Signal Processing And Output Calculation (79.2.3) - Differential Amplifier : Analysis and Numerical Examples (Contd.)
Students

Academic Programs

AI-powered learning for grades 8-12, aligned with major curricula

Professional

Professional Courses

Industry-relevant training in Business, Technology, and Design

Games

Interactive Games

Fun games to boost memory, math, typing, and English skills

Signal Processing and Output Calculation

Signal Processing and Output Calculation

Enroll to start learning

You’ve not yet enrolled in this course. Please enroll for free to listen to audio lessons, classroom podcasts and take practice test.

Practice

Interactive Audio Lesson

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

Differential Amplifiers Basics

🔒 Unlock Audio Lesson

Sign up and enroll to listen to this audio lesson

0:00
--:--
Teacher
Teacher Instructor

Today we will discuss differential amplifiers. These circuits amplify the difference between two input signals while rejecting any common signals. What do you think is the main purpose of using such amplifiers?

Student 1
Student 1

Is it to reduce noise coming from common signals?

Teacher
Teacher Instructor

Exactly! This property is crucial in telecommunications and audio applications. Now, can anyone recall the types of transistors typically used in these amplifiers?

Student 2
Student 2

BJTs and MOSFETs, right?

Teacher
Teacher Instructor

Spot on! We’ll explore both today, focusing on calculating gains and understanding small-signal parameters.

DC Operating Point

🔒 Unlock Audio Lesson

Sign up and enroll to listen to this audio lesson

0:00
--:--
Teacher
Teacher Instructor

Let’s start with the DC operating point. Why do we need to determine this when designing a differential amplifier?

Student 3
Student 3

To ensure the transistors operate in the active region?

Teacher
Teacher Instructor

Exactly! For BJTs, we typically apply a base voltage of 2.6V. Can anyone explain how we calculate the emitter voltage?

Student 4
Student 4

By subtracting the V_BE drop, right? So it would be 2.6V minus 0.6V...

Teacher
Teacher Instructor

Correct! This gives us an emitter voltage of 2.0V. Keep in mind the implications for the collector current calculations too!

Small Signal Parameters

🔒 Unlock Audio Lesson

Sign up and enroll to listen to this audio lesson

0:00
--:--
Teacher
Teacher Instructor

Moving on, let’s find out about small signal parameters. Why are they important for our differential amplifiers?

Student 1
Student 1

They help us understand the behavior of the amplifier for varying input signals, right?

Teacher
Teacher Instructor

Yes! The transconductance 'g_m' and output resistance 'r_o' are key. Can anyone calculate 'g_m' if the collector current is 1 mA?

Student 2
Student 2

It would be approximately 38.5 mS since g_m = I_C / V_T!

Teacher
Teacher Instructor

Wonderful! Now let’s derive the differential mode gain using these parameters.

Output Swing

🔒 Unlock Audio Lesson

Sign up and enroll to listen to this audio lesson

0:00
--:--
Teacher
Teacher Instructor

Now that we have our gains, let’s analyze output swing. What does this term refer to in our amplifier context?

Student 3
Student 3

It’s the maximum voltage output range the amplifier can produce without distortion!

Teacher
Teacher Instructor

Exactly! By calculating the limits based on our DC voltage points, we understand the functional range of our amplifier.

Student 4
Student 4

For example, how would we calculate the positive output swing based on a 12V supply?

Teacher
Teacher Instructor

Great question! It would be 12V minus our collector voltage, which we determined earlier!

Introduction & Overview

Read summaries of the section's main ideas at different levels of detail.

Quick Overview

This section covers the analysis and numerical examples related to differential amplifiers using BJT and MOSFET, focusing on their operating points, output swing, and small-signal parameters.

Standard

The section provides a comprehensive overview of differential amplifiers, illustrating how to analyze circuits using BJTs and MOSFETs. Key points include determining DC operating points, calculating small-signal parameters, and understanding differential and common mode gains through practical numerical examples.

Detailed

Detailed Summary

In this section, we delve deeper into the analysis of differential amplifiers, focusing on configurations using Bipolar Junction Transistors (BJTs) and Metal-Oxide-Semiconductor Field-Effect Transistors (MOSFETs). We begin with the operational amplifier's DC operating point, illustrating how to ensure transistors remain in the active region during operation. By employing numerical examples, we explore the significance of small-signal parameters like differential and common mode gains. The section concludes by emphasizing output swing considerations and identifying the impact of varying resistor values.

Throughout the examples, specific calculations are presented, demonstrating how variations in input signals affect both differential and common mode outputs, reinforcing the fundamental role of differential amplifiers in effective signal processing.

Youtube Videos

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.

Operating Point of Transistors

Chapter 1 of 4

🔒 Unlock Audio Chapter

Sign up and enroll to access the full audio experience

0:00
--:--

Chapter Content

To start with, we do have this DC voltage given to us which is 2.6. In fact, this DC voltage should be sufficiently high, so that Q and Q should be in active region. And on the other hand this DC voltage should not be too high otherwise, Q and Q may enter into saturation region.

Detailed Explanation

The operating point of a transistor is determined by the DC voltage and is crucial for its proper functioning. In this case, we have a DC voltage of 2.6 volts. This voltage needs to be carefully selected; it must be high enough to keep transistors Q1 and Q2 in their active operating region, where they can amplify signals effectively, but not so high that they push the transistors into saturation, where they can no longer operate efficiently. Thus, setting the right DC voltage is essential for optimal performance.

Examples & Analogies

Imagine a car engine that needs to operate within a specific RPM range. If the engine runs too slow (below the active region), it stutters and doesn't perform well, like transistors during low voltage. If it runs too fast (over the saturation region), it might overheat and fail. Just as maintaining the right RPM is critical for the engine, the DC voltage must be just right for the transistors.

Calculation of Emitter and Collector Currents

Chapter 2 of 4

🔒 Unlock Audio Chapter

Sign up and enroll to access the full audio experience

0:00
--:--

Chapter Content

Now, we assume that of course, this is the emitter current 1 mA. So, we assume that the base current is very small. So, we can say that the collector current of transistor 1 as well as transistor 2 both of them we can well approximate by 1 mA.

Detailed Explanation

When analyzing a transistor circuit, the emitter current is identified as 1 mA. In many transistor analyses, the base current is often much smaller than the collector current due to the transistor's amplification capabilities. Therefore, we can approximate the collector current for both transistors, Q1 and Q2, as 1 mA each. This simplification helps us manage our calculations and understand the behavior of the circuit more clearly.

Examples & Analogies

Think of it like a water faucet. The flow rate of water (collector current) depends primarily on how much you open the faucet, while a tiny trickle of water that leaks from the faucet (base current) can be ignored in comparison. The significant flow of water is what matters in determining how much you rely on it for washing or filling a container, just like the collector current is what affects how transistors amplify signals.

Determining DC Voltage at Collector

Chapter 3 of 4

🔒 Unlock Audio Chapter

Sign up and enroll to access the full audio experience

0:00
--:--

Chapter Content

So, the drop across this resistor it is 5.2 V. So, the voltage at the collector DC voltage at the collector it is 12 V ‒ 5.2. So, that = 6.8.

Detailed Explanation

In the process of finding the DC voltage at the collector of a transistor, we first calculate the voltage drop across the load resistor, which is found to be 5.2 volts. The voltage at the collector is then calculated by subtracting this drop from the supply voltage of 12 volts. Thus, we find the collector voltage to be 6.8 volts. This is an important value as it directly influences how the transistor will operate and interact with subsequent stages in the circuit.

Examples & Analogies

Consider a battery powering a series of light bulbs. The battery provides a total voltage, but as it passes through each light bulb (the resistors), some voltage is 'used up' or dropped across each one. What remains is the voltage at the end of the circuit, just like the remaining voltage at the collector after accounting for voltage drops in a transistor circuit.

Small Signal Parameters and Their Importance

Chapter 4 of 4

🔒 Unlock Audio Chapter

Sign up and enroll to access the full audio experience

0:00
--:--

Chapter Content

Now, we obtain the small signal parameters of both the transistors. Next thing is we need to find the small signal gain namely, a differential mode gain and common mode gain.

Detailed Explanation

After establishing the DC operating points, we now turn our attention to the small-signal parameters of the transistors. These parameters are essential for analyzing how the transistors will behave with small alternating signals superimposed on the DC bias. We focus on calculating both the differential mode gain, which measures how well the circuit amplifies differences between two input signals, and the common mode gain, which measures how well the circuit handles input signals common to both inputs. Understanding these gains is critical for designing effective amplifiers.

Examples & Analogies

Think of it like tuning a musical instrument. When a musician adjusts the strings to improve the sound, they focus both on the difference in pitch (the differential tone) and on the overall sound (the common tone). Similarly, in our transistors, we need to understand how well they amplify both the differences (differential gain) and the similarities (common gain) to ensure we achieve the best sound or signal quality.

Key Concepts

  • Operating Point: It ensures transistor operation is stable within active regions.

  • Small Signal Model: Helps analyze amplifier behavior under small voltage changes.

  • Differential Mode vs Common Mode: Differentiates the signals needed for effective amplification.

Examples & Applications

Example 1: Calculating the output voltage swing for a differential amplifier with a supply voltage of 12V and a DC operating point of 6.8V.

Example 2: Finding the differential and common mode gain based on given input signal variations.

Memory Aids

Interactive tools to help you remember key concepts

🎵

Rhymes

In signal processing, we aim to amplify, A differential amp helps noise to defy.

📖

Stories

Imagine two friends at a party, they talk to each other in a loud crowd. The differential amplifier helps hear their conversation above the noise, just as it amplifies differential signals over common noise in circuits.

🧠

Memory Tools

Remember 'D.O.C.' for Differential Operating Conditions: D for differential signal, O for operating point, and C for common mode rejection.

🎯

Acronyms

BJT - 'Bigger Just Talk' indicating the significance of BJTs in amplifying signals.

Flash Cards

Glossary

Differential Amplifier

An amplifier that amplifies the difference between two input voltages while rejecting common mode voltages.

DC Operating Point

The steady-state operating conditions of the amplifier circuits to ensure that the transistors remain in the desired region of operation.

Small Signal Parameters

Parameters (like transconductance and output resistance) that define the behavior of the amplifier under small variations around the DC operating point.

Differential Mode Gain

The gain of the amplifier when a differential signal is applied, calculated as the ratio of differential output voltage to differential input voltage.

Common Mode Gain

The gain of the amplifier when common-mode signals are applied to both inputs, typically much lower than differential mode gain.

Output Swing

The maximum range of output voltage that an amplifier can achieve without distortion.

Reference links

Supplementary resources to enhance your learning experience.