Summary of Numerical Examples on Current Mirror - 89.3.1 | 89. Numerical examples on current mirror and its applications (Part-D) | Analog Electronic Circuits - Vol 4
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Interactive Audio Lesson

Listen to a student-teacher conversation explaining the topic in a relatable way.

Overview of Current Mirrors in Amplifiers

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0:00
Teacher
Teacher

Today, we will explore the use of current mirrors in differential amplifiers. Can someone tell me what a current mirror is?

Student 1
Student 1

Isn't it a way to copy current from one branch of a circuit to another?

Teacher
Teacher

Exactly! Current mirrors are essential for maintaining consistent current in different parts of a circuit. Now, why do you think this is important in amplifiers?

Student 2
Student 2

It helps keep the signal clean and stable, right?

Teacher
Teacher

Correct! A stable signal leads to improved performance in amplifiers. Let's now see how we calculate DC currents in these setups.

Calculating DC Currents

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Teacher
Teacher

Let's look at an example. If the load resistor is 11.4 k ohms and the current is 1mA, how do we determine the current through each transistor?

Student 3
Student 3

If the total current is 1 mA and it splits between two transistors, wouldn't each one carry 0.5 mA?

Teacher
Teacher

Exactly! And that mirrored current flows through both transistors in the mirror. Let’s calculate what happens when we adjust the input signal... What happens to the output?

Student 4
Student 4

The output current would mirror the input signal changes, right?

Teacher
Teacher

Yes! You're understanding the functionality perfectly.

Understanding Gain Calculations

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Teacher
Teacher

Now, with the DC currents established, how do we proceed to calculate the differential mode and common mode gains?

Student 1
Student 1

We use the transconductance and the output resistances, right?

Teacher
Teacher

That's right! A common mistake is overlooking how impedance affects gain. Can anyone summarize the gain calculation for differential mode?

Student 2
Student 2

We compute it using the formula: Gain = gm * (ro1 || ro2) for the differential mode.

Teacher
Teacher

Excellent! Now, how would the common mode gain differ?

Student 3
Student 3

It’s likely lower because of degeneration in the circuit, using the resistor to reduce common mode impact.

Teacher
Teacher

Absolutely correct! Time for some practice!

Practical Applications of Current Mirrors

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Teacher
Teacher

Let’s talk about where we can see current mirrors in real-world applications. Can someone give me an example?

Student 4
Student 4

I remember learning about them in transistor amplifiers, especially in active loads.

Teacher
Teacher

Great example! They also appear in voltage references and biasing circuits. Why do you think they are preferred?

Student 1
Student 1

Because they maintain a constant current, improving the stability over varying conditions.

Teacher
Teacher

Exactly! Stability is key in analog circuits. This is what enables efficient amplification and signal integrity.

Introduction & Overview

Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.

Quick Overview

This section discusses numerical examples of current mirror applications in differential amplifiers, detailing DC current calculations, and differential and common mode gains.

Standard

In Section 3.1, we explore numerical examples related to current mirrors embedded in differential amplifiers. The section explains how to calculate the DC currents in transistors, establish gain measures, and analyzes the implications of these calculations on circuit performance, specifically focusing on differential and common mode gains.

Detailed

Detailed Overview of Current Mirror Numerical Examples

In this section, we elaborate on the applications of current mirrors within differential amplifiers. We begin by examining a differential amplifier that includes an active load, specifically emphasizing the role of current mirrors in ensuring matching currents through connected transistors. The primary calculations aimed at evaluating DC currents through the transistors demonstrate a clear relationship between theoretical values and practical applications.

The section also covers detailed examples of calculating differential and common mode gains. Furthermore, we analyze how the configuration of these circuits affects performance, leading to insights into designing more efficient analog electronic circuits. By the end of this section, the student will understand the operational principles behind current mirrors, appreciate their applications in amplifiers, and be proficient in executing relevant calculations.

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Analog Electronic Circuits _ by Prof. Shanthi Pavan
Analog Electronic Circuits _ by Prof. Shanthi Pavan

Audio Book

Dive deep into the subject with an immersive audiobook experience.

Introduction to Current Mirrors

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In the numerical examples, we will be reestablishing the same thing, to start with we do have to find the DC currents through all the transistors.

Detailed Explanation

This chunk introduces the concept of current mirrors within the context of example problems. The basic task is to determine the direct current (DC) flowing through each transistor in the circuit. Understanding the flow of current in transistors is essential, as it helps in analyzing the performance and behavior of current mirrors.

Examples & Analogies

Imagine a water system where pipes (transistors) carry water (current). To ensure that the right amount of water flows to each pipe, you need to calculate how much is needed at each point. In a circuit, similar calculations ensure each transistor receives the correct amount of current, allowing the circuit to function properly.

Understanding DC Current Calculation

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Again the value of this resistor it is 11.4 k, so that makes this current = 1 mA. So, the current flow here it is also 1 mA...

Detailed Explanation

This section provides specific values that illustrate how to calculate the DC currents through different components in the current mirror circuit. Initially, a resistor is given a value of 11.4 kOhm, which leads to a current of 1 mA. By applying the principles of circuit analysis, we see how this current splits into half (0.5 mA each) across two paths involving more transistors.

Examples & Analogies

Consider two garden hoses splitting water from a single source. If the total flow is 1 mL of water per second and the hoses are identical, each one will carry 0.5 mL per second. This is similar to how the current splits between the transistors in the circuit. Understanding this split helps to ensure that each component operates correctly.

Voltage Setting and Output Conditions

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So, since this is transistor-7 it is diode connected. So, the voltage here it is decided by the 12 V β€’ 0.6 V so the voltage here it is 11.4 V...

Detailed Explanation

The voltage across a diode-connected transistor is crucial for determining the output states within the circuit. By referencing a 12 V supply and the diode forward voltage drop of 0.6 V, a constant voltage of 11.4 V is established at this point. This voltage contributes to the stability required for the current mirror to function effectively.

Examples & Analogies

Think of this scenario like a water tower maintaining pressure by a drop in its spout. The pressure at the bottom (11.4 V) remains consistent regardless of the total amount of water (12 V) in the tank due to the outlet's resistance. This ensures every connected pipe receives a steady flow, just as transistors do in the current mirror design.

Differential and Common Mode Gain Calculations

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So, yeah so now our next calculation it is we need to find differential mode gain and common mode gain...

Detailed Explanation

This part explains how to calculate the differential mode gain and common mode gain, which are key performance metrics for amplifiers. The differential mode gain measures how well the circuit amplifies the difference in input signals, while the common mode gain focuses on signals that are common to both inputs. Understanding these concepts is vital for assessing amplifier performance in practical scenarios.

Examples & Analogies

Imagine you are measuring the height of two plants. If you want to know which plant is taller (differential gain), you compare their heights. But if you want to see if both plants reach a certain mark on a wall (common mode gain), you look at the height relative to that mark. Similarly, these gains in circuits help compare how well signals are amplified.

Conclusion of Numerical Examples

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Now, coming to the conclusion of this entire session it is what we have talked about basically we started with numerical examples of simple current mirror...

Detailed Explanation

In concluding the numerical examples section, the main takeaway is a summary of various current mirror implementations, calculations of quiescent currents, and an analysis of both differential and common mode gains. Competence in these calculations signifies a solid understanding of how current mirrors function within electronic circuits.

Examples & Analogies

Think of wrapping up a project presentation where you summarize your findings on how best to build a model. You outline what you did step by step, similar to retelling the calculations of the current mirrors in your circuit design, ensuring that your audience understands the importance of each part of your process.

Definitions & Key Concepts

Learn essential terms and foundational ideas that form the basis of the topic.

Key Concepts

  • Current Mirror: A circuit used to replicate a current across different transistors.

  • Differential Gain: The amplification of the difference between two input voltages.

  • Common Mode Gain: The response of the amplifier to common signals present at both inputs.

  • Transconductance (gm): A measure of the performance of a transistor, indicating how well it converts input voltage to output current.

Examples & Real-Life Applications

See how the concepts apply in real-world scenarios to understand their practical implications.

Examples

  • In a current mirror circuit, if a reference current of 1 mA is established, each mirrored branch should ideally also carry 1 mA.

  • When analyzing a differential amplifier, if a 12 V supply is used with a load resistor yielding 1 mA, the drop across the resistor confirms the expected outputs.

Memory Aids

Use mnemonics, acronyms, or visual cues to help remember key information more easily.

🎡 Rhymes Time

  • A current mirror shows the flow, keeps it steady, helps us know.

πŸ“– Fascinating Stories

  • Imagine a gardener (the current mirror) that waters two identical plants (transistors) equally from one hose.

🧠 Other Memory Gems

  • CAM - Copying Active Mirroring to remember how current mirrors operate.

🎯 Super Acronyms

DGAIN - Differential Gain Are Important Numbers, to remember the relevance of differential gains.

Flash Cards

Review key concepts with flashcards.

Glossary of Terms

Review the Definitions for terms.

  • Term: Current Mirror

    Definition:

    A circuit configuration used to provide a constant current to a load while replicating the current from a reference branch.

  • Term: Differential Amplifier

    Definition:

    An amplifier that amplifies the difference between two input signals while rejecting any signals that are common to both inputs.

  • Term: Common Mode Gain

    Definition:

    The gain of an amplifier when the same input signal is applied to both inputs.

  • Term: Differential Mode Gain

    Definition:

    The gain of an amplifier when the difference between the input signals is amplified.