Output Terminal And Current Calculation (50.1.2) - Common Base and Common Gate Amplifiers : Analysis (Part B)
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

Output Terminal and Current Calculation

Output Terminal and Current 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.

Understanding Unloaded Condition

🔒 Unlock Audio Lesson

Sign up and enroll to listen to this audio lesson

0:00
--:--
Teacher
Teacher Instructor

Today we will discuss what an unloaded condition means when analyzing common base amplifiers. Can anyone explain why we might want to short the output node to AC ground?

Student 1
Student 1

Because it allows us to measure the output current accurately without affecting the circuit's operating point.

Teacher
Teacher Instructor

Exactly! This is crucial for determining the true performance of the amplifier. Let's remember the acronym 'GROUNDED' to recall that we short the output to ground for accurate signal current analysis. What happens to our current measurements when we do this?

Student 2
Student 2

It gives us a zero impedance at the output, allowing us to isolate the currents.

Teacher
Teacher Instructor

Correct! So, by shorting to ground, we measure the current more precisely, which leads us to analyze the circuit's response effectively.

Student 3
Student 3

Does this mean we should assume input current remains unchanged?

Teacher
Teacher Instructor

Great question! Yes, under small signal conditions, we can assume the input current as a constant for our calculations. This leads us to look at how these currents relate to gain.

Student 4
Student 4

What does 'g_m' stand for in our calculations?

Teacher
Teacher Instructor

'g_m' represents transconductance, which is crucial for relating input current to output current in our gain equation. Let's summarize key points: we analyze unloaded conditions to measure current accurately, understand zero output impedance, and relate currents through transconductance.

Current Gain Derivation

🔒 Unlock Audio Lesson

Sign up and enroll to listen to this audio lesson

0:00
--:--
Teacher
Teacher Instructor

Now, let’s derive the current gain formula. Who can start us off with the basic relationship of input and output currents?

Student 1
Student 1

The output current should be related to input current through transconductance, right?

Teacher
Teacher Instructor

Yes! We start with the simple relationship that output current depends on both transconductance and input signal. So if we denote these as 'i_o = g_m * v_e', what happens next when we consider their ratios?

Student 2
Student 2

We can express current gain as the ratio of 'i_o' to 'i_in'.

Teacher
Teacher Instructor

Exactly! This brings us to the formula for current gain which simplifies to approximately 'g_m'. Can anyone recall what happens if we neglect certain resistances in this context?

Student 3
Student 3

It means we assume they are much smaller compared to g_m, and our gain approaches α.

Teacher
Teacher Instructor

Right again! This shows how reliable the common base configuration is for current mode buffering, as α is close to 1. Let's summarize: we derived that the current gain relates to transconductance and input/output relationships.

Significance of Current Gain in Applications

🔒 Unlock Audio Lesson

Sign up and enroll to listen to this audio lesson

0:00
--:--
Teacher
Teacher Instructor

Let's talk about the significance of current gain in common base and common gate configurations. Why is this important in practical applications?

Student 4
Student 4

It tells us how effectively the circuit can handle current in amplifying signals.

Teacher
Teacher Instructor

Exactly! When we have a current gain close to one, we can use these configurations as buffers between different circuit stages. How would that impact design?

Student 1
Student 1

It allows for better signal integrity because the input and output resistances differ.

Teacher
Teacher Instructor

Spot on! This allows for high input resistance and low output resistance, ideal for maintaining signal quality. Therefore, these configurations play a vital role in multi-stage amplifiers. What conclusion can we draw about our biasing arrangements?

Student 3
Student 3

The biasing must support stable operation to ensure consistent current gain.

Teacher
Teacher Instructor

Well put! In practice, these amplifiers serve significant roles in signal processing, buffering, and current amplification, which we must always consider in circuit design.

Introduction & Overview

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

Quick Overview

This section explains the current gain in common base and common gate configurations and how to determine output current under unloaded conditions.

Standard

In this section, the focus is on deriving the current gain for the common base and common gate amplifier configurations. It discusses how to analyze the circuit in unloaded conditions, define input and output currents, and the significance of the current gain, which is close to one for these configurations, making them suitable for current mode amplification.

Detailed

Detailed Summary

In this section, we delve into the analysis of current gain within common base and common gate amplifier configurations. The main goal is to derive the current gain formula while ensuring that the output terminal remains unloaded. To do this, we start by defining 'unloaded' as the condition wherein the output node is shorted to AC ground, allowing for an accurate assessment of signal current without affecting the transistor's operating point.

We draw attention to the small signal model where the emitter voltage is defined and understanding the various components of currents that flow through the amplifier. The section emphasizes that under unloaded conditions, the input and output currents can be related directly, leading to a simplified representation of current gain, typically expressed in terms of transconductance and emitter voltage gain. By defining output current as a function of input current and using the parameters of the transistor – specifically the alpha (α) and beta (β) parameters – we conclude that the current gain is generally close to one for these amplifiers, validating their effectiveness as current buffers. Throughout the section, we noted the importance of resistance considerations in these calculations, especially in relation to biasing schemes.

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.

Common Base Configuration Overview

Chapter 1 of 6

🔒 Unlock Audio Chapter

Sign up and enroll to access the full audio experience

0:00
--:--

Chapter Content

So, here we do have the common base configuration. We do have the corresponding circuit here and to get the current gain what we have to do? At the output node we have to make their corresponding terminal unloaded. What do you mean by unloaded? We have to basically short this node to ac ground and then we have to find how much the current it is coming from the circuit signal current.

Detailed Explanation

In this chunk, we begin with an overview of the common base configuration used in transistor circuits. To accurately measure the current gain, the output terminal needs to be 'unloaded', meaning we connect it to an AC ground. This connection allows us to measure the actual current flowing through the circuit without the influence of load resistances.

Examples & Analogies

Think of it like tuning a musical instrument. When you tune a guitar, you might mute the strings to hear only the sound generated by the chords you strum. Here, grounding the output node ensures we only measure the signal's direct current without interference.

Understanding Signal Conditions

Chapter 2 of 6

🔒 Unlock Audio Chapter

Sign up and enroll to access the full audio experience

0:00
--:--

Chapter Content

And we know that if the signal it is in current form unloaded condition should be the corresponding impedance or the terminating impedance should be 0. So, small signal model if you see the corresponding situation here it is this node the corresponding collector node it is ground and we are observing the corresponding signal current i , for their input signal it is i .

Detailed Explanation

This chunk discusses the condition needed to analyze the small-signal model in current amplifiers. When we say the terminating impedance should be 0, it means we want the current from the circuit to flow freely without any resistance from a load. In the model, we ground the collector node, which allows us to observe the input signal current and how it behaves without external influences.

Examples & Analogies

Imagine trying to pour water from a pitcher into a glass. If the glass has no bottom (analogous to 0 impedance), the water flows freely. However, if the glass has a bottom (resistance), the flow gets restricted. We want the water (current) to flow unimpeded for accurate measurement.

Signal Current Components

Chapter 3 of 6

🔒 Unlock Audio Chapter

Sign up and enroll to access the full audio experience

0:00
--:--

Chapter Content

Now if you see this circuit again the base node it is grounded, voltage at the emitter we do have v . So, the v it is v it is ‒ v right and part of the current is also flowing here. So, we can say that i , it is having different component; one is this part another is this part right and then we also have this current and this current.

Detailed Explanation

In this part, we analyze what happens at the base node, which is grounded. The voltage at the emitter influences how current flows through various components of the circuit. Different components of the current can be identified, which together show how the total input current behaves.

Examples & Analogies

Think of this as a river with different streams flowing into it from multiple sources. Each stream represents a different component of the current. Just as the total flow of the river is the sum of all its streams, the overall input current is the sum of all its components.

Current Gain Calculation

Chapter 4 of 6

🔒 Unlock Audio Chapter

Sign up and enroll to access the full audio experience

0:00
--:--

Chapter Content

So, in summary we can say that i it is it can be directly written in terms of v . On the other hand if you see the current at the output terminal here. So, if this is the current. In fact, this current of course, this node it is grounded. So, the current here it is actually 0 because this is also ground this is also ground. So, the current here it is 0.

Detailed Explanation

Here, we summarize that the input current can be expressed in terms of voltage, which plays a critical role in determining the overall gain of the circuit. We also notice that at the output terminal, since all nodes are grounded, the output current is zero, which is useful for understanding how the current gain is derived.

Examples & Analogies

Consider this like checking how much water is flowing to a fountain (input current), while the fountain itself has its pipes disconnected (output current). Even though water is supplied to the fountain, if the pipes are disconnected, no water will flow out (output current = 0).

Interpreting Current Gain

Chapter 5 of 6

🔒 Unlock Audio Chapter

Sign up and enroll to access the full audio experience

0:00
--:--

Chapter Content

So, i it is summation of only these two currents we do have this current and we do have this current. So, i it is v × g and then we do have . So, if I take ratio of this two what we are getting here it is the v this is getting cancelled. So, the current gain = .

Detailed Explanation

This chunk explains that the output current (denoted as i) is simply the sum of currents through different paths in the circuit. The ratio of the input voltage multiplied by the transconductance (g) gives us the current gain. Noticing that voltage terms cancel gives us a simplified view of how the gain behaves.

Examples & Analogies

Imagine you are measuring how much power a water pump can push out versus how much power is input to it. If the water levels equalize throughout the system, you can simply compare output and input levels. This cancellation simplifies your calculations, just as it simplifies the gain calculation here.

Current Gain and Transistor Parameters

Chapter 6 of 6

🔒 Unlock Audio Chapter

Sign up and enroll to access the full audio experience

0:00
--:--

Chapter Content

In fact, you may recall that is nothing but the β of the transistor. So, this ( ). In fact, you may recall this is nothing but, α of the transistor that is very obvious.

Detailed Explanation

In this final chunk, we relate the current gain to the parameters of the transistor, β (beta) and α (alpha). The current gain, derived through our calculations, links directly to these parameters, showing how efficient the transistor is at amplifying currents from the emitter to the collector.

Examples & Analogies

This is similar to how efficiency ratings show how well an appliance (like a heater) converts energy input into usable heat output. In this case, knowing the parameters of the transistor lets us predict how efficiently it can amplify signals.

Key Concepts

  • Current Gain: Indicates the ratio of output current to input current, crucial for understanding amplifier performance.

  • Common Base and Common Gate Configurations: Used for amplifying signals and act as buffers.

  • Unloaded Condition: A specific configuration that allows for accurate measurement of output current.

Examples & Applications

In a common base amplifier with a transconductance of 20 mS and an input voltage of 0.1V, the output current can be calculated as i_o = g_m * v_e = 20 mS * 0.1V = 2 mA.

If a common base amplifier has a current gain (α) of approximately 0.98, it means for every 1 mA flowing into the emitter, roughly 0.98 mA flows out of the collector.

Memory Aids

Interactive tools to help you remember key concepts

🎵

Rhymes

In the common base, no current to waste, a ground is what we need, for true gain to proceed!

📖

Stories

Imagine a delivery man (the input current) giving packages (the output current) to a box (the amplifier) that's need to be emptied (grounded). If the box is open (unloaded condition), we can measure how many packages are transferred directly.

🧠

Memory Tools

Remember 'GIVE' for gain in a common base: Grounded Input for Voltage Effect.

🎯

Acronyms

CBA (Common Base Amplifier)

Current Buffering Amplifier - signifying its role in signal buffering.

Flash Cards

Glossary

Common Base Configuration

An amplifier configuration characterized by low input impedance and high output impedance, primarily used for current amplification.

Unloaded Condition

A condition where the output terminal of an amplifier is shorted to ground to accurately measure output current.

Current Gain

The ratio of output current to input current, indicating the effectiveness of the amplifier in amplifying current.

Transconductance (g_m)

A measure of the ability of a transistor to control the output current by varying the input voltage.

Alpha (α)

A parameter indicating the current gain of a bipolar junction transistor, typically close to 1.

Beta (β)

The current gain of a BJT, defined as the ratio of collector current to base current.

Reference links

Supplementary resources to enhance your learning experience.