORMERS

Transformers are not semiconductor devices but are integral to most power supplies. The basic schematic symbol shows two coil-like windings. The transformer consists of two inductors in close proximity but not physically connected. The two windings are called the primary (input) and secondary (output). An alternating voltage applied to the primary induces an alternating voltage in the secondary. The primary and secondary are physically isolated. A transformer can be designed in three modes:

1. STEP UP TRANSFORMER: Provides a secondary voltage greater than the primary voltage (e.g., 240 Vac output from 120 Vac input).
2. STEP DOWN TRANSFORMER: Provides a secondary voltage less than the primary voltage (e.g., 30 Vac output from 120 Vac input).
3. ISOLATION TRANSFORMER: Provides an output voltage equal to the input voltage, used to electrically isolate power supplies from power lines.

TURN RATIO

The turn ratio of a transformer is the ratio of the number of turns in the primary to the number of turns in the secondary. For example, a 4:1 ratio means four turns in the primary for every one turn in the secondary. The turn ratio is equal to the voltage ratio of the two windings.

🔑 Turn Ratio Formula: The relationship is defined as: ( \frac{V_2}{V_1} = \frac{N_2}{N_1} ) Where:

• ( N_2 ) = number of turns in the secondary winding
• ( N_1 ) = number of turns in the primary winding
• ( V_2 ) = voltage in the secondary
• ( V_1 ) = voltage applied to the primary

📐 Formula for Calculating Secondary Voltage: ( V_2 = \left( \frac{N_2}{N_1} \right) V_1 )

📌 Example: For a step-down transformer with a 4:1 turn ratio and a ( V_1 ) of 120 Vac: ( V_2 = \left( \frac{1}{4} \right) \times 120Vac = 30Vac )

CALCULATING SECONDARY CURRENT

Ideally, transformers are 100% efficient, meaning input power equals output power (( P_2 = P_1 )). Since power is the product of voltage and current, we can write: ( V_2 I_2 = V_1 I_1 ) This leads to: ( \frac{I_1}{I_2} = \frac{V_2}{V_1} ) The current ratio is the inverse of the voltage ratio. This implies:

1. For a step-down transformer, ( I_2 > I_1 ) (current increases).
2. For a step-up transformer, ( I_2 < I_1 ) (current decreases).

Since the voltage ratio equals the turn ratio, we can also write: ( \frac{I_1}{I_2} = \frac{N_2}{N_1} ) Therefore: ( I_2 = \left( \frac{N_1}{N_2} \right) I_1 )

📌 Example: A fuse limits the primary current ( ( I_1 ) ) to 1A. What is the limit on the secondary current for a transformer with a turn ratio of 4:1? ( I_2 = \left( \frac{N_1}{N_2} \right) I_1 = \left( \frac{1}{4} \right) \times 1A = 250mA ) If the secondary current tries to exceed 250mA, the primary current will exceed 1A and blow the fuse.

TRANSFORMER INPUT/OUTPUT PHASE RELATIONSHIP

The phase relationship between the primary and secondary voltages is indicated by dots on the schematic symbol.

• If the dots are on the same side (e.g., top of primary and top of secondary), the output voltage is in phase with the input voltage.
• If the dots are on opposite sides (e.g., top of primary and bottom of secondary), the output voltage is 180° out of phase with the input voltage.

HALF WAVE RECTIFIERS

A half-wave rectifier is made up of a diode and a resistor and is used to eliminate either the positive or negative part of the input AC signal.

• Positive Half Wave Rectifier: The diode conducts during the positive half-cycle of the input, producing an output voltage across the load resistor. It is reverse-biased during the negative half-cycle.
• Negative Half Wave Rectifiers: The diode direction is reversed. The diode conducts on the negative half-cycle of the input, and the load voltage equals the secondary voltage. The positive half-cycle of the input is eliminated. The operating principle is the same as the positive half-wave rectifier, with the only difference being the polarity of the output will be reversed.

⭐ Key Takeaways

The transformer is a non-semiconductor device consisting of primary and secondary windings that are physically isolated. Its core function is to transfer energy through electromagnetic induction. The turn ratio ( ( N_2/N_1 ) ) is the fundamental parameter, dictating that the voltage ratio is directly proportional to it ( ( V_2/V_1 = N_2/N_1 ) ), while the current ratio is inversely proportional ( ( I_1/I_2 = N_2/N_1 ) ). The types—step-up, step-down, and isolation—determine whether output voltage is higher, lower, or equal to the input. The dot notation on the schematic indicates whether the secondary voltage is in phase or 180° out of phase with the primary. The half-wave rectifier, a simple circuit using a single diode and resistor, serves to block one half of an AC waveform, resulting in a pulsating DC output.

🧠 Quick Revision Questions

1. What is the fundamental difference between a step-up and a step-down transformer?
2. Write the formula that relates the primary and secondary voltages to the number of turns in the windings.
3. For an ideal transformer with a turn ratio of 10:1 (primary:secondary) and a primary voltage of 120V, what is the secondary voltage? What type of transformer is this?
4. If the secondary current in a step-down transformer is 2A, and the turn ratio is 1:5 (primary:secondary), what is the primary current?
5. Explain what the two dots on a transformer's schematic symbol indicate.

📘 Lecture 36 — Half Wave Rectifiers

📖 Overview: This lecture covers negative half wave rectifiers and provides general rules for determining output polarity based on diode orientation. It then focuses on the critical calculations for load voltage, load current, and average (DC) values in half wave rectifier circuits, including the use of transformer turns ratios and RMS-to-peak conversions.

🗂️ Topics Covered

The lecture begins with negative half wave rectifiers and general polarity rules based on diode direction. It then moves into calculating peak load voltage using transformer turns ratios and RMS-to-peak conversions, demonstrated through multiple examples. Next, it covers calculating peak load current, followed by determining average load voltage and current using the 1/π (0.318) conversion factor.

📝 Lecture Summary

Negative Half Wave Rectifiers

When the diode direction is reversed in a half wave rectifier, the diode conducts on the negative half cycle of the input. During this time, ( V_L = V_2 ). The diode is reverse biased during the positive half cycle, and ( V_D = V_2 ), eliminating the positive half cycle. The operating principle is the same as positive half wave rectifiers; the only difference is the output polarity is reversed.

General Rules for Polarity

Two general rules determine the output polarity of a half wave rectifier:

🔑 Rule 1: When the diode points towards the load (R_L), the output from the rectifier will be positive.

🔑 Rule 2: When the diode points towards the transformer, the output from the rectifier will be negative.

Calculating Load Voltage and Current Values

The load voltage in a half wave rectifier is found as:

📐 Formula: ( V_{L(pk)} = V_{2(pk)} - V_F ) → The peak load voltage equals the peak secondary voltage minus the diode forward voltage drop (typically 0.7V for silicon).

The peak secondary voltage is determined by the transformer turns ratio:

📐 Formula: ( V_{2(pk)} = \frac{N_2}{N_1} \times V_{1(pk)} ) → The peak secondary voltage equals the turns ratio multiplied by the peak primary voltage.

When source voltages are given as RMS values, they are converted to peak values:

📐 Formula: ( V_{pk} = \frac{V_{rms}}{0.707} ) → Peak voltage equals RMS voltage divided by 0.707.

📌 Example 1: For a circuit with 120V RMS primary, turns ratio 1/5, and silicon diode:

• ( V_{1(pk)} = 120 / 0.707 = 169.7V_{pk} )
• ( V_{2(pk)} = (1/5)(169.7) = 33.94V_{pk} )
• ( V_{L(pk)} = 33.94 - 0.7 = 33.24V_{pk} )

📌 Example 2: For a circuit with 25V AC secondary rating:

• ( V_{2(pk)} = 25 / 0.707 = 35.36V_{pk} )
• ( V_{L(pk)} = 35.36 - 0.7 = 34.66V_{pk} )

Calculating Load Current

Once the peak voltage is determined, the peak load current is found using Ohm's law:

📐 Formula: ( I_{L(pk)} = \frac{V_{L(pk)}}{R_L} ) → Peak load current equals peak load voltage divided by load resistance.

📌 Example: For a circuit with 200V RMS primary, turns ratio 1/5, load resistance 10kΩ:

• ( V_{1(pk)} = 200 / 0.707 = 282.9V_{pk} )
• ( V_{2(pk)} = (1/5)(282.9) = 56.6V_{pk} )
• ( V_{L(pk)} = 56.6 - 0.7 = 55.9V_{pk} )
• ( I_{L(pk)} = 55.9V / 10kΩ = 5.59mA_{pk} )

Average Load Voltage and Current

Since rectifiers are used to convert AC to DC, the average (DC) voltage ( V_{ave} ) is a very important value. For a half wave rectifier:

📐 Formula: ( V_{ave} = \frac{V_{pk}}{\pi} ) or ( V_{ave} = 0.318(V_{pk}) ) → The average DC voltage equals the peak voltage divided by π (approximately 0.318 times peak voltage).

💡 Why this matters: This is the DC equivalent value that would be measured by a DC voltmeter at the output of a half wave rectifier.

📌 Example: For a circuit with 75V RMS primary and turns ratio 1/2:

• ( V_{1(pk)} = 75 / 0.707 = 106.1V_{pk} )
• ( V_{2(pk)} = (1/2)(106.1) = 53.04V_{pk} )
• ( V_{L(pk)} = 53.04 - 0.7 = 52.34V_{pk} )
• ( V_{ave} = 52.34 / \pi = 16.66V_{dc} )

Calculating Average Current

The average (DC) load current ( I_{ave} ) can be calculated in two ways:

Method 1: Determine ( V_{ave} ) first, then use Ohm's law: 📐 ( I_{ave} = \frac{V_{ave}}{R_L} )

Method 2: Convert from peak current using the same conversion: 📐 ( I_{ave} = \frac{I_{pk}}{\pi} ) or ( I_{ave} = 0.318(I_{pk}) )

📌 Example: For a circuit with 24V AC secondary rating and 20kΩ load:

• ( V_{2(pk)} = 24 / 0.707 = 33.9V_{pk} )
• ( V_{L(pk)} = 33.9 - 0.7 = 33.2V_{pk} )
• ( I_{L(pk)} = 33.2 / 20kΩ = 1.66mA_{pk} )
• ( I_{ave} = 1.66mA / \pi = 529.13 \mu A )

⭐ Key Takeaways

The most critical concepts from this lecture are: diode direction determines output polarity—pointing toward the load gives positive output, pointing toward the transformer gives negative output. The peak load voltage subtracts the diode forward drop (0.7V for silicon) from the peak secondary voltage. Always convert RMS values to peak by dividing by 0.707 before using them in calculations. The average (DC) value for a half wave rectifier is the peak value divided by π (or multiplied by 0.318), which represents the DC equivalent of the rectified waveform. This average value can be calculated for both voltage and current using the same conversion factor.

🧠 Quick Revision Questions

1. What is the general rule for determining output polarity based on diode orientation in a half wave rectifier?
2. How do you calculate the peak secondary voltage given the RMS primary voltage and transformer turns ratio?
3. What is the formula for average (DC) voltage in a half wave rectifier, and what is the numerical constant used?
4. In the example with 200V RMS primary, 1/5 turns ratio, and 10kΩ load, what are the peak load voltage and peak load current?
5. What are the two methods for calculating average (DC) load current in a half wave rectifier?

📘 Lecture 37 — (Peak Inverse Voltage (PIV) - Full wave rectifier)

📖 Overview: This lecture covers the analysis of negative half wave rectifiers and introduces the concept of Peak Inverse Voltage (PIV). It then transitions to the full wave rectifier, explaining its circuit operation, the role of the center-tapped transformer, and how to calculate load voltage and current values. Understanding PIV is critical for diode selection, and full wave rectification is essential for more efficient DC power conversion.

🗂️ Topics Covered

The lecture begins with a method for analyzing negative half wave rectifiers by adapting positive half wave analysis. It then defines Peak Inverse Voltage (PIV) for half wave rectifiers. The main focus is on the full wave rectifier, including its circuit components, the function of the center-tapped transformer, basic circuit operation during both half cycles, and formulas for calculating peak load voltage and average DC load voltage. Worked examples demonstrate these calculations.

📝 Lecture Summary

Negative Half Wave Rectifiers:

The analysis of a negative half wave rectifier is nearly identical to that of a positive half wave rectifier. The only difference is that the voltage polarities are reversed. A simple method for performing the mathematical analysis is:

1. Analyze the circuit as if it was a positive half wave rectifier.
2. After completing calculations, change all voltage polarity signs from positive to negative.

EXAMPLE: Determine the dc output voltage for the circuit shown in the fig. SOLUTION: We shall start by solving the circuit as if it was a positive half wave rectifier. First, V2(pk) = 48Vrms / 0.707 = 67.9 Vpk And, VL(PK) = V2(pk) – 0.7 = 67.2 Vpk Finally, Vave = Vpk / π = 21.39 Vdc Now we will simply change all positive voltage values to negative voltage values. V2(pk) = -67.9 Vpk VL(pk) = -67.2 Vpk Vave = -21.39 Vdc

Peak Inverse Voltage (PIV):

The maximum amount of reverse bias that a diode will be exposed to in a rectifier is called the peak inverse voltage or PIV of the rectifier. For the half wave rectifier, the value of PIV is found as: 📐 Formula: PIV = V2(pk) → The peak inverse voltage across the diode is equal to the peak secondary voltage of the transformer. The basis for this equation is that when the diode is reversed biased, there is no voltage drop across the load. Therefore, all of V2 is dropped across the diode in the rectifier.

Full Wave Rectifier:

The full wave rectifier consists of two diodes and a resistor, as shown in the fig (a). The result of this change in circuit is illustrated in fig (b), where the output from the full wave rectifier is compared with that of a half wave rectifier. Note that the full wave rectifier has two positive half cycles out for every one produced by the half wave rectifier. The transformer shown in the fig. is a center tapped transformer. This type of transformer has a load connected to the center of the secondary winding. The voltage from the center tap to each of the outer windings is equal to one half of the secondary voltage. For example, a 24V center tapped transformer has 12V from the center tap to each outer terminal. Center tapping plays a major role in the operation of the full wave rectifier. For this reason, the full wave rectifier can not be line operated, meaning it cannot be connected directly to the AC input like the half wave rectifier can.

Basic Circuit Operation

Figure shows the operation of the full wave rectifier during one complete cycle of the input signal. During the positive half cycle of the input, D1 is forward biased. Using ideal operating characteristics of the diode, VL can be found. When the polarity of the input reverses, D2 is forward biased, and D1 is reverse biased. The direction of the current through the load will not change even though the polarity of the transformer secondary will. Thus, another positive half cycle is produced across the load.

Calculating Load Voltage and Current Values:

Using the practical diode model, the peak load voltage for a full wave rectifier is found as: 📐 Formula: VL(pk) = V2(pk)/2 – 0.7 → The peak load voltage is half the peak secondary voltage minus the diode voltage drop. The full wave rectifier will produce twice as many output pulses as the half wave rectifier. For this reason, the average load voltage for the full wave rectifier is found as: 📐 Formula: Vave = 2VL(pk) / π → The average DC voltage is two times the peak load voltage divided by pi. Or, Vave = 0.636 VL(pk)

EXAMPLE: Determine the dc load voltage for the circuit shown in the fig. SOLUTION: Primary voltage is in rms, so we can calculate the peak voltage: V1(pk) = Vrms / 0.707 = 75 / 0.707 = 106.08 Vpk. Primary voltage and turn ratio is known so we can determine the secondary voltage: V2(pk) = (N2/N1) V1pk = 1/4 (106.08) = 26.52 Vpk. Now load voltage can be calculated as: VL(pk) = V2(pk)/2 – 0.7 = 26.52/2 - 0.7 = 12.56 Vpk. Now dc value of the voltage can be calculated as: Vave = 2VL(pk) / π = 0.636 (12.56) = 7.98 Vdc.

EXAMPLE: Determine the values of VL(pk) and Vave for the circuit shown in the fig. SOLUTION: The transformer is rated at 30 Vac. Therefore, the value of V2(pk) is found as: V2(pk) = 30 Vac / 0.707 = 42.2 Vpk. The peak load voltage can now be found as: VL(pk) = V2(pk) / 2 – 0.7 = 21.2 – 0.7 = 20.5 Vpk. Finally, the dc load voltage is found as: Vave = 2VL(pk) / π = 41 Vpk / π = 13.05 Vdc.

⭐ Key Takeaways

The critical takeaways for the exam are: first, you must know the formula and concept of Peak Inverse Voltage (PIV = V2(pk)) for a half-wave rectifier. Second, understand that a center-tapped transformer splits the secondary voltage in half for a full-wave rectifier, so VL(pk) = V2(pk)/2 – 0.7V. Third, the average DC output voltage for a full-wave rectifier is twice that of a half-wave rectifier, calculated as Vave = 2VL(pk)/π or 0.636VL(pk). Fourth, master the step-by-step calculation process of converting RMS primary voltage to peak primary, then using the turns ratio to find peak secondary, and finally using the correct formulas to find peak load and average DC voltages. Finally, remember that in a full-wave rectifier, current always flows in the same direction through the load, providing two output pulses per input cycle.

🧠 Quick Revision Questions

1. What is the formula for the Peak Inverse Voltage (PIV) of a half-wave rectifier?
2. What is the critical difference in the transformer used in a full-wave rectifier compared to a half-wave rectifier, and why is it important?
3. What is the formula for calculating the peak load voltage (VL(pk)) in a full-wave rectifier using the practical diode model?
4. How does the average DC output voltage (Vave) of a full-wave rectifier compare to that of a half-wave rectifier, assuming the same peak voltage at the secondary?
5. A full-wave rectifier uses a center-tapped transformer with a 50Vrms secondary (from end to end). Calculate the peak load voltage and the average DC output voltage.

📘 Lecture 38 — (Negative full wave rectifier - Full wave bridge rectifier)

📖 Overview: This lecture extends the analysis of full-wave rectifiers to include current calculations, negative full-wave rectifiers, and the full-wave bridge rectifier. It explains how to compute peak and average load currents, introduces the concept of peak inverse voltage (PIV) for center-tapped rectifiers, and details the operation and advantages of the bridge rectifier circuit.

🗂️ Topics Covered

The lecture begins with calculating peak and average load currents using Ohm's law through worked examples. It then introduces the negative full-wave rectifier, explaining how reversing diode polarity changes output polarity. Peak Inverse Voltage (PIV) is defined for full-wave rectifiers, including practical considerations for diode drops. Finally, the full-wave bridge rectifier is presented, covering its circuit operation, advantages over the center-tapped design, and calculation of load voltage/current including the effect of two diode drops.

📝 Lecture Summary

CALCULATING CURRENTS:

Once peak and average load voltages are known, Ohm's law is used to find corresponding currents. The formulas are I_L(pk) = V_L(pk) / R_L and I_ave = V_ave / R_L.

🔑 Definition — I_L(pk): Peak load current, the maximum current flowing through the load during one cycle. 🔑 Definition — I_ave: Average (DC) load current, the direct current equivalent delivered to the load.

📐 Formula: I_L(pk) = V_L(pk) / R_L → The peak current equals peak voltage divided by load resistance. 📐 Formula: I_ave = V_ave / R_L → The average current equals average voltage divided by load resistance.

📌 Example: For the circuit from last lecture with V_L(pk) = 20.5V, V_ave = 13.05V, and R_L = 5.1kΩ: I_L(pk) = 20.5V / 5.1kΩ = 4.02 mA I_ave = 13.05V / 5.1kΩ = 2.56 mA

📌 Example: For a circuit with a 24V RMS transformer (V_2(pk) = 24V/0.707 = 33.94V_pk), a center-tap yields V_L(pk) = V_2(pk)/2 – 0.7V = 16.27V_pk. V_ave = 0.636 × V_L(pk) = 10.34V_dc. With R_L = 8kΩ: I_L(pk) = 16.27V / 8kΩ = 2.03 mA; I_ave = 10.34V / 8kΩ = 1.29 mA.

💡 Why this matters: These current values are essential for specifying components like diodes and load resistors, ensuring they can handle the peak and average power.

NEGATIVE FULL WAVE RECTIFIER:

The negative full wave rectifier produces a negative output voltage by reversing the direction of all diodes in a positive full-wave rectifier. The analysis is identical to the negative half-wave rectifier, except both halves of the input cycle are used.

🔑 Definition — Negative full wave rectifier: A full-wave rectifier with reversed diode polarity, producing a negative DC output voltage with respect to ground.

📌 Example: Reversing all diodes in a center-tapped full-wave rectifier circuit (as shown in the lecture fig.) results in the same waveform but negative polarity, with the load voltage negative relative to ground.

PEAK INVERSE VOLTAGE:

For a full-wave center-tapped rectifier, when one diode is reverse biased, the voltage across it is approximately V_2(pk). Since the peak load voltage is V_2(pk)/2, the PIV equals twice the peak load voltage. With diode drops, a more accurate formula subtracts 0.7V.

🔑 Definition — PIV (Peak Inverse Voltage): The maximum reverse voltage a diode must withstand without breaking down.

📐 Formula: PIV = 2 × V_L(pk) (ideal) → The inverse voltage is twice the peak load voltage because the full secondary voltage (minus the load) appears across the off diode. 📐 Formula: PIV = V_2(pk) – 0.7 (practical) → Accounts for the forward drop of the conducting diode in series.

💡 Why this matters: Proper PIV rating prevents diode failure. In center-tapped rectifiers, the PIV is high; bridge rectifiers have lower PIV requirements.

FULL WAVE BRIDGE RECTIFIER:

The full wave bridge rectifier uses four diodes and does not require a center-tapped transformer. Its advantages:

1. Can be coupled directly to the AC power line.
2. Peak output voltage is nearly double that of a center-tapped rectifier using the same secondary voltage, yielding higher DC voltage.

🔑 Definition — Bridge rectifier: A full-wave rectifier using four diodes in a bridge configuration, with two diodes conducting during each half-cycle.

BASIC CIRCUIT OPERATION:

Conduction alternates between two diode pairs: D1 and D3 conduct during the positive half-cycle, while D2 and D4 conduct during the negative half-cycle. Current through the load always flows in the same direction.

📌 Example: During the positive half-cycle of V_2, D1 (anode positive) and D3 (cathode negative) are forward-biased, allowing current through R_L. During the negative half-cycle, D2 and D4 conduct, again producing current in the same direction through R_L.

CALCULATING THE LOAD VOLTAGE AND CURRENT VALUES:

Without a center-tap, the ideal peak load voltage is V_L(pk) = V_2(pk). Considering two diode drops (0.7V each), the practical peak load voltage is V_L(pk) = V_2(pk) – 1.4V. Average voltage is then V_ave = 2V_L(pk)/π.

📐 Formula (ideal): V_L(pk) = V_2(pk) → No voltage division occurs, giving higher output. 📐 Formula (practical): V_L(pk) = V_2(pk) – 1.4 → Accounts for two series diodes conducting.

📌 Example: For a bridge rectifier with a 12V RMS transformer (V_2(pk) = 16.97V_pk) and R_L = 12kΩ: V_L(pk) = 16.97V – 1.4V = 15.57V_pk V_ave = (2 × 15.57V) / π = 9.91V_dc I_ave = 9.91V / 12kΩ = 825.8 μA

⭐ Key Takeaways

The bridge rectifier is the preferred full-wave design because it doubles the output voltage compared to the center-tapped type without requiring a center-tapped transformer, making it more efficient and cost-effective. Currents are found directly from voltages using Ohm's law. PIV for a center-tapped full-wave rectifier is twice the peak load voltage (or V_2(pk)), and must account for the 0.7V drop in practical circuits. The bridge rectifier’s PIV per diode is lower than in the center-tapped design, an additional advantage. Always include diode voltage drops (1.4V for bridge, 0.7V for center-tapped when applicable) for accurate results.

🧠 Quick Revision Questions

1. How is I_ave calculated for a full-wave rectifier given V_ave and R_L?
2. What is the key structural difference between a negative and positive full-wave rectifier?
3. What is the PIV formula for a practical center-tapped full-wave rectifier?
4. Why does the bridge rectifier produce a higher peak output voltage than the center-tapped rectifier with the same transformer?
5. In a bridge rectifier, which diode pairs conduct during the positive half-cycle?

📘 Lecture 39 — (Filters - Clippers - Series clippers)

📖 Overview: This lecture discusses the Peak Inverse Voltage (PIV) in bridge rectifiers and introduces filtering techniques used to smooth rectifier output. It then focuses on diode clippers, particularly series clippers, explaining how they clip portions of an input signal and how adding a DC battery modifies the clipping threshold.

🗂️ Topics Covered

This lecture covers the analysis of Peak Inverse Voltage in bridge rectifiers using the ideal diode model. It introduces filters, primarily the capacitor filter, and explains its charge/discharge action for reducing output voltage variations. The lecture then defines clippers, outlines important points for their analysis, and categorizes them into series and parallel types. A detailed explanation of series clippers is provided, including the effect of adding a DC battery, with supporting examples and calculations for output waveforms.

📝 Lecture Summary

PEAK INVERSE VOLTAGE:

Using the ideal diode model, the Peak Inverse Voltage (PIV) of each diode in the bridge rectifier is equal to V₂. This is the same voltage applied to the diode in the full-wave center-tapped rectifier. To illustrate, in a given figure, two things are done: (1) the conducting diodes (D₁ and D₃) are replaced by straight wires, and (2) the positive side of the secondary is labeled A and the negative side B. Connecting the common A points and B points gives an equivalent circuit. With this circuit, you can see that the two reverse-biased diodes and the secondary of the transformer are all in parallel. Since parallel voltages are equal, the PIV across each diode is equal to V₂. The same situation will exist for D₁ and D₃ when they are reverse-biased.

FILTERS:

Filters are used in power supplies to reduce the variations in the rectifier output signal. The goal is to produce a constant DC output voltage, so it is necessary to remove as much of the rectifier output variations as possible.

The capacitor filter is the most basic and commonly used filter type. This filter is simply a capacitor connected in parallel with the load resistance. The filtering action is based on the charge/discharge action of the capacitor. During the positive half-cycle of the input, D₁ will conduct and the capacitor will charge rapidly. As the input starts to go negative, D₁ will turn off, and the capacitor will slowly discharge through the load resistance. As the input from the rectifier drops below the charged voltage of the capacitor, the capacitor acts as the voltage source for the load. It is the difference between the charge and discharge times of the capacitor that reduces the variation in the rectifier output voltage. This difference is caused by two distinct RC time constants in the circuit.

💡 Why this matters: The capacitor filter is essential for converting a pulsating DC from a rectifier into a smoother, more constant DC voltage suitable for powering electronic circuits.

CLIPPERS:

There are a variety of diode networks called clippers that have the ability to clip off a portion of the input signal without distorting the remaining part of the alternating waveform. The half-wave rectifier is the simplest form of diode clipper. Depending on the orientation of the diode, the positive or negative region of the input signal is "clipped" off.

IMPORTANT POINTS FOR CLIPPERS: (1) Make a sketch in your mind about the response of the network. (2) Determine the applied voltage (Transition Voltage) that causes a change in the diode bias. (3) Be continuously aware of the defined terminal and polarity of V₀. (4) Sketch the input signal on the top and the output at the bottom to determine the output at instantaneous points of the input.

TYPES OF CLIPPERS: There are two general categories of clippers: (1) Series and (2) Parallel.

SERIES CLIPPERS: The series configuration is defined as one where the diode is in series with the load, as in a half-wave rectifier.

ADDITION OF A BATTERY IN THE SERIES CLIPPER CIRCUIT: In the figure, the direction of the diode suggests that the signal Vᵢ must be positive to turn it on. The DC supply further requires that the voltage Vᵢ be greater than V volts to turn the diode on. The negative region of the input signal is pressuring the diode into the off region.

To determine the applied voltage that will cause a change in state for the diode, the ideal diode transition occurs at the point on the characteristic where Vd = 0 and Id = 0. In this case, the transition will occur at Vᵢ = V.

When the diode is short-circuited, the output voltage V₀ can be calculated by applying KVL: Vᵢ – V – V₀ = 0, so V₀ = Vᵢ – V. In this case, V₀ = Vₘ – V.

🔑 Definition — Peak Inverse Voltage (PIV): The maximum reverse voltage that a diode must withstand without breaking down. 📐 Formula for PIV (bridge rectifier): PIV = V₂ (the secondary voltage) → This means each diode in a bridge rectifier must block a reverse voltage equal to the full secondary voltage. 📌 Example: In a bridge rectifier circuit, when D₁ and D₃ are conducting, D₂ and D₄ are reverse-biased. The equivalent circuit shows D₂ and D₄ in parallel with the transformer secondary. Since parallel components have the same voltage, the PIV across each of these diodes is equal to the secondary voltage, V₂.

EXAMPLE: Determine the output waveform for the network shown. SOLUTION: The equivalent circuit is drawn. Applying KVL: V₀ = Vᵢ + 5V. The transition voltage = 5V. When the diode is off, V₀ = VR = iDR = (0)R = 0V (no diode current is flowing).

EXAMPLE: Determine the output for the square wave input shown in the figure. SOLUTION: In the positive half cycle, the diode is in the short circuit condition and by KVL: V₀ = 20 + 5 = 25V. For the negative half cycle, when Vᵢ = -10V, the result is placing the diode in the reverse condition.

⭐ Key Takeaways

For the exam, you must be able to state that the PIV for each diode in a bridge rectifier equals V₂. You need to understand that capacitor filters reduce output voltage variations through their charge and discharge cycles. It is critical to know that clippers are diode networks that clip off a portion of the input signal and are categorized as series or parallel. For series clippers with a DC battery, remember that the transition voltage equals the battery voltage, and the output voltage during conduction is V₀ = Vᵢ – V (or V₀ = Vᵢ + V, depending on battery polarity). Finally, you must be proficient in applying KVL to determine the output waveform for given input signals.

🧠 Quick Revision Questions

1. What is the Peak Inverse Voltage (PIV) across each diode in a bridge rectifier, assuming an ideal diode model?
2. Briefly explain how a capacitor filter smooths the output of a rectifier.
3. What are the two general categories of clippers?
4. In a series clipper circuit with an added DC battery, what defines the transition voltage for the diode?
5. For the example with a square wave input of +20V and -10V, and a 5V battery in series with the diode, what is the output voltage for the negative half-cycle?

📘 Lecture 40 — (Parallel Clippers - Clamper)

📖 Overview: This lecture covers parallel clipper circuits and clamper circuits. It explains how to determine the output voltage for these circuits, including the effects of using ideal versus silicon diodes. The analysis methods for clamping circuits, including the crucial role of the capacitor's charging and discharging, are also thoroughly detailed.

🗂️ Topics Covered

The lecture begins with the definition and analysis of parallel clippers, including an example determining the output voltage for an ideal diode and then a silicon diode. The second half introduces clampers, listing key points for analysis and providing examples of a biased clamper circuit with both ideal and silicon diodes, including the calculation of output waveforms.

📝 Lecture Summary

PRALLEL CLIPPER

The parallel clipper configuration has the diode in a branch parallel to the load.

🔑 Definition — Parallel Clipper: A clipping circuit where the diode is placed in a branch that is parallel to the load.

📌 Example: Determine the output voltage for the network shown in the fig. An input voltage Vi swings from +16V to -20V. The circuit has a 4V battery, a resistor R, and a diode in parallel with the load. When the diode is OFF (positive half cycle), the output voltage (Vo) is taken across the load. When the diode is ON (negative half cycle), it clamps the circuit. Solution:

1. To find the transition voltage, set Id = 0 and Vd = 0.
2. Applying KVL: Vi - Id*R - 4V = 0 => Vi - 0 - 4V = 0 => Vi = 4V is the transition voltage.
3. This means the diode is ON for Vi < 4V and OFF for Vi > 4V.
4. For the positive half cycle (Vi > 4V, diode OFF), the voltage across the load (VL) equals the output voltage (Vo).
5. V0 = VL = 16 Volts.

📌 Example: Repeat the previous example, but assuming the diode is a silicon diode (VT = 0.7V), and draw its output. Solution:

1. The diode will be ON for the negative half cycle.
2. To find the transition voltage, apply Id = 0 and Vd = 0.7V.
3. The transition voltage (VT) is the input voltage at which the diode just starts to conduct. KVL: Vi + Vd - 4V = 0 => Vi = 4V - Vd = 4V - 0.7V = 3.3V.
4. This means the voltage across the diode must be 0.7V for it to conduct. The circuit will reach this state when the input Vi is 3.3V.
5. For the positive half cycle (when the diode is OFF), the output voltage is still the input voltage.
6. V0 = VL = 16V.

💡 Why this matters: The 0.7V drop of the silicon diode shifts the transition point of the clipper. In the previous example with an ideal diode, clipping started at 4V. With a silicon diode, clipping starts at 3.3V.

CLAMPERS

Clampers are circuits that clamp the input signal to a different DC level. The total swing of the output is equal to the total swing of the input signal.

🔑 Definition — Clamper: A circuit that adds a DC level to an AC signal without changing its shape or peak-to-peak swing.

For analyzing clampers, remember these points:

1. The total swing of the output is equal to the total swing of the input signal.
2. Start the analysis by considering that part of the input which will forward bias the diode.
3. During the period the diode is 'ON', assume that the capacitor will charge up instantaneously to a level determined by the network.
4. During the period for which the diode is 'OFF', assume that the capacitor will hold its charge.
5. Throughout the analysis, keep complete awareness of the polarity for V0, so that the proper value of Vo is determined.

📌 Example: Determine v0 for the network shown in the fig. An input voltage Vi swings from +10V to -20V. The circuit has a 5V battery, a resistor R, a capacitor C, and a diode. Solution:

1. Our analysis will begin at time t1 to t2.
2. During this period, the diode is forward biased and 'ON'. The circuit shows the capacitor charging.
3. Apply KVL to the input loop: -20 + Vc - 5 = 0 => Vc = 25V. The capacitor charges to 25V.
4. For the period t2 to t3, the diode is reverse biased and 'OFF'. The capacitor holds its charge of 25V.
5. The time constant of the discharging network is determined by the product of RC, which is T = RC = 0.01S.
6. KVL across the outer loop during this period: 10 + 25 - V0 = 0 => V0 = 35V.
7. The output waveform will have a peak-to-peak swing of 70V (from -35V to +35V) and is shifted up by 25V.

📌 Example: Repeat the previous example using a silicon diode with VT = 0.7V. Solution:

1. For the short circuit state (diode ON, period t1 to t2), the network takes a new appearance.
2. Determine V0 using KVL in the output section: +5 - 0.7 - V0 = 0 => V0 = 4.3V.
3. For the input section, apply KVL: -20 + Vc + 0.7 - 5 = 0 => Vc = 25 - 0.7 => Vc = 24.3V.
4. For the period t2 to t3 (diode OFF), the capacitor holds its charge.
5. By KVL in the outer loop: 10 + 24.3 - V0 = 0 => V0 = 34.3V.

⭐ Key Takeaways

A parallel clipper uses a diode in a branch parallel to the load to remove a portion of the input signal. The transition voltage, where clipping begins, is determined by the battery and the diode's turn-on voltage. Clampers shift an entire AC signal to a different DC level, maintaining its shape and swing. The analysis of clampers hinges on the charging of a capacitor to a specific voltage when the diode is forward biased and the capacitor holding that charge when the diode is reverse biased. The inclusion of a silicon diode's forward voltage drop (0.7V) slightly reduces the capacitor's charge and the final output voltage levels compared to an ideal diode.

🧠 Quick Revision Questions

1. In a parallel clipper with a 5V battery and an ideal diode, at what input voltage does the diode start conducting?
2. How does the transition voltage change when an ideal diode in a parallel clipper is replaced by a silicon diode?
3. What is the key difference between a clipper and a clamper circuit?
4. According to the rules for analyzing clampers, what happens to the capacitor voltage when the diode is OFF?
5. In the clamper example with the silicon diode, why was the capacitor's final charge 24.3V instead of 25V?

📘 Lecture 41 — Diode Logic Gates (OR Gate)

📖 Overview: This lecture covers diode-based logic gates (OR and AND gates), their circuit analysis with worked examples, then introduces voltage multipliers (half-wave doublers) and their operation, and concludes with zener diode characteristics and shunt regulator concepts. These topics bridge digital logic with analog power electronics.

🗂️ Topics Covered

The lecture begins with diode OR gate configuration and example analysis (finding V and I), then covers diode AND gate with similar example. It transitions to voltage multipliers including half-wave voltage doublers with detailed circuit operation during positive and negative half-cycles. The lecture concludes with zener diode modeling (incremental resistance, equivalent circuit) and zener diode shunt regulator parameters (line regulation, load regulation).

📝 Lecture Summary

DIODE LOGIC GATES (OR GATE)

The OR gate using diodes produces an output equal to the logical OR of inputs: Y = A + B + C. When multiple inputs are present, the diode with the highest forward bias voltage dominates and conducts, while others remain off.

🔑 Definition — Diode OR Gate: A logic gate where the output is HIGH if any input is HIGH; implemented using diodes that conduct when their anode is more positive than the cathode.

📐 Formula: Y = A + B + C → output equals the highest input voltage

📌 Example: Find V and I in the OR gate circuit.

• D1 has 3V applied across it (forward biased), making it conduct more than D2 and D3.
• D2 and D3 are OFF.
• Therefore, V = 3V
• By Ohm's Law: I = (3 - 0)/1k = 3mA

DIODE LOGIC GATES (AND GATE)

The AND gate using diodes produces an output equal to the logical AND of inputs: Y = A · B · C. The diode with the lowest input voltage at its cathode will conduct, pulling the output down.

🔑 Definition — Diode AND Gate: A logic gate where the output is HIGH only if ALL inputs are HIGH; implemented using diodes with output taken at the common cathode point.

📌 Example: Find V and I in the AND gate circuit.

• D3 is ON: cathode at +1V, anode at +5V → potential difference 4V (most forward biased).
• D1 and D2: potential difference across D2 is 3V (less than D3, so D3 dominates).
• Due to D3 being fully forward biased, D1 and D2 become reverse-biased and OFF.
• Therefore, V = +1V
• By Ohm's Law: I = (5 - 1)/1k = 4mA

VOLTAGE MULTIPLIERS

Voltage multipliers are circuits that provide a DC output which is a multiple of the peak input voltage. A voltage doubler provides DC output twice the peak input voltage.

💡 Why this matters: Voltage multipliers are not power generators — when voltage increases by factor N, current decreases by approximately the same factor. Used in low current, high voltage applications (e.g., cathode ray tube in televisions).

HALF-WAVE VOLTAGE DOUBLERS

The half-wave doubler uses two diodes and two capacitors. Assume ideal diodes for analysis.

During negative alternation of input:

• D1 is forward-biased (short circuit), D2 is reverse-biased (open circuit).
• C1 charges until its plate-to-plate voltage equals the source voltage.
• C2 discharges through the load resistance RL.

When input polarity reverses (positive half-cycle):

• D1 is OFF (open), D2 is ON (short).
• C1 (charged to peak value Vs) and the source voltage act as series-aiding voltage sources.
• C2 charges to the sum: 2V_s(pk).

Detailed operation:

• During first half cycle: point x is positive relative to y. Charging current flows through C1 and D1 (forward biased). C1 charges to peak value in T/4. D2 remains OFF.
• During negative half cycle: point x is negative relative to y. Voltage of C1 (Vc1) and Vin are in series (additive). Current flows from y through C2, through D2 (conducting), back to x. C2 charges to magnitude 2Vm.

📐 Formula (applying KVL): -Vc1 - Vin = Vo Since Vc1 = Vmax, therefore: Vo = -2Vmax (output waveform is negative twice the peak input)

Dual Power Supply

A dual-polarity power supply provides both positive and negative DC output voltages.

ZENER DIODE

In the zener region, the differential voltage and current relationship is given by:

📐 Formula: ∆Vz = rz ∆Iz Where rz = inverse of the slope (incremental resistance) of zener diode at the operating point.

🔑 Definition — Zener Diode Model: The almost linear i-v characteristic allows modeling as: Vz = Vzo + rz Iz Where Vzo is the knee voltage (practically considered equal to Vzk), valid when Iz > Izk and Vz > Vzo.

ZENER DIODE SHUNT REGULATOR

A regulator must possess two properties:

1. No change in Vo with any change in Vs (line regulation)
2. No change in Vo with any change in IL (load regulation)

📐 Formulas:

• Line regulation = ∆Vo / ∆Vs
• Load regulation = ∆Vo / ∆IL

⭐ Key Takeaways

The most critical concepts from this lecture are: (1) Diode OR gate output equals the highest input voltage, while diode AND gate output equals the lowest input voltage — both rely on which diode is most forward biased. (2) Voltage multipliers, particularly half-wave doublers, charge capacitors sequentially during alternate half-cycles, with the second capacitor charging to twice the peak input voltage through series-aiding sources. (3) Zener diodes operate in the breakdown region with incremental resistance rz, modeled as Vz = Vzo + rz Iz. (4) A shunt regulator's performance is defined by line regulation (∆Vo/∆Vs) and load regulation (∆Vo/∆IL). (5) Voltage multipliers conserve power — increasing voltage proportionally decreases current, limiting them to low-current applications.

🧠 Quick Revision Questions

1. In a diode OR gate with inputs 5V, 3V, and 1V, what is the output voltage and which diodes conduct?
2. Why does D3 dominate in the AND gate example even though D2 also has forward bias?
3. During which half-cycle does C2 charge to 2Vpk in a half-wave voltage doubler?
4. Write the formula for zener diode voltage in terms of Vzo, rz, and Iz.
5. What is the difference between line regulation and load regulation in a zener shunt regulator?

📘 Lecture 42 — (Zener diode-LED-PIN Photodiodes - Tunnel diode)

📖 Overview: This lecture covers the application of Zener diodes in shunt regulators, providing detailed analysis and design examples. It then introduces Light Emitting Diodes (LEDs), their operating principles and characteristics, followed by brief overviews of PIN Photodiodes and the unique negative resistance region of Tunnel Diodes. This is critical for understanding voltage regulation and optoelectronic devices.

🗂️ Topics Covered

The lecture begins with Zener diode shunt regulators, explaining how to calculate regulation, output voltage, and design parameters like series resistance. Several practical design examples are solved, including finding Vzo, line regulation, load regulation, and power dissipation. The second half introduces LEDs, covering their light emission principle, materials, lead identification, and forward/reverse voltage characteristics. Finally, PIN photodiodes and Tunnel diodes are briefly introduced, focusing on the tunnel diode's negative resistance region between its peak and valley points.

📝 Lecture Summary

ZENER DIODE SHUNT REGULATOR:

A Zener diode shunt regulator maintains a constant output voltage across a load by using a series resistor (R) and a Zener diode (Dz) in parallel with the load. The Zener diode operates in its breakdown region, keeping its voltage relatively constant despite changes in supply voltage or load current. The regulation depends on the incremental resistance (rz) of the Zener diode and the series resistor.

🔑 Regulation: The ability of a circuit to maintain a constant output voltage despite changes in input voltage or load current. 📐 Formulas for Regulation:

• Line Regulation = rz / (R + rz) → This measures the change in output voltage (Vo) for a change in supply voltage (Vs).
• Load Regulation = - (rz || RL) → This measures the change in output voltage (Vo) for a change in load current (IL).

The output voltage can be found using the Thevenin equivalent model. Let Vzo be the Zener voltage at the knee of the characteristic curve (zero current intercept). 📐 Output Voltage Formula: Vo = Vzo + (Vs – Vzo) * rz / (R + rz) To design the series resistor R, the worst-case conditions of minimum supply voltage (Vsmin) and maximum load current (ILmax) are considered, ensuring the Zener current (Iz) stays above its minimum (Izmin). 📐 Series Resistor Design Formula: R = (Vsmin – Vzo – rz*Izmin) / (Izmin + ILmax)

📌 Example: A zener diode whose nominal voltage is 10V at 10mA has an incremental resistance of 50Ω. What voltage must Ahmad Tulaib expect if Bilal disturbs the settings to half the diode current? What is the value of Vzo of the zener model? Solution:

1. Half the current: Current change = -5mA. Vz = 10V - (0.05A)(50Ω) = 10V - 0.25V = 9.75V.
2. Double the current: Current change = +10mA. Vz = 10V + (0.01A)(50Ω) = 10V + 0.5V = 10.5V.
3. Finding Vzo: Vzo is the voltage when current decreases by 10mA from the test point. Vzo = 10V - (0.01A)(50Ω) = 10V - 0.5V = 9.5V.

📌 Example: A zener diode exhibits a constant voltage of 5.6V for currents greater than five times the knee current Izk is specified to be 1mA. It is to be used in the design of a shunt regulator fed from a 15V supply. The load current varies over the range 0 to 15mA. Find a suitable value for the resistance R. What is the maximum power dissipation to the zener diode. Solution:

1. Minimum Zener Current: Izmin = 5 * Izk = 5 * (1mA) = 5mA.
2. Total Current through R: With ILmax = 15mA, the current through R is Ir = Izmin + ILmax = 5mA + 15mA = 20mA.
3. Resistance Calculation: R = (Vs – Vz) / Ir = (15V – 5.6V) / 20mA = 9.4V / 0.02A = 470Ω.
4. Maximum Zener Power: This occurs at no load (IL=0), where all 20mA flows through the Zener. Pz = Vz * Iz = 5.6V * 20mA = 112mW.

📌 Example: A shunt regulator utilizes a zener diode whose voltage is 5.1V at a current of 50mA and whose incremental resistance is 7Ω. The diode is fed from a supply of 15 V nominal voltages through a 200 Ω resistor. What is the output voltage at no load? Find the line regulation and load regulation. Solution:

1. Find Vzo: Vzo = Vz – rz*Iz = 5.1V – (7Ω * 0.05A) = 5.1V – 0.35V = 4.75V.
2. Output Voltage (No Load): Vo = Vzo + (Vs – Vzo)rz/(R+rz) = 4.75V + (15V-4.75V)(7Ω)/(200Ω+7Ω) = 4.75V + (10.25*7)/207 = 4.75V + 0.346V = 5.1V.
3. Line Regulation: rz/(R+rz) = 7/(200+7) = 33.8mV/V.
4. Load Regulation: - (rz || R) = - (7 || 200) = - (7*200)/(7+200) = -1400/207 = -6.8 Ω = -6.8 mV/mA.

📌 Example: Ghania is trying to design a shunt regulator, of about 20V... Two kinds of zener diodes are available: 6.8V devices with rz of 10 Ω and 5.1V devices with rz of 30 Ω. For the two choices possible, find load regulation. Solution:

• Choice a): Three 6.8V diodes in series give 20.4V. The total incremental resistance is 3 * 10Ω = 30Ω. Load regulation (neglecting R) = -30Ω.
• Choice b): Four 5.1V diodes in series give 20.4V. The total incremental resistance is 4 * 30Ω = 120Ω. Load regulation (neglecting R) = -120Ω.
  • 💡 Why this matters: A smaller effective rz results in better (smaller numerical value) load regulation, meaning the output voltage is more stable with changing load current.

📌 Example: It is required to design a shunt regulator to provide a regulated voltage of about 10V. The available 10V, 1W zener of type 1N4740 is specified to have a 10V drop at a test current of 25mA. At this current its rz is 7 Ω. The raw supply available has a nominal value of 20V but can vary as much as ±25%. The regulator is required to supply a load current of 0 to 20mA. Design for a minimum zener current of 5mA. Solution (Data: Vz=10V, Iz=25mA, rz=7Ω, Vs=20V±25%, Izmin=5mA, ΔIL=20mA): a). Find Vzo: Vzo = Vz - Izrz = 10V - 25m(7) = 9.825V. b). Calculate R: Minimum Vs (Vsmin) = 20V - 5V = 15V. Izmin = 5mA occurs when IL is maximum (20mA). R = (Vsmin - Vzo - rzIzmin) / (Izmin + ILmax) = (15 - 9.825 - 0.035) / (5m + 20m) = 5.14V / 25mA = 205.6Ω. Use 205Ω. c). Line Regulation: rz/(R+rz) = 7/(205+7) = 33mV/V. A ±25% change in Vs is ±5V. ΔVo = ±5V * 33mV/V = ±165mV. Percentage change = ±0.165V / 10V * 100% = ±1.65%. d). Load Regulation: -rz||R = -7||205 = -6.77 Ω = -6.77 V/A. From no load (0mA) to full load (20mA), ΔVo = -6.77Ω * 20mA = -135.4mV. Percentage change = -0.1354V / 10V * 100% = -1.35%. e). Maximum Zener Current & Power: Occurs at no load (IL=0) and max Vs (Vsmax=25V). Total current = (25V - 9.825V) / (205Ω + 7Ω) = 15.175V / 212Ω = 71.6mA. More precisely, Vz = 9.825V + 71.6m(7) = 10.326V. Pz = 71.6m(10.326) = 739.4mW. This is under the 1W rating.

LIGHT EMITTING DIODES:

LEDs are diodes that emit light when forward-biased. The graphic symbol has two arrows pointing away from the diode, representing emitted light. When electrons in the n-region cross the junction into the p-region, they recombine with holes. This recombination releases energy in the form of a photon of light. The color of the light depends on the semiconductor material (e.g., Gallium, Arsenic, Phosphorus).

• Lead Identification: Since LEDs have clear cases, the leads are often identified by (1) different lengths, (2) a flattened lead (usually the cathode), or (3) a flattened side on the case near the cathode.

LED CHARACTERISTICS:

LEDs have similar characteristic curves to standard pn junction diodes but with different voltage ranges. 🔑 Forward Voltage (VF): The voltage drop across the diode when it is conducting and emitting light. It is typically between +1.2V to +4.3V. 🔑 Reverse Breakdown Voltage: The voltage at which the diode will break down in reverse bias. It is typically much lower than for standard diodes, ranging from -3V to -10V.

🔑 Definition — PIN Photodiode: A photodiode with an intrinsic (i) layer between the p and n layers. The intrinsic layer increases the depletion region width, allowing for faster response times and higher efficiency in detecting light. It is commonly used in fiber optic communications.

TUNNEL DIODE:

A tunnel diode is a heavily doped (approximately 1000 times more than standard diodes) pn junction diode used in ultra-high frequency (UHF) and microwave applications. Its characteristic curve exhibits a negative resistance region.

The key operating points in the forward bias region are:

• Peak Voltage (Vpk): The voltage at which the forward current reaches its maximum, called Peak Current (Ipk).
• Valley Voltage (Vv): The voltage at which the forward current reaches its minimum, called Valley Current (Iv).
• Negative Resistance Region: The region of operation between Vpk and Vv. In this region, as the forward voltage increases, the forward current decreases (dV/dI is negative), which is the opposite of Ohm's law. This property is useful in high-frequency oscillators and switching circuits.

⭐ Key Takeaways

1. A Zener diode shunt regulator uses a series resistor to limit current and a Zener diode in parallel with the load to clamp the output voltage. The Zener's incremental resistance (rz) is a key parameter for calculating regulation.
2. Line Regulation (ΔVo/ΔVs = rz/(R+rz)) measures output stability against input voltage changes, while Load Regulation (ΔVo/ΔIL = -rz||R) measures stability against load current changes. A smaller rz value gives better regulation.
3. For designing the series resistor R, you must consider the worst-case scenario where the supply voltage is minimum and the load current is maximum, ensuring the Zener has its minimum required current (Izmin) to maintain regulation.
4. LEDs convert electrical energy into light when forward-biased, with a forward voltage typically between 1.2V and 4.3V, and a low reverse breakdown voltage. The color of the emitted light depends on the semiconductor material.
5. A Tunnel diode has a unique negative resistance region between its peak and valley voltages, where current decreases as voltage increases, making it useful for high-frequency oscillators.

🧠 Quick Revision Questions

1. What are the two formulas for calculating line regulation and load regulation in a Zener shunt regulator, and what do they physically represent?
2. If a Zener diode has a Vz of 12V at 20mA and an rz of 10Ω, what is its Vzo and what would its voltage be at a current of 35mA?
3. In designing a shunt regulator, why is it important to consider the minimum supply voltage and maximum load current when calculating the value of the series resistor R?
4. What is the fundamental physical process that causes an LED to emit light, and what determines the color of that light?
5. What is the key characteristic of a Tunnel diode's operating curve between the peak voltage (Vpk) and the valley voltage (Vv), and what is this region called?

📘 Lecture 43 — Bipolar Junction Transistor (BJT)

📖 Overview: This lecture introduces the Bipolar Junction Transistor (BJT), a three-terminal semiconductor device. It covers the physical construction and doping of NPN and PNP transistors, their symbols, and the biasing conditions (forward/reverse) for the emitter-base and collector-base junctions. The lecture also explains the three operating configurations (common base, common emitter, common collector) and the three fundamental modes of operation (cut-off, active, and saturation), including their characteristic graphs and biasing conditions.

🗂️ Topics Covered

The lecture begins by explaining the construction of a BJT as a three-layer semiconductor device with heavily doped emitter, lightly doped base, and very lightly doped collector. It then details the two types (NPN and PNP) and their circuit symbols. The operation of the PNP transistor is analyzed by examining the forward-biased emitter-base junction and the reverse-biased collector-base junction. The three operating configurations (common base, common emitter, common collector) are listed, followed by the directions of currents in a BJT. The three modes of operation (cut-off, active, saturation) are defined based on the biasing of the two PN junctions. Finally, graphical presentations are provided, including the input characteristic (exponential curve of collector current vs. VBE) and the output or collector characteristic for the common base configuration.

📝 Lecture Summary

Bipolar Junction Transistor (BJT) Construction

The BJT is a three-layer semiconductor device that consists of either two n-type and one p-type layers (NPN) or two p-type and one n-type layers (PNP). It has three terminals: Emitter, Base, and Collector. The emitter layer is heavily doped to emit electrons (or holes). The base layer is lightly doped and very thin (thickness ratio of about 150:1 compared to other layers); it helps transfer minority carriers from emitter to collector. The collector is very lightly doped and collects the emitted carriers. There are two basic types: the NPN transistor and the PNP transistor.

Bipolar Junction Transistor The Symbol for NPN Transistor

The circuit symbol for the NPN transistor has an arrow on the emitter terminal pointing outward (away from the base), indicating the direction of conventional current flow.

Bipolar Junction Transistor The Symbol for PNP Transistor

The circuit symbol for the PNP transistor has an arrow on the emitter terminal pointing inward (toward the base).

Bipolar Junction Transistor Operation in PNP Transistor Emitter Base Junction

When the emitter-base junction (p-type emitter, n-type base) is forward-biased (positive terminal of battery to emitter, negative to base), the depletion layer shrinks, allowing majority carriers to cross the junction.

Collector Base Junction

When the collector-base junction (p-type collector, n-type base) is reverse-biased (positive terminal of battery to base, negative to collector), the depletion layer widens, and current flows due to minority carriers.

In a typical PNP circuit, the emitter is positive and the base is negative (VEE forward biases the EBJ), while the base is positive and the collector is negative (VCC reverse biases the CBJ). The total emitter current (I_E) flows inward (into the transistor), while the base current (I_B) and collector current (I_C) flow outward from the transistor.

Operating Configurations of BJT

There are three basic operating configurations:

1. Common Base Configuration
2. Common Emitter Configuration
3. Common Collector Configuration

Directions of Currents in BJT

The relative directions of the emitter, base, and collector currents are shown in circuit diagrams. In an NPN transistor, emitter current flows out, base current flows in, and collector current flows in. In a PNP transistor, the directions are opposite.

Modes of Operation of BJT

There are three modes of operation based on the biasing of the base-emitter junction (BEJ) and the base-collector junction (BCJ):

1. Cut Off: Both junctions are reverse-biased. The transistor is “OFF,” and no current flows.
2. Active Mode: The BEJ is forward-biased, and the BCJ is reverse-biased. The transistor operates as an amplifier.
3. Saturation: Both junctions are forward-biased. The transistor is “ON” (fully conducting).

Graphical Presentation of BJT

The input or driving point characteristic is a graph of collector current (I_C) versus base-emitter voltage (V_BE). The curve shows an exponential rise in I_C starting at about 0.5 volts and saturating at 0.7 volts. 📐 Formula: I_C = I_S * e^(V_BE / V_T) → The collector current is equal to the saturation current (I_S) times an exponential function of the base-emitter voltage (V_BE) divided by the thermal voltage (V_T).

Graphical Presentation of BJT Output or Collector Characteristic for Common Base Configuration

This graph plots collector current (I_C) on the y-axis against collector-emitter voltage (V_CE) on the x-axis. Below a certain base current (e.g., 50 μA), it is the cut-off region. The area along the y-axis is the saturation region, where I_C is maximum and V_CE is minimum. The unshaded area between these two regions is the active region, which depends on I_C, base current, and V_CE.

Modes of Operation of BJT (Table)

The lecture provides a summary table for the biasing of the two junctions in each mode:

• Cut Off: EBJ = Reverse, CBJ = Reverse
• Active: EBJ = Forward, CBJ = Reverse
• Saturation: EBJ = Forward, CBJ = Forward

This table is then used to determine the mode of operation for given transistor circuits.

⭐ Key Takeaways

The most critical concepts to remember are the physical construction of a BJT (three layers, three terminals) with specific doping levels (emitter heavily, base lightly, collector very lightly) and the extremely thin base. The two basic types are NPN and PNP, and their circuit symbols show the direction of conventional current (arrow on emitter). The mode of operation (cut-off, active, saturation) is strictly defined by the biasing of the two PN junctions: both reverse = cut-off, one forward one reverse = active, both forward = saturation. The input characteristic shows an exponential relationship between collector current and base-emitter voltage, with conduction starting around 0.5V and saturating at 0.7V. Finally, the three common configurations (common base, common emitter, common collector) are fundamental ways to connect the transistor in a circuit.

🧠 Quick Revision Questions

1. What are the three layers of a BJT, and what are the respective doping levels (heavy, light, very light) for an NPN transistor?
2. Draw the circuit symbols for both NPN and PNP transistors, and indicate the direction of conventional current flow on the emitter terminal.
3. A transistor has a forward-biased base-emitter junction and a reverse-biased base-collector junction. What is its mode of operation?
4. From the input characteristic graph, at approximately what voltage does the collector current begin to rise exponentially?
5. Name the three operating configurations for a BJT.

📘 Lecture 44 — Bipolar Junction Transistor (BJT)

📖 Overview: This lecture introduces the BJT's internal current relationships, focusing on the three fundamental currents (collector, base, emitter) and their mathematical interdependencies. It then transitions into DC biasing techniques, specifically fixed bias and emitter-stabilized bias circuits, which are essential for establishing stable operating points in transistor amplifiers.

🗂️ Topics Covered

The lecture begins by defining the three BJT currents (collector, base, emitter) and their relationships through constants like β and α, including a solved example demonstrating saturation detection. It then covers DC biasing for BJTs, starting with the fixed bias circuit, its DC equivalent, loop equations for base-emitter and collector-emitter loops, and load line analysis with a graphical example. Finally, it introduces the emitter-stabilized bias circuit, deriving equations for the base-emitter and collector-emitter loops, and concludes with the concept of stabilization achieved by adding an emitter resistor.

📝 Lecture Summary

THE CURRENTS OF A BJT

The BJT has three terminal currents: collector current ($I_C$), base current ($I_B$), and emitter current ($I_E$). The collector current is given by $I_C = I_S e^{(V_{BE}/V_T)}$, where $I_S$ is the saturation current (also called the current scale factor). The base current is derived from the collector current using the common emitter current gain $\beta$: $I_B = I_C / \beta$, where $\beta = I_C / I_B$. The emitter current is the sum of base and collector currents: $I_E = I_B + I_C$. Substituting $I_B = I_C / \beta$ gives $I_E = I_C (\beta+1)/\beta$. Alternatively, the common base current gain $\alpha$ is defined as $\alpha = \beta / (\beta+1)$, allowing $I_C = \alpha I_E$ and $I_E = (I_S / \alpha)e^{(V_{BE}/V_T)}$. Also, $\beta = \alpha / (1-\alpha)$.

🔑 Definition — Saturation current ($I_S$): The current scale factor used in the exponential diode equation for the BJT. 📐 Formula: $I_C = I_S e^{(V_{BE}/V_T)}$ → The collector current depends exponentially on the base-emitter voltage. 📌 Example: Show the transistor is in saturation. Given a circuit with $V_{CC}=10V$, $V_{BB}=6V$, $R_E=3.3k\Omega$, $R_C=4.7k\Omega$, $\beta=100$. For the input loop: $6 - V_{BE} - I_E(3.3k)=0$. Rearranging $I_E = (6-0.7)/3.3k = 1.61mA$. Then $I_B = I_E/101 = 1.61m/101 = 15.94\mu A$. Also $V_E = I_E R_E = 1.61m(3.3k) = 5.3V$. $I_C = 15.94\mu(100) = 1.59mA$. For the output loop: $10 - 1.59m(4.7k) - V_C = 0$, so $V_C = 10 - 7.49 = 2.50V$. Since $V_C < V_B$, the transistor is in saturation. 💡 Why this matters: The condition $V_C < V_B$ (collector voltage less than base voltage) confirms the base-collector junction is forward-biased, which defines saturation.

DC BIASING-BJT: Fixed Bias Circuit

The fixed bias circuit is the simplest BJT biasing configuration. Its DC equivalent shows the base-emitter forward bias loop. Applying KVL to the base-emitter loop: $V_{CC} - I_B R_B - V_{BE} = 0$, so $I_B = (V_{CC} - V_{BE})/R_B$. For the collector-emitter loop: $V_{CE} + I_C R_C - V_{CC} = 0$, giving $V_{CE} = V_{CC} - I_C R_C$. Also $V_{CE} = V_C - V_E$; since $V_E=0$ in this case, $V_{CE} = V_C$. Also $V_{BE} = V_B - V_E$ and $V_E=0$, so $V_{BE} = V_B$.

🔑 Definition — Fixed bias circuit: A biasing configuration where a single resistor $R_B$ connects the base to $V_{CC}$, establishing the base current. 📐 Formula: $I_B = (V_{CC} - V_{BE})/R_B$ → Base current is determined by the voltage drop across $R_B$ divided by $R_B$. 📐 Formula: $V_{CE} = V_{CC} - I_C R_C$ → Collector-emitter voltage is the supply voltage minus the drop across the collector resistor. 📌 Example: Find $I_{CQ}$, $I_{BQ}$, $V_{CEQ}$, $V_B$, $V_C$, $V_{BC}$ with $\beta=50$. (Circuit values assumed from context: $V_{CC}=12V$, $R_B=240k\Omega$, $R_C=2.2k\Omega$, $V_{BE}=0.7V$). $I_B = (12-0.7)/240k = 47.08\mu A$. $I_C = \beta I_B = 50(47.08\mu) = 2.35mA$. $V_{CE} = V_{CC} - I_C R_C = 12 - 2.35m(2.2k) = 12 - 5.17 = 6.83V$. $V_B = V_{BE} = 0.7V$. $V_C = V_{CE} = 6.83V$. $V_{BC} = V_B - V_C = 0.7 - 6.83 = -6.13V$ (reverse-biased).

LOAD LINE ANALYSIS

The output equation of the fixed bias circuit is $V_{CE} = V_{CC} - I_C R_C$, which is a straight line. The x-intercept ($I_C=0$) gives $V_{CE} = V_{CC}$. The y-intercept ($V_{CE}=0$) gives $I_C = V_{CC}/R_C$. These two points define the DC load line on the transistor's output characteristics. The slope of the load line is determined by $-1/R_C$.

📐 Formula: $I_C = V_{CC}/R_C$ (when $V_{CE}=0$) → Maximum possible collector current if $V_{CE}$ drops to zero. 📌 Example: Given a curve, find $V_{CC}$, $R_C$, and $R_B$ for the fixed bias circuit. From the graph, when $I_C=0$, $V_{CE}=V_{CC}=20V$. When $V_{CE}=0$, $I_C=V_{CC}/R_C=10mA$, so $R_C = 20/10m = 2k\Omega$. For $I_B=25\mu A$, $R_B = (V_{CC} - V_{BE})/I_B = (20-0.7)/25\mu = 772k\Omega$.

DC BIASING-BJT: Emitter-Stabilized Bias Circuit

The emitter-stabilized bias circuit adds a resistor $R_E$ in the emitter leg for improved stability. For the base-emitter loop: KVL gives $V_{CC} - I_B R_B - V_{BE} - I_E R_E = 0$. Substituting $I_E = (\beta+1)I_B$ yields $V_{CC} - I_B R_B - V_{BE} - (\beta+1)I_B R_E = 0$. Solving: $I_B = (V_{CC} - V_{BE}) / (R_B + (\beta+1)R_E)$. For the collector-emitter loop: $I_E R_E + V_{CE} + I_C R_C - V_{CC} = 0$. Assuming $I_E \approx I_C$, $V_{CE} = V_{CC} - I_C(R_C + R_E)$. The emitter voltage $V_E = I_E R_E$, collector-to-ground $V_C = V_{CE} + V_E = V_{CC} - I_C R_C$. The base voltage $V_B = V_{CC} - I_B R_B = V_{BE} + V_E$.

🔑 Definition — Emitter-stabilized bias circuit: A biasing configuration with an emitter resistor $R_E$ that provides negative feedback for thermal stability. 📐 Formula: $I_B = (V_{CC} - V_{BE}) / (R_B + (\beta+1)R_E)$ → Base current now depends on a larger effective resistance including $(\beta+1)R_E$. 📌 Example: (From lecture) $I_E R_E + V_{CE} + I_C R_C - V_{CC} = 0$ and $V_{CE} = V_{CC} - I_C(R_C + R_E)$.

Stabilization

Stabilization refers to the ability of a biasing circuit to keep DC bias currents and voltages constant despite changes in temperature and transistor beta ($\beta$). The addition of the emitter resistor $R_E$ provides negative feedback: if $I_C$ increases (due to temperature), $I_E$ increases, causing a larger voltage drop across $R_E$. This reduces $V_{BE}$ (since $V_B$ is fixed), which in turn reduces $I_B$, counteracting the original increase in $I_C$.

💡 Why this matters: Stabilization is crucial for reliable circuit operation because transistor parameters (especially $\beta$) vary significantly with temperature and between individual devices.

⭐ Key Takeaways

The three BJT currents are interrelated by the constants $\beta$ and $\alpha$, where $\beta = I_C/I_B$ and $\alpha = \beta/(\beta+1)$, with $\alpha$ being slightly less than 1. The fixed bias circuit, while simple, has poor thermal stability because its base current depends only on $V_{CC}$ and $R_B$. Load line analysis provides a graphical method to determine the quiescent operating point ($Q$-point) by intersecting the transistor's output characteristic curves with the load line defined by $V_{CE} = V_{CC} - I_C R_C$. The emitter-stabilized bias circuit significantly improves stability by incorporating $R_E$, which provides negative feedback that opposes changes in collector current. The condition for saturation is $V_C < V_B$ (base-collector junction forward-biased), which the solved example demonstrated.

🧠 Quick Revision Questions

1. Derive the relationship between $I_E$, $I_C$, and $\beta$ starting from $I_E = I_B + I_C$ and $I_B = I_C/\beta$.
2. In the fixed bias circuit, what are the two intercepts of the DC load line on the transistor output characteristics?
3. Explain how adding an emitter resistor $R_E$ improves the stability of the biasing circuit against temperature changes.
4. For the BJT saturation example, calculate $V_B$ and $V_C$ and state the condition that confirms saturation.
5. Write the KVL equation for the base-emitter loop of the emitter-stabilized bias circuit and derive the expression for $I_B$.

📘 Lecture 45 — Bipolar Junction Transistor BJT-circuit analysis

📖 Overview: This lecture covers the analysis of BJT circuits with different biasing configurations, focusing on emitter-stabilized bias circuits, voltage divider bias, and biasing using two power supplies. Understanding these biasing techniques is essential for designing amplifier circuits that maintain stable operation regardless of temperature variations and transistor parameter variations.

🗂️ Topics Covered

The lecture begins with emitter-stabilized bias circuit load line analysis and includes example problems for finding base voltage limits for active mode operation and determining resistor values. It then covers voltage divider bias circuit analysis using Thévenin's theorem, followed by a complex multi-transistor example with iterative calculations. Finally, biasing using two power supplies is introduced with a common base amplifier design example.

📝 Lecture Summary

Emitter-Stabilized Bias Circuit: Load Line Analysis

The collector-emitter loop equation defining the load line is: VCE = VCC - IC(RC + RE). For IC = 0mA, VCE = VCC. For VCE = 0, IC = VCC/(RC + RE). This load line represents all possible operating points for the transistor.

Example: Finding Maximum Base Voltage for Active Mode

Scenario: Zammad is designing an amplifier circuit. Find the highest voltage to which the base can be raised while the transistor remains in active mode. Assume α=1.

Solution: Let the base voltage for active operation be V. For active mode, consider VB = VC = V.

For the output loop: 10 - 4.7kΩ(IC) - VC = 0 (Equation 1)

For the base-emitter loop: VB - VBE - IE(RE) = 0 (Equation 2)

Since α=1, IC = IE

From Equation (2): IE = (VB - VBE)/RE = (V - 0.7)/3.3kΩ

Substituting into Equation (1): 10 - (V-0.7)(4.7/3.3) - V = 0 3.3(10) - 4.7V - 3.29 - 3.3V = 0 8V = 29.71 V = 3.71V

💡 Why this matters: This calculation ensures the transistor stays in the active region where it can amplify signals linearly without distortion.

Example: Finding RC and RE

Given parameters: IC = 0.5mA, α = 1, VCB = 2V

Solution: For the output loop using KVL: (10-6)/0.5mA = RC RC = 8kΩ

Since α=1, IC = IE = 0.5mA

For the base-emitter loop using KVL: (4-0.7)/0.5mA = RE RE = 6.6kΩ

DC BIASING-BJT: Voltage Divider Bias Circuit

The voltage divider bias circuit uses two resistors R1 and R2 to establish a stable base voltage independent of transistor β.

Applying Thévenin's theorem: RTh = R1 || R2, ETh = VR2 = R2VCC/(R1+R2)

The Thévenin equivalent circuit gives the KVL equation: ETh - IBRTh - VBE - IERE = 0

Putting IE = (β+1)IB and solving for IB: IB = (ETh - VBE)/(RTh + (β+1)RE)

🔑 Definition — Voltage Divider Bias: A biasing method using two resistors to create a stable base voltage that makes the operating point relatively independent of transistor beta variations.

Once IB is known, the remaining problem may be solved exactly as in the previous configurations.

Example: Find Voltage at All Nodes and Current Through All Branches

Given: β = 100

Solution: Using Thévenin's theorem: VBB = 15(RB2)/(RB1+RB2) = 15(50)/150 = 5V

RBB = RB1 || RB2 = 100kΩ || 50kΩ = 33.3kΩ

Writing KVL for loop L: VBB = IBRBB + VBE + IERE

Putting IB = IE/(β+1): IB = 1.29/101 = 0.0128mA

Base voltage: VB = VBE + IERE = 0.7 + 1.29(3) = 4.57V

Assuming active mode, collector current: IC = αIE = 0.99(1.29) = 1.28mA

Collector voltage: VC = 15 - ICRC = 15 - 1.28(5) = 8.6V

Example: Multi-Transistor Circuit Analysis

Find voltages at all nodes and current through all branches.

Solution: Assume Q1 is in active mode: VB1 = 4.57V, IE1 = 1.29mA, IB1 = 0.0128mA, IC1 = 1.28mA

VC1 = 15 - IC1RC1 = 15 - (1.28)5 = 8.6V

Assume Q2 emitter-base junction is forward biased: VE2 = VC1 + VEB = 8.6 + 0.7 = 9.3V

IE2 = (15-VE2)/RE2 = (15-9.3)/2k = 2.85mA

IC2 = αIE2 = 0.99(2.85m) = 2.82mA (β=100)

VC2 = IC2RC = 7.62V

Since VC2 is lower than VB2 = VC1 by 0.98V, Q2 is active.

Iteration for accuracy: IB2 = IE2/(β2+1) = 2.85m/101 = 0.028mA

IC1 = IRc1 - IB2 = 1.25mA VC1 = 15 - 5(1.252m) = 8.74V

VE2 = 8.74 + 0.7 = 9.44V IE2 = (15-9.44)/2k = 2.78mA

IC2 = 0.99(2.78m) = 2.75mA VC2 = 2.75(2.7) = 7.43V

IB2 = 2.78/101 = 0.0275mA

💡 Why this matters: Multi-transistor circuits require iterative analysis because the base current of one transistor affects the collector current calculation of the previous transistor, requiring successive approximations for accuracy.

Biasing Using Two Power Supplies

The loop equation for the loop marked as L is derived for circuits using both positive and negative supply voltages, providing more design flexibility and better stability.

Example: Common Base Amplifier Design

Design the circuit to establish a DC emitter current of 1mA and provide the highest possible gain while allowing for a maximum signal swing of ±2V at the collector. Use +10V and -5V power supplies.

Solution: Since the amplifier is to be used in common base configuration, RB = 0.

Considering the base-emitter junction forward biased: RE = (-0.7 - (-5))/1mA = 4.3kΩ

To allow for ±2V signal swing at the collector while choosing as large a value of RC as possible: Set VC = +2V (so -2V signal would not saturate the BJT)

RC = (10-2)/IC = 8/1mA RC = 8kΩ

📐 Formula: For common base amplifier design: RE = (VEE - VBE)/IE and RC = (VCC - VC)/IC, where VC is chosen to allow maximum signal swing without saturation.

⭐ Key Takeaways

The load line analysis for emitter-stabilized bias circuits defines all possible operating points, with the endpoints being VCE = VCC and IC = VCC/(RC+RE). Voltage divider bias provides stable biasing by using Thévenin's theorem to simplify the base circuit, resulting in base current IB = (ETh - VBE)/(RTh + (β+1)RE). When analyzing multi-transistor circuits, iterative calculations are necessary because the base current of one transistor modifies the collector current of the preceding stage. For common base amplifier design using two power supplies, the emitter resistor establishes the DC emitter current while the collector resistor is chosen to allow maximum signal swing without entering saturation. The key design goal is always to maintain the transistor in the active region where it can provide linear amplification.

🧠 Quick Revision Questions

1. What are the two endpoints of the load line for an emitter-stabilized bias circuit?
2. How do you calculate the Thévenin equivalent voltage and resistance for a voltage divider bias circuit?
3. In the multi-transistor example, why was iteration necessary when calculating IC1 and IB2?
4. What is the formula for finding RC in a common base amplifier design given power supply voltages and desired signal swing?
5. Why does setting VC = +2V in the common base amplifier example prevent the BJT from saturating during -2V signal swings?